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General Questions
Q1
Which of the two Chi Square distributions shown below ($A$ or $B$) has the larger degrees of freedom? How do you know? (relevant section)
Q2
Twelve subjects were each given two flavors of ice cream to taste and then were asked whether they liked them. Two of the subjects liked the first flavor and nine of them liked the second flavor. Is it valid to use the Chi Square test to determine whether this difference in proportions is significant? Why or why not? (relevant section)
Q3
A die is suspected of being biased. It is rolled $25$ times with the following result:
Outcome
Frequency
1
9
2
4
3
1
4
8
5
3
6
0
Conduct a significance test to see if the die is biased.
1. What Chi Square value do you get and how many degrees of freedom does it have?
2. What is the $p$ value? (relevant section)
Q4
A recent experiment investigated the relationship between smoking and urinary incontinence. Of the $322$ subjects in the study who were incontinent, $113$ were smokers, $51$ were former smokers, and $158$ had never smoked. Of the $284$ control subjects who were not incontinent, $68$ were smokers, $23$ were former smokers, and $193$ had never smoked.
1. Create a table displaying this data.
2. What is the expected frequency in each cell?
3. Conduct a significance test to see if there is a relationship between smoking and incontinence. What Chi Square value do you get? What $p$ value do you get?
4. What do you conclude? (relevant section)
Q5
At a school pep rally, a group of sophomore students organized a free raffle for prizes. They claim that they put the names of all of the students in the school in the basket and that they randomly drew $36$ names out of this basket. Of the prize winners, $6$ were freshmen, $14$ were sophomores, $9$ were juniors, and $7$ were seniors. The results do not seem that random to you. You think it is a little fishy that sophomores organized the raffle and also won the most prizes. Your school is composed of $30\%$ freshmen, $25\%$ sophomores, $25\%$ juniors, and $20\%$ seniors.
1. What are the expected frequencies of winners from each class?
2. Conduct a significance test to determine whether the winners of the prizes were distributed throughout the classes as would be expected based on the percentage of students in each group. Report your Chi Square and $p$ values.
3. What do you conclude? (relevant section)
Q6
Some parents of the West Bay little leaguers think that they are noticing a pattern. There seems to be a relationship between the number on the kids' jerseys and their position. These parents decide to record what they see. The hypothetical data appear below. Conduct a Chi Square test to determine if the parents' suspicion that there is a relationship between jersey number and position is right. Report your Chi Square and p values. (relevant section)
Infield
Outfield
Pitcher
Total
0-9
12
5
5
22
10-19
5
10
2
17
20+
4
4
7
15
Total
21
19
14
54
Q7
True/false: A Chi Square distribution with $2\; df$ has a larger mean than a Chi Square distribution with $12\; df$. (relevant section)
Q8
True/false: A Chi Square test is often used to determine if there is a significant relationship between two continuous variables. (relevant section)
Q9
True/false: Imagine that you want to determine if the spinner shown below is biased. You spin it $50$ times and write down how many times the arrow lands in each section. You will reject the null hypothesis at the $0.05$ level and determine that this spinner is biased if you calculate a Chi Square value of $7.82$ or higher. (relevant section)
Questions from Case Studies
The following question uses data from the SAT and GPA (SG) case study.
Q10
Answer these items to determine if the math SAT scores are normally distributed. You may want to first standardize the scores. (relevant section)
1. If these data were normally distributed, how many scores would you expect there to be in each of these brackets:
1. smaller than $1$ $SD$ below the mean
2. in between the mean and $1$ $SD$ below the mean
3. in between the mean and $1$ $SD$ above the mean
4. greater than $1$ $SD$ above the mean?
2. How many scores are actually in each of these brackets?
3. Conduct a Chi Square test to determine if the math SAT scores are normally distributed based on these expected and observed frequencies. (relevant section)
The following questions are from the Diet and Health (DH) case study.
Q11
(DH#3) Conduct a Pearson Chi Square test to determine if there is any relationship between diet and outcome. Report the Chi Square and $p$ values and state your conclusions. (relevant section)
The following questions are from ARTIST (reproduced with permission).
Q12
A study compared members of a medical clinic who filed complaints with a random sample of members who did not complain. The study divided the complainers into two subgroups: those who filed complaints about medical treatment and those who filed nonmedical complaints. Here are the data on the total number in each group and the number who voluntarily left the medical clinic. Set up a two-way table. Analyze these data to see if there is a relationship between complaint (no, yes - medical, yes - nonmedical) and leaving the clinic (yes or no).
No Complaint Medical Complaint Non Medical Complaint
Total 743 199 440
Left 22 26 28
Q13
Imagine that you believe there is a relationship between a person's eye color and where he or she prefers to sit in a large lecture hall. You decide to collect data from a random sample of individuals and conduct a chi-square test of independence. What would your two-way table look like? Use the information to construct such a table, and be sure to label the different levels of each category.
Q14
A geologist collects hand-specimen sized pieces of limestone from a particular area. A qualitative assessment of both texture and color is made with the following results. Is there evidence of association between color and texture for these limestones? Explain your answer.
Color
Texture
Light
Medium
Dark
Fine
4
20
8
Medium
5
23
12
Coarse
21
23
4
Q15
Suppose that college students are asked to identify their preferences in political affiliation (Democrat, Republican, or Independent) and in ice cream (chocolate, vanilla, or strawberry). Suppose that their responses are represented in the following two-way table (with some of the totals left for you to calculate).
Chocolate Vanilla Strawberry Total
Democrat 26 43 13 82
Republican 45 12 8 65
Independent 9 13 4
Total 68 25 173
1. What proportion of the respondents prefer chocolate ice cream?
2. What proportion of the respondents are Independents?
3. What proportion of Independents prefer chocolate ice cream?
4. What proportion of those who prefer chocolate ice cream are Independents?
5. Analyze the data to determine if there is a relationship between political party preference and ice cream preference.
Q16
NCAA collected data on graduation rates of athletes in Division I in the mid $1980s$. Among $2,332$ men, $1,343$ had not graduated from college, and among $959$ women, $441$ had not graduated.
1. Set up a two-way table to examine the relationship between gender and graduation.
2. Identify a test procedure that would be appropriate for analyzing the relationship between gender and graduation. Carry out the procedure and state your conclusion.
Select Answers
S3
1. $\text{Chi Square} = 16.0$, $df = 5$
S4
1. Incontinent/Smoker cell: $96.2$
S5
1. $p = 0.18$
S6
$\text{Chi Square} = 10.2$
S10
1. Scores smaller than $1$ $SD$ below the mean: $24$
S11
$\text{Chi Square} = 16.6$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/17%3A_Chi_Square/17.07%3A_Chi_Square_%28Exercises%29.txt |
Because distribution-free tests do not assume normality, they can be less susceptible to non-normality and extreme values. Therefore, they can be more powerful than the standard tests of means that assume normality.
18: Distribution-Free Tests
Learning Objectives
• State how distribution-free tests can avoid an inflated Type I error rate
• State how how distribution-free tests can affect power
Most tests based on the normal distribution are said to be robust when the assumption of normality is violated. To the extent to which actual probability values differ from nominal probability values, the actual probability values tend to be higher than the nominal \(p\) values. For example, if the probability of a difference as extreme or more extreme were \(0.04\), the test might report that the probability value is \(0.06\). Although this sounds like a good thing because the Type I error rate is lower than the nominal rate, it has a serious downside: reduced power. When the null hypothesis is false, then the probability of rejecting the null hypothesis can be substantially lower than it would have been if the distributions were distributed normally.
Tests assuming normality can have particularly low power when there are extreme values or outliers. A contributing factor is the sensitivity of the mean to extreme values. Although transformations can ameliorate this problem in some situations, they are not a universal solution.
Tests assuming normality often have low power for leptokurtic distributions. Transformations are generally less effective for reducing kurtosis than for reducing skew.
Because distribution-free tests do not assume normality, they can be less susceptible to non-normality and extreme values. Therefore, they can be more powerful than the standard tests of means that assume normality.
18.02: Randomization Tests - Two Conditions
Learning Objectives
• Compute a randomization test of the difference between independent groups
The data in Table $1$ are from a fictitious experiment comparing an experimental group with a control group. The scores in the Experimental Group are generally higher than those in the Control Group with the Experimental Group mean of $14$ being considerably higher than the Control Group mean of $4$. Would a difference this large or larger be likely if the two treatments had identical effects? The approach taken by randomization tests is to consider all possible ways the values obtained in the experiment could be assigned to the two groups. Then, the location of the actual data within the list is used to assess how likely a difference that large or larger would occur by chance.
Table $1$: Fictitious data
Experimental Control
7 0
8 2
11 5
30 9
First, consider all possible ways the $8$ values could be divided into two sets of $4$. We can apply the formula from the section on Permutations and Combinations for the number of combinations of $n$ items taken $r$ at a time and find that there are $70$ ways.
$_{n}\textrm{C}_r = \frac{n!}{(n-r)!r!} = \frac{8!}{(8-4)!4!} = 70$
Of these $70$ ways of dividing the data, how many result in a difference between means of $10$ or larger? From Table $1$you can see that there are two rearrangements that would lead to a bigger difference than $10$:
1. the score of $7$ could have been in the Control Group with the score of $9$ in the Experimental Group and
2. the score of $8$ could have been in the Control Group with the score of $9$ in the Experimental Group
Therefore, including the actual data, there are $3$ ways to produce a difference as large or larger than the one obtained. This means that if assignments to groups were made randomly, the probability of this large or a larger advantage of the Experimental Group is $3/70 = 0.0429$. Since only one direction of difference is considered (Experimental larger than Control), this is a one-tailed probability. The two-tailed probability is $0.0857$ since there are $6/70$ ways to arrange the data so that the absolute value of the difference between groups is as large or larger than the one obtained.
Clearly, this type of analysis would be very time consuming for even moderate sample sizes. Therefore, it is most useful for very small sample sizes.
An alternate approach made practical by computer software is to randomly divide the data into groups thousands of times and count the proportion of times the difference is as big or bigger than that found with the actual data. If the number of times the data are divided randomly is very large, then this proportion will be very close to the proportion you would get if you had listed all possible ways the data could be divided. The link below goes to a web page that can do these calculations.
Statkey | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/18%3A_Distribution-Free_Tests/18.01%3A_Benefits_of_Distribution_Free_Tests.txt |
Learning Objectives
• Compute a randomization test for differences among more than two conditions
The method of randomization for testing differences among more than two means is essentially very similar to the method when there are exactly two means. Table $1$ shows the data from a fictitious experiment with three groups.
Table $1$: Fictitious data
T1 T2 Control
7 14 0
8 19 2
11 21 5
12 122 9
The first step in a randomization test is to decide on a test statistic. Then we compute the proportion of the possible arrangements of the data for which that test statistic is as large as or larger than the arrangement of the actual data. When comparing several means, it is convenient to use the $F$ ratio. The $F$ ratio is computed not to test for significance directly, but as a measure of how different the groups are. For these data, the $F$ ratio for a one-way ANOVA is $2.06$.
The next step is to determine how many arrangements of the data result in as large or larger $F$ ratios. There are $6$ arrangements that lead to the same $F$ of $2.06$: the six arrangements of the three columns. One such arrangement is shown in Table $2$. The six are:
1. $T1$, $T2$, Control
2. $T1$, Control, $T2$
3. $T2$, $T1$, Control
4. $T2$, Control, $T1$
5. Control, $T1$, $T2$
6. Control, $T2$, $T1$
For each of the $6$ arrangements there are two changes that lead to a higher $F$ ratio: swapping the $7$ for the $9$ (which gives an $F$ of $2.08$) and swapping the $8$ for the $9$ (which gives an $F$ of $2.07$). The former of these two is shown in Table $3$.
Table $2$: Fictitious data with data for $T2$ and Control swapped
T1 Control T2
7 14 0
8 19 2
11 21 5
12 122 9
Table $3$: Data from Table $1$ with the $7$ and the $9$ swapped
T1 T2 Control
9 14 0
8 19 2
11 21 5
12 122 7
Thus, there are six arrangements, each with two swaps that lead to a larger $F$ ratio. Therefore, the number of arrangements with an $F$ as large or larger than the actual arrangement is $6$ (for the arrangements with the same $F$) + $12$ (for the arrangements with a larger $F$), which makes $18$ in all.
The next step is to determine the total number of possible arrangements. This can be computed from the following formula:
$\text{Arrangements} = (n!)^k = (4!)^3 = 13,824$
where $n$ is the number of observations in each group (assumed to be the same for all groups), and $k$ is the number of groups. Therefore, the proportion of arrangements with an $F$ as large or larger than the $F$ of $2.06$ obtained with the data is
$\dfrac{18}{13,824} = 0.0013.$
Thus, if there were no treatment effect, it is very unlikely that an $F$ as large or larger than the one obtained would be found.
18.04: Randomization Association
skills to develop
• Compute a randomization test for Pearson's \(r\)
A significance test for Pearson's \(r\) is described in the section inferential statistics for \(b\) and \(r\). The significance test described in that section assumes normality. This section describes a method for testing the significance of \(r\) that makes no distributional assumptions.
Table \(1\): Example data
X 1.0 2.4 3.8 4.0 11.0
Y 1.0 2.0 2.3 3.7 2.5
The approach is to consider the \(X\) variable fixed and compare the correlation obtained in the actual data to the correlations that could be obtained by rearranging the \(Y\) variable. For the data shown in Table \(1\), the correlation between \(X\) and \(Y\) is \(0.385\). There is only one arrangement of \(Y\) that would produce a higher correlation. This arrangement is shown in Table \(2\) and the \(r\) is \(0.945\). Therefore, there are two arrangements of \(Y\) that lead to correlations as high or higher than the actual data.
Table \(1\): The example data arranged to give the highest \(r\)
X Y
1.0 1.0
2.4 2.0
3.8 2.3
4.0 2.5
11.0 3.7
The next step is to calculate the number of possible arrangements of \(Y\). The number is simply \(N!\), where \(N\) is the number of pairs of scores. Here, the number of arrangements is \(5! = 120\). Therefore, the probability value is \(2/120 = 0.017\). Note that this is a one-tailed probability since it is the proportion of arrangements that give an \(r\) as large or larger. For the two-tailed probability, you would also count arrangements for which the value of \(r\) were less than or equal to \(-0.385\). In randomization tests, the two-tailed probability is not necessarily double the one-tailed probability.
18.05: Fisher's Exact Test
Learning Objectives
• State the situation when Fisher's exact test can be used
• Calculate Fisher's exact test
• Describe how conservative the Fisher exact test is relative to a Chi Square test
The chapter on Chi Square showed one way to test the relationship between two nominal variables. A special case of this kind of relationship is the difference between proportions. This section shows how to compute a significance test for a difference in proportions using a randomization test. Suppose, in a fictitious experiment, \(4\) subjects in an Experimental Group and \(4\) subjects in a Control Group are asked to solve an anagram problem. Three of the \(4\) subjects in the Experimental Group and none of the subjects in the Control Group solved the problem. Table \(1\) shows the results in a contingency table.
Table \(1\): Anagram Problem Data
Experimental Control Total
Solved 3 0 3
Did Not Solve 1 4 5
Total 4 4 8
The significance test we are going to perform is called the Fisher Exact Test. The basic idea is to take the row totals and column totals as "given" and add the probability of obtaining the pattern of frequencies obtained in the experiment and the probabilities of all other patterns that reflect a greater difference between conditions. The formula for obtaining any given pattern of frequencies is:
\[\dfrac{ n! (N-n)! R! (N-R)!}{r!(n-r)!(R-r)!(N-n-R+r)!N!}\]
where \(N\) is the total sample size (\(8\)), \(n\) is the sample size for the first group (\(4\)), \(r\) is the number of successes for the first group (\(3\)), and \(R\) is the total number of successes (\(3\)). For this example, the probability is
\[\dfrac{ 4! (8-4)! 3! (8-3)!}{3!(4-3)!(3-3)!(8-4-3+8)!8!} = 0.0714\]
Since more extreme outcomes do not exist given the row and column totals, the \(p\) value is \(0.0714\). This is a one-tailed probability since it only considers outcomes as extreme or more extreme favoring the Experimental Group. An equally extreme outcome favoring the Control Group is shown in Table \(2\), which also has a probability of \(0.0714\). Therefore, the two-tailed probability is \(0.1428\). Note that in the Fisher Exact Test, the two-tailed probability is not necessarily double the one-tailed probability.
Table \(2\): Anagram Problem Favoring Control Group
Experimental Control Total
Solved 0 3 3
Did not Solve 4 1 5
Total 4 4 8
The Fisher Exact Test is "exact" in the sense that it is not based on a statistic that is approximately distributed as, for example, Chi Square. However, because it assumes that both marginal totals are fixed, it can be considerably less powerful than the Chi Square test. Even though the Chi Square test is an approximate test, the approximation is quite good in most cases and tends to have too low a Type I error rate more often than too high a Type I error rate (see for yourself using this simulation). | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/18%3A_Distribution-Free_Tests/18.03%3A_Randomization_Tests_-_Two_or_More_Conditions.txt |
Learning Objectives
• State the difference between a randomization test and a rank randomization test
• Describe why rank randomization tests are more common
• Be able to compute a Mann-Whitney $U$ test
The major problem with randomization tests is that they are very difficult to compute. Rank randomization tests are performed by first converting the scores to ranks and then computing a randomization test. The primary advantage of rank randomization tests is that there are tables that can be used to determine significance. The disadvantage is that some information is lost when the numbers are converted to ranks. Therefore, rank randomization tests are generally less powerful than randomization tests based on the original numbers.
There are several names for rank randomization tests for differences in central tendency. The two most common are the Mann-Whitney $U$ test and the Wilcoxon Rank Sum test.
Consider the data shown in Table $1$ that were used as an example in the section on randomization tests.
Table $1$: Fictitious data
Experimental Control
7 0
8 2
11 5
30 9
A rank randomization test on these data begins by converting the numbers to ranks.
Table $2$: Fictitious data converted to ranks. Rank sum = $24$
Experimental Control
4 1
5 2
7 3
8 6
The probability value is determined by computing the proportion of the possible arrangements of these ranks that result in a difference between ranks as large or larger than those in the actual data (Table $2$). Since the sum of the ranks (the numbers $1-8$) is a constant ($36$ in this case), we can use the computational shortcut of finding the proportion of arrangements for which the sum of the ranks in the Experimental Group is as high or higher than the sum here ($4 + 5 + 7 + 8 = 24$).
First, consider how many ways the $8$ values could be divided into two sets of $4$. We can apply the formula from the section on Permutations and Combinations for the number of combinations of $n$ items taken $r$ at a time ($\text{n = the total number of observations; r = the number of observations in the first group}$) and find that there are $70$ ways.
$_{n}\textrm{C}_r = \frac{n!}{(n-r)!r!} = \frac{8!}{(8-4)!4!} = 70$
Of these $70$ ways of dividing the data, how many result in a sum of ranks of $24$ or more? Tables $3-5$ show three rearrangements that would lead to a rank sum of $24$ or larger.
Table $3$: Rearrangement of data converted to ranks. Rank sum = $26$
Experimental Control
6 1
5 2
7 3
8 4
Table $4$: Rearrangement of data converted to ranks. Rank sum = $25$
Experimental Control
4 1
6 2
7 3
8 5
Table $5$: Rearrangement of data converted to ranks. Rank sum = $24$
Experimental Control
3 1
6 2
7 4
8 5
Therefore, the actual data represent $1$ arrangement with a rank sum of $24$ or more and the $3$ arrangements represent three others. Therefore, there are $4$ arrangements with a rank sum of $24$ or more. This makes the probability equal to $4/70 = 0.057$. Since only one direction of difference is considered (Experimental larger than Control), this is a one-tailed probability. The two-tailed probability is $(2)(0.057) = 0.114$ since there are $8/70$ ways to arrange the data so that the sum of the ranks is either
1. as large or larger or
2. as small or smaller than the sum found for the actual data.
The beginning of this section stated that rank randomization tests were easier to compute than randomization tests because tables are available for rank randomization tests. Table $6$ can be used to obtain the critical values for equal sample sizes of $4-10$.
Table for unequal sample sizes
For the present data, both $n_1$ and $n_2 = 4$ so, as can be determined from the table, the rank sum for the Experimental Group must be at least $25$ for the difference to be significant at the $0.05$ level (one-tailed). Since the sum of ranks equals $24$, the probability value is somewhat above $0.05$. In fact, by counting the arrangements with the sum of ranks greater than or equal to $24$, we found that the probability value is $0.057$. Naturally a table can only give the critical value rather than the $p$ value itself. However, with a larger sample size such as $10$ subjects per group, it becomes very time-consuming to count all arrangements equaling or exceeding the rank sum of the data. Therefore, for practical reasons, the critical value sometimes suffices.
Table $6$: Critical values. One-Tailed Test. Rank Sum for Higher Group
n1 n2 0.20 0.10 0.05 0.025 0.01 0.005
4 4 22 23 25 26 . .
5 5 33 35 36 38 39 40
6 6 45 48 50 52 54 55
7 7 60 64 66 69 71 73
8 8 77 81 85 87 91 93
9 9 96 101 105 109 112 115
10 10 117 123 128 132 136 139
For larger sample sizes than covered in the tables, you can use the following expression that is approximately normally distributed for moderate to large sample sizes.
$Z=\frac{W_a-n_a(n_a+n_b+1)/2}{\sqrt{n_an_b(n_a+n_b+1)/12}}$
where:
• $W_a$ is the sum of the ranks for the first group
• $n_a$ is the sample size for the first group
• $n_b$ is the sample size for the second group
• $Z$ is the test statistic
The probability value can be determined from $Z$ using the normal distribution calculator.
The data from the Stereograms Case Study can be analyzed using this test. For these data, the sum of the ranks for Group 1 ($W_a$) is $1911$, the sample size for Group 1 ($n_a$) is $43$, and the sample size for Group 2 ($n_b$) is $35$. Plugging these values into the formula results in a $Z$ of $2.13$, which has a two-tailed $p$ of $0.033$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/18%3A_Distribution-Free_Tests/18.06%3A_Rank_Randomization_Two_Conditions.txt |
Learning Objectives
• Compute the Kruskal-Wallis test
The Kruskal-Wallis test is a rank-randomization test that extends the Wilcoxon test to designs with more than two groups. It tests for differences in central tendency in designs with one between-subjects variable. The test is based on a statistic $H$ that is approximately distributed as Chi Square. The formula for $H$ is shown below:
$H = -3(N+1) + \frac{12}{N(N+1)} \sum_{i=1}^{k} \frac{T_{i}^{2}}{n_i}$
where
• $N$ is the total number of observations
• $T_i$ is the sum of ranks for the $i^{th}$ group
• $n_i$ is the sample size for the $i^{th}$ group
• $k$ is the number of groups
The first step is to convert the data to ranks (ignoring group membership) and then find the sum of the ranks for each group. Then, compute $H$ using the formula above. Finally, the significance test is done using a Chi Square distribution with $k-1$ degrees of freedom.
For the "Smiles and Leniency" case study, the sums of the ranks for the four conditions are:
• False: $2732.0$
• Felt: $2385.5$
• Miserable: $2424.5$
• Neutral: $1776.0$
Note that since there are "ties" in the data, the mean rank of the ties is used. For example, there were $10$ scores of $2.5$ which tied for ranks $4-13$. The average of the numbers $4, 5, 6, 7, 8, 9, 10, 11, 12,13$ is $8.5$. Therefore, all values of $2.5$ were assigned ranks of $8.5$.
The sample size for each group is $34$.
$H = -3(136+1) + \frac{12}{(136)(137)} \left ( \frac{(2732)^2}{34} + \frac{(2385.5)^2}{34} + \frac{(2424.5)^2}{34} + \frac{(1776)^2}{34} \right ) = 9.28$
Using the Chi Square Calculator for $\text{Chi Square} = 9.28$ with $4-1 = 3\; df$ results in a $p$ value of $0.0258$. Thus the null hypothesis of no leniency effect can be rejected.
18.08: Rank Randomization for Association
Learning Objectives
• Compute Spearman's $ρ$
• Test Spearman's $ρ$ for significance
The rank randomization test for association is equivalent to the randomization test for Pearson's r except that the numbers are converted to ranks before the analysis is done. Table $1$ shows $5$ values of $X$ and $Y$. Table $2$ shows these same data converted to ranks (separately for $X$ and $Y$).
Table $1$: Example data
X Y
1.0 1.0
2.4 2.0
3.8 2.3
4.0 3.7
11.0 2.5
Table $2$: Ranked data
X Y
1 1
2 2
3 3
4 5
5 4
The approach is to consider the $X$ variable fixed and compare the correlation obtained in the actual ranked data to the correlations that could be obtained by rearranging the $Y$ variable ranks. For the ranked data shown in Table $2$, the correlation between $X$ and $Y$ is $0.90$. The correlation of ranks is called "Spearman's $ρ$."
Table $3$: Ranked data with correlation of $1.0$
X Y
1 1
2 2
3 3
4 4
5 5
There is only one arrangement of $Y$ that produces a higher correlation than $0.90$: A correlation of $1.0$ results if the fourth and fifth observations' $Y$ values are switched (see Table $3$). There are also three other arrangements that produce an $r$ of $0.90$ (see Tables $4$, $5$, and $6$). Therefore, there are five arrangements of $Y$ that lead to correlations as high or higher than the actual ranked data (Tables $2$ through $6$).
Table $4$: Ranked data with correlation of $0.90$
X Y
1 1
2 2
3 4
4 3
5 5
Table $5$: Ranked data with correlation of $0.90$
X Y
1 1
2 3
3 2
4 4
5 5
Table $6$: Ranked data with correlation of $0.90$
X Y
1 2
2 1
3 3
4 4
5 5
The next step is to calculate the number of possible arrangements of $Y$. The number is simply $N!$, where $N$ is the number of pairs of scores. Here, the number of arrangements is $5! = 120$. Therefore, the probability value is $5/120 = 0.042$. Note that this is a one-tailed probability since it is the proportion of arrangements that give a correlation as large or larger. The two-tailed probability is $0.084$.
Since it is hard to count up all the possibilities when the sample size is even moderately large, it is convenient to have a table of critical values.
Table of critical values for Spearman's $\rho$
From the table linked to above, you can see that the critical value for a one-tailed test with $5$ observations at the $0.05$ level is $0.90$. Since the correlation for the sample data is $0.90$, the association is significant at the $0.05$ level (one-tailed). As shown above, the probability value is $0.042$. Since the critical value for a two-tailed test is $1.0$, Spearman's $ρ$ is not significant in a two-tailed test.
18.09: Statistical Literacy Standard
Learning Objectives
• Troponin Concentration and Ventricular Strain
Cardiac troponins are markers of myocardial damage. The levels of troponin in subjects with and without signs of right ventricular strain in the electrocardiogram were compared in the experiment described here.
The Wilcoxon rank sum test was used to test for significance. The troponin concentration in patients with signs of right ventricular strain was higher (\(median = 0.03\) ng/ml) than in patients without right ventricular strain (\(median < 0.01\) ng/ml), \(p<0.001\).
Example \(1\): what do you think?
Why might the authors have used the Wilcoxon test rather than a t test? Do you think the conclusions would have been different?
Solution
Perhaps the distributions were very non-normal. Typically a transformation can be done to make a distribution more normal but that is not always the case. It is almost certain the same conclusion would have been reached, although it would have been described in terms of mean differences instead of median differences.
18.10: Distribution Free Tests (Exercises)
General Questions
Q1
For the following data, how many ways could the data be arranged (including the original arrangement) so that the advantage of the Experimental Group mean over the Control Group mean is as large or larger then the original arrangement?
Experimental Control
5 1
10 2
15 3
16 4
17 9
Q2
For the data in Problem 1, how many ways can the data be rearranged?
Q3
What is the one-tailed probability for a test of the difference?
Q4
For the following data, how many ways can the data be rearranged?
T1 T2 Control
7 14 0
8 19 2
11 21 5
Q5
In general, are rank randomization tests or randomization tests more powerful?
Q6
What is the advantage of rank randomization tests over randomization tests?
Q7
Test whether the differences among conditions for the data in Problem 1 is significant (one tailed) at the \(0.01\) level using a rank randomization test.
Questions from Case Studies
The following question uses data from the SAT and GPA case study.
Q8
Compute Spearman's \(ρ\) for the relationship between UGPA and SAT.
The following question uses data from the Stereograms case study.
Q9
Test the difference in central tendency between the two conditions using a rank-randomization test (with the normal approximation) with a one-tailed test. Give the \(Z\) and the \(p\).
The following question uses data from the Smiles and Leniency case study.
Q10
Test the difference in central tendency between the four conditions using a rank-randomization test (with the normal approximation). Give the Chi Square and the \(p\). | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/18%3A_Distribution-Free_Tests/18.07%3A_Rank_Randomization_Two_or_More_Conditions.txt |
Researchers often seek to learn more than whether the variable under investigation has an effect and/or the direction of the effect. This is particularly true for research that has practical applications. For example, an investigation of the efficacy of a pain-relief drug would seek to determine the extent of the relief and not merely whether there was any relief. Similarly, a study of a test-preparation course's efficacy would seek to determine how much the course raises students' test scores. Finally, a study of the relationship between exercise and blood pressure would seek to determine how much blood pressure decreases for a given amount of exercise. In all of these examples, a significance test would not be sufficient since it would only provide the researcher with information about the existence and direction of the effect. It would not provide any information about the size of the effect.
Before we proceed with a discussion of how to measure effect size, it is important to consider that for some research it is the presence or absence of an effect rather than its size that is important. A controversial example is provided by Bem (\(2011\)) who investigated precognition. Bem found statistically significant evidence that subjects' responses are affected by future events. That is, he rejected the null hypothesis that there is no effect. The important question is not the size of the effect but, rather, whether it exists at all. It would be truly remarkable if future events affect present responses even a little. It is important to note that subsequent research (Ritchie, Wiseman, & French, \(2012\)) has failed to replicate Bem's results and the likelihood that the precognition effects he described are real is very low.
Bem, D. J. (201). Feeling the future: Experimental evidence for anomalous retroactive influences on cognition and affect. Journal of Personality and Social Psychology, 100, 407–425.
Ritchie, S. J., Wiseman R., and French, C. C. (2012) Failing the Future: Three Unsuccessful Attempts to Replicate Bem's 'Retroactive Facilitation of Recall' Effect. PLoS ONE 7.
19.02: Proportions
Learning Objectives
• Compute absolute risk reduction
• Compute relative risk reduction
• Compute number needed to treat
Often the interpretation of a proportion is self-evident. For example, the obesity rate for white non-Hispanic adults living in the United States was estimated by a study conducted between $2006$ and $2008$ to be $24\%$. This value of $24\%$ is easily interpretable and indicates the magnitude of the obesity problem in this population.
Often the question of interest involves the comparison of two outcomes. For example, consider the analysis of proportions in the case study "Mediterranean Diet and Health." In this study, one group of people followed the diet recommended by the American Heart Association (AHA), whereas a second group followed the "Mediterranean Diet." One interesting comparison is between the proportions of people who were healthy throughout the study as a function of diet. It turned out that $0.79$ of the people who followed the AHA diet and $0.90$ of those who followed the Mediterranean diet were healthy. How is the effect size of diet best measured?
We will take the perspective that we are assessing the benefits of switching from the AHA diet to the Mediterranean diet. One way to assess the benefits is to compute the difference between the proportion who were not healthy on the AHA diet ($0.21$) with the proportion who were not healthy on the Mediterranean diet ($0.10$). Therefore, the difference in proportions is:
$0.21 - 0.10 = 0.11$
This measure of the benefit is called the Absolute Risk Reduction (ARR).
To define ARR more formally, let $C$ be the proportion of people in the control group with the ailment of interest and $T$ be the proportion in the treatment group. ARR can then be defined as:
$ARR = C - T$
Alternatively, one could measure the difference in terms of percentages. For our example, the proportion of non-healthy people on the Mediterranean diet ($0.10$) is $52\%$ lower than the proportion of non-healthy people on the AHA diet ($0.21$). This value is computed as follows:
$\frac{(0.21 - 0.10)}{0.21} \times 100 = 52\%$
This measure of the benefit is called the Relative Risk Reduction (RRR). The general formula for RRR is:
$RRR = \frac{(C - T)}{C} \times 100$
where $C$ and $T$ are defined as before.
A third commonly used measure is the "odds ratio." For our example, the odds of being healthy on the Mediterranean diet are $90:10 = 9:1$; the odds on the AHA diet are $79:21 = 3.76:1$. The ratio of these two odds is $9/3.76 = 2.39$. Therefore, the odds of being healthy on the Mediterranean diet is $2.39$ times the odds of being healthy on the AHA diet. Note that the odds ratio is the ratio of the odds and not the ratio of the probabilities.
A fourth measure is the number of people who need to be treated in order to prevent one person from having the ailment of interest. In our example, being treated means changing from the AHA diet to the Mediterranean diet. The number who need to be treated can be defined as:
$N = \frac{1}{ARR}$
For our example,
$N = \frac{1}{0.11}=9$
Therefore, one person who would otherwise not be healthy would be expected to stay healthy for every nine people changing from the AHA diet to the Mediterranean diet.
The obvious question is which of these measures is the best one. Although each measure has its proper uses, the $RRR$ measure can exaggerate the importance of an effect, especially when the absolute risks are low. For example, if a drug reduced the risk of a certain disease from $1$ in $1,000,000$ to $1$ in $2,000,000$, the $RRR$ is $50\%$. However, since the $ARR$ is only $0.0000005$, the practical reduction in risk is minimal. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/19%3A_Effect_Size/19.01%3A_Prelude_to_Effect_Size.txt |
Learning Objectives
• State how the inherent meaningfulness of the scales affects the type of measure that should be used
• Compute $g$
• Compute $d$
• State the effect of the variability of subjects on the size of standardized measures
When the units of a measurement scale are meaningful in their own right, then the difference between means is a good and easily interpretable measure of effect size. For example, a study conducted by Holbrook, Crowther, Lotter, Cheng and King in $2000$ investigated the effectiveness of benzodiazepine for the treatment of insomnia. These researchers found that, compared to a placebo, this drug increased total sleep duration by a mean of $61.8$ minutes. This difference in means shows clearly the degree to which benzodiazepine is effective. (It is important to note that the drug was found to sometimes have adverse side effects.)
When the dependent variable is measured on a ratio scale, it is often informative to consider the proportional difference between means in addition to the absolute difference. For example, if in the Holbrook et al. study the mean total sleep time for the placebo group was $120$ minutes, then the $61.8$-minute increase would represent a $51\%$ increase in sleep time. On the other hand, if the mean sleep time for the placebo were $420$ minutes, then the $61.8$-minute increase would represent a $15\%$ increase in sleep time.
It is interesting to note that if a log transformation is applied to the dependent variable, then equal percent changes on the original scale will result in equal absolute changes on the log scale. For example, suppose the mean sleep time increased $10\%$ from $400$ minutes to $440$ in one condition and $10\%$ from $300$ to $330$ minutes in a second condition. If we take the log base $10$ of these values, we find that
$\log_{10}(440) - \log_{10}(400) = 2.643 - 2.602 = 0.041$
Similarly,
$\log_{10}(330) - \log_{10}(300) = 2.518 - 2.477 = 0.041$
Many times the dependent variable is measured on a scale that is not inherently meaningful. For example, in the "Animal Research" case study, attitudes toward animal research were measured on a $7$-point scale. The mean rating of women on whether animal research is wrong was $1.47$ scale units higher than the mean rating of men. However, it is not clear whether this $1.47$-unit difference should be considered a large effect or a small effect, since it is not clear exactly what this difference means.
When the scale of a dependent variable is not inherently meaningful, it is common to consider the difference between means in standardized units. That is, effect size is measured in terms of the number of standard deviations the means differ by. Two commonly used measures are Hedges' $g$ and Cohen's $d$. Both of these measures consist of the difference between means divided by the standard deviation. They differ only in that Hedges' $g$ uses the version of the standard deviation formula in which you divide by $N-1$, whereas Cohen's $d$ uses the version in which you divide by $N$. The two formulas are given below.
$g = \frac{M_1-M_2}{\sqrt{MSE}}$
$d = g \sqrt{\frac{N}{N-2}}$
where $M_1$ is the mean of the first group, $M_2$ is the mean of the second group, $MSE$ is the mean square error, and $N$ is the total number of observations.
Standardized measures such as Cohen's $d$ and Hedges' $g$ have the advantage that they are scale free. That is, since the dependent variable is standardized, the original units are replaced by standardized units and are interpretable even if the original scale units do not have clear meaning. Consider the Animal Research case study in which attitudes were measured on a $7$-point scale. On a rating of whether animal research is wrong, the mean for women was $5.353$, the mean for men was $3.882$, and $MSE$ was $2.864$. Hedges' $g$ can be calculated to be $0.87$. It is more meaningful to say that the means were $0.87$ standard deviations apart than $1.47$ scale units apart, since the scale units are not well-defined.
It is natural to ask what constitutes a large effect. Although there is no objective answer to this question, the guidelines suggested by Cohen ($1988$) stating that an effect size of $0.2$ is a small effect, an effect size of $0.5$ is a medium effect, and an effect size of $0.8$ is a large effect have been widely adopted. Based on these guidelines, the effect size of $0.87$ is a large effect.
It should be noted, however, that these guidelines are somewhat arbitrary and have not been universally accepted. For example, Lenth ($2001$) argued that other important factors are ignored if Cohen's definition of effect size is used to choose a sample size to achieve a given level of power.
Interpretational Issues
It is important to realize that the importance of an effect depends on the context. For example, a small effect can make a big difference if only extreme observations are of interest. Consider a situation in which a test is used to select students for a highly selective program. Assume that there are two types of students (red and blue) and that the mean for the red students is $52$, the mean for the blue students is $50$, both distributions are normal, and the standard deviation for each distribution is $10$. The difference in means is therefore only $0.2$ standard deviations and would generally be considered to be a small difference. Now assume that only students who scored $70$ or higher would be selected for the program. Would there be a big difference between the proportion of blue and red students who would be able to be accepted into the program? It turns out that the proportion of red students who would qualify is $0.036$ and the proportion of blue students is $0.023$. Although this difference is small in absolute terms, the ratio of red to blue students who qualify is $1.6:1$. This means that if $100$ students were to be accepted and if equal numbers of randomly-selected red and blue students applied, $62\%$ would be red and $38\%$ would be blue. In most contexts this would be considered an important difference.
When the effect size is measured in standard deviation units as it is for Hedges' $g$ and Cohen's $d$, it is important to recognize that the variability in the subjects has a large influence on the effect size measure. Therefore, if two experiments both compared the same treatment to a control but the subjects were much more homogeneous in $\text{Experiment 1}$ than in $\text{Experiment 2}$, then a standardized effect size measure would be much larger in the former experiment than in the latter. Consider two hypothetical experiments on the effect of an exercise program on blood pressure. Assume that the mean effect on systolic blood pressure of the program is $10mm\; Hg$ and that, due to differences in the subject populations sampled in the two experiments, the standard deviation was $20$ in $\text{Experiment 1}$ and $30$ in $\text{Experiment 2}$. Under these conditions, the standardized measure of effect size would be $0.50$ in $\text{Experiment 1}$ and $0.33$ in $\text{Experiment 2}$. This standardized difference in effect size occurs even though the effectiveness of the treatment is exactly the same in the two experiments.
Reference
1. Cohen, J. (1988) Statistical Power Analysis for the Behavioral Sciences (second ed.). Lawrence Erlbaum Associates.
2. Lenth, R. V. (2001) Some Practical Guidelines for Effective Sample Size Determination. The American Statistician, 55, 187-193. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/19%3A_Effect_Size/19.03%3A_Difference_Between_Two_Means.txt |
Learning Objectives
• State the difference in bias between $η^2$ and $ω^2$
• Compute $η^2$ Compute $ω^2$
• Distinguish between $ω^2$ and partial $ω^2$
• State the bias in $R^2$ and what can be done to reduce it
Effect sizes are often measured in terms of the proportion of variance explained by a variable. In this section, we discuss this way to measure effect size in both ANOVA designs and in correlational studies.
ANOVA Designs
Responses of subjects will vary in just about every experiment. Consider, for example, the "Smiles and Leniency" case study. A histogram of the dependent variable "leniency" is shown in Figure $1$. It is clear that the leniency scores vary considerably. There are many reasons why the scores differ. One, of course, is that subjects were assigned to four different smile conditions and the condition they were in may have affected their leniency score. In addition, it is likely that some subjects are generally more lenient than others, thus contributing to the differences among scores. There are many other possible sources of differences in leniency ratings including, perhaps, that some subjects were in better moods than other subjects and/or that some subjects reacted more negatively than others to the looks or mannerisms of the stimulus person. You can imagine that there are innumerable other reasons why the scores of the subjects could differ.
One way to measure the effect of conditions is to determine the proportion of the variance among subjects' scores that is attributable to conditions. In this example, the variance of scores is $2.794$. The question is how this variance compares with what the variance would have been if every subject had been in the same treatment condition. We estimate this by computing the variance within each of the treatment conditions and taking the mean of these variances. For this example, the mean of the variances is $2.649$. Since the mean variance within the smile conditions is not that much less than the variance ignoring conditions, it is clear that "Smile Condition" is not responsible for a high percentage of the variance of the scores. The most convenient way to compute the proportion explained is in terms of the sum of squares "conditions" and the sum of squares total. The computations for these sums of squares are shown in the chapter on ANOVA. For the present data, the sum of squares for "Smile Condition" is $27.535$ and the sum of squares total is $377.189$. Therefore, the proportion explained by "Smile Condition" is:
$\frac{27.535}{377.189} = 0.073$
Thus, $0.073$ or $7.3\%$ of the variance is explained by "Smile Condition."
An alternative way to look at the variance explained is as the proportion reduction in error. The sum of squares total ($377.189$) represents the variation when "Smile Condition" is ignored and the sum of squares error ($377.189 - 27.535 = 349.654$) is the variation left over when "Smile Condition" is accounted for. The difference between $377.189$ and $349.654$ is $27.535$. This reduction in error of $27.535$ represents a proportional reduction of $27.535/377.189 = 0.073$, the same value as computed in terms of proportion of variance explained.
This measure of effect size, whether computed in terms of variance explained or in terms of percent reduction in error, is called $η^2$ where $η$ is the Greek letter eta. Unfortunately, $η^2$ tends to overestimate the variance explained and is therefore a biased estimate of the proportion of variance explained. As such, it is not recommended (despite the fact that it is reported by a leading statistics package).
An alternative measure, $ω^2$ (omega squared), is unbiased and can be computed from
$\omega ^2 = \frac{SSQ_{condition}-(k-1)MSE}{SSQ_{total}+MSE}$
where $MSE$ is the mean square error and $k$ is the number of conditions. For this example, $k = 4$ and $ω^2 = 0.052$.
It is important to be aware that both the variability of the population sampled and the specific levels of the independent variable are important determinants of the proportion of variance explained. Consider two possible designs of an experiment investigating the effect of alcohol consumption on driving ability. As can be seen in Table $1$, $\text{Design 1}$ has a smaller range of doses and a more diverse population than $\text{Design 2}$. What are the implications for the proportion of variance explained by Dose? Variation due to Dose would be greater in $\text{Design 2}$ than $\text{Design 1}$ since alcohol is manipulated more strongly than in $\text{Design 1}$. However, the variance in the population should be greater in $\text{Design 1}$ since it includes a more diverse set of drivers. Since with $\text{Design 1}$ the variance due to Dose would be smaller and the total variance would be larger, the proportion of variance explained by Dose would be much less using $\text{Design 1}$ than using $\text{Design 2}$. Thus, the proportion of variance explained is not a general characteristic of the independent variable. Instead, it is dependent on the specific levels of the independent variable used in the experiment and the variability of the population sampled.
Table $1$: Design Parameters
Design
Dose
Population
1
0.00
All Drivers between 16 and 80 Years of Age
0.30
0.60
2
0.00
Experienced Drivers between 25 and 30 Years of Age
0.50
1.00
Factorial Designs
In one-factor designs, the sum of squares total is the sum of squares condition plus the sum of squares error. The proportion of variance explained is defined relative to sum of squares total. In an $A \times B$ design, there are three sources of variation ($A, B, A \times B$) in addition to error. The proportion of variance explained for a variable ($A$, for example) could be defined relative to the sum of squares total ($SSQ_A + SSQ_B + SSQ_{A\times B} + SSQ_{error}$) or relative to $SSQ_A + SSQ_{error}$.
To illustrate with an example, consider a hypothetical experiment on the effects of age ($6$ and $12$ years) and of methods for teaching reading (experimental and control conditions). The means are shown in Table $2$. The standard deviation of each of the four cells ($Age \times Treatment$ combinations) is $5$. (Naturally, for real data, the standard deviations would not be exactly equal and the means would not be whole numbers.) Finally, there were $10$ subjects per cell resulting in a total of $40$ subjects.
Table $2$: Condition Means
Treatment
Age Experimental Control
6 40 42
12 50 56
The sources of variation, degrees of freedom, and sums of squares from the analysis of variance summary table as well as four measures of effect size are shown in Table $3$. Note that the sum of squares for age is very large relative to the other two effects. This is what would be expected since the difference in reading ability between $6$- and $12$-year-olds is very large relative to the effect of condition.
Table $3$: ANOVA Summary Table
Source df SSQ $η^2$ partial
$η^2$
$ω^2$ partial
$ω^2$
Age 1 1440 0.567 0.615 0.552 0.586
Condition 1 160 0.063 0.151 0.053 0.119
A x C 1 40 0.016 0.043 0.006 0.015
Error 36 900
Total 39 2540
First, we consider the two methods of computing $η^2$, labeled $η^2$ and partial $η^2$. The value of $η^2$ for an effect is simply the sum of squares for this effect divided by the sum of squares total. For example, the $η^2$ for Age is $1440/2540 = 0.567$. As in a one-factor design, $η^2$ is the proportion of the total variation explained by a variable. Partial $η^2$ for Age is $SSQ_{Age}$ divided by ($SSQ_{Age} + SSQ_{error}$), which is $1440/2340 = 0.615$.
As you can see, the partial $η^2$ is larger than $η^2$. This is because the denominator is smaller for the partial $η^2$. The difference between $η^2$ and partial $η^2$ is even larger for the effect of condition. This is because $SSQ_{Age}$ is large and it makes a big difference whether or not it is included in the denominator.
As noted previously, it is better to use $ω^2$ than $η^2$ because $η^2$ has a positive bias. You can see that the values for $ω^2$ are smaller than for $η^2$. The calculations for $ω^2$ are shown below:
$\omega ^2 = \frac{SSQ_{effect}-df_{effect}MS_{error}}{SSQ_{total}+MS_{error}}$
$\omega _{partial}^2 = \frac{SSQ_{effect}-df_{effect}MS_{error}}{SSQ_{effect}+(N-df_{effect})MS_{error}}$
where $N$ is the total number of observations.
The choice of whether to use $ω^2$ or the partial $ω^2$ is subjective; neither one is correct or incorrect. However, it is important to understand the difference and, if you are using computer software, to know which version is being computed. (Beware, at least one software package labels the statistics incorrectly).
Correlational Studies
In the section "Partitioning the Sums of Squares" in the Regression chapter, we saw that the sum of squares for $Y$ (the criterion variable) can be partitioned into the sum of squares explained and the sum of squares error. The proportion of variance explained in multiple regression is therefore:
$SSQ_{explained}/SSQ_{total }$
In simple regression, the proportion of variance explained is equal to $r^2$; in multiple regression, it is equal to $R^2$.
In general, $R^2$ is analogous to $η^2$ and is a biased estimate of the variance explained. The following formula for adjusted $R^2$ is analogous to $ω^2$ and is less biased (although not completely unbiased):
$R_{adjusted}^{2} = 1 - \frac{(1-R^2)(N-1)}{N-p-1}$
where $N$ is the total number of observations and $p$ is the number of predictor variables.
19.05: Statistical Literacy
Learning Objectives
• Health Effects of Coffee
This article describes some health effects of drinking coffee. Among the key findings were
1. women who drank four or more cups a day reduced their risk of endometrial cancer by \(25\%\) compared with those who drank less than one cup a day
2. men who drank six or more cups had a \(60\%\) lower risk of developing the most deadly form of prostate cancer than those who drank less than one cup a day
Example \(1\): what do you think?
What is the technical term for the measure of risk reduction reported? What measures of risk reduction cannot be determined from the article? What additional information would have been helpful for assessing risk reduction?
Solution
This is called the "relative risk reduction." The article does not provide information necessary to compute the absolute risk reduction, the odds ratio, or the number needed to treat. It would have been helpful if the article had reported the proportion of women drinking less than one cup a day who developed endometrial cancer as well as the analogous statistic for men and prostate cancer.
19.06: Effect Size (Exercises)
General Questions
Q1
If the probability of a disease is \(0.34\) without treatment and \(0.22\) with treatment then what is the
1. Absolute risk reduction
2. Relative risk reduction
3. Odds ratio
4. Number needed to treat
Q2
When is it meaningful to compute the proportional difference between means?
Q3
The mean for an experimental group is \(12\), the mean for the control group were \(8\), the \(MSE\) from the ANOVA is \(16\), and \(N\), the number of observations is \(20\), compute \(g\) and \(d\).
Q4
Two experiments investigated the same variables but one of the experiment had subject who differed greatly from each other where as the subjects in the other experiment were relatively homogeneous. Which experiment would likely have the larger value of \(g\)?
Q5
Why is \(ω^2\) preferable to \(η^2\)?
Q6
What is the difference between \(η^2\) and partial \(η^2\)?
Questions from Case Studies
The following questions are from the Teacher Ratings case study.
Q7
What are the values of \(d\) and \(g\)?
Q8
What are the values of \(ω^2\) and \(η^2\)?
The following question is from the Smiles and Leniency case study.
Q9
What are the values of \(ω^2\) and \(η^2\)?
The following question is from the Obesity and Bias case study.
Q10
For compute \(ω^2\) and partial \(ω^2\)for the effect of "Weight" in a "Weight x Relatedness" ANOVA. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/19%3A_Effect_Size/19.04%3A_Proportion_of_Variance_Explained.txt |
Learning Objectives
Research conducted by
Emily Zitek and Mindy Ater, Rice University
Emily Zitek
Overview
People have different ways of improving their mood when angry. We have all seen people punch a wall when mad, and indeed, previous research has indicated that some people aggress to improve their mood (Bushman, Baumeister & Phillips, \(2001\)). What do the top athletes do when angry? Striegel (\(1994\)) found that anger often hurts an athlete’s performance and that capability to control anger is what makes good athletes even better. This study adds to the past research and examines the difference in ways to improve an angry mood by gender and sports participation.
The participants were \(78\) Rice University undergraduates, ages \(17\) to \(23\). Of these \(78\) participants, \(48\) were females and \(30\) were males and \(25\) were athletes and \(53\) were non-athletes. People who did not play a varsity or club sport were considered non-athletes. The \(13\) contact sport athletes played soccer, football, rugby, or basketball, and the \(12\) non-contact sport athletes participated in Ultimate Frisbee, baseball, tennis, swimming, volleyball, crew, or dance.
The participants were asked to respond to a questionnaire that asked about what they do to improve their mood when angry or furious. Then they filled out a demographics questionnaire.
Note
This study used the most recent version of the State-Trait Anger Expression Inventory (STAXI-2) (Spielberger, Sydeman, Owen & Marsh, 1999) which was modified to create an Angry Mood Improvement Inventory similar to that created by Bushman et al. (2001).
Questions to Answer
Do athletes and non-athletes deal with anger in the same way? Are there any gender differences? Specifically, are men more likely to believe that aggressive behavior can improve an angry mood?
Design Issues
This study has an extremely unbalanced design. There were a lot more non-athletes than athletes in the sample. In the future, more athletes should be used. This study originally wanted to look at contact and non-contact athletes separately, but there were not enough participants to do this. Future studies could look at this.
Descriptions of Variables
Table \(1\): Description of Variables. Note that the description of the items comes from Spielberger et al. (1999)
Variable Description
Sports 1 = athletes, 2 = non-athletes
Gender 1 = males, 2 = females
Anger-Out (AO) high scores demonstrate that people deal with anger by expressing it in a verbally or physically aggressive fashion
Anger-In
(AI)
high scores demonstrate that people experience anger but do not express it (suppress their anger)
Control-Out (CO) high scores demonstrate that people control the outward expression of angry feelings
Control-In (CI) high scores demonstrate that people control angry feelings by calming down or cooling off
Expression (AE) index of general anger expression:
(Anger-Out) + (Anger-In) - (Control-Out) - (Control-In) + 48
angry_moods.xls
Links
Publisher's description
20.02: Flatulence
Learning Objectives
• Flatulence: Are you embarrassed by your flatus?
Research conducted by
Shannon E. Collins, UH-D undergraduate,
Faculty Advisor: Heidi Ziemer
Case study prepared by
Shannon E. Collins
Overview
The purpose of this study was to find out whether or not people are embarrassed by their flatulence. The participants were \(35\) University of Houston – Downtown students. Flatulence is a normal part of being human, but it can cause an alarming rate of embarrassment in certain situations. How many times have you been subjected to the unpleasant odor emitted from someone around you? Medical research indicates that it is normal to have anywhere from \(7\) to \(20\) episodes of gas in a day.
Would you believe that women produce more of the bad smelling stuff than men do, and that women are more likely to complain to their doctors about the smell of their flatulence? The smell comes from sulfur gasses, the most offensive of which is hydrogen sulfide; it smells like rotten eggs. Still, everybody does it, we just don’t know how embarrassed they are by it.
Questions to Answer
Are people without male siblings more embarrassed by their flatulence than people with one or more male siblings? Do people that come from households where flatulence was acceptable report less embarrassment than people that come from households where it was not acceptable? Are women or men more embarrassed by their flatus?
Design Issues
Embarrassment scores were reported on \(14\) different measures and tallied as a total embarrassment score, then divided into seven categories, producing one number per category. Only the scores on the seven categories are reported here. The data in this research is self report data, and because the topic is sensitive some people may have been less than honest about their reported flatulence.
Descriptions of Variables
Table \(1\): Description of variables
Variable Description
Gender 1 = male, 2 = female
famaccp Household acceptance of flatus
1 to 7, 1=very acceptable and 7=very unacceptable
brother Number of brothers the participant has
howlong How long before farting in front of partner?
1 = 1 year, .5= 6 months, .25=3 months, and smaller decimals represent portions of a year. Numbers larger than 1 indicate longer than 1 year.
perday Number per day
Embarrassing Situations The following variables were rated on this scale: 1=extremely embarrassed and 7= not really embarrassed
mtgwork Meeting at work
talkprof Talking to a professor
romint Romantic interest
flatulence.xls
20.03: Physicians Reactions
Learning Objectives
• Physicians' Reactions to Patient Size
Research conducted by
Mikki Hebl and Jingping Xu
Emily Zitek
Overview
Obese people face discrimination on a daily basis in employment, education, and relationship contexts. Past research has shown that even doctors, who are trained to treat all their patients warmly and have access to literature suggesting uncontrollable and hereditary aspects of obesity, believe obese individuals are undisciplined and suffer from controllability issues. This case study examines how doctors treat overweight as compared to normal weight patients.
Various doctors at one of three major hospitals in the Texas Medical Center of Houston participated in the study. These doctors were sent a packet containing a medical chart similar to the one they view upon seeing a patient. This chart portrayed a patient who was displaying symptoms of a migraine headache but was otherwise healthy. This chart also contained a measure of the patient's weight. Doctors were randomly assigned to receive the chart of a patient who was overweight or the chart of a patient who was of normal weight. After reviewing the chart, the doctors then had to indicate how much time they believed they would spend with the patient.
Questions to Answer
Do doctors discriminate against overweight patients? Specifically, do the doctors who review charts of overweight patients say they would spend the same amount of time with their patients as the doctors who review charts of normal weight patients?
Design Issues
The method and data described here are only a small part of a larger study. See the reference below for a full description of the study.
Descriptions of Variables
Table \(1\): Description of variables
Variable Description
Patient weight 1 = average weight, 2 = overweight
Time represents how long the doctors said they would spend with the patient
Weight.xls | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.01%3A_Angry_Moods.txt |
Learning Objectives
• Teacher Ratings
Research conducted by
Annette Towler and Robert Dipboye
Emily Zitek
Overview
How powerful are rumors? Frequently, students ask friends and/or look at instructor evaluations to decide if a class is worth taking. Kelley (\(1950\)) found that instructor reputation has a profound impact on actual teaching ratings, and Towler and Dipboye (\(1998\)) replicated and extended this study.
Subjects were randomly assigned to one of two conditions. Before viewing the lecture, students were given a summary of the instructors' prior teaching evaluations. There were two conditions: Charismatic instructor and Punitive instructor.
Then all subjects watched the same twenty-minute lecture given by the exact same lecturer. Following the lecture, subjects answered three questions about the leadership qualities of the lecturer. A summary rating score was computed and used as the variable "rating" here.
Questions to Answer
Does an instructor's prior reputation affect student ratings?
Design Issues
The data presented here are part of a larger study. See the references below to learn more.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Condition this represents the content of the description that the students were given about the professor (1 = charismatic, 2 = punitive)
Rating how favorably the subjects rated the professor after hearing the lecture (higher ratings are more favorable)
Ratings.xls
20.05: Diet and Health
Learning Objectives
• Mediterranean Diet and Health
Research conducted by
De Longerill et al
Case study prepared by
David Lane and Emily Zite
Overview
Most doctors would probably agree that a Mediterranean diet, rich in vegetables, fruits, and grains, is healthier than a high-saturated fat diet. Indeed, previous research has found that the diet can lower risk of heart disease. However, there is still considerable uncertainty about whether the Mediterranean diet is superior to a low-fat diet recommended by the American Heart Association. This study is the first to compare these two diets.
The subjects, \(605\) survivors of a heart attack, were randomly assigned follow either
1. a diet close to the "prudent diet step \(1\)" of the American Heart Association (control group) or
2. a Mediterranean-type diet consisting of more bread and cereals, more fresh fruit and vegetables, more grains, more fish, fewer delicatessen foods, less meat.
An experimental canola-oil-based margarine was used instead of butter or cream. The oils recommended for salad and food preparation were canola and olive oils exclusively. Moderate red wine consumption was allowed.
Over a four-year period, patients in the experimental condition were initially seen by the dietician, two months later, and then once a year. Compliance with the dietary intervention was checked by a dietary survey and analysis of plasma fatty acids. Patients in the control group were expected to follow the dietary advice given by their physician.
The researchers collected information on number of deaths from cardiovascular causes e.g., heart attack, strokes, as well as number of nonfatal heart-related episodes. The occurrence of malignant and nonmalignant tumors was also carefully monitored.
Questions to Answer
Is the Mediterranean diet superior to a low-fat diet recommended by the American Heart Association?
Design Issues
The strength of the design is that subjects were randomly assigned to conditions. A possible weakness is that compliance rates depended on reports rather than observation since observation is impractical in this type of research.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Type of diet AHA or Mediterranean
Various outcome measures of health and disease
does the patient have cancer, etc.?
Diet.xls
Links
More on the Mediterranean Diet | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.04%3A_Teacher_Ratings.txt |
Learning Objectives
• To study the research on effects of smiling
Research conducted by
Marianne LaFrance and Marvin Hecht
David Lane
Overview
Dale Carnegie stated that smiling helps win friends and influence people. Research on the effects of smiling has backed this up and shown that a smiling person is judged to be more pleasant, attractive, sincere, sociable, and competent than a non-smiling person.
There is evidence that smiling can attenuate judgments of possible wrongdoing. This phenomenon termed the "smile-leniency effect" was the focus of a study by Marianne LaFrance & Marvin Hecht in 1995.
Questions to Answer
Does smiling increase leniency? Are different types of smiles differentially effective?
Design Issues
There was a single person used for all the conditions. This may limit the generalizeability of the results.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Smile 1 is false smile
2 is felt smile
3 is miserable smile
4 is neutral control
Leniency A measure of how lenient the judgments were.
Leniency.xls
20.07: Animal Research
Learning Objectives
• Gender difference in attitudes toward the use of animals in research
Research conducted by
Nicole Hilliard, Faculty Advisor: Heidi Ziemer
Emily Zitek
Overview
The use of animals in research is a controversial and emotionally charged issue. Personal feelings regarding the use of animals in research vary widely. While many believe that the use of animals in research has been and continues to be essential, others want the practice stopped by cutting off funding or the passing of legislative restrictions. Research on human attitudes toward the use of animals in research has consistently shown systematic differences of opinion with gender differences among the largest.
In this study, a convenience sample of \(34\) University of Houston - Downtown students completed a simple survey that asked their gender and how much they agreed with the following two statements: "The use of animals in research is wrong," and "The use of animals in research is necessary". They rated their agreement with each of these statements on a \(7\)-point scale from strongly disagree (\(1\)) to strongly agree (\(7\)).
Questions to Answer
Is there a gender difference with respect to the belief that animal research is wrong? Is there a gender difference with respect to the belief that animal research is necessary?
Design Issues
This is self-report data. It is possible that the willingness to admit to thinking animal research is wrong or necessary is what differs by gender, not how the participants actually feel.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Gender 1 = female, 2 = male
Wrong high scores indicate that the participant believes that animal research is wrong
Necessary high scores indicate that the participant believes that animal research is necessary
Animals.xls
Links
American Association for the Advancement of Science
20.08: ADHD Treatment
Learning Objectives
Research conducted by
Pearson et al. (2003, see reference below)
Case study prepared by
David Lane and Emily Zitek
Overview
This study investigated the cognitive effects of stimulant medication in children with mental retardation and Attention-Deficit/Hyperactivity Disorder. This case study shows the data for the Delay of Gratification (DOG) task. Children were given various dosages of a drug, methylphenidate (MPH) and then completed this task as part of a larger battery of tests. The order of doses was counterbalanced so that each dose appeared equally often in each position. For example, six children received the lowest dose first, six received it second, etc. The children were on each dose one week before testing.
This task, adapted from the preschool delay task of the Gordon Diagnostic System (Gordon, 1983), measures the ability to suppress or delay impulsive behavioral responses. Children were told that a star would appear on the computer screen if they waited “long enough” to press a response key. If a child responded sooner in less than four seconds after their previous response, they did not earn a star, and the 4-second counter restarted. The DOG differentiates children with and without ADHD of normal intelligence (e.g., Mayes et al., 2001), and is sensitive to MPH treatment in these children (Hall & Kataria, 1992).
Questions to Answer
Does higher dosage lead to higher cognitive performance (measured by the number of correct responses to the DOG task)?
Design Issues
This is a repeated-measures design because each participant performed the task after each dosage.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
d0 Number of correct responses after taking a placebo
d15 Number of correct responses after taking .15 mg/kg of the drug
d30 Number of correct responses after taking .30 mg/kg of the drug
d60 Number of correct responses after taking .60 mg/kg of the drug
ADHD.xls
Methylphenidate
20.09: Weapons and Aggression
Learning Objectives
• Study of the "Weapons" effect
Research conducted by
Anderson, Benjamin, and Bartholow
David Lane
Overview
The "weapons effect" is the finding that the presence of a weapon or even a picture of a weapon can cause people to behave more aggressively. Although once a controversial finding, the weapons effect is now a well-established phenomenon. Based on this, Anderson, Benjamin, and Bartholow (1998) hypothesize that the presence of a weapon-word prime (such as "dagger" or "bullet") should increase the accessibility of an aggressive word (such as "destroy" or "wound"). The accessibility of a word can be measured by the time it takes to name a word presented on computer screen.
The subjects were undergraduate students ranging in age from \(18\) to \(24\) years. They were told that the purpose of this study was to test reading ability of various words. On each of the \(192\) trials, a computer presented a priming stimulus word (either a weapon or non-weapon word) for \(1.25\) seconds, a blank screen for \(0.5\) seconds, and then a target word (aggressive or non-aggressive word). Each subject named both aggressive and non-aggressive words following both weapon and non-weapon "primes." The experimenter instructed the subjects to read the first word to themselves and then to read the second word out loud as quickly as they could. The computer recorded response times and computed mean response times for each participant for each of the four conditions.
Examples of the four types of words
• Weapon word primes: shotgun, grenade
• Non-weapon word primes: rabbit, fish
• Aggressive word: injure, shatter
• Non-aggressive word: consider, relocate
Questions to Answer
Does the mere presence of a weapon increase the accessibility of aggressive thoughts? More specifically, can a person name an aggressive word more quickly if it is preceded by a weapon word prime than if it is preceded by a neutral (non-aggressive) word prime?
Design Issues
This is a within-subjects design, and each participant provided four scores to the analysis.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
gender 1 = female, 2 = male
aw The time in milliseconds (msec) to name aggressive word following a weapon word prime.
an The time in milliseconds (msec) to name aggressive word following a non-weapon word prime.
cw The time in milliseconds (msec) to name a control word following a weapon word prime.
cn The time in milliseconds (msec) to name a control word following a non-weapon word prime.
Guns.xls | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.06%3A_Smiles_and_Leniency.txt |
Learning Objectives
• Predicting college GPA from high school scores
Research conducted by
Thomas W. MacFarland
Emily Zitek
Overview
When deciding whether to admit an applicant, colleges take lots of factors, such as grades, sports, activities, leadership positions, awards, teacher recommendations, and test scores, into consideration. Using SAT scores as a basis of whether to admit a student or not has created some controversy. Among other things, people question whether the SATs are fair and whether they predict college performance.
This study examines the SAT and GPA information of \(105\) students who graduated from a state university with a B.S. in computer science. Using the grades and test scores from high school, can you predict a student's college grades?
Questions to Answer
Can the math and verbal SAT scores be used to predict college GPA? Are the high school and college GPAs related?
Design Issues
The conclusions from this study should not be generalized to students of other majors.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
high_GPA High school grade point average
math_SAT Math SAT score
verb_SAT Verbal SAT score
comp_GPA Computer science grade point average
univ_GPA Overall university grade point average
SAT.xls
Links
Want a job? Hand over your SAT results! Is the SAT a fair test?
20.11: Stereograms
Learning Objectives
• Study to determine the effects of information for an embedded image given ahead of time to a person
Research conducted by
Frisby, J. P. and Clatworthy, J.L.
Case study prepared by
Emily Zitek from DASL story contributed by Michael Friendly
Overview
The rectangles below appear to be composed of random dots. However, if the images are viewed with a stereo viewer, the separate images will fuse and reveal an embedded \(3D\) figure. In this example, fusing the images of these random dot stereograms will reveal a diamond. (Another way for you to fuse the images is to fixate on a point in between them and defocus your eyes. This technique takes practice, but you can try it out with the links below.)
This experiment sought to determine whether giving someone information about the embedded image can help speed up how long it takes to view this image. Seventy-eight participants were given no information, verbal information, and/or visual information (a drawing of the object) about what the embedded image should look like before attempting to fuse the images and actually view the 3D design.
Questions to Answer
Does giving someone information about an embedded image in a stereogram affect the amount of time it takes to see this image? More specifically, does the amount of time it takes to fuse the image in a stereogram differ when the person is given both verbal and visual information about what the image should look like as opposed to when the person is only given verbal information or no information at all?
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Time
Time to produce a fused image of the random dot stereogram
Group Treatment group divided by type of information received:
1 = no information or only verbal information
2 = both verbal and visual information
Fusion.xls
Links
View random dot stereograms. Information about random dot stereograms
20.12: Driving
Learning Objectives
• Driving in inclement weather
Darin Baskin
Emily Zitek
Overview
Many people believe that weather patterns influence driving safety. As a result, there are many web sites and other publications dedicated to giving people tips about how to drive in various weather conditions (see references and links below). Additionally, car accidents are often attributed to bad weather (e.g., see Taylor & Quinn, 1991). This study examines the beliefs and behaviors of people with respect to the important topic of driving in inclement weather.
The participants in this study filled out a questionnaire consisting of some demographic questions and then questions asking about their transportation habits and other beliefs concerning inclement weather. This questionnaire was administered to a convenience sample of \(61\) University of Houston - Downtown students at various locations (i.e., classrooms, hallways, and the food court).
Questions to Answer
Is gender or age related to the likelihood of driving in inclement weather? Does the number of accidents that someone thinks occur during inclement weather relate to how often he or she takes public transportation or chooses to drive during inclement weather?
Design Issues
This is a correlational study, so we cannot infer causation.
Descriptions of Variables
Variable Description
Age The age of the participant in years
Gender 1 = female, 2 = male
Cho2drive How often he or she chooses to drive in inclement weather
1 = always, 3 = sometimes, 5 = never
Pubtran % of travel time spent on public transportation in inclement weather
Accident % of accidents thought to occur from driving in inclement weather
Driving.xls
Links
Driving on Wet Roads. Jokes about Driving in Inclement Weather. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.10%3A_SAT_and_College_GPA.txt |
Learning Objectives
• Strrop Interference demonstration
Statistics Class
David Lane
Overview
Naming the ink color of color words can be difficult. For example, if asked to name the color of the word "blue" is difficult because the answer (red) conflicts with the word "blue." This interference is called "Stroop Interference" after the researcher who first discovered the phenomenon.
This case study is a classroom demonstration. Students in an introductory statistics class were each given three tasks. In the "words" task, students read the names of \(60\) color words written in black ink; in the "color" task, students named the colors of \(60\) rectangles; in the "interference" task, students named the ink color of \(60\) conflicting color words. The times to read the stimuli were recorded. There were \(31\) female and \(16\) male students.
Questions to Answer
Is naming conflicting color names faster or slower than naming color rectangles? Which is faster, naming color rectangles or reading color names? Are there gender differences?
Design Issues
This was not a well-controlled experiment since it was just a classroom demonstration. The order in which the students performed the tasks may not have been counterbalanced or randomized.
Descriptions of Variables
Table \(1\): Description of Variables
Variable
Description
Gender 1 for female, 2 for male
Words Time in seconds to read 60 color words
Colors Time in seconds to name 60 color rectangles
Interfer Time in seconds to name colors of conflicting words
Stroop.xls
Links
Full text of the above reference.
20.14: TV Violence
Learning Objectives
• Does Television Viewing Encourage Aggression in Children?
Research conducted by
Mariana Fernandez, University of Houston-Downtown undergraduate
Nichole Rivera
Overview
How much television is too much for children? Television advocates espouse the educational benefits that children may reap from instructive programming. However, many researchers say that excess television watching may contribute to aggressive behavior in children. Young boys, in particular may be susceptible to this effect. What are the effects, if any, on children’s behavior when television is used as a babysitter?
In a survey of University of Houston-Downtown students, parents reported their children's age, characteristic behavior, and television viewing habits. Convenience sampling was used to gather \(30\) subjects (\(N=30\)).
Questions to Answer
Is there a relationship between hours of television watched and child's obedience? Will a child be more or less aggressive if he/she watches a lot of television?
Design Issues
This survey offered a very limited sample (\(N=30\)), which was further hindered by reporting participants’ filling out an individual survey for each individual child. This contributes to some lack of true variability in responses because participants tended to report similar behavior for each child. This may magnify errors associated with self-reported data. The sample would provide greater reliability if each participant reported on only one child’s behavior.
The survey has broad questions which do not provide much context for reported behaviors. In some instances aggression may be positively rated, but this survey treats all aggression as a negative characteristic. In addition, the instrument itself measures largely nominal data, making in depth analysis difficult.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
TV hours Total number of TV hours watched per day
Obedience How obedient the child is
1 = very obedient, 5 = not obedient
Attitude Attitude while playing with other children
1 = non-aggressive, 5 = very aggressive
TV.xls
Links
TV Guide - Mighty Morphin' Power Rangers v. Teenage Mutant Ninja Turtles | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.13%3A_Stroop_Interference.txt |
Learning Objectives
• Bias Against Associates of the Obese
Research conducted by
Mikki Hebl and Laura Mannix
Emily Zitek
Overview
Obesity is a major stigma in our society. People who are obese face a great deal of prejudice and discrimination. For example, Roehling (1999) showed that obese people experience a lot of discrimination in the workplace (e.g., they are less likely to be hired and get lower wages). We know that people who are obese are stigmatized, but what about people who are somehow associated with an obese person? Neuberg et al. (1994) found that friends of gay men and lesbians suffer from "stigma by association". Perhaps the negative effects of the obesity stigma can also spread to other people. This study seeks to examine how the stigma of obesity can spread to a job applicant of average weight.
As part of a larger study, participants had to rate how qualified a particular job applicant was. This applicant was sitting by a woman. The researchers manipulated the following two variables: the weight of the woman and the relationship between the woman and the applicant. The woman was either obese or of average weight. This woman was also portrayed as being the applicant's girlfriend or a woman simply waiting to participate in a different experiment.
Questions to Answer
Are male applicants who are seated next to an obese woman rated as less qualified for a job? Are applicants who are seated next to their girlfriend rated differently from applicants seated next to a woman with whom they do not have an intimate relationship? Finally, does the effect of the type of relationship differ depending on the weight of the woman?
Design Issues
This study only looked how at how an obese woman seated next to a male job applicant could affect qualification ratings. Future research could address other gender combinations.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Weight The weight of the woman sitting next to the job applicant
1 = obese, 2 = average weight
Relate Type of relationship between the job application and the woman seated next to him
1 = girlfriend, 2 = acquaintance (waiting for another experiment)
Qualified Larger numbers represent higher professional qualification ratings
Weight2.xls
Links
The Obesity Society
20.16: Shaking and Stirring Martinis
Learning Objectives
• To test the difference between shaken and stirred martinis
Research conducted by
This is just made up data.
David Lane
Overview
This is an example to illustrate hypothesis testing and the binomial distribution. The statistician R. Fisher explained the concept of hypothesis testing with a story of a lady tasting tea. Here is an example based on James Bond who insisted that Martinis should be shaken rather than stirred. In this hypothetical experiment to determine whether Mr. Bond could tell the difference between a shaken and a stirred martini, we gave Mr. Bond a series of \(16\) taste tests. In each test, we flipped a fair coin to determine whether to stir or shake the martini. Then we presented the martini to Mr. Bond and asked him to decide whether it was shaken or stirred. Mr. Bond was correct on \(13/16\) trials.
Questions to Answer
Does Mr. Bond have the ability to tell the difference between a Martini that is shaken and one that is stirred?
Design Issues
This is only a made-up study.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Y 0 = incorrect, 1 = correct
Martini.xls
Links
The Lady Tasting Tea
20.17: Adolescent Lifestyle Choices
Learning Objectives
Research conducted by
Ka He, Ellen Kramer, Robert F. Houser, Virginia R. Chomitz, and Karen A. Hacker
Case study prepared by
Robert F. Houser, Alyssa Koomas, and Georgette Baghdady
Overview
Teen pregnancy, sexually transmitted disease, drug abuse, and suicide are some of the behaviorally-mediated negative health outcomes that can occur during adolescence. Identifying the characteristics of adolescents who are able to make healthy lifestyle choices is imperative toward understanding positive health behaviors in this age group. This information could be used to develop targeted interventions that support at-risk adolescents in making healthy lifestyle choices, and hopefully prevent such negative outcomes.
This study collected survey data from \(1487\) high school students in an urban Massachusetts community. The survey assessed health-related behaviors, stressful events, demographics, familial characteristics, perceptions of peer and parental support, and academic performance. In collaboration with community stakeholders and parents, the researchers selected six health-related behaviors and developed two sets of criteria to define positive health behaviors. One set used “strict” definitions, namely, not drinking alcohol in the last \(30\) days, no attempted suicide in the past \(12\) months, and no experience at all with tobacco, hard illegal drugs, marijuana, and sexual partners. The second set used “broad” definitions that allowed for mild use and safe experimentation (except for suicidal behavior). Students who adhered to all six health-related behaviors according to the “strict” definitions formed one subgroup for analysis, and those who reported behaviors in accordance with the “broad” definitions formed another subgroup. These two lifestyle subgroups were analyzed separately in relation to the personal and social-environmental factors assessed by the survey.
Questions to Answer
What personal and social-environmental characteristics are associated with adolescents who practice healthy lifestyle behaviors according to the “strict” definitions? How much more likely are adolescents with these characteristics to be practicing healthy behaviors than adolescents without these characteristics?
Design Issues
The results of this study may not be applicable to adolescents in non-urban schools, as the sample was drawn from a diverse, urban school. As well, the definitions that make up positive health behaviors may vary by region and social group. Adolescents self-reported their health-related behaviors and other information via the survey. Missing responses may have caused bias in the results.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Healthy behaviors based on the “strict” definitions
Whether or not the adolescent practices all 6 health-related behaviors according to the “strict” definitions
Immigration status
Whether the adolescent was born in the US or is an immigrant
Stress score
An index from 0 to 14 assessing 14 possible stressful events in the adolescent’s life, such as failing grades, moving, death in the family, divorce in the family, abuse, and violence
Stress index
Whether the adolescent’s stress score is at or above the median stress score of 2, or below
Academic performance
The adolescent’s average academic letter grade (A, B, C, D, F)
Links
1 in 3 Teens Text While Driving
2011 Youth Risk Behavior Surveillance Survey | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.15%3A_Obesity_and_Bias.txt |
Learning Objectives
• To study chocolate’s healthful metabolic mechanisms
Research conducted by
Beatrice A. Golomb, Sabrina Koperski, and Halbert L. White
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
Recent research has brought to light the beneficial health effects of chocolate. Studies have linked chocolate with lower blood pressure, lower bad cholesterol, improved insulin sensitivity, and reductions in the risks of diabetes, heart disease, and stroke. The authors of this study hypothesized that chocolate’s healthful metabolic mechanisms might also reduce fat deposition in spite of its high caloric content.
This study used the baseline data from a clinical study that examined noncardiac effects of cholesterol-lowering drugs in healthy adults. The baseline data included body mass index (BMI), chocolate consumption frequency, age, sex, physical activity frequency, depression, and some dietary variables. Chocolate consumption frequency was assessed with the question: “How many times a week do you consume chocolate?” Dietary intakes of total calories, fruits and vegetables, and saturated fat were assessed with a validated food frequency questionnaire. A food frequency questionnaire is a limited checklist of foods and beverages with a frequency response section for subjects to report how often each item was consumed over a specified period of time. Depression was measured with a validated scale related to mood. BMI is a measure of body fatness that is associated with many adverse health conditions.
Questions to Answer
What can we conclude from the researchers’ findings that there is an association between consuming chocolate frequently and lower BMI? How do we interpret regression models?
Design Issues
The authors used baseline data from an unrelated clinical study examining noncardiac effects of cholesterol-lowering drugs. That clinical study included men ranging in age from \(20\) to \(85\) years, but only postmenopausal women. The results of the chocolate study cannot, therefore, be generalized to younger adult women. Except for BMI, the data for all of the study variables were “self-reported” by the subjects via questionnaires. The assessment of critical variables, such as chocolate consumption frequency and vigorous physical activity frequency, could differ when using different measurement tools. The study was cross-sectional in nature, precluding conclusions about causation.
Descriptions of Variables
Table \(1\): Description of Variables
VARIABLE
DESCRIPTION
BMI
Body mass index, calculated as: (weight in kilograms) / (height in meters)2
Chocolate consumption frequency
Number of times per week a subject consumed chocolate
Calories
Overall calorie intake of a subject determined via food frequency questionnaire
Age
Range of 20 to 85 years, postmenopausal if female
Sex
68% male, 32% female
Activity
Number of times per 7-day period a subject engaged in vigorous physical activity for at least 20 minutes
Links
Golomb et al. article
Rose et al. article
What is body mass index (BMI)?
20.19: Bedroom TV and Hispanic Children
Learning Objectives
• Study of overweight and obesity in Hispanic children
Research conducted by
Du Feng, Debra B. Reed, M. Christina Esperat, and Mitsue Uchida
Case study prepared by
Robert F. Houser, Alyssa Koomas, and Georgette Baghdady
Overview
The prevalence of overweight and obesity in children in the U.S. is a growing public health concern that disproportionately affects Hispanic youth. As noted by the authors, in \(2005\) to \(2006\), \(15.5\%\) of all U.S. children aged \(2\) to \(19\) years were overweight or obese, compared with \(23.2\%\) for boys and \(18.5\%\) for girls among Mexican-Americans in this age group. Past research has revealed diverse environmental and behavioral factors that may contribute to this disparity. For example, studies have shown that Hispanic children watch more television than white children.
This study examined TV viewing among \(314\) Hispanic children aged \(5\) to \(9\) years in West Texas and the possible effects of having a TV in the child’s bedroom. Children’s weights and heights were measured, body mass indexes (BMI) calculated, and sex- and age-adjusted BMI percentiles obtained. The \(2000\) CDC Growth Charts were used to assess whether or not a child was overweight or at risk for becoming overweight. Their parents completed a family survey assessing demographics, acculturation, parental support of physical activity, dietary practices, the presence of a TV in the participating child’s bedroom, and the child’s TV/DVD viewing time.
Questions to Answer
Do children with a TV in their bedroom spend more time watching TV/DVDs on a daily basis than children without a TV in their bedroom? Do children with a TV in their bedroom have less support from their parents for physical activity than children without a TV in their bedroom? What might account for missing responses to survey questions?
Design Issues
Except for BMI, the data for all of the study variables were “self-reported” by the parents. The study used a cross-sectional design, which cannot be relied upon to provide conclusive evidence of causal relationships.
Descriptions of Variables
Table \(1\): Description of Variables
VARIABLE DESCRIPTION
TVIB, No TVIB Presence or absence of a TV in the participating child’s bedroom
Daily TV/DVD time Average number of hours the child spent watching TV and DVDs per day
Parental support of physical activity Scale score calculated as the average of parent’s responses to 8 survey items assessing the parent’s support of physical activity for the child. Items rated on 4-point Likert scale (0 = never, 3 = always). Research has shown a significant positive relationship between parental support of physical activity and children’s physical activity level
Daily fruit and vegetable intake Average number of cups of fruits and vegetables (fresh, frozen, dried, canned, and 100% juice) consumed by the child per day
Daily sweetened beverages Average number of ounces of soda, fruit drink, sports drink, tea, and lemonade consumed by the child per day
Links
New York Times article
Television and Children information guide | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.18%3A_Chocolate_and_Body_Weight.txt |
Learning Objectives
• Excess Body Weight and Sleep Apnea
Research conducted by
Kari Johansson, Erik Hemmingsson, Richard Harlid, Ylva Trolle Lagerros, Fredrik Granath, Stephan Rössner, and Martin Neovius
Philip Sedgwick
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
In his statistical article, “Standard deviation versus standard error,” UK researcher Philip Sedgwick presents us with an interesting discussion of the proper use of standard deviation (SD) and standard error of the mean (SEM). He uses an example of a weight loss study of \(63\) obese men suffering from obstructive sleep apnea who were being treated with continuous positive airway pressure (CPAP). The weight loss program lasted one year. Outcome measures included change in body weight measured in kilograms (kg).
More than \(60\%\) of people experiencing obstructive sleep apnea are obese. CPAP therapy is the most common treatment. It uses a machine and mask to prevent the airway from collapsing, thus enabling a person to breathe more easily during sleep. Weight loss is an effective treatment for sleep apnea.
Questions to Answer
What is the proper use of the SD? What is the proper use of the SEM?
Design Issues
None for the Sedgwick article.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Weight Body weight at baseline in kg
Weight change Change in body weight at one year from baseline in kg
Links
What Is Sleep Apnea?
t Table (two-tailed) for significance and calculation of confidence interval
Johansson et al. article
20.21: Misusing SEM
Learning Objectives
• Misusing Standard Error of the Mean ($SEM$)
Peter Nagele
Case study prepared by
Robert F. Houser, Georgette Baghdady, and Jennifer E. Konick
Overview
Authors of published research articles often erroneously use the standard error of the mean to describe the variability of their study sample. Nagele demonstrated this misuse of the standard error of the mean as a descriptive statistic by manually searching four leading anesthesia journals in $2001$.
Here are quotes on key points from Nagele’s article and our notes:
“Descriptive statistics aim to describe a given study sample without regard to the entire population.”
“If normally distributed, the study sample can be described entirely by two parameters: the mean and the standard deviation ($SD$).” However, a study sample variable is never exactly normally distributed. When a variable is close to normally distributed, the mean and median are quite similar. Therefore, the mean and $SD$ would be sufficient.
“The $SD$ represents the variability within the sample.” It tells us about “the distribution of individual data points around the mean.” The latter statement, however, is a generalization since the $SD$ cannot tell us exactly where each data point lies relative to the mean.
“[I]nferential statistics generalize about a population on the basis of data from a sample of this population.”
The standard error of the mean ($SEM$) “is used in inferential statistics to give an estimate of how the mean of the sample is related to the mean of the underlying population.” It “informs us how precise our estimate of the [population] mean is.”
Thus, “the $SEM$ estimates the precision and uncertainty [with which] the study sample represents the underlying population.”
The standard error of the mean is calculated by dividing the sample standard deviation by the square root of the sample size ($SEM=SD/\sqrt{n}$).
“[T]he $SEM$ is always smaller than the $SD$.” However, this is only true as long as the sample size is greater than $1$.
“In general, the use of the $SEM$ should be limited to inferential statistics [for which] the author explicitly wants to inform the reader about the precision of the study, and how well the sample truly represents the entire population [of interest].” A sample never truly represents the population.
Questions to Answer
How prevalent is the inappropriate use of the $SEM$ in describing the variability of the study sample in research publications? What is the proper use of the $SEM$?
Design Issues
The author focused on four leading anesthesia journals in his field of expertise. The misapplication of the $SEM$ in descriptive statistics can be found in professional journals of many, if not all, fields of research.
Descriptions of Variables
Table $1$: Description of Variables
Variable Description
Incorrect use of SEM; total Total frequency of misuse of SEM; expressed as number of articles and percent
Laboratory studies using SEM incorrectly A subset of the above variable; expressed as number of articles and percent
Correct use of SD Frequency of correct use of standard deviation; expressed as number of articles and percent
Sem.xls
Nagele article | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.20%3A_Weight_and_Sleep_Apnea.txt |
Learning Objectives
• School garden program benefits
Research conducted by
Michelle M. Ratcliffe, Kathleen A. Merrigan, Beatrice L. Rogers, and Jeanne P. Goldberg
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
School garden programs are gaining popularity because of their numerous benefits for children: outdoor exercise, social skills, connecting with nature, environmental stewardship, active learning, experiential science education, higher academic achievement, and transformed attitudes and habits related to fruits and vegetables. By integrating the regular science class with gardening activities in which students plant, nurture, harvest, prepare, and consume produce grown in the schoolyard, studies are showing that garden-based learning can improve children’s consumption of fruits and vegetables.
This study investigated the impact of participating in a school garden program on the ability to identify, willingness to taste, preference for, and consumption of vegetables. Subjects were \(320\) sixth-grade students aged \(11\) to \(13\) years at two intervention schools and one control school. At the intervention schools, garden-based learning activities were incorporated into the regular science class for a period of four months. The control school did not include a garden program as part of its science class. Two questionnaires – Garden Vegetable Frequency Questionnaire and taste test – assessed the outcome variables using vegetables typically grown in school gardens that were also ethnically and culturally appropriate for the study population. The Garden Vegetable Frequency Questionnaire assessed the types of vegetables consumed the day before as well as usual consumption frequency. The taste test involved tasting five raw vegetables (carrots, string beans, snow peas, broccoli, and Swiss chard). Both questionnaires were administered at the outset and end of the study. Change scores (posttest minus pretest) were compared between the garden (intervention) group and the control group.
Questions to Answer
Do hands-on school garden programs increase vegetable consumption in children? What are some of the potential sources of bias in research studies?
Design Issues
This study used a “quasi-experimental” design, which differs from an experiment in that the students were selected and assigned to the intervention group and control group by a method other than random assignment. With this type of design, there is a greater chance that the intervention and control groups might differ at the outset of the study in ways that could bias the results of the study. Since the study population was middle-school students living in low-income, urban communities, the results of the study cannot be generalized to other settings. The study did not measure the actual amounts of vegetables that students consumed, so no conclusions can be drawn about number or size of servings.
Descriptions of Variables
Tables \(1\): Description of Variables
Variable Description
School garden program group Garden (intervention) and Control groups: Whether or not a student experiences hands-on gardening activities at school
Consumption of vegetables at school Assessed by the taste test, it measures whether or not a student ate each of five specific vegetables at school
Consumption of vegetables at home Assessed by the taste test, it measures whether or not a student also ate each of the five specific vegetables at home
Links
Ratcliffe et al. article
The benefits of school gardens
First Lady Michelle Obama Hosts White House Garden Spring 2011 Planting | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.22%3A_School_Gardens_and_Vegetable_Consumption.txt |
Learning Objectives
• TV viewing time and adverse health
Research conducted by
Perrie E. Pardee, Gregory J. Norman, Robert H. Lustig, Daniel Preud’homme, and Jeffrey B. Schwimmer
Case study prepared by
Robert F. Houser and Andrew Kennedy
Overview
A strong, evidence-based association exists between TV viewing time and the risk of being obese in children and adolescents. Little or no research, however, has explored adverse health outcomes associated with TV viewing among obese children. This study aimed at identifying whether or not time spent watching TV is associated with hypertension (high blood pressure) in obese children.
Obese children aged \(4\) to \(17\) years were recruited and evaluated at three pediatric centers. Obesity was defined as a body mass index (BMI) greater than or equal to the \(95^{th}\) percentile for the child’s age and gender.
Questions to Answer
Is TV watching associated with hypertension in obese children?
Design Issues
The study involved a cross-sectional design, which prevented the determination of possible causality among the associations found. There could be unmeasured factors that play a role in the association between TV viewing and hypertension.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Hypertension Defined as a systolic and/or diastolic blood pressure greater than or equal to the 95th percentile for the child’s age, gender, and height
Age A child’s age in years
BMI A child's body mass index, calculated as: (weight in kilograms) / (height in meters)2
Hours of TV/day An estimate of a child’s average daily time spent watching TV in hours
Links
Pardee et al. article
Luma et al. article
20.24: Dietary Supplements
Learning Objectives
• Dietary supplements and health risk behaviors
Research conducted by
Wen-Bin Chiou, Chao-Chin Yang, and Chin-Sheng Wan
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
Although the dietary-supplement market in the U.S. is enormous, there is no apparent association between the use of dietary supplements and improved public health. The researchers of this study explored this paradox under the hypothesis that taking dietary supplements triggers a phenomenon called the “licensing effect,” namely, the tendency for positive choices to license subsequent self-indulgent, risky or unhealthful choices. The researchers hypothesized that supplement use confers “perceived health credentials,” leading people to feel invulnerable to health hazards and thus more likely to engage in risky, health-related behaviors.
The study involved two experiments. In the first experiment, $82$ participants were randomly assigned to either a vitamin-pill (multivitamin) group or control (placebo) group and were told the kind of pill they would be taking. However, only the control group was given correct information. In actuality, both groups received the placebo pill. After taking the pills, the participants completed a survey on leisure-time activities, rating the desirability of nine hedonic (pleasurable) activities, such as excessive drinking and wild parties, and nine exercise activities, such as yoga and running, on $7$-point scales. The survey also included a general invulnerability scale to assess a participant’s perceived invulnerability to harm and disease. After completing the survey, the participants were offered a free lunch, choosing freely between a buffet and a healthful, organic meal.
The second experiment involved different participants. The vitamin-pill (multivitamin) group again unknowingly took placebo pills. After completing a questionnaire that included the general invulnerability scale and reading a medical report on the health benefits of walking, the distance participants walked in one hour was measured with a pedometer.
Questions to Answer
Does taking dietary supplements disinhibit unhealthy behaviors, such as eating unhealthful meals? Is the study sufficiently powered to detect significant differences between males and females?
Design Issues
The research was conducted in Taiwan, where cultural attitudes and behaviors related to dietary supplements may differ from those in the U.S. It is possible that the results might not generalize to other countries, so more research is needed. Participants in $\text{Experiment 1}$ had a wide range in age, from $18$ to $46$ years, with a mean ($SD$) of $30.9$ ($7.8$) years. It would be helpful to consider age in the analysis, especially if age is associated with invulnerability scores. Leisure-time activities and invulnerability were assessed only post-intervention; future studies should also measure these variables before the intervention to see if the two groups had similar scores at the start of the study. The general invulnerability scale used to assess perceived invulnerability to harm and disease has been validated only for adolescents.
Descriptions of Variables
Table $1$: Description of Variables
VARIABLE DESCRIPTION
Experimental condition Vitamin-pill (multivitamin) condition or Control (placebo) condition
Meal choice Either a buffet meal or a healthful, organic meal
Gender The sex of participants
Links
The licensing effect
No Significant Difference … Says Who?
20.25: Young People and Binge Drinking
Learning Objectives
• Binge drinking and serious public health problems
Research conducted by
Richard O. de Visser and Julian D. Birch
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
Binge drinking is a serious public health problem bringing harm to both the individual and society. It compromises a person's health, increasing the risk of many diseases, injury, and death. It also results in a greater incidence of motor vehicle crashes, violence, the spread of sexually-transmitted diseases, and unintended pregnancies. Binge drinking is prevalent among both young and older adults, men and women, and high and low income levels. Governments have formulated guidelines for moderate or sensible drinking levels. The government of the United Kingdom (UK) issued guidelines for sensible drinking as \(2-3\) alcohol units per day for women and \(3-4\) units per day for men, an alcohol unit being \(10\) milliliters of ethanol. A binge drinking episode is when a person drinks above double the recommended daily guidelines in a short period of time.
Questions to Answer
What can we learn about the binge drinking patterns of university students in England? Do the bingers and non-bingers differ in their knowledge of the sensible drinking guidelines issued by the UK government?
Design Issues
The university students in the sample "self-selected" to participate in the study by responding to recruiting efforts made via email messages and requests in lectures.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Sex Female or male
mo_binge_n Number of times the university students did binge drinking in the last month (using sex-specific definitions)
modrunk Number of times the university students drank in the last month
wk_unit_prop Familiarity with alcohol unit-based guidelines (measured on a 5-point scale)
k_unit_sum Knowledge of alcohol unit-based guidelines (score out of 7)
u_fam Familiarity with alcohol unit-based guidelines (measured on a 5-point scale)
Binge.xls
Links
de Visser et al. article | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.23%3A_TV_and_Hypertension.txt |
Learning Objectives
• Sugar Consumption in the US Diet between 1822 and 2005
Research conducted by
Stephan Guyenet and Jeremy Landen
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
Sugar has many forms: cane sugar, beet sugar, honey, molasses, fruit juice concentrate, glucose, sucrose, fructose, high-fructose corn syrup, maple syrup, brown rice syrup, barley malt syrup, agave nectar, to list a few. High-fructose corn syrup, in particular, was introduced into the US food industry in the early \(1970s\) and has become ubiquitous in processed foods and soft drinks. Many of the added sugars in packaged foods and beverages could be considered "hidden sugar" because, if we do not examine the ingredients list on food labels or know sugar's many aliases, we are most likely unaware of how much sugar we consume each day.
To explore sugar consumption trends in the US, researchers Stephan Guyenet and Jeremy Landen compiled data on caloric sweetener sales spanning \(184\) years. They extracted annual caloric sweetener sales per capita for \(1822\) to \(1908\) from US Department of Commerce and Labor reports, and for \(1909\) to \(2005\) from the US Department of Agriculture (USDA) web site. The researchers adjusted the sales data for post-production losses using the USDA's \(1970-2005\) loss estimate of \(28.8\) percent to obtain reasonable estimates of annual per capita consumption of added sugars. Post-production losses of a food commodity occur at the retail, foodservice and consumer levels from, for example, spoilage, pests, cooking losses and plate waste.
Guyenet presents a striking graph and regression analysis of sugar consumption in the US from \(1822\) to \(2005\) in a blog to promote awareness and discussion.
Questions to Answer
Do different time periods between \(1822\) and \(2005\) reveal different trends in sugar consumption in the US diet? Can a regression graph be used to make predictions outside the range of the study data?
Design Issues
The data represent added caloric sugars such as cane sugar, high-fructose corn syrup and maple syrup, not naturally occurring sugars such as those in fruits and vegetables. Thus the data do not represent total sugar consumption. The data are not direct measures of consumption, but rather estimates derived from sales figures by adjusting for losses before consumption. The adjustment, applied across all years, is based on the USDA loss estimate from \(1970-2005\), which may or may not underestimate sugar consumption in earlier time periods.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
year All years from 1822 to 2005
sugar_consum Estimated consumption of added sugars in the US diet in pounds per year per person
Sugar.xls
Links
By 2606, the US Diet will be 100 Percent Sugar, a blog by Stephan Guyenet
How to Spot Added Sugar on Food Labels
Dietary Sugars Intake and Cardiovascular Health: A Scientific Statement From the American Heart Association
Sugar: The Bitter Truth, a lecture by Robert H. Lustig
60 Minutes: Is Sugar Toxic?
20.27: Nutrition Information Sources and Older Adults
Learning Objectives
• Better educated people and information sources
Research conducted by
Diane L. McKay, Robert F. Houser, Jeffrey B. Blumberg, and Jeanne P. Goldberg
Case study prepared by
Robert F. Houser, Alyssa Koomas, Georgette Baghdady, and Jennifer E. Konick
Overview
Various socioeconomic factors, such as occupation, income, race, and education level, are associated with health outcomes. Prominent among them, education level has proved to be a strong predictor of diet quality, health behavior patterns, and disease risk. Studies have found that better-educated people have healthier diets than those with less education, leading some researchers to hypothesize that better-educated people may obtain nutrition information from more reliable sources than less-educated people.
This study examined that hypothesis among a sample of \(176\) adults aged \(50\) years or older. The participants completed a survey which asked whether or not they primarily relied upon each of the following sources for information about nutrition: doctors, other medical professionals, newspapers, magazines, television, radio, friends, relatives, and neighbors. Analysis involved comparing the sources by education level. Older adults are highly vulnerable to diet-related disease. Knowing which sources they rely on can enable nutrition educators and professionals to target those sources with high-quality nutrition messages, tailored to the needs and education level of the older-adult audience.
Questions to Answer
What sources of nutrition information do older adults rely on? Do these sources differ according to the educational attainment and gender of the adults? Are these sources of nutrition information related to dietary practices, such as taking supplements?
Design Issues
Given that the sample was drawn only from the New England area and that \(93\%\) were Caucasian, the results of this study should not be generalized to older adults in other regions or racial and ethnic groups. The Internet as a source of nutrition information was not included in the survey; it is likely a primary source among today's older adults.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
coll4yrplus Highest level of education completed:
0 = "< 4 years of college" (i.e., secondary school, high school, vocational school, community or junior college)
1 = "≥ 4 years of college" (i.e., four-year college, graduate or professional school)
gender 1 = female, 2 = male
doctor Is your doctor a primary source of information about nutrition?
1 = yes, 2 = no
magazine Are magazines a primary source of information about nutrition?
1 = yes, 2 = no
tv Is TV a primary source of information about nutrition?
1 = yes, 2 = no
friends Are friends a primary source of information about nutrition?
1 = yes, 2 = no
supps Are you taking any dietary supplements?
1 = yes, 2 = no
Data Files
Nutrition_information.xls
Links
Nutrition Information For You
Evaluating Nutrition Information (see pages 36-43)
Nutrition Accuracy in Popular Magazines | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.26%3A_Sugar_Consumption_in_the_US_Diet.txt |
Learning Objectives
• The "placebo" effect
Research conducted by
Alia J. Crum and Ellen J. Langer
Case study prepared by
Robert F. Houser and Alyssa Koomas
Overview
The "placebo effect" is an effect that cannot be attributed to a drug or remedy, but rather to a change in a person's mind-set or perception. The placebo effect is widely accepted in clinical trials and its effects may shock you. For instance, one study found that subjects developed real rashes after being exposed to fake poison ivy (Blakeslee, \(1998\))! This study examined the placebo effect with relation to physical activity and health. Could becoming aware of how much you exercise result in weight loss even if you didn't make any changes to your diet or exercise routine?
The subjects were \(84\) female maids of ages \(19\) to \(65\) years at seven hotels. They were told the purpose of the study was to improve the health and happiness of hotel maids. According to the authors, "[e]ach of seven hotels was randomly assigned to one of two conditions: informed or control" (page \(166\)). "Four hotels were assigned to the informed condition, and three were assigned to the control condition" (pages \(166-167\)). Each subject filled out a questionnaire asking about her perceived amount of exercise during and outside of work. Physiological measurements were taken for weight, body mass index, body-fat percentage, waist-to-hip ratio, and blood pressure. The maids in the informed condition were then given an oral presentation and handouts explaining how their work as hotel maids is good exercise, so good in fact that it meets or exceeds the Surgeon General's recommendations for physical activity. The maids in the control condition were not given this information. After four weeks, the researchers re-administered the questionnaire and took follow-up physiological measurements.
Questions to Answer
Does the placebo effect play a role in the health benefits of exercise? If we alter a person's perception of the exercise she performs, does it result in weight loss?
Design Issues
Instead of assigning individual maids randomly to either the informed or control condition, all of the maids in the same hotel were assigned to the same condition. This was done in an effort to prevent information contamination. This type of study design is known as a "cluster randomized trial," and calls for advanced statistical practices that we will not worry about in this case study.
Simple random sampling with a sufficient number of subjects randomly assigned to intervention and control groups ideally leads to intervention and control groups that are similar with respect to many demographic characteristics. Simple random sampling of individuals and random assignment of individuals to conditions were not used in this study. The authors of this study pointed out that "[s]ubjects in the informed group were significantly younger than subjects in the control group." Consequently, they attempted to control for age differences in their statistical analysis.
The questionnaire asked about self-reported levels of exercise and dietary intake. Future research should use more rigorous methods to assess physical activity and diet.
Descriptions of Variables
Table \(1\): Description of Variables
VARIABLE DESCRIPTION
cond Condition: Either Informed or Control
age Age in years
ex1 Perceived amount of exercise at Time 1 (On a scale from 0 to 10 with 0 = “none” and 10 = “a great deal”)
ex2 Perceived amount of exercise at Time 2 (On a scale from 0 to 10 with 0 = “none” and 10 = “a great deal”)
wt1 Weight in pounds at Time 1
wt2 Weight in pounds at Time 2
aex Change score for exercise equal to the perceived amount of exercise at Time 2 minus the perceived amount of exercise at Time 1
awt Weight change equal to the weight at Time 2 minus the weight at Time 1
Mindset.xls
Links
Crum et al. article
New York Times article | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.28%3A_Mind_Set_-_Exercise_and_the_Placebo_Effect.txt |
Learning Objectives
• To explore the phenomenon of future anhedonia
Research conducted by
Karim S. Kassam, Daniel T. Gilbert, Andrew Boston, and Timothy D. Wilson
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
In Aesop's fable, "The Ant and the Grasshopper," an ant toils all summer to gather food for the winter while a grasshopper sunbathes and enjoys the present abundance of food without concern for the upcoming winter. Consequently when winter arrives, the grasshopper despairs that it has no food. The moral of the fable is that it is best to prepare for the days of necessity. Clearly the grasshopper failed to predict accurately how it would feel in the winter while it sunbathed with a full belly in the summer.
The authors of this study explored the intriguing phenomenon of future anhedonia and its relation to the concept of time discounting in order to understand people's predictions about how they might feel when a future event happens. Time discounting occurs when people put less value on future events than present events. Future anhedonia refers to people's mistaken belief that a future event would elicit a less intense affective reaction than if the same event happened in the present. In six experiments, the authors asked participants to predict how happy they would feel both in the present and in the future upon receiving either \(20\) dollars outright or \(25\) dollars in the form of a Starbucks coffeehouse gift card. The difference between the scores of present and future happiness is a measure of future anhedonia.
Questions to Answer
Do people expect their affective reactions to an event to be less intense in the future than in the present?
Design Issues
The monetary amount (\(\$20, \$25\)) may not have been enough to psychologically engage a large number of participants. The wide age range in Experiment 1b of \(15\) to \(72\) years is unusual for a psychological study. Also in Experiment 1b, several participants reported that they would pay \(\$25\) for a \(\$25\)-gift card that Starbucks was considering selling at a discounted price, which might indicate that they did not fully understand the question.
Descriptions of Variables
Table \(1\): Description of Variables
VARIABLE DESCRIPTION
Gender The sex of a participant
Happiness score A participant’s estimate of his/her affective reaction to an event using a 9-point scale with endpoints 1 = “not at all happy” and 9 = “extremely happy”
diff_happy A difference score equal to a participant’s predicted present happiness score for a present event minus his/her predicted future happiness score for the same event in the future. A positive difference indicates future anhedonia.
diff_money The difference in the maximum amount of money that a participant predicted as his/her willingness to pay in the present minus the predicted amount he/she would pay at a future time for a \$25 Starbucks coffeehouse gift card. A positive difference indicates future anhedonia.
cond Condition: Whether an event was expected or unexpected
today A participant’s predicted present happiness score for a present event
future A participant’s predicted future happiness score for a future event
Predicting.xls
Links
Kassam et al. article
Aesop's Fable: The Ant and the Grasshopper
Prospection: Experiencing the Future
20.30: Exercise and Memory
Learning Objectives
• To study the benefits of exercise on memory
Research conducted by
M. E. Hopkins, F. C. Davis, M. R. Van Tieghem, P. J. Whalen, and D. J. Bucci
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
Physical exercise has many beneficial effects on physiological processes, including those that affect cognition and memory. Exercise increases brain-derived neurotrophic factor (BDNF), which is a protein found in the learning and memory centers of the brain where it supports nerve cell survival and the growth of new neurons and neuronal connections. A polymorphism of BDNF (a variant genotype) alters the release of BDNF during exercise. The researchers of this study sought to compare the effects of a single bout of exercise versus a \(4\)-week exercise regimen on cognition and memory and to determine if BDNF genotype influences the intensity of those effects of exercise.
Questions to Answer
How do regular exercise and/or an acute bout of exercise affect cognitive memory? Does type of BDNF genotype (Val/Val or Met carrier) mediate the effect of exercise on memory? How do we calculate a one-way ANOVA by hand and how do different post-hoc tests compare?
Design Issues
The group sample sizes are small, perhaps limiting the power to detect significant differences between the four exercise/control groups.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Group 0W-: sedentary group
0W+: sedentary group with one bout of exercise at least 2 hours before Visit 2
4W-: regularly exercising group
4W+: regularly exercising group with a bout of exercise at least 2 hours before Visit 2
Accuracy
The percentage of objects each group accurately identified as old or new when performing the novel object recognition task during each study visit
Difference score Accuracy achieved by the subject in the novel object recognition task during Visit 2 minus accuracy during Visit 1, in percent
BDNF genotype Whether a subject's BDNF genotype is Val/Val or Met carrier (Val/Met and Met/Met)
Links
How Exercise Affects the Brain: Age and Genetics Play a Role
BDNF | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.29%3A_Predicting_Present_and_Future_Affect.txt |
Learning Objectives
• To study the parents' perception of their children's weight status
Research conducted by
Debra Etelson, Donald A. Brand, Patricia A. Patrick, and Anushree Shirali
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
With increasing public awareness of child obesity as a major public health problem, studies are showing that it has not translated into an increased awareness of obesity in one’s own child. Dietary patterns and weight status in childhood tend to carry into adolescence and adulthood, promoting the onset of chronic and other diseases. A key ingredient for combating childhood obesity is parental involvement and commitment. However, this is predicated on whether or not parents can recognize overweight and obesity in their children.
This study examined parents’ perceptions of their children’s weight status, their understanding of the health risks of obesity relative to other conditions they may perceive as health risks, and their knowledge of some healthy eating practices. Children’s actual weight status was expressed as their body mass index (BMI) percentile, as determined by the CDC growth charts based on age and sex. According to the CDC growth charts for children, a child with a BMI percentile less than the \(5^{th}\) percentile is underweight; from the \(5^{th}\) to less than the \(85^{th}\), a child is at a healthy weight; from the \(85^{th}\) to less than the \(95^{th}\) percentile, a child is overweight; and a BMI percentile equal to or greater than the \(95^{th}\) percentile, a child is considered to be obese.
A visual analog scale was used to measure parents’ perceptions of their child’s weight. The visual analog scale consisted simply of a \(10\)-cm straight line anchored at the left end by the label “extremely underweight” and at the right end by the label “extremely overweight.” A parent placed a mark along the line to indicate where they perceived their child’s weight to be. The researchers interpreted the marks as percentiles in their analysis.
Questions to Answer
Do parents recognize when their children are overweight or obese? Do parents who make incorrect judgments about healthy food practices also make incorrect judgments about their child’s weight status?
Design Issues
This study defines a parent’s perception of their child’s BMI percentile as “accurate” if their score on a visual analog scale fell within \(30\) points of the child’s true BMI percentile. This wide range defining accuracy potentially allows for misclassification of a child’s weight status among normal, overweight, and obese categories. For example, a parent who perceives their child’s weight status as being at the \(80^{th}\) percentile, i.e., in the normal range, when in reality the child is obese with a BMI percentile of \(98\), the parent’s assessment would be considered accurate by the operational definition used in this study. The authors explain that they chose this definition to give parents as much leeway as possible in assessing their child’s weight on the visual analog scale.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
Sex The sex of the participating parent’s child
Overwt_Obese
Whether or not a child’s body mass index (BMI) is equal to or greater than the 85th percentile for the child’s age and sex, which means that the child is either overweight (85th to less than 95th percentile) or obese (95th percentile or above)
PA_overwt Parental attitude expressing level of concern if their child were overweight, measured on a 4-point Likert Scale. In data analysis, the four categories were condensed into two categories:
0 = “not at all” or “a little” concerned
1 = “quite” or “extremely” concerned
PA-TV Parental attitude expressing level of concern if their child watched >20 hours of TV per week, measured on a 4-point Likert Scale. In data analysis, the four categories were condensed into two categories:
0 = "not at all" or "a little" concerned
1 = "quite" or "extremely" concerned
Accurate Whether or not the parent's perception of their child's weight status was accurate. Parent's perception was considered accurate if the BMI percentile it corresponded to fell within 30 points of the child's actual BMI percentile
Juice_boxes The amount of juice that a parent thinks is healthy for their child to drink each day (a juice box contains eight ounces). We condensed the original four response categories into two categories:
0 = "1 or 2 juice boxes per day"
1 = "3 to 8 juice boxes per day"
Fast_food_meals How often a parent feels it is okay to eat at fast-food restaurants. We condensed the original four response categories into two categories: 0 = "once a month"
Links
Etelson et al. article
BMI percentiles for children | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.31%3A_Parental_Recognition_of_Child_Obesity.txt |
Learning Objectives
• To study large disparities in educational attainment among various racial and ethnic groups
Research conducted by
United States Census Bureau
Case study prepared by
Robert F. Houser, Georgette Baghdady, and Jennifer E. Konick
Overview
The U.S. Census Bureau defines educational attainment as the highest level of education that a person has completed. Large disparities in educational attainment continue to exist among racial and ethnic groups. The gender gap in educational attainment, however, has been undergoing a dramatic social shift in recent decades. In Table \(1\) below, the U.S. Census Bureau tabulated these trends among Whites, Blacks, Asians and Pacific Islanders, and Hispanics between \(1970\) and \(2010\). This case study focuses only on college graduates. The data for "College graduate or more" represent the percentage of adults aged \(25\) years and older that obtained a degree from regular four-year colleges and universities and graduate or professional schools in each racial and ethnic group.
The U.S. Census Bureau defines the racial and ethnic categories in the following manner:
• “White” refers to persons having origins in any of the original peoples of Europe, the Middle East, or North Africa.
• “Black” refers to persons having origins in any of the Black racial groups of Africa.
• “Asian” refers to persons having origins in any of the original peoples of the Far East, Southeast Asia, or the Indian subcontinent.
• “Pacific Islander” refers to persons having origins in any of the original peoples of the Pacific Islands, such as Hawaii, Guam, Samoa, and Tonga.
• “Hispanic” refers to an ethnic group comprised of persons of any race who are of Cuban, Mexican, Puerto Rican, South or Central American, or other Spanish culture or origin.
Educational attainment is strongly associated with future employment, income, and health status.
Questions to Answer
How has the percentage of college graduates changed over time between \(1970\) and \(2010\) among the racial and ethnic groups and between the genders within each group? How might we illustrate these changes graphically?
Design Issues
Beginning with the \(2000\) U.S. Census, respondents were given the option of selecting more than one race category to indicate their racial identities. Therefore, data on race from \(2000\) and beyond are not directly comparable with earlier censuses. The data in Table \(1\) represent persons who selected only one race category and exclude persons who selected more than one race.
In the \(2005\) U.S. Census and beyond, the “Asian and Pacific Islander” category was split into two separate categories, “Asian” and “Native Hawaiian or Other Pacific Islander.” There were several reasons for the split. The combined category was not a homogeneous group because it put together peoples with few social or cultural similarities and who are dissimilar on important demographic characteristics. For example, in \(1990\), about \(11\) percent of Pacific Islanders aged \(25\) years and older obtained a bachelor’s degree compared with about \(40\) percent of Asians. Since Pacific Islanders are numerically a smaller group than Asians (in \(2010\), there were about a half million Pacific Islanders versus about \(14.6\) million Asians), not including them in the data of Table \(1\) starting in \(2005\) biases the percentage of college graduates upwards somewhat, but not strongly.
Descriptions of Variables
Table \(1\): Description of Variables
Variable Description
College graduate or more Obtained a degree from regular four-year colleges and universities and graduate or professional schools
Year Decade years from 1970 to 2010
White_M
White_F
Percentage of college graduates in U.S. subpopulation of White males aged 25 years and over; likewise for White females
Black_M
Black_F
Percentage of college graduates in U.S. subpopulation of Black males aged 25 years and over; likewise for Black females
AsnPac_M
AsnPac_F
Percentage of college graduates in U.S. subpopulation of Asian and Pacific Islander males aged 25 years and over; likewise for Asian and Pacific Islander females
Hispan_M
Hispan_F
Percentage of college graduates in U.S. subpopulation of Hispanic males aged 25 years and over; likewise for Hispanic females
Data Files
Educational_attainment.xls
Links
Overview of Race and Hispanic Origin: 2010
Latinos and Education: Explaining the Attainment Gap
Why Do Women Outnumber Men in College? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/20%3A_Case_Studies/20.32%3A_Educational_Attainment_and_Racial_Ethnic_and_Gender_Disparity.txt |
Learning Objectives
• To study the Analysis Lab calculator
General Instructions
Analysis Lab that can perform basic descriptive and inferential statistics. Descriptive statistics include the graphic methods: box plots, histograms, stem and leaf displays, and normal quantile plots. Inferential statistics include independent-group t tests, Chi Square tests, tests in simple regression/correlation, and Analysis of Variance.
Reading/Entering Data
There are currently two "libraries" of datasets. The default library is "RVLS_data." You can change the library by using the pop-up menu. Just under the pop-up menu for libraries is a pop-up menu of datasets in the library. Choose the dataset you want to analyze. A description of the dataset will be shown on the right side of the display. The data from the case studies will be included in the near future.
To enter your own data, click the button labeled "Enter/Edit User Data." A window will open with an area for you to enter or paste in your data. You may get a warning that you must use the keyboard shortcut to paste because your system may not allow Java programs to read the clipboard. Pasting formatted text will not work. If you have trouble pasting in your data, try pasting them as unformatted text into Word, copying, and then pasting into Analysis Lab.
The first line should contain the names of the variables (separated by spaces or tabs). The remaining lines should contain the data themselves. Missing data cannot be handled so all observations must have valid data for all variables. All variables must be numeric. If one of your variables is to be used as a "Grouping" or "Classification" variable, then values of the grouping variable must be integers ranging from one to the total number of groups. Use grouping variables so that you can do a separate analysis for each level of the variable or to use the variable as an independent variable in an analysis of variance. Once you have entered your data, click on the "Accept data" button. The data will be temporarily saved so that if you click the "Enter/Edit User Data" button again they will be shown
Choosing Variables
To analyze a variable, select it in the "$Y$" pop-up menu and then click on the type of analysis you wish to perform. To see the relationship between two variables, select one in the "$X$" menu and the other in the "$Y$" menu. Then click on the "Correlation/regression" button. Specify a grouping variable to do an analysis separately for each group of observations. You also specify a grouping variable to conduct an analysis of variance or do a t test.
Copying/Printing
Analysis Lab does not have any copying or printing capabilities. However, you can copy and(optionally) print by doing a screen capture. If you are using a Macintosh with OSX, press "$4$" while holding down the command (Apple), shift, and control keys. You will be given the opportunity to select an area of the screen to be copied to the clipboard.
If you are using Windows, make sure the window you want to copy is selected. Then, hold down the ALT key and press the Print Screen Key key. The window will be copied to the clipboard for pasting into another application such as a word processor.
Problems
On occasion, some browsers do not refresh the display. If you get a blank window, try resizing it slightly.
ANOVA Instructions
The following information is a guide to doing Analysis of Variance (ANOVA) with Analysis Lab. Analysis Lab was designed to allow the most common types of designs to be analyzed with a minimum of effort.
One Between-Subjects Variable
This analysis is used when there are two or more independent groups of subjects and one score per subject.
1. The data should be arranged so that there is one variable that indicates group membership and a second variable that contains the dependent variable. The variable indicating group membership must use consecutive integers (e.g., $1, 2, 3$) as group indicators. The data below have two variables; $G$ for group membership and $Y$ for the dependent variable. There are three groups and four observations (subjects) per group.
G Y
1 5
1 7
1 8
1 6
2 3
2 4
2 5
2 2
3 11
3 13
3 12
3 9
1. Paste the data into analysis lab.
1. Copy your data.
2. Click the Enter/Edit data button.
3. Paste the data using keyboard shortcuts (Control-V for Windows, CMD-V for Macs). Then click "Accept Data."
Choose $Y$ as the dependent variable and $G$ as the grouping variable as shown. Then click the ANOVA button. Note that the description says what analysis will be done. $IV$ stands for "Independent Variable" and $DV$ stands for "Dependent Variable."
1. The ANOVA summary table will be presented.
One Within-Subjects Variable
Use this design when you have one group of observations (subjects) with two or more scores per observation.
1. Format your data so that each row contains all the data for one observation. Each score is a separate variable. For the following example, there are five observations with three scores per observation. The variable names are $T1$, $T2$, and $T3$.
T1 T2 T3
2 3 4
3 5 4
2 4 6
1 3 3
6 7 9
1. Paste the data into analysis lab (see the instructions for One-Way Between-Subjects).
2. Click the "Advanced" button next to the ANOVA" button.
3. Select the dependent variables you want to analyze. and click the "Do Anova" button. Here the variables $T1$, and $T2$, and $T3$ are chosen.
1. The Summary Table is presented. The effect $DV$ stands for "Dependent Variable." It is a test of the within-subjects variable.
Two or More Between-Subject Variables
Use this design when you have more than one between-subject variable and only one score per subject. In the following example, the variables are $A$ and $B$. The variable $A$ has two levels and the variable $B$ has three levels. There are $3$ scores per "cell." A cell is a combination of one level of $A$ and one level of $B$.
Note that Analysis Lab uses unweighted means analysis rather than the general linear model. This will give slightly different results from most other programs when there are unequal sample sizes and the degrees of freedom are greater than one.
1. Format the data so that there is one variable indicating the level of each independent variable. The variable must use consecutive integer variables. There is also a dependent variable. In the example, the variable $A$ indicates the level of the variable $A$, $B$ indicates the level of the variable $B$, and $Y$ is the dependent variable.
A B Y
1 1 3
1 1 5
1 1 4
1 2 7
1 2 6
1 2 8
1 3 4
1 3 9
1 3 6
2 1 9
2 1 9
2 1 10
2 2 13
2 2 11
2 2 14
2 3 6
2 3 7
2 3 9
1. Paste the data into analysis lab (see the instructions for One-Way Between-Subjects).
2. Click the "Advanced" button next to the ANOVA" button.
3. Choose the dependent variable and the independent variables. The independent variables are called "group variables." Then click the "Do Anova" button.
1. The Summary table is shown below.
One or More Between-Subject Variables and one Within-Subject Variable
1. Format the data so that a variable indicates each level of each between-subject variable. Include more than one dependent variable. For this example, there is one between-subjects variable ($A$) with two levels and one within-subjects variable ($T$) with four levels.
A T1 T2 T3 T4
1 5 4 3 6
1 7 6 2 9
1 6 5 7 5
2 8 9 12 14
2 6 7 9 11
2 2 4 3 6
1. Paste the data into analysis lab (see the instructions for One-Way Between-Subjects).
2. Click the "Advanced" button next to the ANOVA" button.
3. Choose the dependent variables and the independent variables. The independent variables are called "group variables." In this example there is one independent variable ($A$) and four dependent variables ($T1$, $T2$, $T3$, and $T4$). Then click the "Do Anova" button.
1. The Summary table is shown. $DV$ stands for dependent variable and is the effect of the within-subjects factor. $A \times DV$ is the interaction of the between-subjects variable and the within-subjects variable.
One or More Between-Subjects Variables and Two more More Within-Subjects Variables
1. Format the data so that a variable indicates each level of each between-subject variable. Include more than one dependent variable. For this example, there are two between-subjects variable (Age and Gender) and two within-subjects variable (Day and Trial ). There are four groups of subjects (the combination of two levels of Age and two levels of Gender). Each subject is given two trials a day for two days. For convenience $A$ is used as an abbreviation for age, $G$ for Gender, $D$ for Day, and $T$ for Trial. The variable $D1T1$ is the score for $\text{Day 1 Trial 1}$, $D1T2$ for $\text{Day 1, Trial 2}$, etc.
A B D1T1 D1T2 D2T1 D2T2
1 1 5 7 4 8
1 1 4 7 5 9
1 1 5 5 8 9
1 2 5 4 5 7
1 2 4 8 11 12
1 2 8 7 6 5
2 1 12 13 12 15
2 1 10 9 11 13
2 1 8 8 7 15
2 2 13 14 11 12
2 2 7 5 8 7
2 2 9 11 11 10
1. Paste the data into analysis lab (see the instructions for One-Way Between-Subjects).
2. Click the "Advanced" button next to the ANOVA" button.
3. Choose the dependent variables and the independent variables. The independent variables are called "group variables." In this example there are two independent variable ($A$ and $B$) and four dependent variables ($D1T1$, $D1T2$, $D2T1$, and $D2T2$). Then click the "Do Complex Anova" at the bottom of the window.
1. Next you name the within-subject variables and specify the number of levels of each. Since there are two days and two trials per day, both of the values are two. Analysis Lab allows up to four. In this example there are two ( $D$ and $T$). Also indicate that $A$ and $B$ are the between-subjects (group) variables.
1. Next you will be asked to associate the variable names to the levels of the variables. The way the variables were named in this example makes this a trivial exercise. However, if you used other variable names, it would take a little effort. Click the "Do ANOVA" button when finished.
1. The Summary Table is presented.
21.08: Power Calculator
Power Calculator
David M. Lane
Prerequisites
Power
Difference between means
Standard deviation
Sample size per group
Alpha level
Power (two tailed test)
Fill in the fields and then press the "Caclulate" button. The power for a two-tailed t test will be displayed.
If you change a value you can press enter or the tab key to recalculate.
This calculator is based on jStat from jstat.org and is distributed under the MIT license:
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/21%3A_Calculators/21.01%3A_Analysis_Lab.txt |
Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what "good" data can be distinguished from "bad."
• 1.1: Introduction
Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what "good" data can be distinguished from "bad."
• 1.2: Definitions of Statistics, Probability, and Key Terms
The mathematical theory of statistics is easier to learn when you know the language. This module presents important terms that will be used throughout the text.
• 1.3: Data, Sampling, and Variation in Data and Sampling
Data are individual items of information that come from a population or sample. Data may be classified as qualitative, quantitative continuous, or quantitative discrete. Because it is not practical to measure the entire population in a study, researchers use samples to represent the population. A random sample is a representative group from the population chosen by using a method that gives each individual in the population an equal chance of being included in the sample.
• 1.4: Frequency, Frequency Tables, and Levels of Measurement
Some calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that generated that value were only accurate to the nearest tenth. Round off your final answer to one more decimal place than was present in the original data. This means that if you have data measured to the nearest tenth of a unit, report the final statistic to the nearest hundredth.
• 1.5: Experimental Design and Ethics
A poorly designed study will not produce reliable data. There are certain key components that must be included in every experiment. To eliminate lurking variables, subjects must be assigned randomly to different treatment groups. One of the groups must act as a control group, demonstrating what happens when the active treatment is not applied. Participants in the control group receive a placebo treatment that looks exactly like the active treatments but cannot influence the response variable.
• 1.6: Data Collection Experiment (Worksheet)
A statistics Worksheet: The student will demonstrate the systematic sampling technique. The student will construct relative frequency tables. The student will interpret results and their differences from different data groupings.
• 1.7: Sampling Experiment (Worksheet)
A statistics Worksheet: The student will demonstrate the simple random, systematic, stratified, and cluster sampling techniques. The student will explain the details of each procedure used.
• 1.E: Sampling and Data (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
Contributors and Attributions
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected].
01: Sampling and Data
Learning Objectives
By the end of this chapter, the student should be able to:
• Recognize and differentiate between key terms.
• Apply various types of sampling methods to data collection.
• Create and interpret frequency tables.
You are probably asking yourself the question, "When and where will I use statistics?" If you read any newspaper, watch television, or use the Internet, you will see statistical information. There are statistics about crime, sports, education, politics, and real estate. Typically, when you read a newspaper article or watch a television news program, you are given sample information. With this information, you may make a decision about the correctness of a statement, claim, or "fact." Statistical methods can help you make the "best educated guess."
Since you will undoubtedly be given statistical information at some point in your life, you need to know some techniques for analyzing the information thoughtfully. Think about buying a house or managing a budget. Think about your chosen profession. The fields of economics, business, psychology, education, biology, law, computer science, police science, and early childhood development require at least one course in statistics.
Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what "good" data can be distinguished from "bad." | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.01%3A_Introduction.txt |
The science of statistics deals with the collection, analysis, interpretation, and presentation of data. We see and use data in our everyday lives.
Collaborative Exercise
In your classroom, try this exercise. Have class members write down the average time (in hours, to the nearest half-hour) they sleep per night. Your instructor will record the data. Then create a simple graph (called a dot plot) of the data. A dot plot consists of a number line and dots (or points) positioned above the number line. For example, consider the following data:
5; 5.5; 6; 6; 6; 6.5; 6.5; 6.5; 6.5; 7; 7; 8; 8; 9
The dot plot for this data would be as follows:
• Does your dot plot look the same as or different from the example? Why?
• If you did the same example in an English class with the same number of students, do you think the results would be the same? Why or why not?
• Where do your data appear to cluster? How might you interpret the clustering?
The questions above ask you to analyze and interpret your data. With this example, you have begun your study of statistics.
In this course, you will learn how to organize and summarize data. Organizing and summarizing data is called descriptive statistics. Two ways to summarize data are by graphing and by using numbers (for example, finding an average). After you have studied probability and probability distributions, you will use formal methods for drawing conclusions from "good" data. The formal methods are called inferential statistics. Statistical inference uses probability to determine how confident we can be that our conclusions are correct.
Effective interpretation of data (inference) is based on good procedures for producing data and thoughtful examination of the data. You will encounter what will seem to be too many mathematical formulas for interpreting data. The goal of statistics is not to perform numerous calculations using the formulas, but to gain an understanding of your data. The calculations can be done using a calculator or a computer. The understanding must come from you. If you can thoroughly grasp the basics of statistics, you can be more confident in the decisions you make in life.
Probability
Probability is a mathematical tool used to study randomness. It deals with the chance (the likelihood) of an event occurring. For example, if you toss a fair coin four times, the outcomes may not be two heads and two tails. However, if you toss the same coin 4,000 times, the outcomes will be close to half heads and half tails. The expected theoretical probability of heads in any one toss is $\frac{1}{2}$ or 0.5. Even though the outcomes of a few repetitions are uncertain, there is a regular pattern of outcomes when there are many repetitions. After reading about the English statistician Karl Pearson who tossed a coin 24,000 times with a result of 12,012 heads, one of the authors tossed a coin 2,000 times. The results were 996 heads. The fraction $\frac{996}{2000}$ is equal to 0.498 which is very close to 0.5, the expected probability.
The theory of probability began with the study of games of chance such as poker. Predictions take the form of probabilities. To predict the likelihood of an earthquake, of rain, or whether you will get an A in this course, we use probabilities. Doctors use probability to determine the chance of a vaccination causing the disease the vaccination is supposed to prevent. A stockbroker uses probability to determine the rate of return on a client's investments. You might use probability to decide to buy a lottery ticket or not. In your study of statistics, you will use the power of mathematics through probability calculations to analyze and interpret your data.
Key Terms
In statistics, we generally want to study a population. You can think of a population as a collection of persons, things, or objects under study. To study the population, we select a sample. The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population.
Because it takes a lot of time and money to examine an entire population, sampling is a very practical technique. If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students' grade point averages. In presidential elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country. Manufacturers of canned carbonated drinks take samples to determine if a 16 ounce can contains 16 ounces of carbonated drink.
From the sample data, we can calculate a statistic. A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic. The statistic is an estimate of a population parameter. A parameter is a number that is a property of the population. Since we considered all math classes to be the population, then the average number of points earned per student over all the math classes is an example of a parameter.
One of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The accuracy really depends on how well the sample represents the population. The sample must contain the characteristics of the population in order to be a representative sample. We are interested in both the sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the sample statistic to test the validity of the established population parameter.
A variable, notated by capital letters such as $X$ and $Y$, is a characteristic of interest for each person or thing in a population. Variables may be numerical or categorical. Numerical variables take on values with equal units such as weight in pounds and time in hours. Categorical variables place the person or thing into a category. If we let $X$ equal the number of points earned by one math student at the end of a term, then $X$ is a numerical variable. If we let $Y$ be a person's party affiliation, then some examples of $Y$ include Republican, Democrat, and Independent. $Y$ is a categorical variable. We could do some math with values of $X$ (calculate the average number of points earned, for example), but it makes no sense to do math with values of $Y$ (calculating an average party affiliation makes no sense).
Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single value.
Two words that come up often in statistics are mean and proportion. If you were to take three exams in your math classes and obtain scores of 86, 75, and 92, you would calculate your mean score by adding the three exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is $\frac{22}{40}$ and the proportion of women students is $\frac{18}{40}$. Mean and proportion are discussed in more detail in later chapters.
The words "mean" and "average" are often used interchangeably. The substitution of one word for the other is common practice. The technical term is "arithmetic mean," and "average" is technically a center location. However, in practice among non-statisticians, "average" is commonly accepted for "arithmetic mean."
Example $1$
Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money first year college students spend at ABC College on school supplies that do not include books. We randomly survey 100 first year students at the college. Three of those students spent $150,$200, and $225, respectively. Answer • The population is all first year students attending ABC College this term. • The sample could be all students enrolled in one section of a beginning statistics course at ABC College (although this sample may not represent the entire population). • The parameter is the average (mean) amount of money spent (excluding books) by first year college students at ABC College this term. • The statistic is the average (mean) amount of money spent (excluding books) by first year college students in the sample. • The variable could be the amount of money spent (excluding books) by one first year student. Let $X$ = the amount of money spent (excluding books) by one first year student attending ABC College. • The data are the dollar amounts spent by the first year students. Examples of the data are$150, $200, and$225.
Exercise $1$
Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money spent on school uniforms each year by families with children at Knoll Academy. We randomly survey 100 families with children in the school. Three of the families spent $65,$75, and $95, respectively. Answer • The population is all families with children attending Knoll Academy. • The sample is a random selection of 100 families with children attending Knoll Academy. • The parameter is the average (mean) amount of money spent on school uniforms by families with children at Knoll Academy. • The statistic is the average (mean) amount of money spent on school uniforms by families in the sample. • The variable is the amount of money spent by one family. Let $X$ = the amount of money spent on school uniforms by one family with children attending Knoll Academy. • The data are the dollar amounts spent by the families. Examples of the data are$65, $75, and$95.
Example $2$
Determine what the key terms refer to in the following study.
A study was conducted at a local college to analyze the average cumulative GPA’s of students who graduated last year. Fill in the letter of the phrase that best describes each of the items below.
1._____ Population 2._____ Statistic 3._____ Parameter 4._____ Sample 5._____ Variable 6._____ Data
1. all students who attended the college last year
2. the cumulative GPA of one student who graduated from the college last year
3. 3.65, 2.80, 1.50, 3.90
4. a group of students who graduated from the college last year, randomly selected
5. the average cumulative GPA of students who graduated from the college last year
6. all students who graduated from the college last year
7. the average cumulative GPA of students in the study who graduated from the college last year
Answer
1. f; 2. g; 3. e; 4. d; 5. b; 6. c
Example $3$
Determine what the key terms refer to in the following study.
As part of a study designed to test the safety of automobiles, the National Transportation Safety Board collected and reviewed data about the effects of an automobile crash on test dummies. Here is the criterion they used:
Speed at which Cars Crashed Location of “drive” (i.e. dummies)
35 miles/hour Front Seat
Cars with dummies in the front seats were crashed into a wall at a speed of 35 miles per hour. We want to know the proportion of dummies in the driver’s seat that would have had head injuries, if they had been actual drivers. We start with a simple random sample of 75 cars.
Answer
• The population is all cars containing dummies in the front seat.
• The sample is the 75 cars, selected by a simple random sample.
• The parameter is the proportion of driver dummies (if they had been real people) who would have suffered head injuries in the population.
• The statistic is proportion of driver dummies (if they had been real people) who would have suffered head injuries in the sample.
• The variable $X$ = the number of driver dummies (if they had been real people) who would have suffered head injuries.
• The data are either: yes, had head injury, or no, did not.
Example $4$
Determine what the key terms refer to in the following study.
An insurance company would like to determine the proportion of all medical doctors who have been involved in one or more malpractice lawsuits. The company selects 500 doctors at random from a professional directory and determines the number in the sample who have been involved in a malpractice lawsuit.
Answer
• The population is all medical doctors listed in the professional directory.
• The parameter is the proportion of medical doctors who have been involved in one or more malpractice suits in the population.
• The sample is the 500 doctors selected at random from the professional directory.
• The statistic is the proportion of medical doctors who have been involved in one or more malpractice suits in the sample.
• The variable $X$ = the number of medical doctors who have been involved in one or more malpractice suits.
• The data are either: yes, was involved in one or more malpractice lawsuits, or no, was not.
Collaborative Exercise
Do the following exercise collaboratively with up to four people per group. Find a population, a sample, the parameter, the statistic, a variable, and data for the following study: You want to determine the average (mean) number of glasses of milk college students drink per day. Suppose yesterday, in your English class, you asked five students how many glasses of milk they drank the day before. The answers were 1, 0, 1, 3, and 4 glasses of milk.
Practice
Use the following information to answer the next five exercises. Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS symptoms have revealed themselves. Of interest is the average (mean) length of time in months patients live once they start the treatment. Two researchers each follow a different set of 40 patients with AIDS from the start of treatment until their deaths. The following data (in months) are collected.
Researcher A:
3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher B:
3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
Determine what the key terms refer to in the example for Researcher A.
population
Answer
AIDS patients.
sample
Exercise $4$
parameter
Answer
The average length of time (in months) AIDS patients live after treatment.
statistic
Exercise $6$
variable
Answer
$X =$ the length of time (in months) AIDS patients live after treatment
Glossary
The mathematical theory of statistics is easier to learn when you know the language. This module presents important terms that will be used throughout the text.
Average
also called mean; a number that describes the central tendency of the data
Categorical Variable
variables that take on values that are names or labels
Data
a set of observations (a set of possible outcomes); most data can be put into two groups: qualitative(an attribute whose value is indicated by a label) or quantitative (an attribute whose value is indicated by a number). Quantitative data can be separated into two subgroups: discrete and continuous. Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage)
Numerical Variable
variables that take on values that are indicated by numbers
Parameter
a number that is used to represent a population characteristic and that generally cannot be determined easily
Population
all individuals, objects, or measurements whose properties are being studied
Probability
a number between zero and one, inclusive, that gives the likelihood that a specific event will occur
Proportion
the number of successes divided by the total number in the sample
Representative Sample
a subset of the population that has the same characteristics as the population
Sample
a subset of the population studied
Statistic
a numerical characteristic of the sample; a statistic estimates the corresponding population parameter.
Variable
a characteristic of interest for each person or object in a population | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.02%3A_Definitions_of_Statistics_Probability_and_Key_Terms.txt |
Data may come from a population or from a sample. Small letters like $x$ or $y$ generally are used to represent data values. Most data can be put into the following categories:
• Qualitative
• Quantitative
Qualitative data are the result of categorizing or describing attributes of a population. Hair color, blood type, ethnic group, the car a person drives, and the street a person lives on are examples of qualitative data. Qualitative data are generally described by words or letters. For instance, hair color might be black, dark brown, light brown, blonde, gray, or red. Blood type might be AB+, O-, or B+. Researchers often prefer to use quantitative data over qualitative data because it lends itself more easily to mathematical analysis. For example, it does not make sense to find an average hair color or blood type.
Quantitative data are always numbers. Quantitative data are the result of counting or measuring attributes of a population. Amount of money, pulse rate, weight, number of people living in your town, and number of students who take statistics are examples of quantitative data. Quantitative data may be either discrete or continuous.
All data that are the result of counting are called quantitative discrete data. These data take on only certain numerical values. If you count the number of phone calls you receive for each day of the week, you might get values such as zero, one, two, or three.
All data that are the result of measuring are quantitative continuous data assuming that we can measure accurately. Measuring angles in radians might result in such numbers as $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{4}$, and so on. If you and your friends carry backpacks with books in them to school, the numbers of books in the backpacks are discrete data and the weights of the backpacks are continuous data.
Sample of Quantitative Discrete Data
The data are the number of books students carry in their backpacks. You sample five students. Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. The numbers of books (three, four, two, and one) are the quantitative discrete data.
Exercise $1$
The data are the number of machines in a gym. You sample five gyms. One gym has 12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22 machines, and the other gym has 20 machines. What type of data is this?
Answer
quantitative discrete data
Sample of Quantitative Continuous Data
The data are the weights of backpacks with books in them. You sample the same five students. The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have different weights. Weights are quantitative continuous data because weights are measured.
Exercise $2$
The data are the areas of lawns in square feet. You sample five houses. The areas of the lawns are 144 sq. feet, 160 sq. feet, 190 sq. feet, 180 sq. feet, and 210 sq. feet. What type of data is this?
Answer
quantitative continuous data
Exercise $3$
You go to the supermarket and purchase three cans of soup (19 ounces) tomato bisque, 14.1 ounces lentil, and 19 ounces Italian wedding), two packages of nuts (walnuts and peanuts), four different kinds of vegetable (broccoli, cauliflower, spinach, and carrots), and two desserts (16 ounces Cherry Garcia ice cream and two pounds (32 ounces chocolate chip cookies).
Name data sets that are quantitative discrete, quantitative continuous, and qualitative.
Solution
One Possible Solution:
• The three cans of soup, two packages of nuts, four kinds of vegetables and two desserts are quantitative discrete data because you count them.
• The weights of the soups (19 ounces, 14.1 ounces, 19 ounces) are quantitative continuous data because you measure weights as precisely as possible.
• Types of soups, nuts, vegetables and desserts are qualitative data because they are categorical.
Try to identify additional data sets in this example.
Sample of qualitative data
The data are the colors of backpacks. Again, you sample the same five students. One student has a red backpack, two students have black backpacks, one student has a green backpack, and one student has a gray backpack. The colors red, black, black, green, and gray are qualitative data.
Exercise $4$
The data are the colors of houses. You sample five houses. The colors of the houses are white, yellow, white, red, and white. What type of data is this?
Answer
qualitative data
Collaborative Exercise $1$
Work collaboratively to determine the correct data type (quantitative or qualitative). Indicate whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with the words "the number of."
1. the number of pairs of shoes you own
2. the type of car you drive
3. where you go on vacation
4. the distance it is from your home to the nearest grocery store
5. the number of classes you take per school year.
6. the tuition for your classes
7. the type of calculator you use
8. movie ratings
9. political party preferences
10. weights of sumo wrestlers
11. amount of money (in dollars) won playing poker
12. number of correct answers on a quiz
13. peoples’ attitudes toward the government
14. IQ scores (This may cause some discussion.)
Answer
Items a, e, f, k, and l are quantitative discrete; items d, j, and n are quantitative continuous; items b, c, g, h, i, and m are qualitative.
Exercise $5$
Determine the correct data type (quantitative or qualitative) for the number of cars in a parking lot. Indicate whether quantitative data are continuous or discrete.
Answer
quantitative discrete
Exercise $6$
A statistics professor collects information about the classification of her students as freshmen, sophomores, juniors, or seniors. The data she collects are summarized in the pie chart Figure $1$. What type of data does this graph show?
Answer
This pie chart shows the students in each year, which is qualitative data.
Exercise $7$
The registrar at State University keeps records of the number of credit hours students complete each semester. The data he collects are summarized in the histogram. The class boundaries are 10 to less than 13, 13 to less than 16, 16 to less than 19, 19 to less than 22, and 22 to less than 25.
What type of data does this graph show?
Answer
A histogram is used to display quantitative data: the numbers of credit hours completed. Because students can complete only a whole number of hours (no fractions of hours allowed), this data is quantitative discrete.
Qualitative Data Discussion
Below are tables comparing the number of part-time and full-time students at De Anza College and Foothill College enrolled for the spring 2010 quarter. The tables display counts (frequencies) and percentages or proportions (relative frequencies). The percent columns make comparing the same categories in the colleges easier. Displaying percentages along with the numbers is often helpful, but it is particularly important when comparing sets of data that do not have the same totals, such as the total enrollments for both colleges in this example. Notice how much larger the percentage for part-time students at Foothill College is compared to De Anza College.
De Anza College Foothill College
Table $1$: Fall Term 2007 (Census day)
Number Percent Number Percent
Full-time 9,200 40.9% Full-time 4,059 28.6%
Part-time 13,296 59.1% Part-time 10,124 71.4%
Total 22,496 100% Total 14,183 100%
Tables are a good way of organizing and displaying data. But graphs can be even more helpful in understanding the data. There are no strict rules concerning which graphs to use. Two graphs that are used to display qualitative data are pie charts and bar graphs.
• In a pie chart, categories of data are represented by wedges in a circle and are proportional in size to the percent of individuals in each category.
• In a bar graph, the length of the bar for each category is proportional to the number or percent of individuals in each category. Bars may be vertical or horizontal.
• A Pareto chart consists of bars that are sorted into order by category size (largest to smallest).
Look at Figures $3$ and $4$ and determine which graph (pie or bar) you think displays the comparisons better.
It is a good idea to look at a variety of graphs to see which is the most helpful in displaying the data. We might make different choices of what we think is the “best” graph depending on the data and the context. Our choice also depends on what we are using the data for.
Percentages That Add to More (or Less) Than 100%
Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, the percentages add to more than 100% because students can be in more than one category. A bar graph is appropriate to compare the relative size of the categories. A pie chart cannot be used. It also could not be used if the percentages added to less than 100%.
Characteristic/Category Percent
Table $2$: De Anza College Spring 2010
Full-Time Students 40.9%
Students who intend to transfer to a 4-year educational institution 48.6%
Students under age 25 61.0%
TOTAL 150.5%
Omitting Categories/Missing Data
The table displays Ethnicity of Students but is missing the "Other/Unknown" category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that the frequencies do not add up to the total number of students. In this situation, create a bar graph and not a pie chart.
Table $2$: Ethnicity of Students at De Anza College Fall Term 2007 (Census Day)
Frequency Percent
Asian 8,794 36.1%
Black 1,412 5.8%
Filipino 1,298 5.3%
Hispanic 4,180 17.1%
Native American 146 0.6%
Pacific Islander 236 1.0%
White 5,978 24.5%
TOTAL 22,044 out of 24,382 90.4% out of 100%
The following graph is the same as the previous graph but the “Other/Unknown” percent (9.6%) has been included. The “Other/Unknown” category is large compared to some of the other categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know when we think about what the data are telling us.
This particular bar graph in Figure $4$ can be difficult to understand visually. The graph in Figure $5$ is a Pareto chart. The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret.
Pie Charts: No Missing Data
The following pie charts have the “Other/Unknown” category included (since the percentages must add to 100%). The chart in Figure $6$ is organized by the size of each wedge, which makes it a more visually informative graph than the unsorted, alphabetical graph in Figure $6$.
Sampling
Gathering information about an entire population often costs too much or is virtually impossible. Instead, we use a sample of the population. A sample should have the same characteristics as the population it is representing. Most statisticians use various methods of random sampling in an attempt to achieve this goal. This section will describe a few of the most common methods. There are several different methods of random sampling. In each form of random sampling, each member of a population initially has an equal chance of being selected for the sample. Each method has pros and cons. The easiest method to describe is called a simple random sample. Any group of n individuals is equally likely to be chosen by any other group of n individuals if the simple random sampling technique is used. In other words, each sample of the same size has an equal chance of being selected. For example, suppose Lisa wants to form a four-person study group (herself and three other people) from her pre-calculus class, which has 31 members not including Lisa. To choose a simple random sample of size three from the other members of her class, Lisa could put all 31 names in a hat, shake the hat, close her eyes, and pick out three names. A more technological way is for Lisa to first list the last names of the members of her class together with a two-digit number, as in Table $2$:
Table $3$: Class Roster
ID Name ID Name ID Name
00 Anselmo 11 King 21 Roquero
01 Bautista 12 Legeny 22 Roth
02 Bayani 13 Lundquist 23 Rowell
03 Cheng 14 Macierz 24 Salangsang
04 Cuarismo 15 Motogawa 25 Slade
05 Cuningham 16 Okimoto 26 Stratcher
06 Fontecha 17 Patel 27 Tallai
07 Hong 18 Price 28 Tran
08 Hoobler 19 Quizon 29 Wai
09 Jiao 20 Reyes 30 Wood
10 Khan
Lisa can use a table of random numbers (found in many statistics books and mathematical handbooks), a calculator, or a computer to generate random numbers. For this example, suppose Lisa chooses to generate random numbers from a calculator. The numbers generated are as follows:
0.94360; 0.99832; 0.14669; 0.51470; 0.40581; 0.73381; 0.04399
Lisa reads two-digit groups until she has chosen three class members (that is, she reads 0.94360 as the groups 94, 43, 36, 60). Each random number may only contribute one class member. If she needed to, Lisa could have generated more random numbers.
The random numbers 0.94360 and 0.99832 do not contain appropriate two digit numbers. However the third random number, 0.14669, contains 14 (the fourth random number also contains 14), the fifth random number contains 05, and the seventh random number contains 04. The two-digit number 14 corresponds to Macierz, 05 corresponds to Cuningham, and 04 corresponds to Cuarismo. Besides herself, Lisa’s group will consist of Marcierz, Cuningham, and Cuarismo.
To generate random numbers:
• Press MATH.
• Arrow over to PRB.
• Press 5:randInt(. Enter 0, 30).
• Press ENTER for the first random number.
• Press ENTER two more times for the other 2 random numbers. If there is a repeat press ENTER again.
Note: randInt(0, 30, 3) will generate 3 random numbers.
Besides simple random sampling, there are other forms of sampling that involve a chance process for getting the sample. Other well-known random sampling methods are the stratified sample, the cluster sample, and the systematic sample.
To choose a stratified sample, divide the population into groups called strata and then take a proportionate number from each stratum. For example, you could stratify (group) your college population by department and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of the first department, number each member of the second department, and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department and do the same for each of the remaining departments. Those numbers picked from the first department, picked from the second department, and so on represent the members who make up the stratified sample.
To choose a cluster sample, divide the population into clusters (groups) and then randomly select some of the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly sample four departments from your college population, the four departments make up the cluster sample. Divide your college faculty by department. The departments are the clusters. Number each department, and then choose four different numbers using simple random sampling. All members of the four departments with those numbers are the cluster sample.
To choose a systematic sample, randomly select a starting point and take every nth piece of data from a listing of the population. For example, suppose you have to do a phone survey. Your phone book contains 20,000 residence listings. You must choose 400 names for the sample. Number the population 1–20,000 and then use a simple random sample to pick a number that represents the first name in the sample. Then choose every fiftieth name thereafter until you have a total of 400 names (you might have to go back to the beginning of your phone list). Systematic sampling is frequently chosen because it is a simple method.
A type of sampling that is non-random is convenience sampling. Convenience sampling involves using results that are readily available. For example, a computer software store conducts a marketing study by interviewing potential customers who happen to be in the store browsing through the available software. The results of convenience sampling may be very good in some cases and highly biased (favor certain outcomes) in others.
Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (they may favor a certain group). It is better for the person conducting the survey to select the sample respondents.
True random sampling is done with replacement. That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling is done without replacement. Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low.
In a college population of 10,000 people, suppose you want to pick a sample of 1,000 randomly for a survey. For any particular sample of 1,000, if you are sampling with replacement,
• the chance of picking the first person is 1,000 out of 10,000 (0.1000);
• the chance of picking a different second person for this sample is 999 out of 10,000 (0.0999);
• the chance of picking the same person again is 1 out of 10,000 (very low).
If you are sampling without replacement,
• the chance of picking the first person for any particular sample is 1000 out of 10,000 (0.1000);
• the chance of picking a different second person is 999 out of 9,999 (0.0999);
• you do not replace the first person before picking the next person.
Compare the fractions 999/10,000 and 999/9,999. For accuracy, carry the decimal answers to four decimal places. To four decimal places, these numbers are equivalent (0.0999).
Sampling without replacement instead of sampling with replacement becomes a mathematical issue only when the population is small. For example, if the population is 25 people, the sample is ten, and you are sampling with replacement for any particular sample, then the chance of picking the first person is ten out of 25, and the chance of picking a different second person is nine out of 25 (you replace the first person).
If you sample without replacement, then the chance of picking the first person is ten out of 25, and then the chance of picking the second person (who is different) is nine out of 24 (you do not replace the first person).
Compare the fractions 9/25 and 9/24. To four decimal places, 9/25 = 0.3600 and 9/24 = 0.3750. To four decimal places, these numbers are not equivalent.
When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual process of sampling causes sampling errors. For example, the sample may not be large enough. Factors not related to the sampling process cause nonsampling errors. A defective counting device can cause a nonsampling error.
In reality, a sample will never be exactly representative of the population so there will always be some sampling error. As a rule, the larger the sample, the smaller the sampling error.
In statistics, a sampling bias is created when a sample is collected from a population and some members of the population are not as likely to be chosen as others (remember, each member of the population should have an equally likely chance of being chosen). When a sampling bias happens, there can be incorrect conclusions drawn about the population that is being studied.
Exercise $8$
A study is done to determine the average tuition that San Jose State undergraduate students pay per semester. Each student in the following samples is asked how much tuition he or she paid for the Fall semester. What is the type of sampling in each case?
1. A sample of 100 undergraduate San Jose State students is taken by organizing the students’ names by classification (freshman, sophomore, junior, or senior), and then selecting 25 students from each.
2. A random number generator is used to select a student from the alphabetical listing of all undergraduate students in the Fall semester. Starting with that student, every 50th student is chosen until 75 students are included in the sample.
3. A completely random method is used to select 75 students. Each undergraduate student in the fall semester has the same probability of being chosen at any stage of the sampling process.
4. The freshman, sophomore, junior, and senior years are numbered one, two, three, and four, respectively. A random number generator is used to pick two of those years. All students in those two years are in the sample.
5. An administrative assistant is asked to stand in front of the library one Wednesday and to ask the first 100 undergraduate students he encounters what they paid for tuition the Fall semester. Those 100 students are the sample.
Answer
a. stratified; b. systematic; c. simple random; d. cluster; e. convenience
Example $9$: Calculator
You are going to use the random number generator to generate different types of samples from the data. This table displays six sets of quiz scores (each quiz counts 10 points) for an elementary statistics class.
#1 #2 #3 #4 #5 #6
5 7 10 9 8 3
10 5 9 8 7 6
9 10 8 6 7 9
9 10 10 9 8 9
7 8 9 5 7 4
9 9 9 10 8 7
7 7 10 9 8 8
8 8 9 10 8 8
9 7 8 7 7 8
8 8 10 9 8 7
Instructions: Use the Random Number Generator to pick samples.
1. Create a stratified sample by column. Pick three quiz scores randomly from each column.
• Number each row one through ten.
• On your calculator, press Math and arrow over to PRB.
• For column 1, Press 5:randInt( and enter 1,10). Press ENTER. Record the number. Press ENTER 2 more times (even the repeats). Record these numbers. Record the three quiz scores in column one that correspond to these three numbers.
• Repeat for columns two through six.
• These 18 quiz scores are a stratified sample.
2. Create a cluster sample by picking two of the columns. Use the column numbers: one through six.
• Press MATH and arrow over to PRB.
• Press 5:randInt( and enter 1,6). Press ENTER. Record the number. Press ENTER and record that number.
• The two numbers are for two of the columns.
• The quiz scores (20 of them) in these 2 columns are the cluster sample.
3. Create a simple random sample of 15 quiz scores.
• Use the numbering one through 60.
• Press MATH. Arrow over to PRB. Press 5:randInt( and enter 1, 60).
• Press ENTER 15 times and record the numbers.
• Record the quiz scores that correspond to these numbers.
• These 15 quiz scores are the systematic sample.
4. Create a systematic sample of 12 quiz scores.
• Use the numbering one through 60.
• Press MATH. Arrow over to PRB. Press 5:randInt( and enter 1, 60).
• Press ENTER. Record the number and the first quiz score. From that number, count ten quiz scores and record that quiz score. Keep counting ten quiz scores and recording the quiz score until you have a sample of 12 quiz scores. You may wrap around (go back to the beginning).
Example $10$
Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
1. A soccer coach selects six players from a group of boys aged eight to ten, seven players from a group of boys aged 11 to 12, and three players from a group of boys aged 13 to 14 to form a recreational soccer team.
2. A pollster interviews all human resource personnel in five different high tech companies.
3. A high school educational researcher interviews 50 high school female teachers and 50 high school male teachers.
4. A medical researcher interviews every third cancer patient from a list of cancer patients at a local hospital.
5. A high school counselor uses a computer to generate 50 random numbers and then picks students whose names correspond to the numbers.
6. A student interviews classmates in his algebra class to determine how many pairs of jeans a student owns, on the average.
Answer
a. stratified; b. cluster; c. stratified; d. systematic; e. simple random; f.convenience
Exercise $11$
Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
A high school principal polls 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors regarding policy changes for after school activities.
Answer
stratified
If we were to examine two samples representing the same population, even if we used random sampling methods for the samples, they would not be exactly the same. Just as there is variation in data, there is variation in samples. As you become accustomed to sampling, the variability will begin to seem natural.
Example $12$: Sampling
Suppose ABC College has 10,000 part-time students (the population). We are interested in the average amount of money a part-time student spends on books in the fall term. Asking all 10,000 students is an almost impossible task. Suppose we take two different samples.
First, we use convenience sampling and survey ten students from a first term organic chemistry class. Many of these students are taking first term calculus in addition to the organic chemistry class. The amount of money they spend on books is as follows:
$128;$87; $173;$116; $130;$204; $147;$189; $93;$153
The second sample is taken using a list of senior citizens who take P.E. classes and taking every fifth senior citizen on the list, for a total of ten senior citizens. They spend:
$50;$40; $36;$15; $50;$100; $40;$53; $22;$22
a. Do you think that either of these samples is representative of (or is characteristic of) the entire 10,000 part-time student population?
Answer
a. No. The first sample probably consists of science-oriented students. Besides the chemistry course, some of them are also taking first-term calculus. Books for these classes tend to be expensive. Most of these students are, more than likely, paying more than the average part-time student for their books. The second sample is a group of senior citizens who are, more than likely, taking courses for health and interest. The amount of money they spend on books is probably much less than the average parttime student. Both samples are biased. Also, in both cases, not all students have a chance to be in either sample.
b. Since these samples are not representative of the entire population, is it wise to use the results to describe the entire population?
Answer
b. No. For these samples, each member of the population did not have an equally likely chance of being chosen.
Now, suppose we take a third sample. We choose ten different part-time students from the disciplines of chemistry, math, English, psychology, sociology, history, nursing, physical education, art, and early childhood development. (We assume that these are the only disciplines in which part-time students at ABC College are enrolled and that an equal number of part-time students are enrolled in each of the disciplines.) Each student is chosen using simple random sampling. Using a calculator, random numbers are generated and a student from a particular discipline is selected if he or she has a corresponding number. The students spend the following amounts:
$180;$50; $150;$85; $260;$75; $180;$200; $200;$150
c. Is the sample biased?
Answer
Students often ask if it is "good enough" to take a sample, instead of surveying the entire population. If the survey is done well, the answer is yes.
Exercise $12$
A local radio station has a fan base of 20,000 listeners. The station wants to know if its audience would prefer more music or more talk shows. Asking all 20,000 listeners is an almost impossible task.
The station uses convenience sampling and surveys the first 200 people they meet at one of the station’s music concert events. 24 people said they’d prefer more talk shows, and 176 people said they’d prefer more music.
Do you think that this sample is representative of (or is characteristic of) the entire 20,000 listener population?
Answer
The sample probably consists more of people who prefer music because it is a concert event. Also, the sample represents only those who showed up to the event earlier than the majority. The sample probably doesn’t represent the entire fan base and is probably biased towards people who would prefer music.
Collaborative Exercise $8$
As a class, determine whether or not the following samples are representative. If they are not, discuss the reasons.
1. To find the average GPA of all students in a university, use all honor students at the university as the sample.
2. To find out the most popular cereal among young people under the age of ten, stand outside a large supermarket for three hours and speak to every twentieth child under age ten who enters the supermarket.
3. To find the average annual income of all adults in the United States, sample U.S. congressmen. Create a cluster sample by considering each state as a stratum (group). By using simple random sampling, select states to be part of the cluster. Then survey every U.S. congressman in the cluster.
4. To determine the proportion of people taking public transportation to work, survey 20 people in New York City. Conduct the survey by sitting in Central Park on a bench and interviewing every person who sits next to you.
5. To determine the average cost of a two-day stay in a hospital in Massachusetts, survey 100 hospitals across the state using simple random sampling.
Variation in Data
Variation is present in any set of data. For example, 16-ounce cans of beverage may contain more or less than 16 ounces of liquid. In one study, eight 16 ounce cans were measured and produced the following amount (in ounces) of beverage:
15.8; 16.1; 15.2; 14.8; 15.8; 15.9; 16.0; 15.5
Measurements of the amount of beverage in a 16-ounce can may vary because different people make the measurements or because the exact amount, 16 ounces of liquid, was not put into the cans. Manufacturers regularly run tests to determine if the amount of beverage in a 16-ounce can falls within the desired range.
Be aware that as you take data, your data may vary somewhat from the data someone else is taking for the same purpose. This is completely natural. However, if two or more of you are taking the same data and get very different results, it is time for you and the others to reevaluate your data-taking methods and your accuracy.
Variation in Samples
It was mentioned previously that two or more samples from the same population, taken randomly, and having close to the same characteristics of the population will likely be different from each other. Suppose Doreen and Jung both decide to study the average amount of time students at their college sleep each night. Doreen and Jung each take samples of 500 students. Doreen uses systematic sampling and Jung uses cluster sampling. Doreen's sample will be different from Jung's sample. Even if Doreen and Jung used the same sampling method, in all likelihood their samples would be different. Neither would be wrong, however.
Think about what contributes to making Doreen’s and Jung’s samples different.
If Doreen and Jung took larger samples (i.e. the number of data values is increased), their sample results (the average amount of time a student sleeps) might be closer to the actual population average. But still, their samples would be, in all likelihood, different from each other. This variability in samples cannot be stressed enough.
Size of a Sample
The size of a sample (often called the number of observations) is important. The examples you have seen in this book so far have been small. Samples of only a few hundred observations, or even smaller, are sufficient for many purposes. In polling, samples that are from 1,200 to 1,500 observations are considered large enough and good enough if the survey is random and is well done. You will learn why when you study confidence intervals.
Be aware that many large samples are biased. For example, call-in surveys are invariably biased, because people choose to respond or not.
Collaborative Exercise $8$
Divide into groups of two, three, or four. Your instructor will give each group one six-sided die. Try this experiment twice. Roll one fair die (six-sided) 20 times. Record the number of ones, twos, threes, fours, fives, and sixes you get in the following table (“frequency” is the number of times a particular face of the die occurs):
First Experiment (20 rolls) Second Experiment (20 rolls)
Face on Die Frequency Face on Die Frequency
1
2
3
4
5
6
Did the two experiments have the same results? Probably not. If you did the experiment a third time, do you expect the results to be identical to the first or second experiment? Why or why not?
Which experiment had the correct results? They both did. The job of the statistician is to see through the variability and draw appropriate conclusions.
Critical Evaluation
We need to evaluate the statistical studies we read about critically and analyze them before accepting the results of the studies. Common problems to be aware of include
• Problems with samples: A sample must be representative of the population. A sample that is not representative of the population is biased. Biased samples that are not representative of the population give results that are inaccurate and not valid.
• Self-selected samples: Responses only by people who choose to respond, such as call-in surveys, are often unreliable.
• Sample size issues: Samples that are too small may be unreliable. Larger samples are better, if possible. In some situations, having small samples is unavoidable and can still be used to draw conclusions. Examples: crash testing cars or medical testing for rare conditions
• Undue influence: collecting data or asking questions in a way that influences the response
• Non-response or refusal of subject to participate: The collected responses may no longer be representative of the population. Often, people with strong positive or negative opinions may answer surveys, which can affect the results.
• Causality: A relationship between two variables does not mean that one causes the other to occur. They may be related (correlated) because of their relationship through a different variable.
• Self-funded or self-interest studies: A study performed by a person or organization in order to support their claim. Is the study impartial? Read the study carefully to evaluate the work. Do not automatically assume that the study is good, but do not automatically assume the study is bad either. Evaluate it on its merits and the work done.
• Misleading use of data: improperly displayed graphs, incomplete data, or lack of context
• Confounding: When the effects of multiple factors on a response cannot be separated. Confounding makes it difficult or impossible to draw valid conclusions about the effect of each factor.
References
1. Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/default.asp (accessed May 1, 2013).
2. Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/methodology.asp (accessed May 1, 2013).
3. Gallup-Healthways Well-Being Index. http://www.gallup.com/poll/146822/ga...questions.aspx (accessed May 1, 2013).
4. Data from www.bookofodds.com/Relationsh...-the-President
5. Dominic Lusinchi, “’President’ Landon and the 1936 Literary Digest Poll: Were Automobile and Telephone Owners to Blame?” Social Science History 36, no. 1: 23-54 (2012), ssh.dukejournals.org/content/36/1/23.abstract (accessed May 1, 2013).
6. “The Literary Digest Poll,” Virtual Laboratories in Probability and Statistics http://www.math.uah.edu/stat/data/LiteraryDigest.html (accessed May 1, 2013).
7. “Gallup Presidential Election Trial-Heat Trends, 1936–2008,” Gallup Politics http://www.gallup.com/poll/110548/ga...9362004.aspx#4 (accessed May 1, 2013).
8. The Data and Story Library, lib.stat.cmu.edu/DASL/Datafiles/USCrime.html (accessed May 1, 2013).
9. LBCC Distance Learning (DL) program data in 2010-2011, http://de.lbcc.edu/reports/2010-11/f...hts.html#focus (accessed May 1, 2013).
10. Data from San Jose Mercury News
Review
Data are individual items of information that come from a population or sample. Data may be classified as qualitative, quantitative continuous, or quantitative discrete.
Because it is not practical to measure the entire population in a study, researchers use samples to represent the population. A random sample is a representative group from the population chosen by using a method that gives each individual in the population an equal chance of being included in the sample. Random sampling methods include simple random sampling, stratified sampling, cluster sampling, and systematic sampling. Convenience sampling is a nonrandom method of choosing a sample that often produces biased data.
Samples that contain different individuals result in different data. This is true even when the samples are well-chosen and representative of the population. When properly selected, larger samples model the population more closely than smaller samples. There are many different potential problems that can affect the reliability of a sample. Statistical data needs to be critically analyzed, not simply accepted.
Footnotes
1. lastbaldeagle. 2013. On Tax Day, House to Call for Firing Federal Workers Who Owe Back Taxes. Opinion poll posted online at: www.youpolls.com/details.aspx?id=12328 (accessed May 1, 2013).
2. Scott Keeter et al., “Gauging the Impact of Growing Nonresponse on Estimates from a National RDD Telephone Survey,” Public Opinion Quarterly 70 no. 5 (2006), http://poq.oxfordjournals.org/content/70/5/759.full (accessed May 1, 2013).
3. Frequently Asked Questions, Pew Research Center for the People & the Press, www.people-press.org/methodol...wer-your-polls (accessed May 1, 2013).
Glossary
Cluster Sampling
a method for selecting a random sample and dividing the population into groups (clusters); use simple random sampling to select a set of clusters. Every individual in the chosen clusters is included in the sample.
Continuous Random Variable
a random variable (RV) whose outcomes are measured; the height of trees in the forest is a continuous RV.
Convenience Sampling
a nonrandom method of selecting a sample; this method selects individuals that are easily accessible and may result in biased data.
Discrete Random Variable
a random variable (RV) whose outcomes are counted
Nonsampling Error
an issue that affects the reliability of sampling data other than natural variation; it includes a variety of human errors including poor study design, biased sampling methods, inaccurate information provided by study participants, data entry errors, and poor analysis.
Qualitative Data
See Data.
Quantitative Data
See Data.
Random Sampling
a method of selecting a sample that gives every member of the population an equal chance of being selected.
Sampling Bias
not all members of the population are equally likely to be selected
Sampling Error
the natural variation that results from selecting a sample to represent a larger population; this variation decreases as the sample size increases, so selecting larger samples reduces sampling error.
Sampling with Replacement
Once a member of the population is selected for inclusion in a sample, that member is returned to the population for the selection of the next individual.
Sampling without Replacement
A member of the population may be chosen for inclusion in a sample only once. If chosen, the member is not returned to the population before the next selection.
Simple Random Sampling
a straightforward method for selecting a random sample; give each member of the population a number. Use a random number generator to select a set of labels. These randomly selected labels identify the members of your sample.
Stratified Sampling
a method for selecting a random sample used to ensure that subgroups of the population are represented adequately; divide the population into groups (strata). Use simple random sampling to identify a proportionate number of individuals from each stratum.
Systematic Sampling
a method for selecting a random sample; list the members of the population. Use simple random sampling to select a starting point in the population. Let k = (number of individuals in the population)/(number of individuals needed in the sample). Choose every kth individual in the list starting with the one that was randomly selected. If necessary, return to the beginning of the population list to complete your sample. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.03%3A_Data_Sampling_and_Variation_in_Data_and_Sampling.txt |
Once you have a set of data, you will need to organize it so that you can analyze how frequently each datum occurs in the set. However, when calculating the frequency, you may need to round your answers so that they are as precise as possible.
Answers and Rounding Off
A simple way to round off answers is to carry your final answer one more decimal place than was present in the original data. Round off only the final answer. Do not round off any intermediate results, if possible. If it becomes necessary to round off intermediate results, carry them to at least twice as many decimal places as the final answer. For example, the average of the three quiz scores four, six, and nine is 6.3, rounded off to the nearest tenth, because the data are whole numbers. Most answers will be rounded off in this manner.
It is not necessary to reduce most fractions in this course. Especially in Probability Topics, the chapter on probability, it is more helpful to leave an answer as an unreduced fraction.
Levels of Measurement
The way a set of data is measured is called its level of measurement. Correct statistical procedures depend on a researcher being familiar with levels of measurement. Not every statistical operation can be used with every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest level):
• Nominal scale level
• Ordinal scale level
• Interval scale level
• Ratio scale level
Data that is measured using a nominal scale is qualitative. Categories, colors, names, labels and favorite foods along with yes or no responses are examples of nominal level data. Nominal scale data are not ordered. For example, trying to classify people according to their favorite food does not make any sense. Putting pizza first and sushi second is not meaningful.
Smartphone companies are another example of nominal scale data. Some examples are Sony, Motorola, Nokia, Samsung and Apple. This is just a list and there is no agreed upon order. Some people may favor Apple but that is a matter of opinion. Nominal scale data cannot be used in calculations.
Data that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. An example of ordinal scale data is a list of the top five national parks in the United States. The top five national parks in the United States can be ranked from one to five but we cannot measure differences between the data.
Another example of using the ordinal scale is a cruise survey where the responses to questions about the cruise are “excellent,” “good,” “satisfactory,” and “unsatisfactory.” These responses are ordered from the most desired response to the least desired. But the differences between two pieces of data cannot be measured. Like the nominal scale data, ordinal scale data cannot be used in calculations.
Data that is measured using the interval scale is similar to ordinal level data because it has a definite ordering but there is a difference between data. The differences between interval scale data can be measured though the data does not have a starting point.
Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In both temperature measurements, 40° is equal to 100° minus 60°. Differences make sense. But 0 degrees does not because, in both scales, 0 is not the absolute lowest temperature. Temperatures like -10° F and -15° C exist and are colder than 0.
Interval level data can be used in calculations, but one type of comparison cannot be done. 80° C is not four times as hot as 20° C (nor is 80° F four times as hot as 20° F). There is no meaning to the ratio of 80 to 20 (or four to one).
Data that is measured using the ratio scale takes care of the ratio problem and gives you the most information. Ratio scale data is like interval scale data, but it has a 0 point and ratios can be calculated. For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a possible 100 points). The exams are machine-graded.
The data can be put in order from lowest to highest: 20, 68, 80, 92.
The differences between the data have meaning. The score 92 is more than the score 68 by 24 points. Ratios can be calculated. The smallest score is 0. So 80 is four times 20. The score of 80 is four times better than the score of 20.
Frequency
Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows:
5; 6; 3; 3; 2; 4; 7; 5; 2; 3; 5; 6; 5; 4; 4; 3; 5; 2; 5; 3.
Table lists the different data values in ascending order and their frequencies.
Table $1$: Frequency Table of Student Work Hours
DATA VALUE FREQUENCY
2 3
3 5
4 3
5 6
6 2
7 1
Definition: Relative Frequency
A frequency is the number of times a value of the data occurs. According to Table Table $1$, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample.
Definition: Relative frequencies
A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals.
Table $2$: Frequency Table of Student Work Hours with Relative Frequencies
DATA VALUE FREQUENCY RELATIVE FREQUENCY
2 3 $\frac{3}{20}$ or 0.15
3 5 $\frac{5}{20}$ or 0.25
4 3 $\frac{3}{20}$ or 0.15
5 6 $\frac{6}{20}$ or 0.30
6 2 $\frac{2}{20}$ or 0.10
7 1 $\frac{1}{20}$ or 0.05
The sum of the values in the relative frequency column of Table $2$ is $\frac{20}{20}$, or 1.
Definition: Cumulative Relative Frequency
Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in Table $3$.
Table $3$: Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies
DATA VALUE FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY
2 3 320320 or 0.15 0.15
3 5 520520 or 0.25 0.15 + 0.25 = 0.40
4 3 320320 or 0.15 0.40 + 0.15 = 0.55
5 6 620620 or 0.30 0.55 + 0.30 = 0.85
6 2 220220 or 0.10 0.85 + 0.10 = 0.95
7 1 120120 or 0.05 0.95 + 0.05 = 1.00
The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated.
Because of rounding, the relative frequency column may not always sum to one, and the last entry in the cumulative relative frequency column may not be one. However, they each should be close to one.
Table $4$ represents the heights, in inches, of a sample of 100 male semiprofessional soccer players.
Table $4$: Frequency Table of Soccer Player Height
HEIGHTS (INCHES) FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY
59.95–61.95 5 $\frac{5}{100} = 0.05$ $0.05$
61.95–63.95 3 $\frac{3}{100} = 0.03$ $0.05 + 0.03 = 0.08$
63.95–65.95 15 $\frac{15}{100} = 0.15$ $0.08 + 0.15 = 0.23$
65.95–67.95 40 $\frac{40}{100} = 0.40$ $0.23 + 0.40 = 0.63$
67.95–69.95 17 $\frac{17}{100} = 0.17$ $0.63 + 0.17 = 0.80$
69.95–71.95 12 $\frac{12}{100} = 0.12$ $0.80 + 0.12 = 0.92$
71.95–73.95 7 $\frac{7}{100} = 0.07$ $0.92 + 0.07 = 0.99$
73.95–75.95 1 $\frac{1}{100} = 0.01$ $0.99 + 0.01 = 1.00$
Total = 100 Total = 1.00
The data in this table have been grouped into the following intervals:
• 61.95 to 63.95 inches
• 63.95 to 65.95 inches
• 65.95 to 67.95 inches
• 67.95 to 69.95 inches
• 69.95 to 71.95 inches
• 71.95 to 73.95 inches
• 73.95 to 75.95 inches
This example is used again in Descriptive Statistics, where the method used to compute the intervals will be explained.
In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints.
Exercise $1$
1. From the Table $4$, find the percentage of heights that are less than 65.95 inches.
2. Find the percentage of heights that fall between 61.95 and 65.95 inches.
Answer
1. If you look at the first, second, and third rows, the heights are all less than 65.95 inches. There are $5 + 3 + 15 = 23$ players whose heights are less than 65.95 inches. The percentage of heights less than 65.95 inches is then $\frac{23}{100}$ or 23%. This percentage is the cumulative relative frequency entry in the third row.
2. Add the relative frequencies in the second and third rows: $0.03 + 0.15 = 0.18$ or 18%.
Exercise $2$
Table $5$ shows the amount, in inches, of annual rainfall in a sample of towns.
1. Find the percentage of rainfall that is less than 9.01 inches.
2. Find the percentage of rainfall that is between 6.99 and 13.05 inches.
Table $5$
Rainfall (Inches) Frequency Relative Frequency Cumulative Relative Frequency
2.95–4.97 6 $\frac{6}{50} = 0.12$ $0.12$
4.97–6.99 7 $\frac{7}{50} = 0.14$ $0.12 + 0.14 = 0.26$
6.99–9.01 15 $\frac{15}{50} = 0.30$ $0.26 + 0.30 = 0.56$
9.01–11.03 8 $\frac{8}{50} = 0.16$ $0.56 + 0.16 = 0.72$
11.03–13.05 9 $\frac{9}{50} = 0.18$ $0.72 + 0.18 = 0.90$
13.05–15.07 5 $\frac{5}{50} = 0.10$ $0.90 + 0.10 = 1.00$
Total = 50 Total = 1.00
Answer
1. $0.56$ or $56%$
2. $0.30 + 0.16 + 0.18 = 0.64$ or $64%$
Exercise $3$
Use the heights of the 100 male semiprofessional soccer players in Table $4$. Fill in the blanks and check your answers.
1. The percentage of heights that are from 67.95 to 71.95 inches is: ____.
2. The percentage of heights that are from 67.95 to 73.95 inches is: ____.
3. The percentage of heights that are more than 65.95 inches is: ____.
4. The number of players in the sample who are between 61.95 and 71.95 inches tall is: ____.
5. What kind of data are the heights?
6. Describe how you could gather this data (the heights) so that the data are characteristic of all male semiprofessional soccer players.
Remember, you count frequencies. To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row.
Answer
1. 29%
2. 36%
3. 77%
4. 87
5. quantitative continuous
6. get rosters from each team and choose a simple random sample from each
Exercise $4$
From Table $5$, find the number of towns that have rainfall between 2.95 and 9.01 inches.
Answer
$6 + 7 + 15 = 28$ towns
Collaborative Exercise $7$
In your class, have someone conduct a survey of the number of siblings (brothers and sisters) each student has. Create a frequency table. Add to it a relative frequency column and a cumulative relative frequency column. Answer the following questions:
1. What percentage of the students in your class have no siblings?
2. What percentage of the students have from one to three siblings?
3. What percentage of the students have fewer than three siblings?
Example $7$
Nineteen people were asked how many miles, to the nearest mile, they commute to work each day. The data are as follows: 2; 5; 7; 3; 2; 10; 18; 15; 20; 7; 10; 18; 5; 12; 13; 12; 4; 5; 10. Table $6$ was produced:
Table $6$: Frequency of Commuting Distances
DATA FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY
3 3 $\frac{3}{19}$ 0.1579
4 1 $\frac{1}{19}$ 0.2105
5 3 $\frac{3}{19}$ 0.1579
7 2 $\frac{2}{19}$ 0.2632
10 3 $\frac{3}{19}$ 0.4737
12 2 $\frac{2}{19}$ 0.7895
13 1 $\frac{1}{19}$ 0.8421
15 1 $\frac{1}{19}$ 0.8948
18 1 $\frac{1}{19}$ 0.9474
20 1 $\frac{1}{19}$ 1.0000
1. Is the table correct? If it is not correct, what is wrong?
2. True or False: Three percent of the people surveyed commute three miles. If the statement is not correct, what should it be? If the table is incorrect, make the corrections.
3. What fraction of the people surveyed commute five or seven miles?
4. What fraction of the people surveyed commute 12 miles or more? Less than 12 miles? Between five and 13 miles (not including five and 13 miles)?
Answer
1. No. The frequency column sums to 18, not 19. Not all cumulative relative frequencies are correct.
2. False. The frequency for three miles should be one; for two miles (left out), two. The cumulative relative frequency column should read: 0.1052, 0.1579, 0.2105, 0.3684, 0.4737, 0.6316, 0.7368, 0.7895, 0.8421, 0.9474, 1.0000.
3. $\frac{5}{19}$
4. $\frac{7}{19}$, $\frac{12}{19}$, $\frac{7}{19}$
Exercise $8$
Table $5$ represents the amount, in inches, of annual rainfall in a sample of towns. What fraction of towns surveyed get between 11.03 and 13.05 inches of rainfall each year?
Answer
$\frac{9}{50}$
Example $9$
Table $7$ contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012.
Table $7$: Total Number of Deaths Worldwide as a Result of Earthquakes
Year Total Number of Deaths
2000 231
2001 21,357
2002 11,685
2003 33,819
2004 228,802
2005 88,003
2006 6,605
2007 712
2008 88,011
2009 1,790
2010 320,120
2011 21,953
2012 768
Total 823,356
Answer the following questions.
1. What is the frequency of deaths measured from 2006 through 2009?
2. What percentage of deaths occurred after 2009?
3. What is the relative frequency of deaths that occurred in 2003 or earlier?
4. What is the percentage of deaths that occurred in 2004?
5. What kind of data are the numbers of deaths?
6. The Richter scale is used to quantify the energy produced by an earthquake. Examples of Richter scale numbers are 2.3, 4.0, 6.1, and 7.0. What kind of data are these numbers?
Answer
1. 97,118 (11.8%)
2. 41.6%
3. 67,092/823,356 or 0.081 or 8.1 %
4. 27.8%
5. Quantitative discrete
6. Quantitative continuous
Exercise $10$
Table $8$ contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011.
Table $8$:
Year Total Number of Crashes Year Total Number of Crashes
1994 36,254 2004 38,444
1995 37,241 2005 39,252
1996 37,494 2006 38,648
1997 37,324 2007 37,435
1998 37,107 2008 34,172
1999 37,140 2009 30,862
2000 37,526 2010 30,296
2001 37,862 2011 29,757
2002 38,491 Total 653,782
2003 38,477
Answer the following questions.
1. What is the frequency of deaths measured from 2000 through 2004?
2. What percentage of deaths occurred after 2006?
3. What is the relative frequency of deaths that occurred in 2000 or before?
4. What is the percentage of deaths that occurred in 2011?
5. What is the cumulative relative frequency for 2006? Explain what this number tells you about the data.
Answer
1. 190,800 (29.2%)
2. 24.9%
3. 260,086/653,782 or 39.8%
4. 4.6%
5. 75.1% of all fatal traffic crashes for the period from 1994 to 2011 happened from 1994 to 2006.
Review
Some calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that generated that value were only accurate to the nearest tenth. Round off your final answer to one more decimal place than was present in the original data. This means that if you have data measured to the nearest tenth of a unit, report the final statistic to the nearest hundredth.
In addition to rounding your answers, you can measure your data using the following four levels of measurement.
• Nominal scale level: data that cannot be ordered nor can it be used in calculations
• Ordinal scale level: data that can be ordered; the differences cannot be measured
• Interval scale level: data with a definite ordering but no starting point; the differences can be measured, but there is no such thing as a ratio.
• Ratio scale level: data with a starting point that can be ordered; the differences have meaning and ratios can be calculated.
When organizing data, it is important to know how many times a value appears. How many statistics students study five hours or more for an exam? What percent of families on our block own two pets? Frequency, relative frequency, and cumulative relative frequency are measures that answer questions like these.
Exercise $11$
What type of measure scale is being used? Nominal, ordinal, interval or ratio.
1. High school soccer players classified by their athletic ability: Superior, Average, Above average
2. Baking temperatures for various main dishes: 350, 400, 325, 250, 300
3. The colors of crayons in a 24-crayon box
4. Social security numbers
5. Incomes measured in dollars
6. A satisfaction survey of a social website by number: 1 = very satisfied, 2 = somewhat satisfied, 3 = not satisfied
7. Political outlook: extreme left, left-of-center, right-of-center, extreme right
8. Time of day on an analog watch
9. The distance in miles to the closest grocery store
10. The dates 1066, 1492, 1644, 1947, and 1944
11. The heights of 21–65 year-old women
12. Common letter grades: A, B, C, D, and F
Answer
1. ordinal
2. interval
3. nominal
4. nominal
5. ratio
6. ordinal
7. nominal
8. interval
9. ratio
10. interval
11. ratio
12. ordinal
Glossary
Cumulative Relative Frequency
The term applies to an ordered set of observations from smallest to largest. The cumulative relative frequency is the sum of the relative frequencies for all values that are less than or equal to the given value.
Frequency
the number of times a value of the data occurs
Relative Frequency
the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes to the total number of outcomes | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.04%3A_Frequency_Frequency_Tables_and_Levels_of_Measurement.txt |
Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol? Questions like these are answered using randomized experiments. In this module, you will learn important aspects of experimental design. Proper study design ensures the production of reliable, accurate data.
The purpose of an experiment is to investigate the relationship between two variables. When one variable causes change in another, we call the first variable the explanatory variable. The affected variable is called the response variable. In a randomized experiment, the researcher manipulates values of the explanatory variable and measures the resulting changes in the response variable. The different values of the explanatory variable are called treatments. An experimental unit is a single object or individual to be measured.
You want to investigate the effectiveness of vitamin E in preventing disease. You recruit a group of subjects and ask them if they regularly take vitamin E. You notice that the subjects who take vitamin E exhibit better health on average than those who do not. Does this prove that vitamin E is effective in disease prevention? It does not. There are many differences between the two groups compared in addition to vitamin E consumption. People who take vitamin E regularly often take other steps to improve their health: exercise, diet, other vitamin supplements, choosing not to smoke. Any one of these factors could be influencing health. As described, this study does not prove that vitamin E is the key to disease prevention.
Additional variables that can cloud a study are called lurking variables. In order to prove that the explanatory variable is causing a change in the response variable, it is necessary to isolate the explanatory variable. The researcher must design her experiment in such a way that there is only one difference between groups being compared: the planned treatments. This is accomplished by the random assignment of experimental units to treatment groups. When subjects are assigned treatments randomly, all of the potential lurking variables are spread equally among the groups. At this point the only difference between groups is the one imposed by the researcher. Different outcomes measured in the response variable, therefore, must be a direct result of the different treatments. In this way, an experiment can prove a cause-and-effect connection between the explanatory and response variables.
The power of suggestion can have an important influence on the outcome of an experiment. Studies have shown that the expectation of the study participant can be as important as the actual medication. In one study of performance-enhancing drugs, researchers noted:
Results showed that believing one had taken the substance resulted in [performance] times almost as fast as those associated with consuming the drug itself. In contrast, taking the drug without knowledge yielded no significant performance increment.1
When participation in a study prompts a physical response from a participant, it is difficult to isolate the effects of the explanatory variable. To counter the power of suggestion, researchers set aside one treatment group as a control group. This group is given a placebo treatment–a treatment that cannot influence the response variable. The control group helps researchers balance the effects of being in an experiment with the effects of the active treatments. Of course, if you are participating in a study and you know that you are receiving a pill which contains no actual medication, then the power of suggestion is no longer a factor. Blinding in a randomized experiment preserves the power of suggestion. When a person involved in a research study is blinded, he does not know who is receiving the active treatment(s) and who is receiving the placebo treatment. A double-blind experiment is one in which both the subjects and the researchers involved with the subjects are blinded.
Example \(1\)
Researchers want to investigate whether taking aspirin regularly reduces the risk of heart attack. Four hundred men between the ages of 50 and 84 are recruited as participants. The men are divided randomly into two groups: one group will take aspirin, and the other group will take a placebo. Each man takes one pill each day for three years, but he does not know whether he is taking aspirin or the placebo. At the end of the study, researchers count the number of men in each group who have had heart attacks.
Identify the following values for this study: population, sample, experimental units, explanatory variable, response variable, treatments.
Answer
• The population is men aged 50 to 84.
• The sample is the 400 men who participated.
• The experimental units are the individual men in the study.
• The explanatory variable is oral medication.
• The treatments are aspirin and a placebo.
• The response variable is whether a subject had a heart attack.
Example \(2\)
The Smell & Taste Treatment and Research Foundation conducted a study to investigate whether smell can affect learning. Subjects completed mazes multiple times while wearing masks. They completed the pencil and paper mazes three times wearing floral-scented masks, and three times with unscented masks. Participants were assigned at random to wear the floral mask during the first three trials or during the last three trials. For each trial, researchers recorded the time it took to complete the maze and the subject’s impression of the mask’s scent: positive, negative, or neutral.
1. Describe the explanatory and response variables in this study.
2. What are the treatments?
3. Identify any lurking variables that could interfere with this study.
4. Is it possible to use blinding in this study?
Answer
1. The explanatory variable is scent, and the response variable is the time it takes to complete the maze.
2. There are two treatments: a floral-scented mask and an unscented mask.
3. All subjects experienced both treatments. The order of treatments was randomly assigned so there were no differences between the treatment groups. Random assignment eliminates the problem of lurking variables.
4. Subjects will clearly know whether they can smell flowers or not, so subjects cannot be blinded in this study. Researchers timing the mazes can be blinded, though. The researcher who is observing a subject will not know which mask is being worn.
Example \(3\)
A researcher wants to study the effects of birth order on personality. Explain why this study could not be conducted as a randomized experiment. What is the main problem in a study that cannot be designed as a randomized experiment?
Answer
The explanatory variable is birth order. You cannot randomly assign a person’s birth order. Random assignment eliminates the impact of lurking variables. When you cannot assign subjects to treatment groups at random, there will be differences between the groups other than the explanatory variable.
Exercise \(4\)
You are concerned about the effects of texting on driving performance. Design a study to test the response time of drivers while texting and while driving only. How many seconds does it take for a driver to respond when a leading car hits the brakes?
1. Describe the explanatory and response variables in the study.
2. What are the treatments?
3. What should you consider when selecting participants?
4. Your research partner wants to divide participants randomly into two groups: one to drive without distraction and one to text and drive simultaneously. Is this a good idea? Why or why not?
5. Identify any lurking variables that could interfere with this study.
6. How can blinding be used in this study?
Answer
1. Explanatory: presence of distraction from texting; response: response time measured in seconds
2. Driving without distraction and driving while texting
3. Answers will vary. Possible responses: Do participants regularly send and receive text messages? How long has the subject been driving? What is the age of the participants? Do participants have similar texting and driving experience?
4. This is not a good plan because it compares drivers with different abilities. It would be better to assign both treatments to each participant in random order.
5. Possible responses include: texting ability, driving experience, type of phone.
6. The researchers observing the trials and recording response time could be blinded to the treatment being applied.
Ethics
The widespread misuse and misrepresentation of statistical information often gives the field a bad name. Some say that “numbers don’t lie,” but the people who use numbers to support their claims often do.
A recent investigation of famous social psychologist, Diederik Stapel, has led to the retraction of his articles from some of the world’s top journals including Journal of Experimental Social Psychology, Social Psychology, Basic and Applied Social Psychology, British Journal of Social Psychology, and the magazine Science. Diederik Stapel is a former professor at Tilburg University in the Netherlands. Over the past two years, an extensive investigation involving three universities where Stapel has worked concluded that the psychologist is guilty of fraud on a colossal scale. Falsified data taints over 55 papers he authored and 10 Ph.D. dissertations that he supervised.
Stapel did not deny that his deceit was driven by ambition. But it was more complicated than that, he told me. He insisted that he loved social psychology but had been frustrated by the messiness of experimental data, which rarely led to clear conclusions. His lifelong obsession with elegance and order, he said, led him to concoct sexy results that journals found attractive. “It was a quest for aesthetics, for beauty—instead of the truth,” he said. He described his behavior as an addiction that drove him to carry out acts of increasingly daring fraud, like a junkie seeking a bigger and better high.2
The committee investigating Stapel concluded that he is guilty of several practices including:
• creating datasets, which largely confirmed the prior expectations,
• altering data in existing datasets,
• changing measuring instruments without reporting the change, and
• misrepresenting the number of experimental subjects.
Clearly, it is never acceptable to falsify data the way this researcher did. Sometimes, however, violations of ethics are not as easy to spot.
Researchers have a responsibility to verify that proper methods are being followed. The report describing the investigation of Stapel’s fraud states that, “statistical flaws frequently revealed a lack of familiarity with elementary statistics.”3 Many of Stapel’s co-authors should have spotted irregularities in his data. Unfortunately, they did not know very much about statistical analysis, and they simply trusted that he was collecting and reporting data properly.
Many types of statistical fraud are difficult to spot. Some researchers simply stop collecting data once they have just enough to prove what they had hoped to prove. They don’t want to take the chance that a more extensive study would complicate their lives by producing data contradicting their hypothesis.
Professional organizations, like the American Statistical Association, clearly define expectations for researchers. There are even laws in the federal code about the use of research data.
When a statistical study uses human participants, as in medical studies, both ethics and the law dictate that researchers should be mindful of the safety of their research subjects. The U.S. Department of Health and Human Services oversees federal regulations of research studies with the aim of protecting participants. When a university or other research institution engages in research, it must ensure the safety of all human subjects. For this reason, research institutions establish oversight committees known as Institutional Review Boards (IRB). All planned studies must be approved in advance by the IRB. Key protections that are mandated by law include the following:
• Risks to participants must be minimized and reasonable with respect to projected benefits.
• Participants must give informed consent. This means that the risks of participation must be clearly explained to the subjects of the study. Subjects must consent in writing, and researchers are required to keep documentation of their consent.
• Data collected from individuals must be guarded carefully to protect their privacy.
These ideas may seem fundamental, but they can be very difficult to verify in practice. Is removing a participant’s name from the data record sufficient to protect privacy? Perhaps the person’s identity could be discovered from the data that remains. What happens if the study does not proceed as planned and risks arise that were not anticipated? When is informed consent really necessary? Suppose your doctor wants a blood sample to check your cholesterol level. Once the sample has been tested, you expect the lab to dispose of the remaining blood. At that point the blood becomes biological waste. Does a researcher have the right to take it for use in a study?
It is important that students of statistics take time to consider the ethical questions that arise in statistical studies. How prevalent is fraud in statistical studies? You might be surprised—and disappointed. There is a website (www.retractionwatch.com) dedicated to cataloging retractions of study articles that have been proven fraudulent. A quick glance will show that the misuse of statistics is a bigger problem than most people realize.
Vigilance against fraud requires knowledge. Learning the basic theory of statistics will empower you to analyze statistical studies critically.
Example \(5\)
Describe the unethical behavior in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected.
A researcher is collecting data in a community.
1. She selects a block where she is comfortable walking because she knows many of the people living on the street.
2. No one seems to be home at four houses on her route. She does not record the addresses and does not return at a later time to try to find residents at home.
3. She skips four houses on her route because she is running late for an appointment. When she gets home, she fills in the forms by selecting random answers from other residents in the neighborhood.
Answer
1. By selecting a convenient sample, the researcher is intentionally selecting a sample that could be biased. Claiming that this sample represents the community is misleading. The researcher needs to select areas in the community at random.
2. Intentionally omitting relevant data will create bias in the sample. Suppose the researcher is gathering information about jobs and child care. By ignoring people who are not home, she may be missing data from working families that are relevant to her study. She needs to make every effort to interview all members of the target sample.
3. It is never acceptable to fake data. Even though the responses she uses are “real” responses provided by other participants, the duplication is fraudulent and can create bias in the data. She needs to work diligently to interview everyone on her route.
Exercise \(6\)
Describe the unethical behavior, if any, in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected.
A study is commissioned to determine the favorite brand of fruit juice among teens in California.
1. The survey is commissioned by the seller of a popular brand of apple juice.
2. There are only two types of juice included in the study: apple juice and cranberry juice.
3. Researchers allow participants to see the brand of juice as samples are poured for a taste test.
4. Twenty-five percent of participants prefer Brand X, 33% prefer Brand Y and 42% have no preference between the two brands. Brand X references the study in a commercial saying “Most teens like Brand X as much as or more than Brand Y.”
Answer
1. This is not necessarily a problem. The study should be monitored carefully, however, to ensure that the company is not pressuring researchers to return biased results.
2. If the researchers truly want to determine the favorite brand of juice, then researchers should ask teens to compare different brands of the same type of juice. Choosing a sweet juice to compare against a sharp-flavored juice will not lead to an accurate comparison of brand quality.
3. Participants could be biased by the knowledge. The results may be different from those obtained in a blind taste test.
4. The commercial tells the truth, but not the whole truth. It leads consumers to believe that Brand X was preferred by more participants than Brand Y while the opposite is true.
Review
A poorly designed study will not produce reliable data. There are certain key components that must be included in every experiment. To eliminate lurking variables, subjects must be assigned randomly to different treatment groups. One of the groups must act as a control group, demonstrating what happens when the active treatment is not applied. Participants in the control group receive a placebo treatment that looks exactly like the active treatments but cannot influence the response variable. To preserve the integrity of the placebo, both researchers and subjects may be blinded. When a study is designed properly, the only difference between treatment groups is the one imposed by the researcher. Therefore, when groups respond differently to different treatments, the difference must be due to the influence of the explanatory variable.
“An ethics problem arises when you are considering an action that benefits you or some cause you support, hurts or reduces benefits to others, and violates some rule.”4 Ethical violations in statistics are not always easy to spot. Professional associations and federal agencies post guidelines for proper conduct. It is important that you learn basic statistical procedures so that you can recognize proper data analysis.
Exercise \(7\)
Design an experiment. Identify the explanatory and response variables. Describe the population being studied and the experimental units. Explain the treatments that will be used and how they will be assigned to the experimental units. Describe how blinding and placebos may be used to counter the power of suggestion.
Exercise \(7\)
Discuss potential violations of the rule requiring informed consent.
1. Inmates in a correctional facility are offered good behavior credit in return for participation in a study.
2. A research study is designed to investigate a new children’s allergy medication.
3. Participants in a study are told that the new medication being tested is highly promising, but they are not told that only a small portion of participants will receive the new medication. Others will receive placebo treatments and traditional treatments.
Answer
1. Inmates may not feel comfortable refusing participation, or may feel obligated to take advantage of the promised benefits. They may not feel truly free to refuse participation.
2. Parents can provide consent on behalf of their children, but children are not competent to provide consent for themselves.
3. All risks and benefits must be clearly outlined. Study participants must be informed of relevant aspects of the study in order to give appropriate consent.
Footnotes
1 McClung, M. Collins, D. “Because I know it will!”: placebo effects of an ergogenic aid on athletic performance. Journal of Sport & Exercise Psychology. 2007 Jun. 29(3):382-94. Web. April 30, 2013.
2 Y.udhijit Bhattacharjee, “The Mind of a Con Man,” Magazine, New York Times, April 26, 2013. Available online at: http://www.nytimes.com/2013/04/28/ma...src=dayp&_r=2& (accessed May 1, 2013).
3 “Flawed Science: The Fraudulent Research Practices of Social Psychologist Diederik Stapel,” Tillburg University, November 28, 2012, www.tilburguniversity.edu/upl...012_UK_web.pdf (accessed May 1, 2013).
4 Andrew Gelman, “Open Data and Open Methods,” Ethics and Statistics, http://www.stat.columbia.edu/~gelman...nceEthics1.pdf (accessed May 1, 2013).
Glossary
Explanatory Variable
the independent variable in an experiment; the value controlled by researchers
Treatments
different values or components of the explanatory variable applied in an experiment
Response Variable
the dependent variable in an experiment; the value that is measured for change at the end of an experiment
Experimental Unit
any individual or object to be measured
Lurking Variable
a variable that has an effect on a study even though it is neither an explanatory variable nor a response variable
Random Assignment
the act of organizing experimental units into treatment groups using random methods
Control Group
a group in a randomized experiment that receives an inactive treatment but is otherwise managed exactly as the other groups
Informed Consent
Any human subject in a research study must be cognizant of any risks or costs associated with the study. The subject has the right to know the nature of the treatments included in the study, their potential risks, and their potential benefits. Consent must be given freely by an informed, fit participant.
Institutional Review Board
a committee tasked with oversight of research programs that involve human subjects
Placebo
an inactive treatment that has no real effect on the explanatory variable
Blinding
not telling participants which treatment a subject is receiving
Double-blinding
the act of blinding both the subjects of an experiment and the researchers who work with the subjects | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.05%3A_Experimental_Design_and_Ethics.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will demonstrate the systematic sampling technique.
• The student will construct relative frequency tables.
• The student will interpret results and their differences from different data groupings.
Movie Survey
Ask five classmates from a different class how many movies they saw at the theater last month. Do not include rented movies.
1. Record the data.
2. In class, randomly pick one person. On the class list, mark that person’s name. Move down four names on the class list. Mark that person’s name. Continue doing this until you have marked 12 names. You may need to go back to the start of the list. For each marked name record the five data values. You now have a total of 60 data values.
3. For each name marked, record the data.
___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___
___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___
___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___
___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___
___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___
Order the Data
Complete the two relative frequency tables below using your class data.
Frequency of Number of Movies Viewed
Number of Movies Frequency Relative Frequency Cumulative Relative Frequency
0
1
2
3
4
5
6
7+
Frequency of Number of Movies Viewed
Number of Movies Frequency Relative Frequency Cumulative Relative Frequency
0–1
2–3
4–5
6–7+
1. Using the tables, find the percent of data that is at most two. Which table did you use and why?
2. Using the tables, find the percent of data that is at most three. Which table did you use and why?
3. Using the tables, find the percent of data that is more than two. Which table did you use and why?
4. Using the tables, find the percent of data that is more than three. Which table did you use and why?
Discussion Questions
1. Is one of the tables “more correct” than the other? Why or why not?
2. In general, how could you group the data differently? Are there any advantages to either way of grouping the data?
3. Why did you switch between tables, if you did, when answering the question above?
1.07: Sampling Experiment (Worksheet)
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will demonstrate the simple random, systematic, stratified, and cluster sampling techniques.
• The student will explain the details of each procedure used.
In this lab, you will be asked to pick several random samples of restaurants. In each case, describe your procedure briefly, including how you might have used the random number generator, and then list the restaurants in the sample you obtained.
Note 1.7.1
The following section contains restaurants stratified by city into columns and grouped horizontally by entree cost (clusters).
Restaurants Stratified by City and Entree Cost
Restaurants Used in Sample
Entree Cost Under \$10 \$10 to under \$15 \$15 to under \$20 Over \$20
San Jose El Abuelo Taq, Pasta Mia, Emma’s Express, Bamboo Hut Emperor’s Guard, Creekside Inn Agenda, Gervais, Miro’s Blake’s, Eulipia, Hayes Mansion, Germania
Palo Alto Senor Taco, Olive Garden, Taxi’s Ming’s, P.A. Joe’s, Stickney’s Scott’s Seafood, Poolside Grill, Fish Market Sundance Mine, Maddalena’s, Spago’s
Los Gatos Mary’s Patio, Mount Everest, Sweet Pea’s, Andele Taqueria Lindsey’s, Willow Street Toll House Charter House, La Maison Du Cafe
Mountain View Maharaja, New Ma’s, Thai-Rific, Garden Fresh Amber Indian, La Fiesta, Fiesta del Mar, Dawit Austin’s, Shiva’s, Mazeh Le Petit Bistro
Cupertino Hobees, Hung Fu, Samrat, Panda Express Santa Barb. Grill, Mand. Gourmet, Bombay Oven, Kathmandu West Fontana’s, Blue Pheasant Hamasushi, Helios
Sunnyvale Chekijababi, Taj India, Full Throttle, Tia Juana, Lemon Grass Pacific Fresh, Charley Brown’s, Cafe Cameroon, Faz, Aruba’s Lion & Compass, The Palace, Beau Sejour
Santa Clara Rangoli, Armadillo Willy’s, Thai Pepper, Pasand Arthur’s, Katie’s Cafe, Pedro’s, La Galleria Birk’s, Truya Sushi, Valley Plaza Lakeside, Mariani’s
A Simple Random Sample
Pick a simple random sample of 15 restaurants.
1. Describe your procedure.
2. Complete the table with your sample.
1. __________ 6. __________ 11. __________
2. __________ 7. __________ 12. __________
3. __________ 8. __________ 13. __________
4. __________ 9. __________ 14. __________
5. __________ 10. __________ 15. __________
A Systematic Sample
Pick a systematic sample of 15 restaurants.
1. Describe your procedure.
2. Complete the table with your sample.
1. __________ 6. __________ 11. __________
2. __________ 7. __________ 12. __________
3. __________ 8. __________ 13. __________
4. __________ 9. __________ 14. __________
5. __________ 10. __________ 15. __________
A Stratified Sample
Pick a stratified sample, by city, of 20 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number.
1. Describe your procedure.
2. Complete the table with your sample.
1. __________ 6. __________ 11. __________ 16. __________
2. __________ 7. __________ 12. __________ 17. __________
3. __________ 8. __________ 13. __________ 18. __________
4. __________ 9. __________ 14. __________ 19. __________
5. __________ 10. __________ 15. __________ 20. __________
A Stratified Sample
Pick a stratified sample, by entree cost, of 21 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number.
1. Describe your procedure.
2. Complete the table with your sample.
1. __________ 6. __________ 11. __________ 16. __________
2. __________ 7. __________ 12. __________ 17. __________
3. __________ 8. __________ 13. __________ 18. __________
4. __________ 9. __________ 14. __________ 19. __________
5. __________ 10. __________ 15. __________ 20. __________
21. __________
A Cluster Sample
Pick a cluster sample of restaurants from two cities. The number of restaurants will vary.
1. Describe your procedure.
2. Complete the table with your sample.
1. ________ 6. ________ 11. ________ 16. ________ 21. ________
2. ________ 7. ________ 12. ________ 17. ________ 22. ________
3. ________ 8. ________ 13. ________ 18. ________ 23. ________
4. ________ 9. ________ 14. ________ 19. ________ 24. ________
5. ________ 10. ________ 15. ________ 20. ________ 25. ________ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.06%3A_Data_Collection_Experiment_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
1.2: Definitions of Statistics, Probability, and Key Terms
For each of the following eight exercises, identify: a. the population, b. the sample, c. the parameter, d. the statistic, e. the variable, and f. the data. Give examples where appropriate.
Q 1.2.1
A fitness center is interested in the mean amount of time a client exercises in the center each week.
Q. 1.2.2
Ski resorts are interested in the mean age that children take their first ski and snowboard lessons. They need this information to plan their ski classes optimally.
S 1.2.2
1. all children who take ski or snowboard lessons
2. a group of these children
3. the population mean age of children who take their first snowboard lesson
4. the sample mean age of children who take their first snowboard lesson
5. $X =$ the age of one child who takes his or her first ski or snowboard lesson
6. values for $X$, such as 3, 7, and so on
Q 1.2.3
A cardiologist is interested in the mean recovery period of her patients who have had heart attacks.
Q 1.2.4
Insurance companies are interested in the mean health costs each year of their clients, so that they can determine the costs of health insurance.
S 1.2.5
1. the clients of the insurance companies
2. a group of the clients
3. the mean health costs of the clients
4. the mean health costs of the sample
5. $X =$ the health costs of one client
6. values for $X$, such as 34, 9, 82, and so on
Q 1.2.6
A politician is interested in the proportion of voters in his district who think he is doing a good job.
Q 1.2.7
A marriage counselor is interested in the proportion of clients she counsels who stay married.
S 1.2.7
1. all the clients of this counselor
2. a group of clients of this marriage counselor
3. the proportion of all her clients who stay married
4. the proportion of the sample of the counselor’s clients who stay married
5. $X =$ the number of couples who stay married
6. yes, no
Q 1.2.8
Political pollsters may be interested in the proportion of people who will vote for a particular cause.
Q 1.2.9
A marketing company is interested in the proportion of people who will buy a particular product.
S 1.2.9
1. all people (maybe in a certain geographic area, such as the United States)
2. a group of the people
3. the proportion of all people who will buy the product
4. the proportion of the sample who will buy the product
5. $X =$ the number of people who will buy it
6. buy, not buy
Use the following information to answer the next three exercises: A Lake Tahoe Community College instructor is interested in the mean number of days Lake Tahoe Community College math students are absent from class during a quarter.
Q 1.2.10
What is the population she is interested in?
1. all Lake Tahoe Community College students
2. all Lake Tahoe Community College English students
3. all Lake Tahoe Community College students in her classes
4. all Lake Tahoe Community College math students
Q 1.2.11
Consider the following:
$X =$ number of days a Lake Tahoe Community College math student is absent
In this case, $X$ is an example of a:
1. variable.
2. population.
3. statistic.
4. data.
a
Q 1.2.12
The instructor’s sample produces a mean number of days absent of 3.5 days. This value is an example of a:
1. parameter.
2. data.
3. statistic.
4. variable.
1.3: Data, Sampling, and Variation in Data and Sampling
Practice
Exercise 1.3.11
“Number of times per week” is what type of data?
1. qualitative
2. quantitative discrete
3. quantitative continuous
Use the following information to answer the next four exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Antonio, Texas. The first house in the neighborhood around the park was selected randomly, and then the resident of every eighth house in the neighborhood around the park was interviewed.
Exercise 1.3.12
The sampling method was
1. simple random
2. systematic
3. stratified
4. cluster
Answer
b
Exercise 1.3.13
“Duration (amount of time)” is what type of data?
1. qualitative
2. quantitative discrete
3. quantitative continuous
Exercise 1.3.14
The colors of the houses around the park are what kind of data?
1. qualitative
2. quantitative discrete
3. quantitative continuous
Answer
a
Exercise 1.3.15
The population is ______________________
Exercise 1.3.16
Table contains the total number of deaths worldwide as a result of earthquakes from 2000 to 2012.
Year Total Number of Deaths
2000 231
2001 21,357
2002 11,685
2003 33,819
2004 228,802
2005 88,003
2006 6,605
2007 712
2008 88,011
2009 1,790
2010 320,120
2011 21,953
2012 768
Total 823,856
Use Table to answer the following questions.
1. What is the proportion of deaths between 2007 and 2012?
2. What percent of deaths occurred before 2001?
3. What is the percent of deaths that occurred in 2003 or after 2010?
4. What is the fraction of deaths that happened before 2012?
5. What kind of data is the number of deaths?
6. Earthquakes are quantified according to the amount of energy they produce (examples are 2.1, 5.0, 6.7). What type of data is that?
7. What contributed to the large number of deaths in 2010? In 2004? Explain.
Answer
1. 0.5242
2. 0.03%
3. 6.86%
4. $\frac{823,088}{823,856}$
5. quantitative discrete
6. quantitative continuous
7. In both years, underwater earthquakes produced massive tsunamis.
For the following four exercises, determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
Exercise 1.3.17
A group of test subjects is divided into twelve groups; then four of the groups are chosen at random.
Exercise 1.3.18
A market researcher polls every tenth person who walks into a store.
Answer
systematic
Exercise 1.3.19
The first 50 people who walk into a sporting event are polled on their television preferences.
Exercise 1.3.20
A computer generates 100 random numbers, and 100 people whose names correspond with the numbers on the list are chosen.
Answer
simple random
Use the following information to answer the next seven exercises: Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS symptoms have revealed themselves. Of interest is the average (mean) length of time in months patients live once starting the treatment. Two researchers each follow a different set of 40 AIDS patients from the start of treatment until their deaths. The following data (in months) are collected.
3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
Exercise 1.3.21
Complete the tables using the data provided:
Researcher A
Survival Length (in months) Frequency Relative Frequency Cumulative Relative Frequency
0.5–6.5
6.5–12.5
12.5–18.5
18.5–24.5
24.5–30.5
30.5–36.5
36.5–42.5
42.5–48.5
Researcher B
Survival Length (in months) Frequency Relative Frequency Cumulative Relative Frequency
0.5–6.5
6.5–12.5
12.5–18.5
18.5–24.5
24.5–30.5
30.5–36.5
36.5-45.5
Exercise 1.3.22
Determine what the key term data refers to in the above example for Researcher A.
Answer
values for $X$, such as 3, 4, 11, and so on
Exercise 1.3.23
List two reasons why the data may differ.
Exercise 1.3.24
Can you tell if one researcher is correct and the other one is incorrect? Why?
Answer
No, we do not have enough information to make such a claim.
Exercise 1.3.25
Would you expect the data to be identical? Why or why not?
Exercise 1.3.26
How might the researchers gather random data?
Answer
Take a simple random sample from each group. One way is by assigning a number to each patient and using a random number generator to randomly select patients.
Exercise 1.3.27
Suppose that the first researcher conducted his survey by randomly choosing one state in the nation and then randomly picking 40 patients from that state. What sampling method would that researcher have used?
Exercise 1.3.28
Suppose that the second researcher conducted his survey by choosing 40 patients he knew. What sampling method would that researcher have used? What concerns would you have about this data set, based upon the data collection method?
Answer
This would be convenience sampling and is not random.
Use the following data to answer the next five exercises: Two researchers are gathering data on hours of video games played by school-aged children and young adults. They each randomly sample different groups of 150 students from the same school. They collect the following data.
Researcher A
Hours Played per Week Frequency Relative Frequency Cumulative Relative Frequency
0–2 26 0.17 0.17
2–4 30 0.20 0.37
4–6 49 0.33 0.70
6–8 25 0.17 0.87
8–10 12 0.08 0.95
10–12 8 0.05 1
Researcher B
Hours Played per Week Frequency Relative Frequency Cumulative Relative Frequency
0–2 48 0.32 0.32
2–4 51 0.34 0.66
4–6 24 0.16 0.82
6–8 12 0.08 0.90
8–10 11 0.07 0.97
10–12 4 0.03 1
Exercise 1.3.29
Give a reason why the data may differ.
Exercise 1.3.30
Would the sample size be large enough if the population is the students in the school?
Answer
Yes, the sample size of 150 would be large enough to reflect a population of one school.
Exercise 1.3.31
Would the sample size be large enough if the population is school-aged children and young adults in the United States?
Exercise 1.3.32
Researcher A concludes that most students play video games between four and six hours each week. Researcher B concludes that most students play video games between two and four hours each week. Who is correct?
Answer
Even though the specific data support each researcher’s conclusions, the different results suggest that more data need to be collected before the researchers can reach a conclusion.
Exercise 1.3.33
As part of a way to reward students for participating in the survey, the researchers gave each student a gift card to a video game store. Would this affect the data if students knew about the award before the study?
Use the following data to answer the next five exercises: A pair of studies was performed to measure the effectiveness of a new software program designed to help stroke patients regain their problem-solving skills. Patients were asked to use the software program twice a day, once in the morning and once in the evening. The studies observed 200 stroke patients recovering over a period of several weeks. The first study collected the data in the first Table. The second study collected the data in the second Table.
Group Showed improvement No improvement Deterioration
Used program 142 43 15
Did not use program 72 110 18
Group Showed improvement No improvement Deterioration
Used program 105 74 19
Did not use program 89 99 12
Exercise 1.3.34
Given what you know, which study is correct?
Answer
There is not enough information given to judge if either one is correct or incorrect.
Exercise 1.3.35
The first study was performed by the company that designed the software program. The second study was performed by the American Medical Association. Which study is more reliable?
Exercise 1.3.36
Both groups that performed the study concluded that the software works. Is this accurate?
Answer
The software program seems to work because the second study shows that more patients improve while using the software than not. Even though the difference is not as large as that in the first study, the results from the second study are likely more reliable and still show improvement.
Exercise 1.3.37
The company takes the two studies as proof that their software causes mental improvement in stroke patients. Is this a fair statement?
Exercise 1.3.38
Patients who used the software were also a part of an exercise program whereas patients who did not use the software were not. Does this change the validity of the conclusions from Exercise?
Answer
Yes, because we cannot tell if the improvement was due to the software or the exercise; the data is confounded, and a reliable conclusion cannot be drawn. New studies should be performed.
Exercise 1.3.39
Is a sample size of 1,000 a reliable measure for a population of 5,000?
Exercise 1.3.40
Is a sample of 500 volunteers a reliable measure for a population of 2,500?
Answer
No, even though the sample is large enough, the fact that the sample consists of volunteers makes it a self-selected sample, which is not reliable.
Exercise 1.3.41
A question on a survey reads: "Do you prefer the delicious taste of Brand X or the taste of Brand Y?" Is this a fair question?
Exercise 1.3.42
Is a sample size of two representative of a population of five?
Answer
No, even though the sample is a large portion of the population, two responses are not enough to justify any conclusions. Because the population is so small, it would be better to include everyone in the population to get the most accurate data.
Exercise 1.3.43
Is it possible for two experiments to be well run with similar sample sizes to get different data?
Bringing It Together
Exercise 1.3.44
Seven hundred and seventy-one distance learning students at Long Beach City College responded to surveys in the 2010-11 academic year. Highlights of the summary report are listed below.
LBCC Distance Learning Survey Results
Have computer at home 96%
Unable to come to campus for classes 65%
Age 41 or over 24%
Would like LBCC to offer more DL courses 95%
Took DL classes due to a disability 17%
Live at least 16 miles from campus 13%
Took DL courses to fulfill transfer requirements 71%
1. What percent of the students surveyed do not have a computer at home?
2. About how many students in the survey live at least 16 miles from campus?
3. If the same survey were done at Great Basin College in Elko, Nevada, do you think the percentages would be the same? Why?
Exercise 1.3.45
Several online textbook retailers advertise that they have lower prices than on-campus bookstores. However, an important factor is whether the Internet retailers actually have the textbooks that students need in stock. Students need to be able to get textbooks promptly at the beginning of the college term. If the book is not available, then a student would not be able to get the textbook at all, or might get a delayed delivery if the book is back ordered.
A college newspaper reporter is investigating textbook availability at online retailers. He decides to investigate one textbook for each of the following seven subjects: calculus, biology, chemistry, physics, statistics, geology, and general engineering. He consults textbook industry sales data and selects the most popular nationally used textbook in each of these subjects. He visits websites for a random sample of major online textbook sellers and looks up each of these seven textbooks to see if they are available in stock for quick delivery through these retailers. Based on his investigation, he writes an article in which he draws conclusions about the overall availability of all college textbooks through online textbook retailers.
Write an analysis of his study that addresses the following issues: Is his sample representative of the population of all college textbooks? Explain why or why not. Describe some possible sources of bias in this study, and how it might affect the results of the study. Give some suggestions about what could be done to improve the study.
Answer
Answers will vary. Sample answer: The sample is not representative of the population of all college textbooks. Two reasons why it is not representative are that he only sampled seven subjects and he only investigated one textbook in each subject. There are several possible sources of bias in the study. The seven subjects that he investigated are all in mathematics and the sciences; there are many subjects in the humanities, social sciences, and other subject areas, (for example: literature, art, history, psychology, sociology, business) that he did not investigate at all. It may be that different subject areas exhibit different patterns of textbook availability, but his sample would not detect such results.
He also looked only at the most popular textbook in each of the subjects he investigated. The availability of the most popular textbooks may differ from the availability of other textbooks in one of two ways:
• the most popular textbooks may be more readily available online, because more new copies are printed, and more students nationwide are selling back their used copies OR
• the most popular textbooks may be harder to find available online, because more student demand exhausts the supply more quickly.
In reality, many college students do not use the most popular textbook in their subject, and this study gives no useful information about the situation for those less popular textbooks.
He could improve this study by:
• expanding the selection of subjects he investigates so that it is more representative of all subjects studied by college students, and
• expanding the selection of textbooks he investigates within each subject to include a mixed representation of both the most popular and less popular textbooks.
For the following exercises, identify the type of data that would be used to describe a response (quantitative discrete, quantitative continuous, or qualitative), and give an example of the data.
Q 1.3.1
number of tickets sold to a concert
S 1.3.1
quantitative discrete, 150
Q 1.3.2
percent of body fat
Q 1.3.3
favorite baseball team
S 1.3.3
qualitative, Oakland A’s
Q 1.3.4
time in line to buy groceries
Q 1.3.5
number of students enrolled at Evergreen Valley College
S 1.3.5
quantitative discrete, 11,234 students
Q 1.3.6
most-watched television show
Q 1.3.7
brand of toothpaste
S 1.3.7
qualitative, Crest
Q 1.3.8
distance to the closest movie theater
Q 1.3.9
age of executives in Fortune 500 companies
S 1.3.9
quantitative continuous, 47.3 years
Q 1.3.10
number of competing computer spreadsheet software packages
Use the following information to answer the next two exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of resident use of a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every 8th house in the neighborhood around the park was interviewed.
Q 1.3.11
“Number of times per week” is what type of data?
1. qualitative
2. quantitative discrete
3. quantitative continuous
b
Q 1.3.12
“Duration (amount of time)” is what type of data?
1. qualitative
2. quantitative discrete
3. quantitative continuous
Q 1.3.13
Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. Suppose an airline conducts a survey. Over Thanksgiving weekend, it surveys six flights from Boston to Salt Lake City to determine the number of babies on the flights. It determines the amount of safety equipment needed by the result of that study.
1. Using complete sentences, list three things wrong with the way the survey was conducted.
2. Using complete sentences, list three ways that you would improve the survey if it were to be repeated.
S 1.3.13
1. The survey was conducted using six similar flights.
The survey would not be a true representation of the entire population of air travelers.
Conducting the survey on a holiday weekend will not produce representative results.
2. Conduct the survey during different times of the year.
Conduct the survey using flights to and from various locations.
Conduct the survey on different days of the week.
Q 1.3.14
Suppose you want to determine the mean number of students per statistics class in your state. Describe a possible sampling method in three to five complete sentences. Make the description detailed.
Q 1.3.15
Suppose you want to determine the mean number of cans of soda drunk each month by students in their twenties at your school. Describe a possible sampling method in three to five complete sentences. Make the description detailed.
S 1.3.15
Answers will vary. Sample Answer: You could use a systematic sampling method. Stop the tenth person as they leave one of the buildings on campus at 9:50 in the morning. Then stop the tenth person as they leave a different building on campus at 1:50 in the afternoon.
Q 1.3.16
List some practical difficulties involved in getting accurate results from a telephone survey.
Q 1.3.17
List some practical difficulties involved in getting accurate results from a mailed survey.
S 1.3.17
Answers will vary. Sample Answer: Many people will not respond to mail surveys. If they do respond to the surveys, you can’t be sure who is responding. In addition, mailing lists can be incomplete.
Q 1.3.18
With your classmates, brainstorm some ways you could overcome these problems if you needed to conduct a phone or mail survey.
Q 1.3.19
The instructor takes her sample by gathering data on five randomly selected students from each Lake Tahoe Community College math class. The type of sampling she used is
1. cluster sampling
2. stratified sampling
3. simple random sampling
4. convenience sampling
b
Q 1.3.20
A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every eighth house in the neighborhood around the park was interviewed. The sampling method was:
1. simple random
2. systematic
3. stratified
4. cluster
Q 1.3.21
Name the sampling method used in each of the following situations:
1. A woman in the airport is handing out questionnaires to travelers asking them to evaluate the airport’s service. She does not ask travelers who are hurrying through the airport with their hands full of luggage, but instead asks all travelers who are sitting near gates and not taking naps while they wait.
2. A teacher wants to know if her students are doing homework, so she randomly selects rows two and five and then calls on all students in row two and all students in row five to present the solutions to homework problems to the class.
3. The marketing manager for an electronics chain store wants information about the ages of its customers. Over the next two weeks, at each store location, 100 randomly selected customers are given questionnaires to fill out asking for information about age, as well as about other variables of interest.
4. The librarian at a public library wants to determine what proportion of the library users are children. The librarian has a tally sheet on which she marks whether books are checked out by an adult or a child. She records this data for every fourth patron who checks out books.
5. A political party wants to know the reaction of voters to a debate between the candidates. The day after the debate, the party’s polling staff calls 1,200 randomly selected phone numbers. If a registered voter answers the phone or is available to come to the phone, that registered voter is asked whom he or she intends to vote for and whether the debate changed his or her opinion of the candidates.
S 1.3.21
1. convenience
2. cluster
3. stratified
4. systematic
5. simple random
A “random survey” was conducted of 3,274 people of the “microprocessor generation” (people born since 1971, the year the microprocessor was invented). It was reported that 48% of those individuals surveyed stated that if they had $2,000 to spend, they would use it for computer equipment. Also, 66% of those surveyed considered themselves relatively savvy computer users. 1. Do you consider the sample size large enough for a study of this type? Why or why not? 2. Based on your “gut feeling,” do you believe the percents accurately reflect the U.S. population for those individuals born since 1971? If not, do you think the percents of the population are actually higher or lower than the sample statistics? Why? Additional information: The survey, reported by Intel Corporation, was filled out by individuals who visited the Los Angeles Convention Center to see the Smithsonian Institute's road show called “America’s Smithsonian.” 3. With this additional information, do you feel that all demographic and ethnic groups were equally represented at the event? Why or why not? 4. With the additional information, comment on how accurately you think the sample statistics reflect the population parameters. Q 1.3.23 The Gallup-Healthways Well-Being Index is a survey that follows trends of U.S. residents on a regular basis. There are six areas of health and wellness covered in the survey: Life Evaluation, Emotional Health, Physical Health, Healthy Behavior, Work Environment, and Basic Access. Some of the questions used to measure the Index are listed below. Identify the type of data obtained from each question used in this survey: qualitative, quantitative discrete, or quantitative continuous. 1. Do you have any health problems that prevent you from doing any of the things people your age can normally do? 2. During the past 30 days, for about how many days did poor health keep you from doing your usual activities? 3. In the last seven days, on how many days did you exercise for 30 minutes or more? 4. Do you have health insurance coverage? S 1.3.23 1. qualitative 2. quantitative discrete 3. quantitative discrete 4. qualitative Q 1.3.24 In advance of the 1936 Presidential Election, a magazine titled Literary Digest released the results of an opinion poll predicting that the republican candidate Alf Landon would win by a large margin. The magazine sent post cards to approximately 10,000,000 prospective voters. These prospective voters were selected from the subscription list of the magazine, from automobile registration lists, from phone lists, and from club membership lists. Approximately 2,300,000 people returned the postcards. 1. Think about the state of the United States in 1936. Explain why a sample chosen from magazine subscription lists, automobile registration lists, phone books, and club membership lists was not representative of the population of the United States at that time. 2. What effect does the low response rate have on the reliability of the sample? 3. Are these problems examples of sampling error or nonsampling error? 4. During the same year, George Gallup conducted his own poll of 30,000 prospective voters. His researchers used a method they called "quota sampling" to obtain survey answers from specific subsets of the population. Quota sampling is an example of which sampling method described in this module? Q 1.3.25 Crime-related and demographic statistics for 47 US states in 1960 were collected from government agencies, including the FBI's Uniform Crime Report. One analysis of this data found a strong connection between education and crime indicating that higher levels of education in a community correspond to higher crime rates. Which of the potential problems with samples discussed in [link] could explain this connection? S 1.3.26 Causality: The fact that two variables are related does not guarantee that one variable is influencing the other. We cannot assume that crime rate impacts education level or that education level impacts crime rate. Confounding: There are many factors that define a community other than education level and crime rate. Communities with high crime rates and high education levels may have other lurking variables that distinguish them from communities with lower crime rates and lower education levels. Because we cannot isolate these variables of interest, we cannot draw valid conclusions about the connection between education and crime. Possible lurking variables include police expenditures, unemployment levels, region, average age, and size. Q 1.3.27 YouPolls is a website that allows anyone to create and respond to polls. One question posted April 15 asks: “Do you feel happy paying your taxes when members of the Obama administration are allowed to ignore their tax liabilities?”1 As of April 25, 11 people responded to this question. Each participant answered “NO!” Which of the potential problems with samples discussed in this module could explain this connection? Q 1.3.28 A scholarly article about response rates begins with the following quote: “Declining contact and cooperation rates in random digit dial (RDD) national telephone surveys raise serious concerns about the validity of estimates drawn from such research.”2 The Pew Research Center for People and the Press admits: “The percentage of people we interview – out of all we try to interview – has been declining over the past decade or more.”3 1. What are some reasons for the decline in response rate over the past decade? 2. Explain why researchers are concerned with the impact of the declining response rate on public opinion polls. S 1.3.28 1. Possible reasons: increased use of caller id, decreased use of landlines, increased use of private numbers, voice mail, privacy managers, hectic nature of personal schedules, decreased willingness to be interviewed 2. When a large number of people refuse to participate, then the sample may not have the same characteristics of the population. Perhaps the majority of people willing to participate are doing so because they feel strongly about the subject of the survey. 1.4: Frequency, Frequency Tables, and Levels of Measurement Q 1.4.1 Fifty part-time students were asked how many courses they were taking this term. The (incomplete) results are shown below: Part-time Student Course Loads # of Courses Frequency Relative Frequency Cumulative Relative Frequency 1 30 0.6 2 15 3 1. Fill in the blanks in Table. 2. What percent of students take exactly two courses? 3. What percent of students take one or two courses? Q 1.4.2 Sixty adults with gum disease were asked the number of times per week they used to floss before their diagnosis. The (incomplete) results are shown in Table. Flossing Frequency for Adults with Gum Disease # Flossing per Week Frequency Relative Frequency Cumulative Relative Freq. 0 27 0.4500 1 18 3 0.9333 6 3 0.0500 7 1 0.0167 1. Fill in the blanks in Table. 2. What percent of adults flossed six times per week? 3. What percent flossed at most three times per week? S 1.4.2 1. # Flossing per Week Frequency Relative Frequency Cumulative Relative Frequency 0 27 0.4500 0.4500 1 18 0.3000 0.7500 3 11 0.1833 0.9333 6 3 0.0500 0.9833 7 1 0.0167 1 2. 5.00% 3. 93.33% Q 1.4.3 Nineteen immigrants to the U.S were asked how many years, to the nearest year, they have lived in the U.S. The data are as follows: 2; 5; 7; 2; 2; 10; 20; 15; 0; 7; 0; 20; 5; 12; 15; 12; 45; 10 . Table was produced. Frequency of Immigrant Survey Responses Data Frequency Relative Frequency Cumulative Relative Frequency 0 2 219219 0.1053 2 3 319319 0.2632 4 1 119119 0.3158 5 3 319319 0.4737 7 2 219219 0.5789 10 2 219219 0.6842 12 2 219219 0.7895 15 1 119119 0.8421 20 1 119119 1.0000 1. Fix the errors in Table. Also, explain how someone might have arrived at the incorrect number(s). 2. Explain what is wrong with this statement: “47 percent of the people surveyed have lived in the U.S. for 5 years.” 3. Fix the statement in b to make it correct. 4. What fraction of the people surveyed have lived in the U.S. five or seven years? 5. What fraction of the people surveyed have lived in the U.S. at most 12 years? 6. What fraction of the people surveyed have lived in the U.S. fewer than 12 years? 7. What fraction of the people surveyed have lived in the U.S. from five to 20 years, inclusive? Q 1.4.4 How much time does it take to travel to work? Table shows the mean commute time by state for workers at least 16 years old who are not working at home. Find the mean travel time, and round off the answer properly. 24.0 24.3 25.9 18.9 27.5 17.9 21.8 20.9 16.7 27.3 18.2 24.7 20.0 22.6 23.9 18.0 31.4 22.3 24.0 25.5 24.7 24.6 28.1 24.9 22.6 23.6 23.4 25.7 24.8 25.5 21.2 25.7 23.1 23.0 23.9 26.0 16.3 23.1 21.4 21.5 27.0 27.0 18.6 31.7 23.3 30.1 22.9 23.3 21.7 18.6 S 1.4.4 The sum of the travel times is 1,173.1. Divide the sum by 50 to calculate the mean value: 23.462. Because each state’s travel time was measured to the nearest tenth, round this calculation to the nearest hundredth: 23.46. Q 1.4.5 Forbes magazine published data on the best small firms in 2012. These were firms which had been publicly traded for at least a year, have a stock price of at least$5 per share, and have reported annual revenue between $5 million and$1 billion. Table shows the ages of the chief executive officers for the first 60 ranked firms.
Age Frequency Relative Frequency Cumulative Relative Frequency
40–44 3
45–49 11
50–54 13
55–59 16
60–64 10
65–69 6
70–74 1
1. What is the frequency for CEO ages between 54 and 65?
2. What percentage of CEOs are 65 years or older?
3. What is the relative frequency of ages under 50?
4. What is the cumulative relative frequency for CEOs younger than 55?
5. Which graph shows the relative frequency and which shows the cumulative relative frequency?
Use the following information to answer the next two exercises: Table contains data on hurricanes that have made direct hits on the U.S. Between 1851 and 2004. A hurricane is given a strength category rating based on the minimum wind speed generated by the storm.
Frequency of Hurricane Direct Hits
Category Number of Direct Hits Relative Frequency Cumulative Frequency
Total = 273
1 109 0.3993 0.3993
2 72 0.2637 0.6630
3 71 0.2601
4 18 0.9890
5 3 0.0110 1.0000
Q 1.4.6
What is the relative frequency of direct hits that were category 4 hurricanes?
1. 0.0768
2. 0.0659
3. 0.2601
4. Not enough information to calculate
b
Q 1.4.7
What is the relative frequency of direct hits that were AT MOST a category 3 storm?
1. 0.3480
2. 0.9231
3. 0.2601
4. 0.3370
1.5: Experimental Design and Ethics
Q 1.5.1
How does sleep deprivation affect your ability to drive? A recent study measured the effects on 19 professional drivers. Each driver participated in two experimental sessions: one after normal sleep and one after 27 hours of total sleep deprivation. The treatments were assigned in random order. In each session, performance was measured on a variety of tasks including a driving simulation.
Use key terms from this module to describe the design of this experiment.
S 1.5.1
Explanatory variable: amount of sleep
Response variable: performance measured in assigned tasks
Treatments: normal sleep and 27 hours of total sleep deprivation
Experimental Units: 19 professional drivers
Lurking variables: none – all drivers participated in both treatments
Random assignment: treatments were assigned in random order; this eliminated the effect of any “learning” that may take place during the first experimental session
Control/Placebo: completing the experimental session under normal sleep conditions
Blinding: researchers evaluating subjects’ performance must not know which treatment is being applied at the time
Q 1.5.2
An advertisement for Acme Investments displays the two graphs in Figures to show the value of Acme’s product in comparison with the Other Guy’s product. Describe the potentially misleading visual effect of these comparison graphs. How can this be corrected?
As the graphs show, Acme consistently outperforms the Other Guys!
Q 1.5.3
The graph in Figure shows the number of complaints for six different airlines as reported to the US Department of Transportation in February 2013. Alaska, Pinnacle, and Airtran Airlines have far fewer complaints reported than American, Delta, and United. Can we conclude that American, Delta, and United are the worst airline carriers since they have the most complaints?
S 1.5.3
You cannot assume that the numbers of complaints reflect the quality of the airlines. The airlines shown with the greatest number of complaints are the ones with the most passengers. You must consider the appropriateness of methods for presenting data; in this case displaying totals is misleading. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/01%3A_Sampling_and_Data/1.E%3A_Sampling_and_Data_%28Exercises%29.txt |
In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called "Descriptive Statistics." You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs.
• 2.1: Prelude to Descriptive Statistics
In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called "Descriptive Statistics." You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs. Our emphasis will be on histograms and box plots.
• 2.2: Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
A stem-and-leaf plot is a way to plot data and look at the distribution, where all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories.
• 2.3: Histograms, Frequency Polygons, and Time Series Graphs
A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat.
• 2.4: Measures of the Location of the Data
The values that divide a rank-ordered set of data into 100 equal parts are called percentiles and are used to compare and interpret data. For example, an observation at the 50th percentile would be greater than 50 % of the other obeservations in the set. Quartiles divide data into quarters. The first quartile is the 25th percentile, the second quartile is 50th percentile, and the third quartile is the the 75th percentile. The interquartile range is the range of the middle 50 % of the data values
• 2.5: Box Plots
Box plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data.
• 2.6: Measures of the Center of the Data
The mean and the median can be calculated to help you find the "center" of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data.
• 2.7: Skewness and the Mean, Median, and Mode
Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A right (or positive) skewed distribution, a left (or negative) skewed distribution and a symmetrical distribution.
• 2.8: Measures of the Spread of the Data
An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.
• 2.9: Descriptive Statistics (Worksheet)
A statistics Worksheet: The student will construct a histogram and a box plot. The student will calculate univariate statistics. The student will examine the graphs to interpret what the data implies.
• 2.E: Descriptive Statistics (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
02: Descriptive Statistics
Learning Objectives
By the end of this chapter, the student should be able to:
• Display data graphically and interpret graphs: stemplots, histograms, and box plots.
• Recognize, describe, and calculate the measures of location of data: quartiles and percentiles.
• Recognize, describe, and calculate the measures of the center of data: mean, median, and mode.
• Recognize, describe, and calculate the measures of the spread of data: variance, standard deviation, and range.
Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data.
In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called "Descriptive Statistics." You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs.
A statistical graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied.
Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs. Our emphasis will be on histograms and box plots.
This book contains instructions for constructing a histogram and a box plot for the TI-83+ and TI-84 calculators. The Texas Instruments (TI) website provides additional instructions for using these calculators. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.01%3A_Prelude_to_Descriptive_Statistics.txt |
One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.
Example $1$
For Susan Dean's spring pre-calculus class, scores for the first exam were as follows (smallest to largest):
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
Stem-and-Leaf Graph
Stem Leaf
3 3
4 2 9 9
5 3 5 5
6 1 3 7 8 8 9 9
7 2 3 4 8
8 0 3 8 8 8
9 0 2 4 4 4 4 6
10 0
The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% $\left(\frac{8}{31}\right)$ were in the 90s or 100, a fairly high number of As.
Exercise $2$
For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):
32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61
Construct a stem plot for the data.
Answer
Stem Leaf
3 2 2 3 4 8
4 0 2 2 3 4 6 7 7 8 8 8 9
5 0 0 1 2 2 2 3 4 6 7 7
6 0 1
The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later.
Example $3$
The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data:
1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3
Do the data seem to have any concentration of values?
HINT: The leaves are to the right of the decimal.
Answer
The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers.
Stem Leaf
1 1 5
2 3 5 7
3 2 3 3 5 8
4 0 2 5 5 7 8
5 5 6
6 5 7
7
8
9
10
11
12 3
Exercise $4$
The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers:
0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0
Answer
Stem Leaf
0 5 7
1 1 2 2 3 3 5 5 7 7 8 9
2 0 2 5 6 8 8 8
3 5 8
4 4 8 9
5 2 5 7 8
6
7
8 0
The value 8.0 may be an outlier. Values appear to concentrate at one and two miles.
Example $5$: Side-by-Side Stem-and-Leaf plot
A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. Tables $1$ and $2$ show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data.
Table $1$: Presidential Ages at Inauguration
President Ageat Inauguration President Age President Age
Pierce 48 Harding 55 Obama 47
Polk 49 T. Roosevelt 42 G.H.W. Bush 64
Fillmore 50 Wilson 56 G. W. Bush 54
Tyler 51 McKinley 54 Reagan 69
Van Buren 54 B. Harrison 55 Ford 61
Washington 57 Lincoln 52 Hoover 54
Jefferson 57 Grant 46 Truman 60
Madison 57 Hayes 54 Eisenhower 62
J. Q. Adams 57 Arthur 51 L. Johnson 55
Monroe 58 Garfield 49 Kennedy 43
J. Adams 61 A. Johnson 56 F. Roosevelt 51
Jackson 61 Cleveland 47 Nixon 56
Taylor 64 Taft 51 Clinton 47
Buchanan 65 Coolidge 51 Trump 70
W. H. Harrison 68 Cleveland 55 Carter 52
$2$ Presidential Age at Death
President Age President Age President Age
Washington 67 Lincoln 56 Hoover 90
J. Adams 90 A. Johnson 66 F. Roosevelt 63
Jefferson 83 Grant 63 Truman 88
Madison 85 Hayes 70 Eisenhower 78
Monroe 73 Garfield 49 Kennedy 46
J. Q. Adams 80 Arthur 56 L. Johnson 64
Jackson 78 Cleveland 71 Nixon 81
Van Buren 79 B. Harrison 67 Ford 93
W. H. Harrison 68 Cleveland 71 Reagan 93
Tyler 71 McKinley 58
Polk 53 T. Roosevelt 60
Taylor 65 Taft 72
Fillmore 74 Wilson 67
Pierce 64 Harding 57
Buchanan 77 Coolidge 60
Answer
Ages at Inauguration Ages at Death
9 9 8 7 7 7 6 3 2 4 6 9
8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 1 1 1 1 1 0 5 3 6 6 7 7 8
9 5 4 4 2 1 1 1 0 6 0 0 3 3 4 4 5 6 7 7 7 8
7 0 0 1 1 1 3 4 7 8 8 9
8 0 1 3 5 8
9 0 0 3 3
Exercise $6$
The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stem-and-leaf plot of these wins and losses.
Losses Wins Year Losses Wins Year
34 48 1968–1969 41 41 1989–1990
34 48 1969–1970 39 43 1990–1991
46 36 1970–1971 44 38 1991–1992
46 36 1971–1972 39 43 1992–1993
36 46 1972–1973 25 57 1993–1994
47 35 1973–1974 40 42 1994–1995
51 31 1974–1975 36 46 1995–1996
53 29 1975–1976 26 56 1996–1997
51 31 1976–1977 32 50 1997–1998
41 41 1977–1978 19 31 1998–1999
36 46 1978–1979 54 28 1999–2000
32 50 1979–1980 57 25 2000–2001
51 31 1980–1981 49 33 2001–2002
40 42 1981–1982 47 35 2002–2003
39 43 1982–1983 54 28 2003–2004
42 40 1983–1984 69 13 2004–2005
48 34 1984–1985 56 26 2005–2006
32 50 1985–1986 52 30 2006–2007
25 57 1986–1987 45 37 2007–2008
32 50 1987–1988 35 47 2008–2009
30 52 1988–1989 29 53 2009–2010
Answer
Table $3$: Atlanta Hawks Wins and Losses
Number of Wins Number of Losses
3 1 9
9 8 8 6 5 2 5 5 9
8 7 6 6 5 5 4 3 1 1 1 1 0 3 0 2 2 2 2 4 4 5 6 6 6 9 9 9
8 8 7 6 6 6 3 3 3 2 2 1 1 0 4 0 0 1 1 2 4 5 6 6 7 7 8 9
7 7 6 3 2 0 0 0 0 5 1 1 1 2 3 4 4 6 7
6 9
Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in Example, the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists of frequency points. The frequency points are connected using line segments.
Example $7$
In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in Table and in Figure.
Number of times teenager is reminded Frequency
0 2
1 5
2 8
3 14
4 7
5 4
Answer
Exercise $8$
In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in Table. Construct a line graph.
Number of times in shop Frequency
0 7
1 10
2 14
3 9
Answer
Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in Example $9$ has age groups represented on the x-axis and proportions on the y-axis.
Example $9$
By the end of 2011, Facebook had over 146 million users in the United States. Table shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph using this data.
Age groups Number of Facebook users Proportion (%) of Facebook users
13–25 65,082,280 45%
26–44 53,300,200 36%
45–64 27,885,100 19%
Answer
Exercise $10$
The population in Park City is made up of children, working-age adults, and retirees. Table shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions.
Age groups Number of people Proportion of population
Children 67,059 19%
Working-age adults 152,198 43%
Retirees 131,662 38%
Answer
Example $11$
The columns in Table contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the x-axis, and the Advanced Placement examinee population percentages on the y-axis.
Race/Ethnicity AP Examinee Population Overall Student Population
1 = Asian, Asian American or Pacific Islander 10.3% 5.7%
2 = Black or African American 9.0% 14.7%
3 = Hispanic or Latino 17.0% 17.6%
4 = American Indian or Alaska Native 0.6% 1.1%
5 = White 57.1% 59.2%
6 = Not reported/other 6.0% 1.7%
Solution
Exercise $12$
Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district.
District Registered voter population Overall city population
1 15.5% 19.4%
2 12.2% 15.6%
3 9.8% 9.0%
4 17.4% 18.5%
5 22.8% 20.7%
6 22.3% 16.8%
Answer
Summary
A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.02%3A_Stem-and-Leaf_Graphs_%28Stemplots%29_Line_Graphs_and_Bar_Graphs.txt |
For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more.
A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data.
The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample.(Remember, frequency is defined as the number of times an answer occurs.) If:
• $f$ is frequency
• $n$ is total number of data values (or the sum of the individual frequencies), and
• $RF$ is relative frequency,
then:
$RF=\dfrac{f}{n} \label{2.3.1}$
For example, if three students in Mr. Ahab's English class of 40 students received from 90% to 100%, then, f = 3, n = 40, and RF = fn = 340 = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures.
To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is $6.05 (6.1 – 0.05 = 6.05)$. We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is $1.495 (1.5 – 0.005 = 1.495)$. If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is $0.9995 (1.0 – 0.0005 = 0.9995)$. If all the data happen to be integers and the smallest value is two, then a convenient starting point is $1.5 (2 - 0.5 = 1.5)$. Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data.
Example $1$
The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured.
60; 60.5; 61; 61; 61.5
63.5; 63.5; 63.5
64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5
66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5
68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5
70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71
72; 72; 72; 72.5; 72.5; 73; 73.5
74
The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.
60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95.
The largest value is 74, so 74 + 0.05 = 74.05 is the ending value.
Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars.
$\dfrac{74.05−59.95}{8}=1.76$
We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals.
The boundaries are:
• 59.95
• 59.95 + 2 = 61.95
• 61.95 + 2 = 63.95
• 63.95 + 2 = 65.95
• 65.95 + 2 = 67.95
• 67.95 + 2 = 69.95
• 69.95 + 2 = 71.95
• 71.95 + 2 = 73.95
• 73.95 + 2 = 75.95
The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.
The following histogram displays the heights on the x-axis and relative frequency on the y-axis.
Exercise $1$
The following data are the shoe sizes of 50 male students. The sizes are discrete data since shoe size is measured in whole and half units only. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.
9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5
11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5
12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14
Answer
Smallest value: 9
Largest value: 14
Convenient starting value: 9 – 0.05 = 8.95
Convenient ending value: 14 + 0.05 = 14.05
$\frac{14.05-8.95}{6}$ = 0.85
The calculations suggests using 0.85 as the width of each bar or class interval. You can also use an interval with a width equal to one.
Example $2$
The following data are the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data, since books are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
2; 2; 2; 2; 2; 2; 2; 2; 2; 2
3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3
4; 4; 4; 4; 4; 4
5; 5; 5; 5; 5
6; 6
Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books.
Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.
Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ .
Answer
Calculate the number of bars as follows:
$\dfrac{6.5 - 0.5}{\text{number of bars}}$ = 1
where 1 is the width of a bar. Therefore, bars = 6.
The following histogram displays the number of books on the x-axis and the frequency on the y-axis.
Note
Go to [link]. There are calculator instructions for entering data and for creating a customized histogram. Create the histogram for Example.
• Press Y=. Press CLEAR to delete any equations.
• Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1, press CLEAR and then arrow down. If necessary, do the same for L2.
• Into L1, enter 1, 2, 3, 4, 5, 6.
• Into L2, enter 11, 10, 16, 6, 5, 2.
• Press WINDOW. Set Xmin = .5, Xscl = (6.5 – .5)/6, Ymin = –1, Ymax = 20, Yscl = 1, Xres = 1.
• Press 2nd Y=. Start by pressing 4:Plotsoff ENTER.
• Press 2nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE. Arrow to the 3rdpicture (histogram). Press ENTER.
• Arrow down to Xlist: Enter L1 (2nd 1). Arrow down to Freq. Enter L2 (2nd 2).
• Press GRAPH.
• Use the TRACE key and the arrow keys to examine the histogram.
Exercise $2$
The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2
3; 3; 3; 3; 3; 3; 3; 3
20 student athletes play one sport. 22 student athletes play two sports. Eight student athletes play three sports.
Fill in the blanks for the following sentence. Since the data consist of the numbers 1, 2, 3, and the starting point is 0.5, a width of one places the 1 in the middle of the interval 0.5 to _____, the 2 in the middle of the interval from _____ to _____, and the 3 in the middle of the interval from _____ to _____.
Answer
1.5
1.5 to 2.5
2.5 to 3.5
Example $3$
Using this data set, construct a histogram.
Number of Hours My Classmates Spent Playing Video Games on Weekends
9.95 10 2.25 16.75 0
19.5 22.5 7.5 15 12.75
5.5 11 10 20.75 17.5
23 21.9 24 23.75 18
20 15 22.9 18.8 20.5
Answer
Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram.
Exercise $3$
The following data represent the number of employees at various restaurants in New York City. Using this data, create a histogram.
22; 35; 15; 26; 40; 28; 18; 20; 25; 34; 39; 42; 24; 22; 19; 27; 22; 34; 40; 20; 38 and 28
Use 10–19 as the first interval.
Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals.
Frequency Polygons
Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons. To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x-axis and y-axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them.
Example $4$
A frequency polygon was constructed from the frequency table below.
Frequency Distribution for Calculus Final Test Scores
Lower Bound Upper Bound Frequency Cumulative Frequency
49.5 59.5 5 5
59.5 69.5 10 15
69.5 79.5 30 45
79.5 89.5 40 85
89.5 99.5 15 100
The first label on the x-axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.
Exercise $4$
Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in the Table.
Age at Inauguration Frequency
41.5–46.5 4
46.5–51.5 11
51.5–56.5 14
56.5–61.5 9
61.5–66.5 4
66.5–71.5 2
Answer
The first label on the x-axis is 39. This represents an interval extending from 36.5 to 41.5. Since there are no ages less than 41.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 44 represents the next interval, or the first “real” interval from the table, and contains four scores. This reasoning is followed for each of the remaining intervals with the point 74 representing the interval from 71.5 to 76.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.
Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets.
Example $5$
We will construct an overlay frequency polygon comparing the scores from Example with the students’ final numeric grade.
Frequency Distribution for Calculus Final Test Scores
Lower Bound Upper Bound Frequency Cumulative Frequency
49.5 59.5 5 5
59.5 69.5 10 15
69.5 79.5 30 45
79.5 89.5 40 85
89.5 99.5 15 100
Frequency Distribution for Calculus Final Grades
Lower Bound Upper Bound Frequency Cumulative Frequency
49.5 59.5 10 10
59.5 69.5 10 20
69.5 79.5 30 50
79.5 89.5 45 95
89.5 99.5 5 100
Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with this data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected.
One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph.
Constructing a Time Series Graph
To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur.
Example $6$
The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only.
Year Jan Feb Mar Apr May Jun Jul
2003 181.7 183.1 184.2 183.8 183.5 183.7 183.9
2004 185.2 186.2 187.4 188.0 189.1 189.7 189.4
2005 190.7 191.8 193.3 194.6 194.4 194.5 195.4
2006 198.3 198.7 199.8 201.5 202.5 202.9 203.5
2007 202.416 203.499 205.352 206.686 207.949 208.352 208.299
2008 211.080 211.693 213.528 214.823 216.632 218.815 219.964
2009 211.143 212.193 212.709 213.240 213.856 215.693 215.351
2010 216.687 216.741 217.631 218.009 218.178 217.965 218.011
2011 220.223 221.309 223.467 224.906 225.964 225.722 225.922
2012 226.665 227.663 229.392 230.085 229.815 229.478 229.104
Year Aug Sep Oct Nov Dec Annual
2003 184.6 185.2 185.0 184.5 184.3 184.0
2004 189.5 189.9 190.9 191.0 190.3 188.9
2005 196.4 198.8 199.2 197.6 196.8 195.3
2006 203.9 202.9 201.8 201.5 201.8 201.6
2007 207.917 208.490 208.936 210.177 210.036 207.342
2008 219.086 218.783 216.573 212.425 210.228 215.303
2009 215.834 215.969 216.177 216.330 215.949 214.537
2010 218.312 218.439 218.711 218.803 219.179 218.056
2011 226.545 226.889 226.421 226.230 225.672 224.939
2012 230.379 231.407 231.317 230.221 229.601 229.594
Answer
Exercise $5$
The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO2 emissions for the United States.
CO2 Emissions
Ukraine United Kingdom United States
2003 352,259 540,640 5,681,664
2004 343,121 540,409 5,790,761
2005 339,029 541,990 5,826,394
2006 327,797 542,045 5,737,615
2007 328,357 528,631 5,828,697
2008 323,657 522,247 5,656,839
2009 272,176 474,579 5,299,563
Uses of a Time Series Graph
Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot.
Review
A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y-axis with the frequency being graphed on the x-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time.Glossary
Frequency
the number of times a value of the data occurs
Histogram
a graphical representation in $x-y$ form of the distribution of data in a data set; $x$ represents the data and $y$ represents the frequency, or relative frequency. The graph consists of contiguous rectangles.
Relative Frequency
the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.03%3A_Histograms_Frequency_Polygons_and_Time_Series_Graphs.txt |
The common measures of location are quartiles and percentiles. Quartiles are special percentiles. The first quartile, Q1, is the same as the 25th percentile, and the third quartile, Q3, is the same as the 75th percentile. The median, M, is called both the second quartile and the 50th percentile.
To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.
Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75th percentile. That translates into a score of at least 1220.
Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.
The median is a number that measures the "center" of the data. You can think of the median as the "middle value," but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data.
1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1
Ordered from smallest to largest:
1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5
Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.
$\dfrac{6.8+7.2}{2} = 7$
The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.
Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, Q1, is the middle value of the lower half of the data, and the third quartile, Q3, is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set:
1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5
The median or second quartile is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two.
1; 1; 2; 2; 4; 6; 6.8
The number two, which is part of the data, is the first quartile. One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two.
The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine.
The third quartile, Q3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example.
The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile (Q3) and the first quartile (Q1).
$IQR = Q_3 – Q_1 \tag{2.4.1}$
The IQR can help to determine potential outliers. A value is suspected to be a potential outlier if it is less than (1.5)(IQR) below the first quartile or more than (1.5)(IQR) above the third quartile. Potential outliers always require further investigation.
Definition: Outliers
A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data.
Example 2.4.1
For the following 13 real estate prices, calculate the IQR and determine if any prices are potential outliers. Prices are in dollars.
389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000
Answer
Order the data from smallest to largest.
114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000
$M = 488,800 \nonumber$
$Q_{1} = \dfrac{230,500 + 387,000}{2} = 308,750\nonumber$
$Q_{3} = \dfrac{639,000 + 659,000}{2} = 649,000\nonumber$
$IQR = 649,000 - 308,750 = 340,250\nonumber$
$(1.5)(IQR) = (1.5)(340,250) = 510,375\nonumber$
$Q_{1} - (1.5)(IQR) = 308,750 - 510,375 = –201,625\nonumber$
$Q_{3} + (1.5)(IQR) = 649,000 + 510,375 = 1,159,375\nonumber$
No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier.
Exercise $1$
For the following 11 salaries, calculate the IQR and determine if any salaries are outliers. The salaries are in dollars.
$33,000;$64,500; $28,000;$54,000; $72,000;$68,500; $69,000;$42,000; $54,000;$120,000; $40,500 Answer Order the data from smallest to largest.$28,000; $33,000;$40,500; $42,000;$54,000; $54,000;$64,500; $68,500;$69,000; $72,000;$120,000
Median = $54,000 $Q_{1} = 40,500\nonumber$ $Q_{3} = 69,000\nonumber$ $IQR = 69,000 - 40,500 = 28,500\nonumber$ $(1.5)(IQR) = (1.5)(28,500) = 42,750\nonumber$ $Q_{1} - (1.5)(IQR) = 40,500 - 42,750 = -2,250\nonumber$ $Q_{3} + (1.5)(IQR) = 69,000 + 42,750 = 111,750\nonumber$ No salary is less than –$2,250. However, $120,000 is more than$11,750, so \$120,000 is a potential outlier.
Example 2.4.2
For the two data sets in the test scores example, find the following:
1. The interquartile range. Compare the two interquartile ranges.
2. Any outliers in either set.
Answer
The five number summary for the day and night classes is
Minimum Q1 Median Q3 Maximum
Day 32 56 74.5 82.5 99
Night 25.5 78 81 89 98
1. The IQR for the day group is $Q_{3} - Q_{1} = 82.5 - 56 = 26.5$
The IQR for the night group is $Q_{3} - Q_{1} = 89 - 78 = 11$
The interquartile range (the spread or variability) for the day class is larger than the night class IQR. This suggests more variation will be found in the day class’s class test scores.
2. Day class outliers are found using the IQR times 1.5 rule. So,
• $Q_{1} - IQR(1.5) = 56 – 26.5(1.5) = 16.25$
• $Q_{3} + IQR(1.5) = 82.5 + 26.5(1.5) = 122.25$
Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers.
Night class outliers are calculated as:
• $Q_{1} - IQR (1.5) = 78 – 11(1.5) = 61.5$
• $Q_{3} + IQR(1.5) = 89 + 11(1.5) = 105.5$
For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier.
Exercise $2$
Find the interquartile range for the following two data sets and compare them.
Test Scores for Class A
69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94
Test Scores for Class B
90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100
Answer
Class A
Order the data from smallest to largest.
65; 66; 67; 69; 69; 76; 77; 77; 79; 80; 81; 83; 85; 89; 90; 91; 94; 96; 98; 99
$Median = \dfrac{80 + 81}{2}$ = 80.5
$Q_{1} = \dfrac{69 + 76}{2} = 72.5$
$Q_{3} = \dfrac{90 + 91}{2} = 90.5$
$IQR = 90.5 - 72.5 = 18$
Class B
Order the data from smallest to largest.
68; 68; 70; 71; 72; 73; 75; 78; 79; 80; 80; 90; 90; 92; 92; 95; 95; 97; 99; 100
$Median = \dfrac{80 + 80}{2} = 80$
$Q_{1} = \dfrac{72 + 73}{2} = 72.5$
$Q_{3} = \dfrac{92 + 95}{2} = 93.5$
$IQR = 93.5 - 72.5 = 21$
The data for Class B has a larger IQR, so the scores between Q3 and Q1 (middle 50%) for the data for Class B are more spread out and not clustered about the median.
Example $3$
Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were:
AMOUNT OF SLEEP PER SCHOOL NIGHT (HOURS) FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY
4 2 0.04 0.04
5 5 0.10 0.14
6 7 0.14 0.28
7 12 0.24 0.52
8 14 0.28 0.80
9 7 0.14 0.94
10 3 0.06 1.00
Find the 28th percentile. Notice the 0.28 in the "cumulative relative frequency" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28th percentile is between the last six and the first seven. The 28th percentile is 6.5.
Find the median. Look again at the "cumulative relative frequency" column and find 0.52. The median is the 50th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50th percentile is between the 25th, or seven, and 26th, or seven, values. The median is seven.
Find the third quartile. The third quartile is the same as the 75th percentile. You can "eyeball" this answer. If you look at the "cumulative relative frequency" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75th percentile, then, must be an eight. Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q3, is the 38th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.)
Exercise $3$
Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65th percentile.
Amount of time spent on route (hours) Frequency Relative Frequency Cumulative Relative Frequency
2 12 0.30 0.30
3 14 0.35 0.65
4 10 0.25 0.90
5 4 0.10 1.00
Answer
The 65th percentile is between the last three and the first four.
The 65th percentile is 3.5.
Example 2.4.4
Using the table above in Example $3$
1. Find the 80th percentile.
2. Find the 90th percentile.
3. Find the first quartile. What is another name for the first quartile?
Solution
Using the data from the frequency table, we have:
1. The 80th percentile is between the last eight and the first nine in the table (between the 40th and 41st values). Therefore, we need to take the mean of the 40th an 41st values. The 80th percentile $= \dfrac{8+9}{2} = 8.5$
2. The 90th percentile will be the 45th data value (location is $0.90(50) = 45$) and the 45th data value is nine.
3. Q1 is also the 25th percentile. The 25th percentile location calculation: $P_{25} = 0.25(50) = 12.5 \approx 13$ the 13th data value. Thus, the 25th percentile is six.
Exercise $4$
Refer to the table above in Exercise $3$. Find the third quartile. What is another name for the third quartile?
Answer
The third quartile is the 75th percentile, which is four. The 65th percentile is between three and four, and the 90th percentile is between four and 5.75. The third quartile is between 65 and 90, so it must be four.
COLLABORATIVE STATISTICS
Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions:
1. How many students were surveyed?
2. What kind of sampling did you do?
3. Construct two different histograms. For each, starting value = _____ ending value = ____.
4. Find the median, first quartile, and third quartile.
5. Construct a table of the data to find the following:
1. the 10th percentile
2. the 70th percentile
3. the percent of students who own less than four sweaters
A Formula for Finding the kth Percentile
If you were to do a little research, you would find several formulas for calculating the kth percentile. Here is one of them.
• $k =$ the kth percentile. It may or may not be part of the data.
• $i =$ the index (ranking or position of a data value)
• $n =$ the total number of data
Order the data from smallest to largest.
Calculate $i = \dfrac{k}{100}(n + 1)$
If $i$ is an integer, then the $k^{th}$ percentile is the data value in the $i^{th}$ position in the ordered set of data.
If $i$ is not an integer, then round $i$ up and round $i$ down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example.
Example 2.4.5
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
1. Find the 70th percentile.
2. Find the 83rd percentile.
Solution
• $k = 70$
• $i$ = the index
• $n = 29$
$i = \dfrac{k}{100}(n + 1) = \dfrac{70}{100}(29 + 1) = 21$. Twenty-one is an integer, and the data value in the 21st position in the ordered data set is 64. The 70th percentile is 64 years.
• $k$ = 83rd percentile
• $i$= the index
• $n = 29$
$i = \dfrac{k}{100}(n + 1) = (\dfrac{83}{100})(29 + 1) = 24.9$, which is NOT an integer. Round it down to 24 and up to 25. The age in the 24th position is 71 and the age in the 25th position is 72. Average 71 and 72. The 83rd percentile is 71.5 years.
Exercise $5$
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Calculate the 20th percentile and the 55th percentile.
Answer
$k = 20$. Index $= i = \dfrac{k}{100}(n+1) = \dfrac{20}{100}(29 + 1) = 6$. The age in the sixth position is 27. The 20th percentile is 27 years.
$k = 55$. Index $= i = \dfrac{k}{100}(n+1) = \dfrac{55}{100}(29 + 1) = 16.5$. Round down to 16 and up to 17. The age in the 16th position is 52 and the age in the 17th position is 55. The average of 52 and 55 is 53.5. The 55th percentile is 53.5 years.
Note 2.4.2
You can calculate percentiles using calculators and computers. There are a variety of online calculators.
A Formula for Finding the Percentile of a Value in a Data Set
• Order the data from smallest to largest.
• $x =$ the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile.
• $y =$ the number of data values equal to the data value for which you want to find the percentile.
• $n =$ the total number of data.
• Calculate $\dfrac{x + 0.5y}{n}(100)$. Then round to the nearest integer.
Example 2.4.6
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
1. Find the percentile for 58.
2. Find the percentile for 25.
Solution
1. Counting from the bottom of the list, there are 18 data values less than 58. There is one value of 58.
$x = 18$ and $y = 1$. $\dfrac{x + 0.5y}{n}(100) = \dfrac{18 + 0.5(1)}{29}(100) = 63.80$. 58 is the 64th percentile.
2. Counting from the bottom of the list, there are three data values less than 25. There is one value of 25.
$x = 3$ and $y = 1$. $\dfrac{x + 0.5y}{n}(100) = \dfrac{3 + 0.5(1)}{29}(100) = 12.07$. Twenty-five is the 12thpercentile.
Exercise $6$
Listed are 30 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Find the percentiles for 47 and 31.
Answer
Percentile for 47: Counting from the bottom of the list, there are 15 data values less than 47. There is one value of 47.
$x = 15$ and $y = 1$. $\dfrac{x + 0.5y}{n}(100) = \dfrac{15 + 0.5(1)}{30}(100) = 51.67$. 47 is the 52nd percentile.
Percentile for 31: Counting from the bottom of the list, there are eight data values less than 31. There are two values of 31.
$x = 8$ and $y = 2$. $\dfrac{x + 0.5y}{n}(100) = \dfrac{8 + 0.5(2)}{30}(100) = 30$. 31 is the 30th percentile.
Interpreting Percentiles, Quartiles, and Median
A percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the pth percentile. For example, 15% of data values are less than or equal to the 15th percentile.
• Low percentiles always correspond to lower data values.
• High percentiles always correspond to higher data values.
A percentile may or may not correspond to a value judgment about whether it is "good" or "bad." The interpretation of whether a certain percentile is "good" or "bad" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered "good;" in other contexts a high percentile might be considered "good". In many situations, there is no value judgment that applies.
Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text.
GUIDELINE
When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information.
• information about the context of the situation being considered
• the data value (value of the variable) that represents the percentile
• the percent of individuals or items with data values below the percentile
• the percent of individuals or items with data values above the percentile.
Example 2.4.7
On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation.
Answer
• Twenty-five percent of students finished the exam in 35 minutes or less.
• Seventy-five percent of students finished the exam in 35 minutes or more.
• A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)
Exercise $7$
For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation.
Answer
Twenty-five percent of runners finished the race in 11.5 seconds or more. Seventy-five percent of runners finished the race in 11.5 seconds or less. A lower percentile is good because finishing a race more quickly is desirable.
Example 2.4.8
On a 20 question math test, the 70th percentile for number of correct answers was 16. Interpret the 70th percentile in the context of this situation.
Answer
• Seventy percent of students answered 16 or fewer questions correctly.
• Thirty percent of students answered 16 or more questions correctly.
• A higher percentile could be considered good, as answering more questions correctly is desirable.
Exercise $8$
On a 60 point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.
Answer
Eighty percent of students earned 49 points or fewer. Twenty percent of students earned 49 or more points. A higher percentile is good because getting more points on an assignment is desirable.
Example 2.4.9
At a community college, it was found that the 30th percentile of credit units that students are enrolled for is seven units. Interpret the 30th percentile in the context of this situation.
Answer
• Thirty percent of students are enrolled in seven or fewer credit units.
• Seventy percent of students are enrolled in seven or more credit units.
• In this example, there is no "good" or "bad" value judgment associated with a higher or lower percentile. Students attend community college for varied reasons and needs, and their course load varies according to their needs.
Exercise $9$
During a season, the 40th percentile for points scored per player in a game is eight. Interpret the 40th percentile in the context of this situation.
Answer
Forty percent of players scored eight points or fewer. Sixty percent of players scored eight points or more. A higher percentile is good because getting more points in a basketball game is desirable.
Example 2.4.10
Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown.
0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes
10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes;
30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes
Determine the following five values.
• Min = 0
• Q1 = 20
• Med = 40
• Q3 = 60
• Max = 300
If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.
However, the principal needs to be careful. The value 300 appears to be a potential outlier.
$Q_{3} + 1.5(IQR) = 60 + (1.5)(40) = 120$.
The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values:
• Min = 0
• Q1 = 20
• Q3 = 60
• Max = 120
We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results.
Review
The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50th percentile would be greater than 50 percent of the other obeservations in the set. Quartiles divide data into quarters. The first quartile (Q1) is the 25th percentile,the second quartile (Q2 or median) is 50th percentile, and the third quartile (Q3) is the the 75th percentile. The interquartile range, or IQR, is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q1 from Q3, and can help determine outliers by using the following two expressions.
• $Q_{3} + IQR(1.5)$
• $Q_{1} - IQR(1.5)$
Formula Review
$i = \dfrac{k}{100}(n+1) \nonumber$
where $i$ = the ranking or position of a data value,
• $k$ = the kth percentile,
• $n$ = total number of data.
Expression for finding the percentile of a data value: $\left(\dfrac{x + 0.5y}{n}\right)(100)$
where $x =$ the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,
$y =$ the number of data values equal to the data value for which you want to find the percentile,
$n =$ total number of data
Glossary
Interquartile Range
or IQR, is the range of the middle 50 percent of the data values; the IQR is found by subtracting the first quartile from the third quartile.
Outlier
an observation that does not fit the rest of the data
Percentile
a number that divides ordered data into hundredths; percentiles may or may not be part of the data. The median of the data is the second quartile and the 50th percentile. The first and third quartiles are the 25th and the 75th percentiles, respectively.
Quartiles
the numbers that separate the data into quarters; quartiles may or may not be part of the data. The second quartile is the median of the data.
2.04: Measures of the Location of the Data
Exercise \(10\)
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
1. Find the 40th percentile.
2. Find the 78th percentile.
Answer
1. The 40th percentile is 37 years.
2. The 78th percentile is 70 years.
Exercise \(11\)
Listed are 32 ages for Academy Award winning best actors in order from smallest to largest.
18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
1. Find the percentile of 37.
2. Find the percentile of 72.
Exercise \(12\)
Jesse was ranked 37th in his graduating class of 180 students. At what percentile is Jesse’s ranking?
Answer
Jesse graduated 37th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37.
\(x = 143\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{143 + 0.5(1)}{180}(100) = 79.72\). Jesse’s rank of 37 puts him at the 80th percentile.
Exercise \(13\)
1. For runners in a race, a low time means a faster run. The winners in a race have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race?
2. The 20th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20th percentile in the context of the situation.
3. A bicyclist in the 90th percentile of a bicycle race completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90th percentile in the context of the situation.
Exercise \(14\)
1. For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race?
2. The 40th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40th percentile in the context of the situation.
Answer
1. For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a higher speed which is faster.
2. 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour or more (faster).
Exercise \(15\)
On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain.
Exercise \(16\)
Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85th percentile in the context of this situation.
Answer
When waiting in line at the DMV, the 85th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer.
Exercise \(17\)
In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78th percentile. Should Li be pleased or upset by this result? Explain.
Exercise \(18\)
In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had \$1,700 in damage and was in the 90th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain and write a sentence that interprets the 90th percentile in the context of this problem.
Answer
The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of \$1700 or less; only 10% had damage repair costs of \$1700 or more.
Exercise \(19\)
The University of California has two criteria used to set admission standards for freshman to be admitted to a college in the UC system:
1. Students' GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an "admissions index" score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students in the state. In this context, what percentile does the top 12% represent?
2. Students whose GPAs are at or above the 96th percentile of all students at their high school are eligible (called eligible in the local context), even if they are not in the top 12% of all students in the state. What percentage of students from each high school are "eligible in the local context"?
Exercise \(20\)
Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34th percentile. The 34th percentile of housing prices is \$240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses?
Answer
You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost \$240,000 or less. 66% of houses cost \$240,000 or more.
Use Exercise to calculate the following values:
Exercise \(21\)
First quartile = _______
Exercise \(22\)
Second quartile = median = 50th percentile = _______
Answer
4
Exercise \(23\)
Third quartile = _______
Exercise \(24\)
Interquartile range (IQR) = _____ – _____ = _____
Answer
\(6 - 4 = 2\)
Exercise \(25\)
10th percentile = _______
Exercise \(26\)
70th percentile = _______
Answer
6 | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.04%3A_Measures_of_the_Location_of_the_Data/2.4E%3A_Measures_of_the_Location_of_the_Data_%28Exercises%29.txt |
Box plots (also called box-and-whisker plots or box-whisker plots) give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. A box plot is constructed from five values: the minimum value, the first quartile, the median, the third quartile, and the maximum value. We use these values to compare how close other data values are to them.
To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The "whiskers" extend from the ends of the box to the smallest and largest data values. The median or second quartile can be between the first and third quartiles, or it can be one, or the other, or both. The box plot gives a good, quick picture of the data.
You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values.
Consider, again, this dataset.
1; 1; 2; 2; 4; 6; 6;.8; 7.2; 8; 8.3; 9; 10; 10; 11.5
The first quartile is two, the median is seven, and the third quartile is nine. The smallest value is one, and the largest value is 11.5. The following image shows the constructed box plot.
See the calculator instructions on the TI web site or in the appendix.
The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line.
It is important to start a box plot with a scaled number line. Otherwise the box plot may not be useful.
Example $1$
The following data are the heights of 40 students in a statistics class.
59; 60; 61; 62; 62; 63; 63; 64; 64; 64; 65; 65; 65; 65; 65; 65; 65; 65; 65; 66; 66; 67; 67; 68; 68; 69; 70; 70; 70; 70; 70; 71; 71; 72; 72; 73; 74; 74; 75; 77
Construct a box plot with the following properties; the calculator instructions for the minimum and maximum values as well as the quartiles follow the example.
• Minimum value = 59
• Maximum value = 77
• Q1: First quartile = 64.5
• Q2: Second quartile or median= 66
• Q3: Third quartile = 70
1. Each quarter has approximately 25% of the data.
2. The spreads of the four quarters are 64.5 – 59 = 5.5 (first quarter), 66 – 64.5 = 1.5 (second quarter), 70 – 66 = 4 (third quarter), and 77 – 70 = 7 (fourth quarter). So, the second quarter has the smallest spread and the fourth quarter has the largest spread.
3. $\text{Range} = \text{maximum value} - \text{the minimum value} = 77 - 59 = 18$
4. Interquartile Range: $IQR = Q_{3} – Q_{1} = 70 - 64.5 = 5.5$.
5. The interval 59–65 has more than 25% of the data so it has more data in it than the interval 66 through 70 which has 25% of the data.
6. The middle 50% (middle half) of the data has a range of 5.5 inches.
Calculator
To find the minimum, maximum, and quartiles:
Enter data into the list editor (Pres STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, and then arrow down.
Put the data values into the list L1.
Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.
Press ENTER.
Use the down and up arrow keys to scroll.
Smallest value = 59.
Largest value = 77.
Q1: First quartile = 64.5.
Q2: Second quartile or median = 66.
Q3: Third quartile = 70.
To construct the box plot:
Press 4: Plotsoff. Press ENTER.
Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER.
Arrow down to Xlist: Press 2nd 1 for L1
Arrow down to Freq: Press ALPHA. Press 1.
Press Zoom. Press 9: ZoomStat.
Press TRACE, and use the arrow keys to examine the box plot.
Exercise $1$
The following data are the number of pages in 40 books on a shelf. Construct a box plot using a graphing calculator, and state the interquartile range.
136; 140; 178; 190; 205; 215; 217; 218; 232; 234; 240; 255; 270; 275; 290; 301; 303; 315; 317; 318; 326; 333; 343; 349; 360; 369; 377; 388; 391; 392; 398; 400; 402; 405; 408; 422; 429; 450; 475; 512
Answer
$IQR = 158$
For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like:
In this case, at least 25% of the values are equal to one. Twenty-five percent of the values are between one and five, inclusive. At least 25% of the values are equal to five. The top 25% of the values fall between five and seven, inclusive.
Example $2$
Test scores for a college statistics class held during the day are:
99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90
Test scores for a college statistics class held during the evening are:
98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5
1. Find the smallest and largest values, the median, and the first and third quartile for the day class.
2. Find the smallest and largest values, the median, and the first and third quartile for the night class.
3. For each data set, what percentage of the data is between the smallest value and the first quartile? the first quartile and the median? the median and the third quartile? the third quartile and the largest value? What percentage of the data is between the first quartile and the largest value?
4. Create a box plot for each set of data. Use one number line for both box plots.
5. Which box plot has the widest spread for the middle 50% of the data (the data between the first and third quartiles)? What does this mean for that set of data in comparison to the other set of data?
Answer
• Min = 32
• Q1 = 56
• M = 74.5
• Q3 = 82.5
• Max = 99
• Min = 25.5
• Q1 = 78
• M = 81
• Q3 = 89
• Max = 98
1. Day class: There are six data values ranging from 32 to 56: 30%. There are six data values ranging from 56 to 74.5: 30%. There are five data values ranging from 74.5 to 82.5: 25%. There are five data values ranging from 82.5 to 99: 25%. There are 16 data values between the first quartile, 56, and the largest value, 99: 75%. Night class:
2. The first data set has the wider spread for the middle 50% of the data. The IQRfor the first data set is greater than the IQR for the second set. This means that there is more variability in the middle 50% of the first data set.
Exercise $2$
The following data set shows the heights in inches for the boys in a class of 40 students.
66; 66; 67; 67; 68; 68; 68; 68; 68; 69; 69; 69; 70; 71; 72; 72; 72; 73; 73; 74
The following data set shows the heights in inches for the girls in a class of 40 students.
61; 61; 62; 62; 63; 63; 63; 65; 65; 65; 66; 66; 66; 67; 68; 68; 68; 69; 69; 69
Construct a box plot using a graphing calculator for each data set, and state which box plot has the wider spread for the middle 50% of the data.
Answer
IQR for the boys = 4
IQR for the girls = 5
The box plot for the heights of the girls has the wider spread for the middle 50% of the data.
Example $3$
Graph a box-and-whisker plot for the data values shown.
10; 10; 10; 15; 35; 75; 90; 95; 100; 175; 420; 490; 515; 515; 790
The five numbers used to create a box-and-whisker plot are:
• Min: 10
• Q1: 15
• Med: 95
• Q3: 490
• Max: 790
The following graph shows the box-and-whisker plot.
Exercise $3$
Follow the steps you used to graph a box-and-whisker plot for the data values shown.
0; 5; 5; 15; 30; 30; 45; 50; 50; 60; 75; 110; 140; 240; 330
Answer
The data are in order from least to greatest. There are 15 values, so the eighth number in order is the median: 50. There are seven data values written to the left of the median and 7 values to the right. The five values that are used to create the boxplot are:
• Min: 0
• Q1: 15
• Med: 50
• Q3: 110
• Max: 330
Review
Box plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data.
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars.
Exercise 2.5.4
Construct a box plot below. Use a ruler to measure and scale accurately.
Exercise 2.5.5
Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or concentrated in some areas, but not in others? How can you tell?
Answer
More than 25% of salespersons sell four cars in a typical week. You can see this concentration in the box plot because the first quartile is equal to the median. The top 25% and the bottom 25% are spread out evenly; the whiskers have the same length.
Bringing It Together
Exercise 2.5.6
Santa Clara County, CA, has approximately 27,873 Japanese-Americans. Their ages are as follows:
Age Group Percent of Community
0–17 18.9
18–24 8.0
25–34 22.8
35–44 15.0
45–54 13.1
55–64 11.9
65+ 10.3
1. Construct a histogram of the Japanese-American community in Santa Clara County, CA. The bars will not be the same width for this example. Why not? What impact does this have on the reliability of the graph?
2. What percentage of the community is under age 35?
3. Which box plot most resembles the information above?
Answer
1. For graph, check student's solution.
2. 49.7% of the community is under the age of 35.
3. Based on the information in the table, graph (a) most closely represents the data.
Glossary
Box plot
a graph that gives a quick picture of the middle 50% of the data
First Quartile
the value that is the median of the of the lower half of the ordered data set
Frequency Polygon
looks like a line graph but uses intervals to display ranges of large amounts of data
Interval
also called a class interval; an interval represents a range of data and is used when displaying large data sets
Paired Data Set
two data sets that have a one to one relationship so that:
• both data sets are the same size, and
• each data point in one data set is matched with exactly one point from the other set.
Skewed
used to describe data that is not symmetrical; when the right side of a graph looks “chopped off” compared the left side, we say it is “skewed to the left.” When the left side of the graph looks “chopped off” compared to the right side, we say the data is “skewed to the right.” Alternatively: when the lower values of the data are more spread out, we say the data are skewed to the left. When the greater values are more spread out, the data are skewed to the right. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.05%3A_Box_Plots.txt |
The "center" of a data set is also a way of describing location. The two most widely used measures of the "center" of the data are the mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center.
Note
The ideal gas law is easy to remember and apply in solving problems, as long as you get the proper values aThe words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. However, in practice among non-statisticians, “average" is commonly accepted for “arithmetic mean.”
When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an $x$ with a bar over it (pronounced “$x$ bar”): $\overline{x}$.
The Greek letter $\mu$ (pronounced "mew") represents the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random.
To see that both ways of calculating the mean are the same, consider the sample:
1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4
$\bar{x} = \dfrac{1+1+1+2+2+3+4+4+4+4+4}{11} = 2.7$
$\bar{x} = \dfrac{3(1) + 2(2) + 1(3) + 5(4)}{11} = 2.7$
In the second calculation, the frequencies are 3, 2, 1, and 5.
You can quickly find the location of the median by using the expression
$\dfrac{n+1}{2}$
The letter $n$ is the total number of data values in the sample. If $n$ is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If $n$ is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered. For example, if the total number of data values is 97, then
$\dfrac{n+1}{2} = \dfrac{97+1}{2} = 49.$
The median is the 49th value in the ordered data. If the total number of data values is 100, then
$\dfrac{n+1}{2} = \dfrac{100+1}{2} = 50.5.$
The median occurs midway between the 50th and 51st values. The location of the median and the value of the median are not the same. The upper case letter $M$ is often used to represent the median. The next example illustrates the location of the median and the value of the median.
Example $1$
AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest):
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47
Calculate the mean and the median.
Answer
The calculation for the mean is:
$\bar{x} = \dfrac{[3+4+(8)(2)+10+11+12+13+14+(15)(2)+(16)(2)+...+35+37+40+(44)(2)+47]}{40} = 23.6$
To find the median, $M$, first use the formula for the location. The location is:
$\dfrac{n+1}{2} = \dfrac{40+1}{2} = 20.5$
Starting at the smallest value, the median is located between the 20th and 21st values (the two 24s):
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47
$M = \dfrac{24+24}{2} = 24$
Calculator
To find the mean and the median:
Clear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER.
Enter data into the list editor. Press STAT 1:EDIT.
Put the data values into list L1.
Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.
Press the down and up arrow keys to scroll.
$\bar{x}$ = 23.6, M = 24
Exercise $1$
The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median.
3; 4; 5; 7; 7; 7; 7; 8; 8; 9; 9; 10; 10; 10; 10; 10; 11; 12; 12; 13; 14; 14; 15; 15; 17; 17; 18; 19; 19; 19; 21; 21; 22; 22; 23; 24; 24; 24; 24
Answer
Mean: $3 + 4 + 5 + 7 + 7 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 11 + 12 + 12 + 13 + 14 + 14 + 15 + 15 + 17 + 17 + 18 + 19 + 19 + 19 + 21 + 21 + 22 + 22 + 23 + 24 + 24 + 24 = 544$
$\dfrac{544}{39} = 13.95$
Median: Starting at the smallest value, the median is the 20th term, which is 13.
Example $2$
Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn$30,000. Which is the better measure of the "center": the mean or the median?
Solution
$\bar{x} = \dfrac{5,000,000+49(30,000)}{50} = 129,400$
$M = 30,000$
(There are 49 people who earn $30,000 and one person who earns$5,000,000.)
The median is a better measure of the "center" than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data.
Exercise $2$
In a sample of 60 households, one house is worth $2,500,000. Half of the rest are worth$280,000, and all the others are worth $315,000. Which is the better measure of the “center”: the mean or the median? Answer The median is the better measure of the “center” than the mean because 59 of the values are$280,000 and one is $2,500,000. The$2,500,000 is an outlier. Either $280,000 or$315,000 gives us a better sense of the middle of the data.
Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal.
Example $3$
Statistics exam scores for 20 students are as follows:
50; 53; 59; 59; 63; 63; 72; 72; 72; 72; 72; 76; 78; 81; 83; 84; 84; 84; 90; 93
Find the mode.
Answer
The most frequent score is 72, which occurs five times. Mode = 72.
Exercise $3$
The number of books checked out from the library from 25 students are as follows:
0; 0; 0; 1; 2; 3; 3; 4; 4; 5; 5; 7; 7; 7; 7; 8; 8; 8; 9; 10; 10; 11; 11; 12; 12
Find the mode.
Answer
The most frequent number of books is 7, which occurs four times. Mode = 7.
Example $4$
Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice.
When is the mode the best measure of the "center"? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing.
The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red.
Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software.
Exercise $4$
Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is$50,000 and the mean is $47,500. What would be the best measure of the “center”? Answer Because$25,000 occurs nearly half the time, the mode would be the best measure of the center because the median and mean don’t represent what most people make at the factory.
The Law of Large Numbers and the Mean
The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean $\bar{x}$ of the sample is very likely to get closer and closer to $\mu$. This is discussed in more detail later in the text.
Sampling Distributions and Statistic of a Sampling Distribution
You can think of a sampling distribution as a relative frequency distribution with a great many samples. (See Sampling and Data for a review of relative frequency). Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below.
# of movies Relative Frequency
0
$\dfrac{5}{30}$
1
$\dfrac{15}{30}$
2
$\dfrac{6}{30}$
3
$\dfrac{3}{30}$
4
$\dfrac{1}{30}$
If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.
A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean $\bar{x}$ is an example of a statistic which estimates the population mean $\mu$.
Calculating the Mean of Grouped Frequency Tables
When only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean:
$mean = \dfrac{\text{data sum}}{\text{number of data values}}.$
We simply need to modify the definition to fit within the restrictions of a frequency table.
Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is
$\dfrac{\text{lower boundary+upper boundary}}{2}.$
We can now modify the mean definition to be
$\text{Mean of Frequency Table} = \dfrac{\sum{fm}}{\sum{f}}$
where $f$ is the frequency of the interval and $m$ is the midpoint of the interval.
Example $5$
A frequency table displaying professor Blount’s last statistic test is shown. Find the best estimate of the class mean.
Grade Interval Number of Students
50–56.5 1
56.5–62.5 0
62.5–68.5 4
68.5–74.5 4
74.5–80.5 2
80.5–86.5 3
86.5–92.5 4
92.5–98.5 1
Solution
• Find the midpoints for all intervals
Grade Interval Midpoint
50–56.5 53.25
56.5–62.5 59.5
62.5–68.5 65.5
68.5–74.5 71.5
74.5–80.5 77.5
80.5–86.5 83.5
86.5–92.5 89.5
92.5–98.5 95.5
• Calculate the sum of the product of each interval frequency and midpoint. $53.25(1) + 59.5(0) + 65.5(4 )+ 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.25$
• $\mu = \dfrac{\sum{fm}}{\sum{f}} = \dfrac{1460.25}{19} = 76.86$
Exercise $5$
Maris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data:
Hours Teenagers Spend on Video Games Number of Teenagers
0–3.5 3
3.5–7.5 7
7.5–11.5 12
11.5–15.5 7
15.5–19.5 9
What is the best estimate for the mean number of hours spent playing video games?
Answer
Find the midpoint of each interval, multiply by the corresponding number of teenagers, add the results and then divide by the total number of teenagers
The midpoints are 1.75, 5.5, 9.5, 13.5,17.5.
$Mean = (1.75)(3) + (5.5)(7) + (9.5)(12) + (13.5)(7) + (17.5)(9) = 409.75$
Review
The mean and the median can be calculated to help you find the "center" of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occuring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set.
Formula Review
$\mu = \dfrac{\sum{fm}}{\sum{f}}$
where $f$ = interval frequencies and $m$ = interval midpoints.
Exercise 2.6.6
Find the mean for the following frequency tables.
1. Grade Frequency
49.5–59.5 2
59.5–69.5 3
69.5–79.5 8
79.5–89.5 12
89.5–99.5 5
2. Daily Low Temperature Frequency
49.5–59.5 53
59.5–69.5 32
69.5–79.5 15
79.5–89.5 1
89.5–99.5 0
3. Points per Game Frequency
49.5–59.5 14
59.5–69.5 32
69.5–79.5 15
79.5–89.5 23
89.5–99.5 2
Use the following information to answer the next three exercises: The following data show the lengths of boats moored in a marina. The data are ordered from smallest to largest: 16; 17; 19; 20; 20; 21; 23; 24; 25; 25; 25; 26; 26; 27; 27; 27; 28; 29; 30; 32; 33; 33; 34; 35; 37; 39; 40
Exercise 2.6.7
Calculate the mean.
Answer
Mean: $16 + 17 + 19 + 20 + 20 + 21 + 23 + 24 + 25 + 25 + 25 + 26 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 32 + 33 + 33 + 34 + 35 + 37 + 39 + 40 = 738$;
$\dfrac{738}{27} = 27.33$
Exercise 2.6.8
Identify the median.
Exercise 2.6.9
Identify the mode.
Answer
The most frequent lengths are 25 and 27, which occur three times. Mode = 25, 27
Use the following information to answer the next three exercises: Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Calculate the following:
Exercise 2.6.10
sample mean = $\bar{x}$ = _______
median = _______
Answer
4
Bringing It Together
Exercise 2.6.12
Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information.
Javier Ercilia
$\bar{x}$ 6.0 miles 6.0 miles
s 4.0 miles 7.0 miles
1. How can you determine which survey was correct ?
2. Explain what the difference in the results of the surveys implies about the data.
3. If the two histograms depict the distribution of values for each supervisor, which one depicts Ercilia's sample? How do you know?
1. If the two box plots depict the distribution of values for each supervisor, which one depicts Ercilia’s sample? How do you know?
Use the following information to answer the next three exercises: We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section.
Number of years Frequency Number of years Frequency
Total = 20
7 1 22 1
14 3 23 1
15 1 26 1
18 1 40 2
19 4 42 2
20 3
What is the IQR?
1. 8
2. 11
3. 15
4. 35
Answer
a
Exercise 2.6.14
What is the mode?
1. 19
2. 19.5
3. 14 and 20
4. 22.65
Exercise 2.6.15
Is this a sample or the entire population?
1. sample
2. entire population
3. neither
Answer
b
Glossary
Frequency Table
a data representation in which grouped data is displayed along with the corresponding frequencies
Mean
a number that measures the central tendency of the data; a common name for mean is 'average.' The term 'mean' is a shortened form of 'arithmetic mean.' By definition, the mean for a sample (denoted by $\bar{x}$) is $\bar{x} = \dfrac{\text{Sum of all values in the sample}}{\text{Number of values in the sample}}$, and the mean for a population (denoted by $\mu$) is $\mu = \dfrac{\text{Sum of all values in the population}}{\text{Number of values in the population}}$.
Median
a number that separates ordered data into halves; half the values are the same number or smaller than the median and half the values are the same number or larger than the median. The median may or may not be part of the data.
Midpoint
the mean of an interval in a frequency table
Mode
the value that appears most frequently in a set of data | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.06%3A_Measures_of_the_Center_of_the_Data.txt |
Consider the following data set.
4; 5; 6; 6; 6; 7; 7; 7; 7; 7; 7; 8; 8; 8; 9; 10
This data set can be represented by following histogram. Each interval has width one, and each value is located in the middle of an interval.
The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median.
The histogram for the data: 4; 5; 6; 6; 6; 7; 7; 7; 7; 8 is not symmetrical. The right-hand side seems "chopped off" compared to the left side. A distribution of this type is called skewed to the left because it is pulled out to the left.
The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so.
The histogram for the data: 6; 7; 7; 7; 7; 8; 8; 8; 9; 10, is also not symmetrical. It is skewed to the right.
The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, the mean is the largest, while the mode is the smallest. Again, the mean reflects the skewing the most.
Generally, if the distribution of data is skewed to the left, the mean is less than the median, which is often less than the mode. If the distribution of data is skewed to the right, the mode is often less than the median, which is less than the mean.
Skewness and symmetry become important when we discuss probability distributions in later chapters.
Example \(1\)
Statistics are used to compare and sometimes identify authors. The following lists shows a simple random sample that compares the letter counts for three authors.
• Terry: 7; 9; 3; 3; 3; 4; 1; 3; 2; 2
• Davis: 3; 3; 3; 4; 1; 4; 3; 2; 3; 1
• Maris: 2; 3; 4; 4; 4; 6; 6; 6; 8; 3
1. Make a dot plot for the three authors and compare the shapes.
2. Calculate the mean for each.
3. Calculate the median for each.
4. Describe any pattern you notice between the shape and the measures of center.
Solution
a.
• Terry’s mean is 3.7, Davis’ mean is 2.7, Maris’ mean is 4.6.
• Terry’s median is three, Davis’ median is three. Maris’ median is four.
• It appears that the median is always closest to the high point (the mode), while the mean tends to be farther out on the tail. In a symmetrical distribution, the mean and the median are both centrally located close to the high point of the distribution.
Exercise \(1\)
Discuss the mean, median, and mode for each of the following problems. Is there a pattern between the shape and measure of the center?
a.
b.
The Ages Former U.S Presidents Died
4 6 9
5 3 6 7 7 7 8
6 0 0 3 3 4 4 5 6 7 7 7 8
7 0 1 1 2 3 4 7 8 8 9
8 0 1 3 5 8
9 0 0 3 3
Key: 8|0 means 80.
c.
Review
Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A left (or negative) skewed distribution has a shape like Figure \(2\). A right (or positive) skewed distribution has a shape like Figure \(3\). A symmetrical distribution looks like Figure \(1\).
Use the following information to answer the next three exercises: State whether the data are symmetrical, skewed to the left, or skewed to the right.
Exercise 2.7.2
1; 1; 1; 2; 2; 2; 2; 3; 3; 3; 3; 3; 3; 3; 3; 4; 4; 4; 5; 5
Answer
The data are symmetrical. The median is 3 and the mean is 2.85. They are close, and the mode lies close to the middle of the data, so the data are symmetrical.
Exercise 2.7.3
16; 17; 19; 22; 22; 22; 22; 22; 23
Exercise 2.7.4
87; 87; 87; 87; 87; 88; 89; 89; 90; 91
Answer
The data are skewed right. The median is 87.5 and the mean is 88.2. Even though they are close, the mode lies to the left of the middle of the data, and there are many more instances of 87 than any other number, so the data are skewed right.
Exercise 2.7.5
When the data are skewed left, what is the typical relationship between the mean and median?
Exercise 2.7.6
When the data are symmetrical, what is the typical relationship between the mean and median?
Answer
When the data are symmetrical, the mean and median are close or the same.
Exercise 2.7.7
What word describes a distribution that has two modes?
Exercise 2.7.8
Describe the shape of this distribution.
Answer
The distribution is skewed right because it looks pulled out to the right.
Exercise 2.7.9
Describe the relationship between the mode and the median of this distribution.
Exercise 2.7.10
Describe the relationship between the mean and the median of this distribution.
Answer
The mean is 4.1 and is slightly greater than the median, which is four.
Exercise 2.7.11
Describe the shape of this distribution.
Exercise 2.7.12
Describe the relationship between the mode and the median of this distribution.
Answer
The mode and the median are the same. In this case, they are both five.
Exercise 2.7.13
Are the mean and the median the exact same in this distribution? Why or why not?
Exercise 2.7.14
Describe the shape of this distribution.
Answer
The distribution is skewed left because it looks pulled out to the left.
Exercise 2.7.15
Describe the relationship between the mode and the median of this distribution.
Exercise 2.7.16
Describe the relationship between the mean and the median of this distribution.
Answer
The mean and the median are both six.
Exercise 2.7.17
The mean and median for the data are the same.
3; 4; 5; 5; 6; 6; 6; 6; 7; 7; 7; 7; 7; 7; 7
Is the data perfectly symmetrical? Why or why not?
Exercise 2.7.18
Which is the greatest, the mean, the mode, or the median of the data set?
11; 11; 12; 12; 12; 12; 13; 15; 17; 22; 22; 22
Answer
The mode is 12, the median is 12.5, and the mean is 15.1. The mean is the largest.
Exercise 2.7.19
Which is the least, the mean, the mode, and the median of the data set?
56; 56; 56; 58; 59; 60; 62; 64; 64; 65; 67
Exercise 2.7.20
Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median? Why?
Answer
The mean tends to reflect skewing the most because it is affected the most by outliers.
Exercise 2.7.21
In a perfectly symmetrical distribution, when would the mode be different from the mean and median? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.07%3A_Skewness_and_the_Mean_Median_and_Mode.txt |
An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.
The standard deviation
• provides a numerical measure of the overall amount of variation in a data set, and
• can be used to determine whether a particular data value is close to or far from the mean.
The standard deviation provides a measure of the overall variation in a data set
The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.
Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B. the average wait time at both supermarkets is five minutes. At supermarket A, the standard deviation for the wait time is two minutes; at supermarket B the standard deviation for the wait time is four minutes.
Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B. Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average.
The standard deviation can be used to determine whether a data value is close to or far from the mean.
Suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A, the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.
Rosa waits for seven minutes:
• Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.
• Rosa's wait time of seven minutes is two minutes longer than the average of five minutes.
• Rosa's wait time of seven minutes is one standard deviation above the average of five minutes.
Binh waits for one minute.
• One is four minutes less than the average of five; four minutes is equal to two standard deviations.
• Binh's wait time of one minute is four minutes less than the average of five minutes.
• Binh's wait time of one minute is two standard deviations below the average of five minutes.
• A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)
The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because $5 + (1)(2) = 7$.
If one were also part of the data set, then one is two standard deviations to the left of five because $5 + (-2)(2) = 1$.
• In general, a value = mean + (#ofSTDEV)(standard deviation)
• where #ofSTDEVs = the number of standard deviations
• #ofSTDEV does not need to be an integer
• One is two standard deviations less than the mean of five because: $1 = 5 + (-2)(2)$.
The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population.
• sample: $x = \bar{x} + \text{(#ofSTDEV)(s)}$
• Population: $x = \mu + \text{(#ofSTDEV)(s)}$
The lower case letter s represents the sample standard deviation and the Greek letter $\sigma$ (sigma, lower case) represents the population standard deviation.
The symbol $\bar{x}$ is the sample mean and the Greek symbol $\mu$ is the population mean.
Calculating the Standard Deviation
If $x$ is a number, then the difference "$x$ – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is $x - \mu$. For sample data, in symbols a deviation is $x - \bar{x}$.
The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter $\sigma$ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of $\sigma$.
To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the $x - \bar{x}$ values for a sample, or the $x - \mu$ values for a population). The symbol $\sigma^{2}$ represents the population variance; the population standard deviation $\sigma$ is the square root of the population variance. The symbol $s^{2}$ represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.
If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by $N$, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample.
Formulas for the Sample Standard Deviation
$s = \sqrt{\dfrac{\sum(x-\bar{x})^{2}}{n-1}} \label{eq1}$
or
$s = \sqrt{\dfrac{\sum f (x-\bar{x})^{2}}{n-1}} \label{eq2}$
For the sample standard deviation, the denominator is $n - 1$, that is the sample size MINUS 1.
Formulas for the Population Standard Deviation
$\sigma = \sqrt{\dfrac{\sum(x-\mu)^{2}}{N}} \label{eq3}$
or
$\sigma = \sqrt{\dfrac{\sum f (x-\mu)^{2}}{N}} \label{eq4}$
For the population standard deviation, the denominator is $N$, the number of items in the population.
In Equations \ref{eq2} and \ref{eq4}, $f$ represents the frequency with which a value appears. For example, if a value appears once, $f$ is one. If a value appears three times in the data set or population, $f$ is three.
Sampling Variability of a Statistic
The statistic of a sampling distribution was discussed in Section 2.6. How much the statistic varies from one sample to another is known as the sampling variability of a statistic. You typically measure the sampling variability of a statistic by its standard error.
The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean in Chapter 7. The notation for the standard error of the mean is $\dfrac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation of the population and $n$ is the size of the sample.
In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation $\sigma_{x}$ or $s_{x}$ from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)
Example $1$
In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of n = 20 fifth grade students. The ages are rounded to the nearest half year:
9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5;
$\bar{x} = \dfrac{9+9.5(2)+10(4)+10.5(4)+11(6)+11.5(3)}{20} = 10.525 \nonumber$
The average age is 10.53 years, rounded to two places.
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s.
Data Freq. Deviations Deviations2 (Freq.)(Deviations2)
x f (x – $\bar{x}$) (x – $\bar{x}$)2 (f)(x – $\bar{x}$)2
9 1 9 – 10.525 = –1.525 (–1.525)2 = 2.325625 1 × 2.325625 = 2.325625
9.5 2 9.5 – 10.525 = –1.025 (–1.025)2 = 1.050625 2 × 1.050625 = 2.101250
10 4 10 – 10.525 = –0.525 (–0.525)2 = 0.275625 4 × 0.275625 = 1.1025
10.5 4 10.5 – 10.525 = –0.025 (–0.025)2 = 0.000625 4 × 0.000625 = 0.0025
11 6 11 – 10.525 = 0.475 (0.475)2 = 0.225625 6 × 0.225625 = 1.35375
11.5 3 11.5 – 10.525 = 0.975 (0.975)2 = 0.950625 3 × 0.950625 = 2.851875
The total is 9.7375
The sample variance, $s^{2}$, is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1):
$s^{2} = \dfrac{9.7375}{20-1} = 0.5125 \nonumber$
The sample standard deviation s is equal to the square root of the sample variance:
$s = \sqrt{0.5125} = 0.715891 \nonumber$
and this is rounded to two decimal places, $s = 0.72$.
Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy.
• For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation). Verify the mean and standard deviation or a calculator or computer.
• For a sample: $x$ = $\bar{x}$ + (#ofSTDEVs)(s)
• For a population: $x$ = $\mu$ + (#ofSTDEVs)$\sigma$
• For this example, use x = $\bar{x}$ + (#ofSTDEVs)(s) because the data is from a sample
1. Verify the mean and standard deviation on your calculator or computer.
2. Find the value that is one standard deviation above the mean. Find ($\bar{x}$ + 1s).
3. Find the value that is two standard deviations below the mean. Find ($\bar{x}$ – 2s).
4. Find the values that are 1.5 standard deviations from (below and above) the mean.
Solution
• Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.
• Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.
• Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.
• Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.
• $\bar{x}$ = 10.525
• Use Sx because this is sample data (not a population): Sx=0.715891
1. ($\bar{x} + 1s) = 10.53 + (1)(0.72) = 11.25$
2. $(\bar{x} - 2s) = 10.53 – (2)(0.72) = 9.09$
• $(\bar{x} - 1.5s) = 10.53 – (1.5)(0.72) = 9.45$
• $(\bar{x} + 1.5s) = 10.53 + (1.5)(0.72) = 11.61$
Exercise 2.8.1
On a baseball team, the ages of each of the players are as follows:
21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40
Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean.
Answer
$\mu$ = 30.68
$s = 6.09$
($\bar{x} + 2s = 30.68 + (2)(6.09) = 42.86$.
Explanation of the standard deviation calculation shown in the table
The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. (For Example $1$, there are $n = 20$ deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.
The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.
Notice that instead of dividing by $n = 20$, the calculation divided by $n - 1 = 20 - 1 = 19$ because the data is a sample. For the sample variance, we divide by the sample size minus one ($n - 1$). Why not divide by $n$? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by ($n - 1$) gives a better estimate of the population variance.
Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.
The standard deviation, $s$ or $\sigma$, is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make $s$ or $\sigma$ very large.
The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot.
Example $2$
Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
1. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
2. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:
1. The sample mean
2. The sample standard deviation
3. The median
4. The first quartile
5. The third quartile
6. IQR
3. Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.
Answer
1. See Table
1. The sample mean = 73.5
2. The sample standard deviation = 17.9
3. The median = 73
4. The first quartile = 61
5. The third quartile = 90
6. IQR = 90 – 61 = 29
2. The $x$-axis goes from 32.5 to 100.5; $y$-axis goes from -2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is ($100.5 - 32.5$) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, $32.5 + 13.6 = 46.1$, $46.1 + 13.6 = 59.7$, $59.7 + 13.6 = 73.3$, $73.3 + 13.6 = 86.9$, $86.9 + 13.6 = 100.5 =$ the ending value; No data values fall on an interval boundary.
The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater ($73 - 33 = 40$) than the spread in the upper 50% ($100 - 73 = 27$). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.
Data Frequency Relative Frequency Cumulative Relative Frequency
33 1 0.032 0.032
42 1 0.032 0.064
49 2 0.065 0.129
53 1 0.032 0.161
55 2 0.065 0.226
61 1 0.032 0.258
63 1 0.032 0.29
67 1 0.032 0.322
68 2 0.065 0.387
69 2 0.065 0.452
72 1 0.032 0.484
73 1 0.032 0.516
74 1 0.032 0.548
78 1 0.032 0.580
80 1 0.032 0.612
83 1 0.032 0.644
88 3 0.097 0.741
90 1 0.032 0.773
92 1 0.032 0.805
94 4 0.129 0.934
96 1 0.032 0.966
100 1 0.032 0.998 (Why isn't this value 1?)
Exercise $2$
The following data show the different types of pet food stores in the area carry.
6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;
Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.
Answer
$\mu = 9.3$ and $s = 2.2$
Standard deviation of Grouped Frequency Tables
Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:
$\text{Mean of Frequency Table} = \dfrac{\sum fm}{\sum f}$
where $f$ interval frequencies and $m =$ interval midpoints.
Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.
Example $3$
Find the standard deviation for the data in Table $3$.
Table $3$
Class Frequency, f Midpoint, m m2 $\bar{x}$ fm2 Standard Deviation
0–2 1 1 1 7.58 1 3.5
3–5 6 4 16 7.58 96 3.5
6–8 10 7 49 7.58 490 3.5
9–11 7 10 100 7.58 700 3.5
12–14 0 13 169 7.58 0 3.5
15–17 2 16 256 7.58 512 3.5
For this data set, we have the mean, $\bar{x}$ = 7.58 and the standard deviation, $s_{x}$ = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated, $s_{x} = \sqrt{\dfrac{f(m - \bar{x})^{2}}{n-1}}$ where $s_{x}$ = sample standard deviation, $\bar{x}$ = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.
Find the standard deviation for the data from the previous example
Class 0-2 3-5 6-8 9–11 12–14 15–17
Frequency, f 1 6 10 7 0 2
First, press the STAT key and select 1:Edit
Input the midpoint values into L1 and the frequencies into L2
Select STAT, CALC, and 1: 1-Var Stats
Select 2nd then 1 then , 2nd then 2 Enter
You will see displayed both a population standard deviation, $\sigma_{x}$, and the sample standard deviation, $s_{x}$.
Comparing Values from Different Data Sets
The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.
• For each data value, calculate how many standard deviations away from its mean the value is.
• Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.
• $\text{#ofSTDEVs} = \dfrac{\text{value-mean}}{\text{standard deviation}}$
• Compare the results of this calculation.
#ofSTDEVs is often called a "z-score"; we can use the symbol $z$. In symbols, the formulas become:
Sample $x = \bar{x} + zs$ $z = \dfrac{x - \bar{x}}{s}$
Population $x = \mu + z\sigma$ $z = \dfrac{x - \mu}{\sigma}$
Example $4$
Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?
Student GPA School Mean GPA School Standard Deviation
John 2.85 3.0 0.7
Ali 77 80 10
Answer
For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.
$z = \text{#ofSTDEVs} = \left(\dfrac{\text{value-mean}}{\text{standard deviation}}\right) = \left(\dfrac{x + \mu}{\sigma}\right) \nonumber$
For John,
$z = \text{#ofSTDEVs} = \left(\dfrac{2.85-3.0}{0.7}\right) = -0.21 \nonumber$
For Ali,
$z = \text{#ofSTDEVs} = (\dfrac{77-80}{10}) = -0.3 \nonumber$
John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.
John's z-score of –0.21 is higher than Ali's z-score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.
Exercise $4$
Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?
Swimmer Time (seconds) Team Mean Time Team Standard Deviation
Angie 26.2 27.2 0.8
Beth 27.3 30.1 1.4
Answer
For Angie:
$z = \left(\dfrac{26.2-27.2}{0.8}\right) = -1.25 \nonumber$
For Beth:
$z = \left(\dfrac{27.3-30.1}{1.4}\right) = -2 \nonumber$
The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.
For ANY data set, no matter what the distribution of the data is:
• At least 75% of the data is within two standard deviations of the mean.
• At least 89% of the data is within three standard deviations of the mean.
• At least 95% of the data is within 4.5 standard deviations of the mean.
• This is known as Chebyshev's Rule.
For data having a distribution that is BELL-SHAPED and SYMMETRIC:
• Approximately 68% of the data is within one standard deviation of the mean.
• Approximately 95% of the data is within two standard deviations of the mean.
• More than 99% of the data is within three standard deviations of the mean.
• This is known as the Empirical Rule.
• It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.
Review
The standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population.
• The Standard Deviation allows us to compare individual data or classes to the data set mean numerically.
• $s = \sqrt{\dfrac{\sum(x-\bar{x})^{2}}{n-1}}$ or $s = \sqrt{\dfrac{\sum f (x-\bar{x})^{2}}{n-1}}$ is the formula for calculating the standard deviation of a sample. To calculate the standard deviation of a population, we would use the population mean, $\mu$, and the formula $\sigma = \sqrt{\dfrac{\sum(x-\mu)^{2}}{N}}$ or $\sigma = \sqrt{\dfrac{\sum f (x-\mu)^{2}}{N}}$.f(xμ)2N.
Formula Review
$s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n} - \bar{x}^2}$
where $s_{x} =\text{sample standard deviation}$ and $\bar{x} = \text{sample mean}$
Use the following information to answer the next two exercises: The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles.
29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150
Exercise 2.8.4
Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth.
Answer
$s$ = 34.5
Exercise 2.8.5
Find the value that is one standard deviation below the mean.
Exercise 2.8.6
Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team?
Baseball Player Batting Average Team Batting Average Team Standard Deviation
Fredo 0.158 0.166 0.012
Karl 0.177 0.189 0.015
Answer
For Fredo:
$z$ = $\dfrac{0.158-0.166}{0.012}$ = –0.67
For Karl:
$z$ = $\dfrac{0.177-0.189}{0.015}$ = –0.8
Fredo’s z-score of –0.67 is higher than Karl’s z-score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.
Exercise 2.8.7
Use Table to find the value that is three standard deviations:
• above the mean
• below the mean
Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84.
Exercise 2.8.5
Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84.
1. Grade Frequency
49.5–59.5 2
59.5–69.5 3
69.5–79.5 8
79.5–89.5 12
89.5–99.5 5
2. Daily Low Temperature Frequency
49.5–59.5 53
59.5–69.5 32
69.5–79.5 15
79.5–89.5 1
89.5–99.5 0
3. Points per Game Frequency
49.5–59.5 14
59.5–69.5 32
69.5–79.5 15
79.5–89.5 23
89.5–99.5 2
Answer
1. $s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n} - \bar{x}^{2}} = \sqrt{\dfrac{193157.45}{30} - 79.5^{2}} = 10.88$
2. $s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n} - \bar{x}^{2}} = \sqrt{\dfrac{380945.3}{101} - 60.94^{2}} = 7.62$
3. $s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n} - \bar{x}^{2}} = \sqrt{\dfrac{440051.5}{86} - 70.66^{2}} = 11.14$
Bringing It Together
Exercise 2.8.7
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
# of movies Frequency
0 5
1 9
2 6
3 4
4 1
1. Find the sample mean $\bar{x}$.
2. Find the approximate sample standard deviation, $s$.
Answer
1. 1.48
2. 1.12
Exercise 2.8.8
Forty randomly selected students were asked the number of pairs of sneakers they owned. Let $X =$ the number of pairs of sneakers owned. The results are as follows:
$X$ Frequency
1 2
2 5
3 8
4 12
5 12
6 0
7 1
1. Find the sample mean $\bar{x}$
2. Find the sample standard deviation, s
3. Construct a histogram of the data.
4. Complete the columns of the chart.
5. Find the first quartile.
6. Find the median.
7. Find the third quartile.
8. Construct a box plot of the data.
9. What percent of the students owned at least five pairs?
10. Find the 40th percentile.
11. Find the 90th percentile.
12. Construct a line graph of the data
13. Construct a stemplot of the data
Exercise 2.8.9
Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year.
177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265
1. Organize the data from smallest to largest value.
2. Find the median.
3. Find the first quartile.
4. Find the third quartile.
5. Construct a box plot of the data.
6. The middle 50% of the weights are from _______ to _______.
7. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why?
8. If our population included every team member who ever played for the San Francisco 49ers, would the above data be a sample of weights or the population of weights? Why?
9. Assume the population was the San Francisco 49ers. Find:
1. the population mean, $\mu$.
2. the population standard deviation, $\sigma$.
3. the weight that is two standard deviations below the mean.
4. When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he?
10. That same year, the mean weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer?
Answer
1. 174; 177; 178; 184; 185; 185; 185; 185; 188; 190; 200; 205; 205; 206; 210; 210; 210; 212; 212; 215; 215; 220; 223; 228; 230; 232; 241; 241; 242; 245; 247; 250; 250; 259; 260; 260; 265; 265; 270; 272; 273; 275; 276; 278; 280; 280; 285; 285; 286; 290; 290; 295; 302
2. 241
3. 205.5
4. 272.5
5. 205.5, 272.5
6. sample
7. population
1. 236.34
2. 37.50
3. 161.34
4. 0.84 std. dev. below the mean
8. Young
Exercise 2.8.10
One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The 12 change scores are as follows:
3; 8; –1; 2; 0; 5; –3; 1; –1; 6; 5; –2
1. What is the mean change score?
2. What is the standard deviation for this population?
3. What is the median change score?
4. Find the change score that is 2.2 standard deviations below the mean.
Exercise 2.8.11
Refer to Figure determine which of the following are true and which are false. Explain your solution to each part in complete sentences.
<figure >
</figure>
1. The medians for all three graphs are the same.
2. We cannot determine if any of the means for the three graphs is different.
3. The standard deviation for graph b is larger than the standard deviation for graph a.
4. We cannot determine if any of the third quartiles for the three graphs is different.
Answer
1. True
2. True
3. True
4. False
Exercise 2.8.12
In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference.
1. Organize the data in a chart.
2. Find the median, the first quartile, and the third quartile.
3. Find the 65th percentile.
4. Find the 10th percentile.
5. Construct a box plot of the data.
6. The middle 50% of the conferences last from _______ days to _______ days.
7. Calculate the sample mean of days of engineering conferences.
8. Calculate the sample standard deviation of days of engineering conferences.
9. Find the mode.
10. If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice.
11. Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences.
Exercise 2.8.13
A survey of enrollment at 35 community colleges across the United States yielded the following figures:
6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622
1. Organize the data into a chart with five intervals of equal width. Label the two columns "Enrollment" and "Frequency."
2. Construct a histogram of the data.
3. If you were to build a new community college, which piece of information would be more valuable: the mode or the mean?
4. Calculate the sample mean.
5. Calculate the sample standard deviation.
6. A school with an enrollment of 8000 would be how many standard deviations away from the mean?
Answer
1. Enrollment Frequency
1000-5000 10
5000-10000 16
10000-15000 3
15000-20000 3
20000-25000 1
25000-30000 2
2. Check student’s solution.
3. mode
4. 8628.74
5. 6943.88
6. –0.09
Use the following information to answer the next two exercises. $X =$ the number of days per week that 100 clients use a particular exercise facility.
$x$ Frequency
0 3
1 12
2 33
3 28
4 11
5 9
6 4
Exercise 2.8.14
The 80th percentile is _____
1. 5
2. 80
3. 3
4. 4
Exercise 2.8.15
The number that is 1.5 standard deviations BELOW the mean is approximately _____
1. 0.7
2. 4.8
3. –2.8
4. Cannot be determined
Answer
a
Exercise 2.8.16
Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the Table.
# of books Freq. Rel. Freq.
0 18
1 24
2 24
3 22
4 15
5 10
7 5
9 1
1. Are there any outliers in the data? Use an appropriate numerical test involving the IQRto identify outliers, if any, and clearly state your conclusion.
2. If a data value is identified as an outlier, what should be done about it?
3. Are any data values further than two standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.)
4. Do parts a and c of this problem give the same answer?
5. Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data?
6. Based on the shape of the data which is the most appropriate measure of center for this data: mean, median or mode?
Glossary
Standard Deviation
a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation.
Contributors and Attributions
Variance
mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as $x$ – $\bar{x}$ where $x$ is a value of the data and $\bar{x}$ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.08%3A_Measures_of_the_Spread_of_the_Data.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will construct a histogram and a box plot.
• The student will calculate univariate statistics.
• The student will examine the graphs to interpret what the data implies.
Collect the Data
Record the number of pairs of shoes you own.
1. Randomly survey 30 classmates about the number of pairs of shoes they own. Record their values.
Survey Results
_____ _____ _____ _____ _____
_____ _____ _____ _____ _____
_____ _____ _____ _____ _____
_____ _____ _____ _____ _____
_____ _____ _____ _____ _____
_____ _____ _____ _____ _____
2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil and scale the axes.
3. Calculate the following values.
1. $\bar{x}$ = _____
2. $s$ = _____
4. Are the data discrete or continuous? How do you know?
5. In complete sentences, describe the shape of the histogram.
6. Are there any potential outliers? List the value(s) that could be outliers. Use a formula to check the end values to determine if they are potential outliers.
Analyze the Data
1. Determine the following values.
1. Min = _____
2. M = _____
3. Max = _____
4. Q1 = _____
5. Q3 = _____
6. IQR = _____
2. Construct a box plot of data
3. What does the shape of the box plot imply about the concentration of data? Use complete sentences.
4. Using the box plot, how can you determine if there are potential outliers?
5. How does the standard deviation help you to determine concentration of the data and whether or not there are potential outliers?
6. What does the IQR represent in this problem?
7. Show your work to find the value that is 1.5 standard deviations:
1. above the mean.
2. below the mean.
2.E: Descriptive Statistics (Exercises)
2.2: Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
Q 2.2.1
Student grades on a chemistry exam were: 77, 78, 76, 81, 86, 51, 79, 82, 84, 99
1. Construct a stem-and-leaf plot of the data.
2. Are there any potential outliers? If so, which scores are they? Why do you consider them outliers?
Q 2.2.2
The table below contains the 2010 obesity rates in U.S. states and Washington, DC.
State Percent (%) State Percent (%) State Percent (%)
Alabama 32.2 Kentucky 31.3 North Dakota 27.2
Alaska 24.5 Louisiana 31.0 Ohio 29.2
Arizona 24.3 Maine 26.8 Oklahoma 30.4
Arkansas 30.1 Maryland 27.1 Oregon 26.8
California 24.0 Massachusetts 23.0 Pennsylvania 28.6
Colorado 21.0 Michigan 30.9 Rhode Island 25.5
Connecticut 22.5 Minnesota 24.8 South Carolina 31.5
Delaware 28.0 Mississippi 34.0 South Dakota 27.3
Washington, DC 22.2 Missouri 30.5 Tennessee 30.8
Florida 26.6 Montana 23.0 Texas 31.0
Georgia 29.6 Nebraska 26.9 Utah 22.5
Hawaii 22.7 Nevada 22.4 Vermont 23.2
Idaho 26.5 New Hampshire 25.0 Virginia 26.0
Illinois 28.2 New Jersey 23.8 Washington 25.5
Indiana 29.6 New Mexico 25.1 West Virginia 32.5
Iowa 28.4 New York 23.9 Wisconsin 26.3
Kansas 29.4 North Carolina 27.8 Wyoming 25.1
1. Use a random number generator to randomly pick eight states. Construct a bar graph of the obesity rates of those eight states.
2. Construct a bar graph for all the states beginning with the letter "A."
3. Construct a bar graph for all the states beginning with the letter "M."
S 2.2.2
1. Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows.
• Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically)
• Press MATH
• Arrow over to PRB
• Press 5:randInt(
• Enter 51,1,8)
Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}.
Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.
For each of the following data sets, create a stem plot and identify any outliers.
Exercise 2.2.7
The miles per gallon rating for 30 cars are shown below (lowest to highest).
19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43
Answer
Stem Leaf
1 9 9 9
2 0 1 1 5 5 5 6 6 8 9
3 1 1 2 2 3 4 5 6 7 7 8 8 8 8
4 1 3 3
The height in feet of 25 trees is shown below (lowest to highest).
25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54
The data are the prices of different laptops at an electronics store. Round each value to the nearest ten.
249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610
Answer
Stem Leaf
2 5 5 6 7 7 8
3 0 0 1 2 3 3 5 5 5 7 7 9
4 1 6 9
5 6 7 7
6 1
The data are daily high temperatures in a town for one month.
61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95
For the next three exercises, use the data to construct a line graph.
Exercise 2.2.8
In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in the Table below.
Number of times in store Frequency
1 4
2 10
3 16
4 6
5 4
Answer
Exercise 2.2.9
In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table.
Years since last purchase Frequency
0 2
1 8
2 13
3 22
4 16
5 9
Exercise 2.2.10
Several children were asked how many TV shows they watch each day. The results of the survey are shown in the Table below.
Number of TV Shows Frequency
0 12
1 18
2 36
3 7
4 2
Answer
Exercise 2.2.11
The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. Table shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.
Seasons Number of students Proportion of population
Spring 8 24%
Summer 9 26%
Autumn 11 32%
Winter 6 18%
Using the data from Mrs. Ramirez’s math class supplied in the table above, construct a bar graph showing the percentages.
Answer
Exercise 2.2.12
David County has six high schools. Each school sent students to participate in a county-wide science competition. Table shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.
High School Science competition population Overall student population
Alabaster 28.9% 8.6%
Concordia 7.6% 23.2%
Genoa 12.1% 15.0%
Mocksville 18.5% 14.3%
Tynneson 24.2% 10.1%
West End 8.7% 28.8%
Use the data from the David County science competition supplied in Exercise. Construct a bar graph that shows the county-wide population percentage of students at each school.
Answer
2.3: Histograms, Frequency, Polygons, and Time Series Graphs
Q 2.3.1
Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows:
Publisher A
# of books Freq. Rel. Freq.
0 10
1 12
2 16
3 12
4 8
5 6
6 2
8 2
Publisher B
# of books Freq. Rel. Freq.
0 18
1 24
2 24
3 22
4 15
5 10
7 5
9 1
Publisher C
# of books Freq. Rel. Freq.
0–1 20
2–3 35
4–5 12
6–7 2
8–9 1
1. Find the relative frequencies for each survey. Write them in the charts.
2. Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two.
3. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical.
4. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not?
5. Make new histograms for Publisher A and Publisher B. This time, make bar widths of two.
6. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer.
Q 2.3.2
Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group.
Singles
Amount($) Frequency Rel. Frequency 51–100 5 101–150 10 151–200 15 201–250 15 251–300 10 301–350 5 Couples Amount($) Frequency Rel. Frequency
100–150 5
201–250 5
251–300 5
301–350 5
351–400 10
401–450 10
451–500 10
501–550 10
551–600 5
601–650 5
1. Fill in the relative frequency for each group.
2. Construct a histogram for the singles group. Scale the x-axis by $50 widths. Use relative frequency on the y-axis. 3. Construct a histogram for the couples group. Scale the x-axis by$50 widths. Use relative frequency on the y-axis.
4. Compare the two graphs:
1. List two similarities between the graphs.
2. List two differences between the graphs.
3. Overall, are the graphs more similar or different?
5. Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x-axis by $50, scale it by$100. Use relative frequency on the y-axis.
6. Compare the graph for the singles with the new graph for the couples:
1. List two similarities between the graphs.
2. Overall, are the graphs more similar or different?
7. How did scaling the couples graph differently change the way you compared it to the singles graph?
8. Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences.
S 2.3.2
Singles
Amount($) Frequency Relative Frequency 51–100 5 0.08 101–150 10 0.17 151–200 15 0.25 201–250 15 0.25 251–300 10 0.17 301–350 5 0.08 Couples Amount($) Frequency Relative Frequency
100–150 5 0.07
201–250 5 0.07
251–300 5 0.07
301–350 5 0.07
351–400 10 0.14
401–450 10 0.14
451–500 10 0.14
501–550 10 0.14
551–600 5 0.07
601–650 5 0.07
1. See the tables above
2. In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included).
3. In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included).
4. Compare the two graphs:
1. Answers may vary. Possible answers include:
• Both graphs have a single peak.
The percentage of people who own at most three t-shirts costing more than $19 each is approximately: 1. 21 2. 59 3. 41 4. Cannot be determined S 2.3.4 c Q 2.3.5 If the data were collected by asking the first 111 people who entered the store, then the type of sampling is: 1. cluster 2. simple random 3. stratified 4. convenience Q 2.3.6 Following are the 2010 obesity rates by U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the $x$-axis with the states. S 2.3.7 Answers will vary. Exercise 2.3.6 Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table. Data Value (# cars) Frequency Relative Frequency Cumulative Relative Frequency Exercise 2.3.7 What does the frequency column in the Table above sum to? Why? Answer 65 Exercise 2.3.8 What does the relative frequency column in in the Table above sum to? Why? Exercise 2.3.9 What is the difference between relative frequency and frequency for each data value in in the Table above ? Answer The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value. Exercise 2.3.10 What is the difference between cumulative relative frequency and relative frequency for each data value? Exercise 2.3.11 To construct the histogram for the data in in the Table above , determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling. Answer Answers will vary. One possible histogram is shown: Exercise 2.3.12 Construct a frequency polygon for the following: 1. Pulse Rates for Women Frequency 60–69 12 70–79 14 80–89 11 90–99 1 100–109 1 110–119 0 120–129 1 2. Actual Speed in a 30 MPH Zone Frequency 42–45 25 46–49 14 50–53 7 54–57 3 58–61 1 3. Tar (mg) in Nonfiltered Cigarettes Frequency 10–13 1 14–17 0 18–21 15 22–25 7 26–29 2 Exercise 2.3.13 Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger. Depth of Hunger Frequency 230–259 21 260–289 13 290–319 5 320–349 7 350–379 1 380–409 1 410–439 1 Answer Find the midpoint for each class. These will be graphed on the x-axis. The frequency values will be graphed on the y-axis values. Exercise 2.3.14 Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men? Life Expectancy at Birth – Women Frequency 49–55 3 56–62 3 63–69 1 70–76 3 77–83 8 84–90 2 Life Expectancy at Birth – Men Frequency 49–55 3 56–62 3 63–69 1 70–76 1 77–83 7 84–90 5 Exercise 2.3.15 Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births. Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 Male 47,804 52,239 53,158 53,694 54,628 54,409 54,606 Total 93,349 101,821 103,415 104,018 106,543 105,629 107,009 Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 Male 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 Total 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354 Sex/Year 1871 1870 1872 1871 1872 1827 1874 1875 Female 56,099 56,431 57,472 56,099 57,472 58,233 60,109 60,146 Male 60,029 58,959 61,293 60,029 61,293 61,467 63,602 63,432 Total 116,128 115,390 118,765 116,128 118,765 119,700 123,711 123,578 Answer Exercise 2.3.16 The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973. Year 1961 1962 1963 1964 1965 1966 1967 Police 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Homicides 8.6 8.9 8.52 8.89 13.07 14.57 21.36 Year 1968 1969 1970 1971 1972 1973 Police 295.99 319.87 341.43 356.59 376.69 390.19 Homicides 28.03 31.49 37.39 46.26 47.24 52.33 1. Construct a double time series graph using a common x-axis for both sets of data. 2. Which variable increased the fastest? Explain. 3. Did Detroit’s increase in police officers have an impact on the murder rate? Explain. 2.4: Measures of the Location of the Data Q 2.4.1 The median age for U.S. blacks currently is 30.9 years; for U.S. whites it is 42.3 years. 1. Based upon this information, give two reasons why the black median age could be lower than the white median age. 2. Does the lower median age for blacks necessarily mean that blacks die younger than whites? Why or why not? 3. How might it be possible for blacks and whites to die at approximately the same age, but for the median age for whites to be higher? Q 2.4.2 Six hundred adult Americans were asked by telephone poll, "What do you think constitutes a middle-class income?" The results are in the Table below. Also, include left endpoint, but not the right endpoint. Salary ($) Relative Frequency
< 20,000 0.02
20,000–25,000 0.09
25,000–30,000 0.19
30,000–40,000 0.26
40,000–50,000 0.18
50,000–75,000 0.17
75,000–99,999 0.02
100,000+ 0.01
1. What percentage of the survey answered "not sure"?
2. What percentage think that middle-class is from $25,000 to$50,000?
3. Construct a histogram of the data.
1. Should all bars have the same width, based on the data? Why or why not?
2. How should the <20,000 and the 100,000+ intervals be handled? Why?
4. Find the 40th and 80th percentiles
5. Construct a bar graph of the data
S 2.4.2
1. $1 - (0.02 + 0.09 + 0.19 + 0.26 + 0.18 + 0.17 + 0.02 + 0.01) = 0.06$
2. $0.19 + 0.26 + 0.18 = 0.63$
3. Check student’s solution.
4. 40th percentile will fall between 30,000 and 40,000
80th percentile will fall between 50,000 and 75,000
5. Check student’s solution.
Q 2.4.3
Given the following box plot:
1. which quarter has the smallest spread of data? What is that spread?
2. which quarter has the largest spread of data? What is that spread?
3. find the interquartile range (IQR).
4. are there more data in the interval 5–10 or in the interval 10–13? How do you know this?
5. which interval has the fewest data in it? How do you know this?
1. 0–2
2. 2–4
3. 10–12
4. 12–13
5. need more information
Q 2.4.4
The following box plot shows the U.S. population for 1990, the latest available year.
1. Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)? How do you know?
2. 12.6% are age 65 and over. Approximately what percentage of the population are working age adults (above age 17 to age 65)?
S 2.4.4
1. more children; the left whisker shows that 25% of the population are children 17 and younger. The right whisker shows that 25% of the population are adults 50 and older, so adults 65 and over represent less than 25%.
2. 62.4%
2.5: Box Plots
Q 2.5.1
In a survey of 20-year-olds in China, Germany, and the United States, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results.
1. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected.
2. Have more Americans or more Germans surveyed been to over eight foreign countries?
3. Compare the three box plots. What do they imply about the foreign travel of 20-year-old residents of the three countries when compared to each other?
Q 2.5.2
Given the following box plot, answer the questions.
1. Think of an example (in words) where the data might fit into the above box plot. In 2–5 sentences, write down the example.
2. What does it mean to have the first and second quartiles so close together, while the second to third quartiles are far apart?
S 2.5.2
1. Answers will vary. Possible answer: State University conducted a survey to see how involved its students are in community service. The box plot shows the number of community service hours logged by participants over the past year.
2. Because the first and second quartiles are close, the data in this quarter is very similar. There is not much variation in the values. The data in the third quarter is much more variable, or spread out. This is clear because the second quartile is so far away from the third quartile.
Q 2.5.3
Given the following box plots, answer the questions.
1. In complete sentences, explain why each statement is false.
1. Data 1 has more data values above two than Data 2 has above two.
2. The data sets cannot have the same mode.
3. For Data 1, there are more data values below four than there are above four.
2. For which group, Data 1 or Data 2, is the value of “7” more likely to be an outlier? Explain why in complete sentences.
Q 2.5.4
A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results.
1. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected for that car series.
2. Which group is most likely to have an outlier? Explain how you determined that.
3. Compare the three box plots. What do they imply about the age of purchasing a BMW from the series when compared to each other?
4. Look at the BMW 5 series. Which quarter has the smallest spread of data? What is the spread?
5. Look at the BMW 5 series. Which quarter has the largest spread of data? What is the spread?
6. Look at the BMW 5 series. Estimate the interquartile range (IQR).
7. Look at the BMW 5 series. Are there more data in the interval 31 to 38 or in the interval 45 to 55? How do you know this?
8. Look at the BMW 5 series. Which interval has the fewest data in it? How do you know this?
1. 31–35
2. 38–41
3. 41–64
S 2.5.4
1. Each box plot is spread out more in the greater values. Each plot is skewed to the right, so the ages of the top 50% of buyers are more variable than the ages of the lower 50%.
2. The BMW 3 series is most likely to have an outlier. It has the longest whisker.
3. Comparing the median ages, younger people tend to buy the BMW 3 series, while older people tend to buy the BMW 7 series. However, this is not a rule, because there is so much variability in each data set.
4. The second quarter has the smallest spread. There seems to be only a three-year difference between the first quartile and the median.
5. The third quarter has the largest spread. There seems to be approximately a 14-year difference between the median and the third quartile.
6. IQR ~ 17 years
7. There is not enough information to tell. Each interval lies within a quarter, so we cannot tell exactly where the data in that quarter is concentrated.
8. The interval from 31 to 35 years has the fewest data values. Twenty-five percent of the values fall in the interval 38 to 41, and 25% fall between 41 and 64. Since 25% of values fall between 31 and 38, we know that fewer than 25% fall between 31 and 35.
Q 2.5.5
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
# of movies Frequency
0 5
1 9
2 6
3 4
4 1
Construct a box plot of the data.
2.6: Measures of the Center of the Data
Q 2.6.1
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table.
Percent of Population Obese Number of Countries
11.4–20.45 29
20.45–29.45 13
29.45–38.45 4
38.45–47.45 0
47.45–56.45 2
56.45–65.45 1
65.45–74.45 0
74.45–83.45 1
1. What is the best estimate of the average obesity percentage for these countries?
2. The United States has an average obesity rate of 33.9%. Is this rate above average or below?
3. How does the United States compare to other countries?
Q 2.6.2
The table below gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children?
Percent of Underweight Children Number of Countries
16–21.45 23
21.45–26.9 4
26.9–32.35 9
32.35–37.8 7
37.8–43.25 6
43.25–48.7 1
S 2.6.2
The mean percentage, $\bar{x} = \frac{1328.65}{50} = 26.75$
2.7: Skewness and the Mean, Median, and Mode
Q 2.7.1
The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years.
1. What does it mean for the median age to rise?
2. Give two reasons why the median age could rise.
3. For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not?
2.8: Measures of the Spread of the Data
Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005.
• $\mu = 1000$ FTES
• median = 1,014 FTES
• $\sigma = 474$ FTES
• first quartile = 528.5 FTES
• third quartile = 1,447.5 FTES
• $n = 29$ years
Q 2.8.1
A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer.
S 2.8.1
The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or below the median.
Q 2.8.2
75% of all years have an FTES:
1. at or below: _____
2. at or above: _____
Q 2.8.3
The population standard deviation = _____
474 FTES
Q 2.8.4
What percent of the FTES were from 528.5 to 1447.5? How do you know?
Q 2.8.5
What is the IQR? What does the IQR represent?
919
Q 2.8.6
How many standard deviations away from the mean is the median?
Additional Information: The population FTES for 2005–2006 through 2010–2011 was given in an updated report. The data are reported here.
Year 2005–06 2006–07 2007–08 2008–09 2009–10 2010–11
Total FTES 1,585 1,690 1,735 1,935 2,021 1,890
Q 2.8.7
Calculate the mean, median, standard deviation, the first quartile, the third quartile and theIQR. Round to one decimal place.
S 2.8.7
• mean = 1,809.3
• median = 1,812.5
• standard deviation = 151.2
• first quartile = 1,690
• third quartile = 1,935
• IQR = 245
Q 2.8.8
Construct a box plot for the FTES for 2005–2006 through 2010–2011 and a box plot for the FTES for 1976–1977 through 2004–2005.
Q 2.8.9
Compare the IQR for the FTES for 1976–77 through 2004–2005 with the IQR for the FTES for 2005-2006 through 2010–2011. Why do you suppose the IQRs are so different?
S 2.8.10
Hint: Think about the number of years covered by each time period and what happened to higher education during those periods.
Q 2.8.11
Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer.
Student GPA School Average GPA School Standard Deviation
Thuy 2.7 3.2 0.8
Vichet 87 75 20
Kamala 8.6 8 0.4
Q 2.8.12
A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing$550, and a drum set costing $600. The mean cost for a piano is$4,000 with a standard deviation of $2,500. The mean cost for a guitar is$500 with a standard deviation of $200. The mean cost for drums is$700 with a standard deviation of \$100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer.
S 2.8.12
For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.
Q 2.8.13
An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.
1. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?
2. Who is the fastest runner with respect to his or her class? Explain why.
Q 2.8.14
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the table belo2
Percent of Population Obese Number of Countries
11.4–20.45 29
20.45–29.45 13
29.45–38.45 4
38.45–47.45 0
47.45–56.45 2
56.45–65.45 1
65.45–74.45 0
74.45–83.45 1
What is the best estimate of the average obesity percentage for these countries? What is the standard deviation for the listed obesity rates? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ obesity rate compared to the average rate? Explain.
S 2.8.14
• $\bar{x} = 23.32$
• Using the TI 83/84, we obtain a standard deviation of: $s_{x} = 12.95$.
• The obesity rate of the United States is 10.58% higher than the average obesity rate.
• Since the standard deviation is 12.95, we see that $23.32 + 12.95 = 36.27$ is the obesity percentage that is one standard deviation from the mean. The United States obesity rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, while 34% obese, does not have an unusually high percentage of obese people.
Q 2.8.15
The Table below gives the percent of children under five considered to be underweight.
Percent of Underweight Children Number of Countries
16–21.45 23
21.45–26.9 4
26.9–32.35 9
32.35–37.8 7
37.8–43.25 6
43.25–48.7 1
What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/02%3A_Descriptive_Statistics/2.09%3A_Descriptive_Statistics_%28Worksheet%29.txt |
Probability theory is concerned with probability, the analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either be single occurrences or evolve over time in an apparently random fashion.
• 3.1: Introduction
You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.
• 3.2: Terminology
In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.
• 3.3: Independent and Mutually Exclusive Events
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If they are not independent, then they are dependent. In sampling with replacement, with selecting each member with the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutu
• 3.4: Two Basic Rules of Probability
The multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A, B. In sampling with replacement each member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member may be chosen only once, and the events are not independent. The events A and B are mutually exclusive events when they have no common outcomes.
• 3.5: Contingency Tables
There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables.
• 3.6: Tree and Venn Diagrams
A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabi
• 3.7: Probability Topics (Worksheet)
The student will use theoretical and empirical methods to estimate probabilities. The student will appraise the differences between the two estimates. The student will demonstrate an understanding of long-term relative frequencies.
• 3.E: Probability Topics (Exericses)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
03: Probability Topics
CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Understand and use the terminology of probability.
• Determine whether two events are mutually exclusive and whether two events are independent.
• Calculate probabilities using the Addition Rules and Multiplication Rules.
• Construct and interpret Contingency Tables.
• Construct and interpret Venn Diagrams.
• Construct and interpret Tree Diagrams.
It is often necessary to "guess" about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.
You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.
Collaborative Exercise
Your instructor will survey your class. Count the number of students in the class today.
• Raise your hand if you have any change in your pocket or purse. Record the number of raised hands.
• Raise your hand if you rode a bus within the past month. Record the number of raised hands.
• Raise your hand if you answered "yes" to BOTH of the first two questions. Record the number of raised hands.
Use the class data as estimates of the following probabilities. $P(\text{change})$ means the probability that a randomly chosen person in your class has change in his/her pocket or purse. $P(\text{bus})$ means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.
• Find $P(\text{change})$.
• Find $P(\text{bus})$.
• Find $P(\text{change AND bus})$. Find the probability that a randomly chosen student in your class has change in his/her pocket or purse and rode a bus within the last month.
• Find $P(\text{change|bus})$. Find the probability that a randomly chosen student has change given that he or she rode a bus within the last month. Count all the students that rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.01%3A_Introduction.txt |
Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment.
A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, $S = \{\text{H, T}\}$ where $\text{H} =$ heads and $\text{T} =$ tails are the outcomes.
An event is any combination of outcomes. Upper case letters like $\text{A}$ and $\text{B}$ represent events. For example, if the experiment is to flip one fair coin, event $\text{A}$ might be getting at most one head. The probability of an event $\text{A}$ is written $P(\text{A})$.
Definition: Probability
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values).
• $P(\text{A}) = 0$ means the event $\text{A}$ can never happen.
• $P(\text{A}) = 1$ means the event $\text{A}$ always happens.
• $P(\text{A}) = 0.5$ means the event $\text{A}$ is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).
Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head ($\text{H}$) and a Tail ($\text{T}$) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.
To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event $\text{A}$ and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is $\{\text{HH, TH, HT,TT}\}$ where $\text{T} =$ tails and $\text{H} =$ heads. The sample space has four outcomes. $\text{A} =$ getting one head. There are two outcomes that meet this condition $\text{\{HT, TH\}}$, so $P(\text{A}) = \frac{2}{4} = 0.5$.
Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event $\text{E} =$ rolling a number that is at least five. There are two outcomes {5, 6}. $P(\text{E}) = \frac{2}{6}$. If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, $\frac{2}{6}$ of the rolls would result in an outcome of "at least five". You would not expect exactly $\frac{2}{6}$. The long-term relative frequency of obtaining this result would approach the theoretical probability of $\frac{2}{6}$ as the number of repetitions grows larger and larger.
Definition: Law of Large Numbers
This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)
It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.
The "OR" Event
An outcome is in the event $\text{A OR B}$ if the outcome is in $\text{A}$ or is in $\text{B}$ or is in both $\text{A}$ and $\text{B}$. For example, let $\text{A} = \{1, 2, 3, 4, 5\}$ and $\text{B} = \{4, 5, 6, 7, 8\}$. $\text{A OR B} = \{1, 2, 3, 4, 5, 6, 7, 8\}$. Notice that 4 and 5 are NOT listed twice.
The "AND" Event
An outcome is in the event $\text{A AND B}$ if the outcome is in both $\text{A}$ and $\text{B}$ at the same time. For example, let $\text{A}$ and $\text{B}$ be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then $\text{A AND B} = {4, 5}$.
The complement of event $\text{A}$ is denoted $\text{A'}$ (read "A prime"). $\text{A'}$ consists of all outcomes that are NOT in $\text{A}$. Notice that
$P(\text{A}) + P(\text{A′}) = 1. \nonumber$
For example, let $\text{S} = \{1, 2, 3, 4, 5, 6\}$ and let $\text{A} = {1, 2, 3, 4}$. Then, $\text{A′} = {5, 6}$ and $P(A) = \frac{4}{6}$, $P(\text{A′}) = \frac{2}{6}$, and
$P(\text{A}) + P(\text{A′}) = \frac{4}{6} + \frac{2}{6} = 1. \nonumber$
The conditional probability of $\text{A}$ given $\text{B}$ is written $P(\text{A|B})$. $P(\text{A|B})$ is the probability that event $\text{A}$ will occur given that the event $\text{B}$ has already occurred. A conditional reduces the sample space. We calculate the probability of $\text{A}$ from the reduced sample space $\text{B}$. The formula to calculate $P(\text{A|B})$ is
$P(\text{A|B}) = \frac{\text{P(A AND B)}}{\text{P(B)}} \nonumber$
where $P(\text{B})$ is greater than zero.
For example, suppose we toss one fair, six-sided die. The sample space $\text{S} = \{1, 2, 3, 4, 5, 6\}$. Let $\text{A} =$ face is 2 or 3 and $\text{B} =$ face is even (2, 4, 6). To calculate $P(\text{A|B})$, we count the number of outcomes 2 or 3 in the sample space $\text{B} = \{2, 4, 6\}$. Then we divide that by the number of outcomes $\text{B}$ (rather than $\text{S}$).
We get the same result by using the formula. Remember that $\text{S}$ has six outcomes.
\begin{align*} P(\text{A|B}) &= \dfrac{ \text{ P(A AND B) } } {P(\text{B})} \[4pt] &= \dfrac{\dfrac{\text{the number of outcomes that are 2 or 3 and even in S}}{6}}{\dfrac{\text{the number of outcomes that are even in S}}{6}} \[4pt] &= \dfrac{\frac{1}{6}}{\frac{3}{6}} = \dfrac{1}{3} \end{align*}
Understanding Terminology and Symbols
It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.
Example $1$
The sample space $S$ is the whole numbers starting at one and less than 20.
1. $S =$ _____________________________
Let event $A =$ the even numbers and event $B =$ numbers greater than 13.
2. $A =$ _____________________, $B =$ _____________________
3. $P(\text{A}) =$ _____________, $P(\text{B}) =$ ________________
4. $\text{A AND B} =$ ____________________, $\text{A OR B} =$ ________________
5. $P(\text{A AND B}) =$ _________, $P(\text{A OR B}) =$ _____________
6. $\text{A′} =$ _____________, $P(\text{A′}) =$ _____________
7. $P(\text{A}) + P(\text{A′}) =$ ____________
8. $P(\text{A|B}) =$ ___________, $P(\text{B|A}) =$ _____________; are the probabilities equal?
Answer
1. $\text{S} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19\}$
2. $\text{A} = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}, \text{B} = \{14, 15, 16, 17, 18, 19\}$
3. $P(\text{A}) = \frac{9}{19}$, $P(\text{B}) = \frac{6}{19}$
4. $\text{A AND B} = \{14,16,18\}$, $\text{A OR B} = \{2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19\}$
5. $P(\text{A AND B}) = \frac{3}{19}$, $P(\text{A OR B}) = \frac{12}{19}$
6. $\text{A′} = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19$; $P(\text{A′}) = \frac{10}{19}$
7. $P(\text{A}) + P(\text{A′}) = 1\left((\frac{9}{19} + \frac{10}{19} = 1\right)$
8. $P(\text{A|B}) = \frac{\text{P(A AND B)}}{\text{P(B)}} = \frac{3}{6}, P(\text{B|A}) = \frac{\text{P(A AND B)}}{\text{P(A)}} = \frac{3}{9}$, No
Exercise $1$
The sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).
1. $S =$ _____________________________
Let event $A =$ the sum is even and event $B =$ the first number is prime.
2. $A =$ _____________________, $B =$ _____________________
3. $P(\text{A}) =$ _____________, $P(\text{B}) =$ ________________
4. $\text{A AND B} =$ ____________________, $\text{A OR B} =$ ________________
5. $P(\text{A AND B}) =$ _________, $P(\text{A OR B}) =$ _____________
6. $\text{B′} =$ _____________, $P(\text{B′)} =$ _____________
7. $P(\text{A}) + P(\text{A′}) =$ ____________
8. $P(\text{A|B}) =$ ___________, $P(\text{B|A}) =$ _____________; are the probabilities equal?
Answer
1. $\text{S} = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$
2. $\text{A} = \{(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)\}$
$\text{B} = \{(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$
3. $P(\text{A}) = \frac{1}{2}$, $P(\text{B}) = \frac{2}{3}$
4. $\text{A AND B} = \{(2,2), (2,4), (3,1), (3,3)\}$
$\text{A OR B} = \{(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$
5. $P(\text{A AND B}) = \frac{1}{3}, P(\text{A OR B}) = \frac{5}{6}$
6. $\text{B′} = \{(1,1), (1,2), (1,3), (1,4)\}, P(\text{B′}) = \frac{1}{3}$
7. $P(\text{B}) + P(\text{B′}) = 1$
8. $P(\text{A|B}) = \frac{P(\text{A AND B})}{P(\text{B})} = \frac{1}{2}, P(\text{B|A}) = \frac{P(\text{A AND B})}{P(\text{B})} = \frac{2}{3}$, No.
Example $\PageIndex{2A}$
A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).
1. Event $\text{T} =$ the outcome is two.
2. Event $\text{A} =$ the outcome is an even number.
3. Event $\text{B} =$ the outcome is less than four.
4. The complement of $\text{A}$.
5. $\text{A GIVEN B}$
6. $\text{B GIVEN A}$
7. $\text{A AND B}$
8. $\text{A OR B}$
9. $\text{A OR B′}$
10. Event $\text{N} =$ the outcome is a prime number.
11. Event $\text{I} =$ the outcome is seven.
Solution
1. $\text{T} = \{2\}$, $P(\text{T}) = \frac{1}{6}$
2. $A = \{2, 4, 6\}$, $P(\text{A}) = \frac{1}{2}$
3. $\text{B} = \{1, 2, 3\}$, $P(\text{B}) = \frac{1}{2}$
4. $\text{A′} = \{1, 3, 5\}, P(\text{A′}) = \frac{1}{2}$
5. $\text{A|B} = \{2\}$, $P(\text{A|B}) = \frac{1}{3}$
6. $\text{B|A} = \{2\}$, $P(\text{B|A}) = \frac{1}{3}$
7. $\text{A AND B} = {2}, P(\text{A AND B}) = \frac{1}{6}$
8. $\text{A OR B} = \{1, 2, 3, 4, 6\}$, $P(\text{A OR B}) = \frac{5}{6}$
9. $\text{A OR B′} = \{2, 4, 5, 6\}$, $P(\text{A OR B′}) = \frac{2}{3}$
10. $\text{N} = \{2, 3, 5\}$, $P(\text{N}) = \frac{1}{2}$
11. A six-sided die does not have seven dots. $P(7) = 0$.
Example $\PageIndex{2B}$
Table describes the distribution of a random sample $S$ of 100 individuals, organized by gender and whether they are right- or left-handed.
Right-handed Left-handed
Males 43 9
Females 44 4
Let’s denote the events $M =$ the subject is male, $F =$ the subject is female, $R =$ the subject is right-handed, $L =$ the subject is left-handed. Compute the following probabilities:
1. $P(\text{M})$
2. $P(\text{F})$
3. $P(\text{R})$
4. $P(\text{L})$
5. $P(\text{M AND R})$
6. $P(\text{F AND L})$
7. $P(\text{M OR F})$
8. $P(\text{M OR R})$
9. $P(\text{F OR L})$
10. $P(\text{M'})$
11. $P(\text{R|M})$
12. $P(\text{F|L})$
13. $P(\text{L|F})$
Answer
1. $P(\text{M}) = 0.52$
2. $P(\text{F}) = 0.48$
3. $P(\text{R}) = 0.87$
4. $P(\text{L}) = 0.13$
5. $P(\text{M AND R}) = 0.43$
6. $P(\text{F AND L}) = 0.04$
7. $P(\text{M OR F}) = 1$
8. $P(\text{M OR R}) = 0.96$
9. $P(\text{F OR L}) = 0.57$
10. $P(\text{M'}) = 0.48$
11. $P(\text{R|M}) = 0.8269$ (rounded to four decimal places)
12. $P(\text{F|L}) = 0.3077$ (rounded to four decimal places)
13. $P(\text{L|F}) = 0.0833$
Review
In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.
Formula Review
$\text{A}$ and $\text{B}$ are events
$P(\text{S}) = 1$ where $\text{S}$ is the sample space
$0 \leq P(\text{A}) \leq 1$
$P(\text{A|B}) = \frac{\text{P(A AND B)}}{\text{P(B)}}$
Glossary
Conditional Probability
the likelihood that an event will occur given that another event has already occurred
Equally Likely
Each outcome of an experiment has the same probability.
Event
a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by $S$. An event is an arbitrary subset in $S$. It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as $A, B, C$, and so on.
Experiment
a planned activity carried out under controlled conditions
Outcome
a particular result of an experiment
Probability
a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let $S$ denote the sample space and $A$ and $B$ are two events in S. Then:
• $0 \leq P(\text{A}) \leq 1$
• If $\text{A}$ and $\text{B}$ are any two mutually exclusive events, then $\text{P}(\text{A OR B}) = P(\text{A}) + P(\text{B})$.
• $P(\text{S}) = 1$
Sample Space
the set of all possible outcomes of an experiment
The AND Event
An outcome is in the event $\text{A AND B}$ if the outcome is in both $\text{A AND B}$ at the same time.
The Complement Event
The complement of event $\text{A}$ consists of all outcomes that are NOT in $\text{A}$.
The Conditional Probability of A GIVEN B
$P(\text{A|B})$ is the probability that event $\text{A}$ will occur given that the event $\text{B}$ has already occurred.
The Or Event
An outcome is in the event $\text{A OR B}$ if the outcome is in $\text{A}$ or is in $\text{B}$ or is in both $\text{A}$ and $\text{B}$.
Exercise 3.2.2
In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)
• Let $\text{F}$ be the event that a student is female.
• Let $\text{M}$ be the event that a student is male.
• Let $\text{S}$ be the event that a student has short hair.
• Let $\text{L}$ be the event that a student has long hair.
1. The probability that a student does not have long hair.
2. The probability that a student is male or has short hair.
3. The probability that a student is a female and has long hair.
4. The probability that a student is male, given that the student has long hair.
5. The probability that a student has long hair, given that the student is male.
6. Of all the female students, the probability that a student has short hair.
7. Of all students with long hair, the probability that a student is female.
8. The probability that a student is female or has long hair.
9. The probability that a randomly selected student is a male student with short hair.
10. The probability that a student is female.
Answer
1. $P(\text{L′)} = P(\text{S})$
2. $P(\text{M OR S})$
3. $P(\text{F AND L})$
4. $P(\text{M|L})$
5. $P(\text{L|M})$
6. $P(\text{S|F})$
7. $P(\text{F|L})$
8. $P(\text{F OR L})$
9. $P(\text{M AND S})$
10. $P(\text{F})$
Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti.
Let $H =$ the event of getting a hat.
Let $N =$ the event of getting a noisemaker.
Let $F =$ the event of getting a finger trap.
Let $C =$ the event of getting a bag of confetti.
Exercise 3.2.3
Find $P(\text{H})$.
Exercise 3.2.4
Find $P(\text{N})$.
Answer
$P(\text{N}) = \frac{15}{42} = \frac{5}{14} = 0.36$
Exercise 3.2.5
Find $P(\text{F})$.
Exercise 3.2.6
Find $P(\text{C})$.
Answer
$P(\text{C}) = \frac{5}{42} = 0.12$
Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange.
Let $B =$ the event of getting a blue jelly bean
Let $G =$ the event of getting a green jelly bean.
Let $O =$ the event of getting an orange jelly bean.
Let $P =$ the event of getting a purple jelly bean.
Let $R =$ the event of getting a red jelly bean.
Let $Y =$ the event of getting a yellow jelly bean.
Exercise 3.2.7
Find $P(\text{B})$.
Exercise 3.2.8
Find $P(\text{G})$.
Answer
$P(\text{G}) = \frac{20}{150} = \frac{2}{15} = 0.13$
Exercise 3.2.9
Find $P(\text{P})$.
Exercise 3.2.10
Find $P(\text{R})$.
Answer
$P(\text{R}) = \frac{22}{150} = \frac{11}{75} = 0.15$
Exercise 3.2.11
Find $P(\text{Y})$.
Exercise 3.2.12
Find $P(\text{O})$.
Answer
$P(text{O}) = \frac{150-22-38-20-28-26}{150} = \frac{16}{150} = \frac{8}{75} = 0.11$
Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region).
Let $\text{A} =$ the event that a country is in Asia.
Let $\text{E} =$ the event that a country is in Europe.
Let $\text{F} =$ the event that a country is in Africa.
Let $\text{N} =$ the event that a country is in North America.
Let $\text{O} =$ the event that a country is in Oceania.
Let $\text{S} =$ the event that a country is in South America.
Exercise 3.2.13
Find $P(\text{A})$.
Exercise 3.2.14
Find $P(\text{E})$.
Answer
$P(\text{E}) = \frac{47}{194} = 0.24$
Exercise 3.2.15
Find $P(\text{F})$.
Exercise 3.2.16
Find $P(\text{N})$.
Answer
$P(\text{N}) = \frac{23}{194} = 0.12$
Exercise 3.2.17
Find $P(\text{O})$.
Exercise 3.2.18
Find $P(\text{S})$.
Answer
$P(\text{S}) = \frac{12}{194} = \frac{6}{97} = 0.06$
Exercise 3.2.19
What is the probability of drawing a red card in a standard deck of 52 cards?
Exercise 3.2.20
What is the probability of drawing a club in a standard deck of 52 cards?
Answer
$\frac{13}{52} = \frac{1}{4} = 0.25$
Exercise 3.2.21
What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six?
Exercise 3.2.22
What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six?
Answer
$\frac{3}{6} = \frac{1}{2} = 0.5$
Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area.
Let $\text{B} =$ the event of landing on blue.
Let $\text{R} =$ the event of landing on red.
Let $\text{G} =$ the event of landing on green.
Let $\text{Y} =$ the event of landing on yellow.
Exercise 3.2.23
If you land on $\text{Y}$, you get the biggest prize. Find $P(\text{Y})$.
Exercise 3.2.24
If you land on red, you don’t get a prize. What is $P(\text{R})$?
Answer
$\text{P}(R) = \frac{4}{8} = 0.5$
Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters.
Let $\text{I} =$ the event that a player in an infielder.
Let $\text{O} =$ the event that a player is an outfielder.
Let $\text{H} =$ the event that a player is a great hitter.
Let $\text{N} =$ the event that a player is not a great hitter.
Exercise 3.2.25
Write the symbols for the probability that a player is not an outfielder.
Exercise 3.2.26
Write the symbols for the probability that a player is an outfielder or is a great hitter.
Answer
$P(\text{O OR H})$
Exercise 3.2.27
Write the symbols for the probability that a player is an infielder and is not a great hitter.
Exercise 3.2.28
Write the symbols for the probability that a player is a great hitter, given that the player is an infielder.
Answer
$P(\text{H|I})$
Exercise 3.2.29
Write the symbols for the probability that a player is an infielder, given that the player is a great hitter.
Exercise 3.2.30
Write the symbols for the probability that of all the outfielders, a player is not a great hitter.
Answer
$P(\text{N|O})$
Exercise 3.2.31
Write the symbols for the probability that of all the great hitters, a player is an outfielder.
Exercise 3.2.32
Write the symbols for the probability that a player is an infielder or is not a great hitter.
Answer
$P(\text{I OR N})$
Exercise 3.2.33
Write the symbols for the probability that a player is an outfielder and is a great hitter.
Exercise 3.2.34
Write the symbols for the probability that a player is an infielder.
Answer
$P(\text{I})$
Exercise 3.2.35
What is the word for the set of all possible outcomes?
Exercise 3.2.36
What is conditional probability?
Answer
The likelihood that an event will occur given that another event has already occurred.
Exercise 3.2.37
A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book
Let $\text{F} =$ event that book is fiction
Let $\text{N} =$ event that book is nonfiction
What is the sample space?
Exercise 3.2.38
What is the sum of the probabilities of an event and its complement?
Answer
1
Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let $\text{E} =$ the event that it lands on an even number. Let $\text{M} =$ the event that it lands on a multiple of three.
Exercise 3.2.39
What does $P(\text{E|M})$ mean in words?
Exercise 3.2.40
What does $P(\text{E OR M})$ mean in words?
Answer
the probability of landing on an even number or a multiple of three | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.02%3A_Terminology.txt |
Independent and mutually exclusive do not mean the same thing.
Independent Events
Two events are independent if the following are true:
• $P(\text{A|B}) = P(\text{A})$
• $P(\text{B|A}) = P(\text{B})$
• $P(\text{A AND B}) = P(\text{A})P(\text{B})$
Two events $\text{A}$ and $\text{B}$ are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.
Sampling a population
Sampling may be done with replacement or without replacement (Figure $1$):
• With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
• Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
If it is not known whether $\text{A}$ and $\text{B}$ are independent or dependent, assume they are dependent until you can show otherwise.
Example $1$: Sampling with and without replacement
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, $\text{J}$ (jack), $\text{Q}$ (queen), $\text{K}$ (king) of that suit.
a. Sampling with replacement:
Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the $\text{Q}$ of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the $\text{Q}$ of spades again. Your picks are {$\text{Q}$ of spades, ten of clubs, $\text{Q}$ of spades}. You have picked the $\text{Q}$ of spades twice. You pick each card from the 52-card deck.
b. Sampling without replacement:
Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the $\text{K}$ of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the $\text{J}$ of spades. Your picks are {$\text{K}$ of hearts, three of diamonds, $\text{J}$ of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.
Exercise $1$
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, $\text{J}$ (jack), $\text{Q}$ (queen), $\text{K}$ (king) of that suit. Three cards are picked at random.
1. Suppose you know that the picked cards are $\text{Q}$ of spades, $\text{K}$ of hearts and $\text{Q}$ of spades. Can you decide if the sampling was with or without replacement?
2. Suppose you know that the picked cards are $\text{Q}$ of spades, $\text{K}$ of hearts, and $\text{J}$ of spades. Can you decide if the sampling was with or without replacement?
Answer a
With replacement
Answer b
No
Example $2$
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, $\text{J}$ (jack), $\text{Q}$ (queen), and $\text{K}$ (king) of that suit. $\text{S} =$ spades, $\text{H} =$ Hearts, $\text{D} =$ Diamonds, $\text{C} =$ Clubs.
1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are $\text{QS}, 1\text{D}, 1\text{C}, \text{QD}$.
2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are $\text{KH}, 7\text{D}, 6\text{D}, \text{KH}$.
Which of a. or b. did you sample with replacement and which did you sample without replacement?
Answer a
Without replacement
Answer b
With replacement
Exercise $2$
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, $\text{J}$ (jack), $\text{Q}$ (queen), and $\text{K}$ (king) of that suit. $\text{S} =$ spades, $\text{H} =$ Hearts, $\text{D} =$ Diamonds, $\text{C} =$ Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.
1. $\text{QS}, 1\text{D}, 1\text{C}, \text{QD}$
2. $\text{KH}, 7\text{D}, 6\text{D}, \text{KH}$
3. $\text{QS}, 7\text{D}, 6\text{D}, \text{KS}$
Answer - without replacement
a. Possible; b. Impossible, c. Possible
Answer - with replacement
a. Possible; c. Possible, c. Possible
Mutually Exclusive Events
$\text{A}$ and $\text{B}$ are mutually exclusive events if they cannot occur at the same time. This means that $\text{A}$ and $\text{B}$ do not share any outcomes and $P(\text{A AND B}) = 0$.
For example, suppose the sample space
$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. \nonumber$
Let $\text{A} = \{1, 2, 3, 4, 5\}, \text{B} = \{4, 5, 6, 7, 8\}$, and $\text{C} = \{7, 9\}$. $\text{A AND B} = \{4, 5\}$.
$P(\text{A AND B}) = \dfrac{2}{10} \nonumber$
and is not equal to zero. Therefore, $\text{A}$ and $\text{B}$ are not mutually exclusive. $\text{A}$ and $\text{C}$ do not have any numbers in common so $P(\text{A AND C}) = 0$. Therefore, $\text{A}$ and $\text{C}$ are mutually exclusive.
If it is not known whether $\text{A}$ and $\text{B}$ are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.
Example $3$
Flip two fair coins.
The sample space is $\{HH, HT, TH, TT\}$ where $T =$ tails and $H =$ heads. The outcomes are $HH,HT, TH$, and $TT$. The outcomes $HT$ and $TH$ are different. The $HT$ means that the first coin showed heads and the second coin showed tails. The $TH$ means that the first coin showed tails and the second coin showed heads.
• Let $\text{A} =$ the event of getting at most one tail. (At most one tail means zero or one tail.) Then $\text{A}$ can be written as $\{HH, HT, TH\}$. The outcome $HH$ shows zero tails. $HT$ and $TH$ each show one tail.
• Let $\text{B} =$ the event of getting all tails. $\text{B}$ can be written as $\{TT\}$. $\text{B}$ is the complement of $\text{A}$, so $\text{B} = \text{A′}$. Also, $P(\text{A}) + P(\text{B}) = P(\text{A}) + P(\text{A′}) = 1$.
• The probabilities for $\text{A}$ and for $\text{B}$ are $P(\text{A}) = \dfrac{3}{4}$ and $P(\text{B}) = \dfrac{1}{4}$.
• Let $\text{C} =$ the event of getting all heads. $\text{C} = \{HH\}$. Since $\text{B} = \{TT\}$, $P(\text{B AND C}) = 0$. $\text{B}$ and Care mutually exclusive. $\text{B}$ and $\text{C}$ have no members in common because you cannot have all tails and all heads at the same time.)
• Let $\text{D} =$ event of getting more than one tail. $\text{D} = \{TT\}$. $P(\text{D}) = \dfrac{1}{4}$
• Let $\text{E} =$ event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) $\text{E} = \{HT, HH\}$. $P(\text{E}) = \dfrac{2}{4}$
• Find the probability of getting at least one (one or two) tail in two flips. Let $\text{F} =$ event of getting at least one tail in two flips. $\text{F} = \{HT, TH, TT\}$. $P(\text{F}) = \dfrac{3}{4}$
Exercise $3$
Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.
Answer
The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is $\{BB, BR, RB, RR\}$.
Event $A =$ Getting at least one black card $= \{BB, BR, RB\}$
$P(\text{A}) = \dfrac{3}{4} = 0.75$
Example $4$
Flip two fair coins. Find the probabilities of the events.
1. Let $\text{F} =$ the event of getting at most one tail (zero or one tail).
2. Let $\text{G} =$ the event of getting two faces that are the same.
3. Let $\text{H} =$ the event of getting a head on the first flip followed by a head or tail on the second flip.
4. Are $\text{F}$ and $\text{G}$ mutually exclusive?
5. Let $\text{J} =$ the event of getting all tails. Are $\text{J}$ and $\text{H}$ mutually exclusive?
Solution
Look at the sample space in Example $3$.
1. Zero (0) or one (1) tails occur when the outcomes $HH, TH, HT$ show up. $P(\text{F}) = \dfrac{3}{4}$
2. Two faces are the same if $HH$ or $TT$ show up. $P(\text{G}) = \dfrac{2}{4}$
3. A head on the first flip followed by a head or tail on the second flip occurs when $HH$ or $HT$ show up. $P(\text{H}) = \dfrac{2}{4}$
4. $\text{F}$ and $\text{G}$ share $HH$ so $P(\text{F AND G})$ is not equal to zero (0). $\text{F}$ and $\text{G}$ are not mutually exclusive.
5. Getting all tails occurs when tails shows up on both coins ($TT$). $\text{H}$’s outcomes are $HH$ and $HT$.
$\text{J}$ and $\text{H}$ have nothing in common so $P(\text{J AND H}) = 0$. $\text{J}$ and $\text{H}$ are mutually exclusive.
Exercise $4$
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
1. Let $\text{F} =$ the event of getting the white ball twice.
2. Let $\text{G} =$ the event of getting two balls of different colors.
3. Let $\text{H} =$ the event of getting white on the first pick.
4. Are $\text{F}$ and $\text{G}$ mutually exclusive?
5. Are $\text{G}$ and $\text{H}$ mutually exclusive?
Answer
1. $P(\text{F}) = \dfrac{1}{4}$
2. $P(\text{G}) = \dfrac{1}{2}$
3. $P(\text{H}) = \dfrac{1}{2}$
4. Yes
5. No
Example $5$
Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event $\text{A} =$ a face is odd. Then $\text{A} = \{1, 3, 5\}$. Let event $\text{B} =$ a face is even. Then $\text{B} = \{2, 4, 6\}$.
• Find the complement of $\text{A}$, $\text{A′}$. The complement of $\text{A}$, $\text{A′}$, is $\text{B}$ because $\text{A}$ and $\text{B}$ together make up the sample space. $P(\text{A}) + P(\text{B}) = P(\text{A}) + P(\text{A′}) = 1$. Also, $P(\text{A}) = \dfrac{3}{6}$ and $P(\text{B}) = \dfrac{3}{6}$.
• Let event $\text{C} =$ odd faces larger than two. Then $\text{C} = \{3, 5\}$. Let event $\text{D} =$ all even faces smaller than five. Then $\text{D} = \{2, 4\}$. $P(\text{C AND D}) = 0$ because you cannot have an odd and even face at the same time. Therefore, $\text{C}$ and $\text{D}$ are mutually exclusive events.
• Let event $\text{E} =$ all faces less than five. $\text{E} = \{1, 2, 3, 4\}$.
Are $\text{C}$ and $\text{E}$ mutually exclusive events? (Answer yes or no.) Why or why not?
Answer
No. $\text{C} = \{3, 5\}$ and $\text{E} = \{1, 2, 3, 4\}$. $P(\text{C AND E}) = \dfrac{1}{6}$. To be mutually exclusive, $P(\text{C AND E})$ must be zero.
• Find $P(\text{C|A})$. This is a conditional probability. Recall that the event $\text{C}$ is {3, 5} and event $\text{A}$ is {1, 3, 5}. To find $P(\text{C|A})$, find the probability of $\text{C}$ using the sample space $\text{A}$. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, $P(\text{C|A}) = \dfrac{2}{3}$.
Exercise $5$
Let event $\text{A} =$ learning Spanish. Let event $\text{B}$ = learning German. Then $\text{A AND B}$ = learning Spanish and German. Suppose $P(\text{A}) = 0.4$ and $P(\text{B}) = 0.2$. $P(\text{A AND B}) = 0.08$. Are events $\text{A}$ and $\text{B}$ independent? Hint: You must show ONE of the following:
• $P(\text{A|B}) = P(\text{A})$
• $P(\text{B|A})$
• $P(\text{A AND B}) = P(\text{A})P(\text{B})$
Answer
$P(\text{A|B}) = \dfrac{\text{P(A AND B)}}{P(\text{B})} = \dfrac{0.08}{0.2} = 0.4 = P(\text{A})$
The events are independent because $P(\text{A|B}) = P(\text{A})$.
Example $6$
Let event $\text{G} =$ taking a math class. Let event $\text{H} =$ taking a science class. Then, $\text{G AND H} =$ taking a math class and a science class. Suppose $P(\text{G}) = 0.6$, $P(\text{H}) = 0.5$, and $P(\text{G AND H}) = 0.3$. Are $\text{G}$ and $\text{H}$ independent?
If $\text{G}$ and $\text{H}$ are independent, then you must show ONE of the following:
• $P(\text{G|H}) = P(\text{G})$
• $P(\text{H|G}) = P(\text{H})$
• $P(\text{G AND H}) = P(\text{G})P(\text{H})$
The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.
1. a. Show that $P(\text{G|H}) = P(\text{G})$.
2. b. Show $P(\text{G AND H}) = P(\text{G})P(\text{H})$.
Solution
1. $P(\text{G|H}) = \dfrac{P(\text{G AND H})}{P(\text{H})} = \dfrac{0.3}{0.5} = 0.6 = P(\text{G})$
2. $P(\text{G})P(\text{H}) = (0.6)(0.5) = 0.3 = P(\text{G AND H})$
Since $\text{G}$ and $\text{H}$ are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that $P(\text{H|G}) = P(\text{H})$ to show that $\text{G}$ and $\text{H}$ are independent events.
Exercise $6$
In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.
• $\text{R} =$ a red marble
• $\text{G} =$ a green marble
• $\text{O} =$ an odd-numbered marble
• The sample space is $\text{S} = \{R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\}$.
$\text{S}$ has ten outcomes. What is $P(\text{G AND O})$?
Answer
Event $\text{G}$ and $\text{O} = \{G1, G3\}$
$P(\text{G and O}) = \dfrac{2}{10} = 0.2$
Example $7$
Let event $\text{C} =$ taking an English class. Let event $\text{D} =$ taking a speech class.
Suppose $P(\text{C}) = 0.75$, $P(\text{D}) = 0.3$, $P(\text{C|D}) = 0.75$ and $P(\text{C AND D}) = 0.225$.
Justify your answers to the following questions numerically.
1. Are $\text{C}$ and $\text{D}$ independent?
2. Are $\text{C}$ and $\text{D}$ mutually exclusive?
3. What is $P(\text{D|C})$?
Solution
1. Yes, because $P(\text{C|D}) = P(\text{C})$.
2. No, because $P(\text{C AND D})$ is not equal to zero.
3. $P(\text{D|C}) = \dfrac{P(\text{C AND D})}{P(\text{C})} = \dfrac{0.225}{0.75} = 0.3$
Exercise $7$
A student goes to the library. Let events $\text{B} =$ the student checks out a book and $\text{D} =$ the student checks out a DVD. Suppose that $P(\text{B}) = 0.40$, $P(\text{D}) = 0.30$ and $P(\text{B AND D}) = 0.20$.
1. Find $P(\text{B|D})$.
2. Find $P(\text{D|B})$.
3. Are $\text{B}$ and $\text{D}$ independent?
4. Are $\text{B}$ and $\text{D}$ mutually exclusive?
Answer
1. $P(\text{B|D}) = 0.6667$
2. $P(\text{D|B}) = 0.5$
3. No
4. No
Example $8$
In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.
Let
• $\text{R =}$ red card is drawn,
• $\text{B} =$ blue card is drawn,
• $\text{E} =$ even-numbered card is drawn.
The sample space $S = R1, R2, R3, B1, B2, B3, B4, B5$.
$S$ has eight outcomes.
• $P(\text{R}) = \dfrac{3}{8}$. $P(\text{B}) = \dfrac{5}{8}$. $P(\text{R AND B}) = 0$. (You cannot draw one card that is both red and blue.)
• $P(\text{E}) = \dfrac{3}{8}$. (There are three even-numbered cards, $R2, B2$, and $B4$.)
• $P(\text{E|B}) = \dfrac{2}{5}$. (There are five blue cards: $B1, B2, B3, B4$, and $B5$. Out of the blue cards, there are two even cards; $B2$ and $B4$.)
• $P(\text{B|E}) = \dfrac{2}{3}$. (There are three even-numbered cards: $R2, B2$, and $B4$. Out of the even-numbered cards, to are blue; $B2$ and $B4$.)
• The events $\text{R}$ and $\text{B}$ are mutually exclusive because $P(\text{R AND B}) = 0$.
• Let $\text{G} =$ card with a number greater than 3. $\text{G} = \{B4, B5\}$. $P(\text{G}) = \dfrac{2}{8}$. Let $\text{H} =$ blue card numbered between one and four, inclusive. $\text{H} = \{B1, B2, B3, B4\}$. $P(\text{G|H}) = \frac{1}{4}$. (The only card in $\text{H}$ that has a number greater than three is B4.) Since $\dfrac{2}{8} = \dfrac{1}{4}$, $P(\text{G}) = P(\text{G|H})$, which means that $\text{G}$ and $\text{H}$ are independent.
Exercise $8$
In a basketball arena,
• 70% of the fans are rooting for the home team.
• 25% of the fans are wearing blue.
• 20% of the fans are wearing blue and are rooting for the away team.
• Of the fans rooting for the away team, 67% are wearing blue.
Let $\text{A}$ be the event that a fan is rooting for the away team.
Let $\text{B}$ be the event that a fan is wearing blue.
Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive?
Answer
• $P(\text{B|A}) = 0.67$
• $P(\text{B}) = 0.25$
So $P(\text{B})$ does not equal $P(\text{B|A})$ which means that $\text{B} and \text{A}$ are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because $P(\text{B AND A}) = 0.20$, not $0$.
Example $9$
In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let $\text{F}$ be the event that a student is female. Let $\text{L}$ be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?
• The following probabilities are given in this example:
• $P(\text{F}) = 0.60$; $P(\text{L}) = 0.50$
• $P(\text{F AND L}) = 0.45$
• $P(\text{L|F}) = 0.75$
The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know $P(\text{F|L})$ yet, so you cannot use the second condition.
Solution 1
Check whether $P(\text{F AND L}) = P(\text{F})P(\text{L})$. We are given that $P(\text{F AND L}) = 0.45$, but $P(\text{F})P(\text{L}) = (0.60)(0.50) = 0.30$. The events of being female and having long hair are not independent because $P(\text{F AND L})$ does not equal $P(\text{F})P(\text{L})$.
Solution 2
Check whether $P(\text{L|F})$ equals $P(\text{L})$. We are given that $P(\text{L|F}) = 0.75$, but $P(\text{L}) = 0.50$; they are not equal. The events of being female and having long hair are not independent.
Interpretation of Results
The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.
Exercise $9$
Mark is deciding which route to take to work. His choices are $\text{I} = \text{the Interstate}$ and $\text{F} = \text{Fifth Street}$
• $P(\text{I}) = 0.44$ and $P(\text{F}) = 0.55$
• $P(\text{I AND F}) = 0$ because Mark will take only one route to work.
What is the probability of $P(\text{I OR F})$?
Answer
Because $P(\text{I AND F}) = 0$,
$P(\text{I OR F}) = P(\text{I}) + P(\text{F}) - P(\text{I AND F}) = 0.44 + 0.56 - 0 = 1$
Example $10$
1. Toss one fair coin (the coin has two sides, $\text{H}$ and $\text{T}$). The outcomes are ________. Count the outcomes. There are ____ outcomes.
2. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes.
3. Multiply the two numbers of outcomes. The answer is _______.
4. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in three is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are $H1$ and $T6$.)
5. Event $\text{A} =$ heads ($\text{H}$) on the coin followed by an even number (2, 4, 6) on the die.
$\text{A}$ = {_________________}. Find $P(\text{A})$.
6. Event $\text{B} =$ heads on the coin followed by a three on the die. $\text{B} =$ {________}. Find $P(\text{B})$.
7. Are $\text{A}$ and $\text{B}$ mutually exclusive? (Hint: What is $P(\text{A AND B})$? If $P(\text{A AND B}) = 0$, then $\text{A}$ and $\text{B}$ are mutually exclusive.)
8. Are $\text{A}$ and $\text{B}$ independent? (Hint: Is $P(\text{A AND B}) = P(\text{A})P(\text{B})$? If $P(\text{A AND B})\ = P(\text{A})P(\text{B})$, then $\text{A}$ and $\text{B}$ are independent. If not, then they are dependent).
Solution
1. $\text{H}$ and $\text{T}$; 2
2. 1, 2, 3, 4, 5, 6; 6
3. 2(6) = 12
4. $T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6$
5. $\text{A} = \{H2, H4, H6\}$; $P(\text{A}) = \dfrac{3}{12}$
6. $\text{B} = \{H3\}$; $P(\text{B}) = \dfrac{1}{12}$
7. Yes, because $P(\text{A AND B}) = 0$
8. $P(\text{A AND B}) = 0$. $P(\text{A})P(\text{B}) = \left(\dfrac{3}{12}\right)\left(\dfrac{1}{12}\right)$. $P(\text{A AND B})$ does not equal $P(\text{A})P(\text{B})$, so $\text{A}$ and $\text{B}$ are dependent.
Exercise $10$
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let $\text{T}$ be the event of getting the white ball twice, $\text{F}$ the event of picking the white ball first, $\text{S}$ the event of picking the white ball in the second drawing.
1. Compute $P(\text{T})$.
2. Compute $P(\text{T|F})$.
3. Are $\text{T}$ and $\text{F}$ independent?.
4. Are $\text{F}$ and $\text{S}$ mutually exclusive?
5. Are $\text{F}$ and $\text{S}$ independent?
Answer
1. $P(\text{T}) = \dfrac{1}{4}$
2. $P(\text{T|F}) = \dfrac{1}{2}$
3. No
4. No
5. Yes
Review
Two events $\text{A}$ and $\text{B}$ are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.
In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.
Formula Review
• If $\text{A}$ and $\text{B}$ are independent, $P(\text{A AND B}) = P(\text{A})P(\text{B}), P(\text{A|B}) = P(\text{A})$ and $P(\text{B|A}) = P(\text{B})$.
• If $\text{A}$ and $\text{B}$ are mutually exclusive, $P(\text{A OR B}) = P(\text{A}) + P(\text{B}) and P(\text{A AND B}) = 0$.
Exercise $11$
$\text{E}$ and $\text{F}$ are mutually exclusive events. $P(\text{E}) = 0.4$; $P(\text{F}) = 0.5$. Find $P(\text{E∣F})$.
Exercise $12$
$\text{J}$ and $\text{K}$ are independent events. $P(\text{J|K}) = 0.3$. Find $P(\text{J})$.
Answer
$P(\text{J}) = 0.3$
Exercise $13$
$\text{U}$ and $\text{V}$ are mutually exclusive events. $P(\text{U}) = 0.26$; $P(\text{V}) = 0.37$. Find:
1. $P(\text{U AND V}) =$
2. $P(\text{U|V}) =$
3. $P(\text{U OR V}) =$
Exercise $14$
$\text{Q}$ and $\text{R}$ are independent events. $P(\text{Q}) = 0.4$ and $P(\text{Q AND R}) = 0.1$. Find $P(\text{R})$.
Answer
$P(\text{Q AND R}) = P(\text{Q})P(\text{R})$
$0.1 = (0.4)P(\text{R})$
$P(\text{R}) = 0.25$
Bringing It Together
Exercise $16$
A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into Table.
Shirt# ≤ 210 211–250 251–290 290≤
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5
For the following, suppose that you randomly select one player from the 49ers or Cowboys.
If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about $P(\text{Shirt} \#1–33|\leq 210 \text{ pounds})$?
Exercise $17$
The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let $\text{C} =$ a man develops cancer in his lifetime and $\text{P} =$ man has at least one false positive.
1. $P(\text{C}) =$ ______
2. $P(\text{P|C}) =$ ______
3. $P(\text{P|C'}) =$ ______
4. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not.
Answer
1. $P(\text{C}) = 0.4567$
2. not enough information
3. not enough information
4. No, because over half (0.51) of men have at least one false positive text
Exercise $18$
Given events $\text{G}$ and $\text{H}: P(\text{G}) = 0.43$; $P(\text{H}) = 0.26$; $P(\text{H AND G}) = 0.14$
1. Find $P(\text{H OR G})$.
2. Find the probability of the complement of event ($\text{H AND G}$).
3. Find the probability of the complement of event ($\text{H OR G}$).
Exercise $19$
Given events $\text{J}$ and $\text{K}: P(\text{J}) = 0.18$; $P(\text{K}) = 0.37$; $P(\text{J OR K}) = 0.45$
1. Find $P(\text{J AND K})$.
2. Find the probability of the complement of event ($\text{J AND K}$).
3. Find the probability of the complement of event ($\text{J AND K}$).
Answer
1. $P(\text{J OR K}) = P(\text{J}) + P(\text{K}) − P(\text{J AND K}); 0.45 = 0.18 + 0.37 - P(\text{J AND K})$; solve to find $P(\text{J AND K}) = 0.10$
2. $P(\text{NOT (J AND K)}) = 1 - P(\text{J AND K}) = 1 - 0.10 = 0.90$
3. $P(\text{NOT (J OR K)}) = 1 - P(\text{J OR K}) = 1 - 0.45 = 0.55$
Glossary
Dependent Events
If two events are NOT independent, then we say that they are dependent.
Sampling with Replacement
If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once.
Sampling without Replacement
When sampling is done without replacement, each member of a population may be chosen only once.
The Conditional Probability of One Event Given Another Event
P(A|B) is the probability that event A will occur given that the event B has already occurred.
The OR of Two Events
An outcome is in the event A OR B if the outcome is in A, is in B, or is in both A and B. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.03%3A_Independent_and_Mutually_Exclusive_Events.txt |
When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.
The Multiplication Rule
If A and B are two events defined on a sample space, then:
$P(A \text{ AND } B) = P(B)P(A|B) \label{eq1}$
This rule may also be written as:
$P(A|B) = \dfrac{P(A \text{ AND } B)}{P(B)} \nonumber$
(The probability of $A$ given $B$ equals the probability of $A$ and $B$ divided by the probability of $B$.)
If $A$ and $B$ are independent, then
$P(A|B) = P(A). \nonumber$
and Equation \ref{eq1} becomes
$P(A \text{ AND } B) = P(A)P(B). \nonumber$
The Addition Rule
If A and B are defined on a sample space, then:
$P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) \label{eq5}$
If A and B are mutually exclusive, then
$P(A \text{ AND } B) = 0. \nonumber$
and Equation \ref{eq5} becomes
$P(A \text{ OR } B) = P(A) + P(B). \nonumber$
Example $1$
Klaus is trying to choose where to go on vacation. His two choices are: $\text{A} = \text{New Zealand}$ and $\text{B} = \text{Alaska}$.
• Klaus can only afford one vacation. The probability that he chooses $\text{A}$ is $P(\text{A}) = 0.6$ and the probability that he chooses $\text{B}$ is $P(\text{B}) = 0.35$.
• $P(\text{A AND B}) = 0$ because Klaus can only afford to take one vacation
• Therefore, the probability that he chooses either New Zealand or Alaska is $P(\text{A OR B}) = P(\text{A}) + P(\text{B}) = 0.6 + 0.35 = 0.95$. Note that the probability that he does not choose to go anywhere on vacation must be 0.05.
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. $\text{A} =$ the event Carlos is successful on his first attempt. $P(\text{A}) = 0.65$. $\text{B} =$ the event Carlos is successful on his second attempt. $P(\text{B}) = 0.65$. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.
1. What is the probability that he makes both goals?
2. What is the probability that Carlos makes either the first goal or the second goal?
3. Are $\text{A}$ and $\text{B}$ independent?
4. Are $\text{A}$ and $\text{B}$ mutually exclusive?
Solutions
a. The problem is asking you to find $P(\text{A AND B}) = P(\text{B AND A})$. Since $P(\text{B|A}) = 0.90: P(\text{B AND A}) = P(\text{B|A}) P(\text{A}) = (0.90)(0.65) = 0.585$
Carlos makes the first and second goals with probability 0.585.
b. The problem is asking you to find $P(\text{A OR B})$.
$P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B}) = 0.65 + 0.65 - 0.585 = 0.715$
Carlos makes either the first goal or the second goal with probability 0.715.
c. No, they are not, because $P(\text{B AND A}) = 0.585$.
$P(\text{B})P(\text{A}) = (0.65)(0.65) = 0.423$
$0.423 \neq 0.585 = P(\text{B AND A})$
So, $P(\text{B AND A})$ is not equal to $P(\text{B})P(\text{A})$.
d. No, they are not because $P(\text{A and B}) = 0.585$.
To be mutually exclusive, $P(\text{A AND B})$ must equal zero.
Exercise $1$
Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. $\text{C} =$ the event that Helen makes the first shot. $P(\text{C}) = 0.75$. $\text{D} =$ the event Helen makes the second shot. $P(\text{D}) = 0.75$. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?
Answer
$P(\text{D|C}) = 0.85$
$P(\text{C AND D}) = P(\text{D AND C})$
$P(\text{D AND C}) = P(\text{D|C})P(\text{C}) = (0.85)(0.75) = 0.6375$
Helen makes the first and second free throws with probability 0.6375.
Example $2$
A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.
1. What is the probability that the member is a novice swimmer?
2. What is the probability that the member practices four times a week?
3. What is the probability that the member is an advanced swimmer and practices four times a week?
4. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?
5. Are being a novice swimmer and practicing four times a week independent events? Why or why not?
Answer
1. $\dfrac{28}{150}$
2. $\dfrac{80}{150}$
3. $\dfrac{40}{150}$
4. $P(\text{advanced AND intermediate}) = 0$, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.
5. No, these are not independent events. $P(\text{novice AND practices four times per week}) = 0.0667$P(\text{novice})P(\text{practices four times per week}) = 0.0996$ $0.0667 \neq 0.0996$
Exercise $2$
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?
Answer
$P = \dfrac{200-140-40}{200} = \dfrac{20}{200} = 0.1$
Example $3$
Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.
Let: $\text{M} =$ math class, $\text{S} =$ speech class, $\text{M|S} =$ math given speech
1. What is the probability that Felicity enrolls in math and speech?
Find $P(\text{M AND S}) = P(\text{M|S})P(\text{S})$.
2. What is the probability that Felicity enrolls in math or speech classes?
Find $P(\text{M OR S}) = P(\text{M}) + P(\text{S}) - P(\text{M AND S})$.
3. Are $\text{M}$ and $\text{S}$ independent? Is $P(\text{M|S}) = P(\text{M})$?
4. Are $\text{M}$ and $\text{S}$ mutually exclusive? Is $P(\text{M AND S}) = 0$?
Answer
a. 0.1625, b. 0.6875, c. No, d. No
Exercise $3$
A student goes to the library. Let events $\text{B} =$ the student checks out a book and $\text{D} =$ the student check out a DVD. Suppose that $P(\text{B}) = 0.40, P(\text{D}) = 0.30$ and $P(\text{D|B}) = 0.5$.
1. Find $P(\text{B AND D})$.
2. Find $P(\text{B OR D})$.
Answer
1. $P(\text{B AND D}) = P(\text{D|B})P(\text{B}) = (0.5)(0.4) = 0.20$.
2. $P(\text{B OR D}) = P(\text{B}) + P(\text{D}) − P(\text{B AND D}) = 0.40 + 0.30 − 0.20 = 0.50$
Example $4$
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let $\text{B} =$ woman develops breast cancer and let $\text{N} =$ tests negative. Suppose one woman is selected at random.
1. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
2. Given that the woman has breast cancer, what is the probability that she tests negative?
3. What is the probability that the woman has breast cancer AND tests negative?
4. What is the probability that the woman has breast cancer or tests negative?
5. Are having breast cancer and testing negative independent events?
6. Are having breast cancer and testing negative mutually exclusive?
Answers
1. $P(\text{B}) = 0.143; P(\text{N}) = 0.85$
2. $P(\text{N|B}) = 0.02$
3. $P(\text{B AND N}) = P(\text{B})P(\text{N|B}) = (0.143)(0.02) = 0.0029$
4. $P(\text{B OR N}) = P(\text{B}) + P(\text{N}) - P(\text{B AND N}) = 0.143 + 0.85 - 0.0029 = 0.9901$
5. No. $P(\text{N}) = 0.85; P(\text{N|B}) = 0.02$. So, $P(\text{N|B})$ does not equal $P(\text{N})$.
6. No. $P(\text{B AND N}) = 0.0029$. For $\text{B}$ and $\text{N}$ to be mutually exclusive, $P(\text{B AND N})$ must be zero
Exercise $4$
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?
Answer
Let $\text{A} =$ student is a senior going to college.
Let $\text{B} =$ student plays sports.
$P(\text{B}) = \dfrac{140}{200}$
$P(\text{B|A}) = \dfrac{50}{140}$
$P(\text{A AND B}) = P(\text{B|A})P(\text{A})$
$P(\text{A AND B}) = (\dfrac{140}{200}$)($\dfrac{50}{140}) = \dfrac{1}{4}$
Example $5$
Refer to the information in Example $4$. $\text{P} =$ tests positive.
1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find $P(\text{P|B}) = 1 - P(\text{N|B})$.
2. What is the probability that a woman develops breast cancer and tests positive. Find $P(\text{B AND P}) = P(\text{P|B})P(\text{B})$.
3. What is the probability that a woman does not develop breast cancer. Find $P(\text{B′}) = 1 - P(\text{B})$.
4. What is the probability that a woman tests positive for breast cancer. Find $P(\text{P}) = 1 - P(\text{N})$.
Answer
a. 0.98; b. 0.1401; c. 0.857; d. 0.15
Exercise $5$
A student goes to the library. Let events $\text{B} =$ the student checks out a book and $\text{D} =$ the student checks out a DVD. Suppose that $P(\text{B}) = 0.40, P(\text{D}) = 0.30$ and $P(\text{D|B}) = 0.5$.
1. Find $P(\text{B′})$.
2. Find $P(\text{D AND B})$.
3. Find $P(\text{B|D})$.
4. Find $P(\text{D AND B′})$.
5. Find $P(\text{D|B′})$.
Answer
1. $P(\text{B′}) = 0.60$
2. $P(\text{D AND B}) = P(\text{D|B})P(\text{B}) = 0.20$
3. $P(\text{B|D}) = \dfrac{P(\text{B AND D})}{P(\text{D})} = \dfrac{(0.20)}{(0.30)} = 0.66$
4. $P(\text{D AND B′}) = P(\text{D}) - P(\text{D AND B}) = 0.30 - 0.20 = 0.10$
5. $P(\text{D|B′}) = P(\text{D AND B′})P(\text{B′}) = (P(\text{D}) - P(\text{D AND B}))(0.60) = (0.10)(0.60) = 0.06$
Review
The multiplication rule and the addition rule are used for computing the probability of $\text{A}$ and $\text{B}$, as well as the probability of $\text{A}$ or $\text{B}$ for two given events $\text{A}$, $\text{B}$ defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events $\text{A}$ and $\text{B}$ are mutually exclusive events when they do not have any outcomes in common.
Formula Review
The multiplication rule: $P(\text{A AND B}) = P(\text{A|B})P(\text{B})$
The addition rule: $P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B})$
Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino.
In this problem, let:
• $\text{C} =$ Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder.
• $\text{L} =$ Latino Californians
Suppose that one Californian is randomly selected.
Exercise $5$
Find $P(\text{C})$.
Exercise $6$
Find $P(\text{L})$.
Answer
0.376
Exercise $7$
Find $P(\text{C|L})$.
Exercise $8$
In words, what is $\text{C|L}$?
Answer
$\text{C|L}$ means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.
Exercise $9$
Find $P(\text{L AND C})$
Exercise $10$
In words, what is $\text{L AND C}$?
Answer
$\text{L AND C}$ is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.
Exercise $11$
Are $\text{L}$ and $\text{C}$ independent events? Show why or why not.
Exercise $12$
Find $P(\text{L OR C})$.
Answer
0.6492
Exercise $13$
In words, what is $\text{L OR C}$?
Exercise $14$
Are $\text{L}$ and $\text{C}$ mutually exclusive events? Show why or why not.
Answer
No, because $P(\text{L AND C})$ does not equal 0.
Glossary
Independent Events
The occurrence of one event has no effect on the probability of the occurrence of another event. Events $\text{A}$ and $\text{B}$ are independent if one of the following is true:
1. $P(\text{A|B}) = P(\text{A})$
2. $P(\text{B|A}) = P(\text{B})$
3. $P(\text{A AND B}) = P(\text{A})P(\text{B})$
Mutually Exclusive
Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events $\text{A}$ and $\text{B}$ are mutually exclusive, then $P(\text{A AND B}) = 0$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.04%3A_Two_Basic_Rules_of_Probability.txt |
A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.
Example $1$
Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:
Speeding violation in the last year No speeding violation in the last year Total
Cell phone user 25 280 305
Not a cell phone user 45 405 450
Total 70 685 755
The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.
Calculate the following probabilities using the table.
1. Find $P(\text{Person is a cell phone user})$.
2. Find $P(\text{person had no violation in the last year})$.
3. Find $P(\text{Person had no violation in the last year AND was a cell phone user})$.
4. Find $P(\text{Person is a cell phone user OR person had no violation in the last year})$.
5. Find $P(\text{Person is a cell phone user GIVEN person had a violation in the last year})$.
6. Find $P(\text{Person had no violation last year GIVEN person was not a cell phone user})$
Answer
1. $\dfrac{\text{number of cell phone users}}{\text{total number in study}}$ = $\dfrac{305}{755}$
2. $\dfrac{\text{number that had no violation}}{\text{total number in study}} = \dfrac{685}{755}$
3. $\dfrac{280}{755}$
4. $\left(\dfrac{305}{755} + \dfrac{685}{755}\right) - \dfrac{280}{755} = \dfrac{710}{755}$
5. $\dfrac{25}{70}$ (The sample space is reduced to the number of persons who had a violation.)
6. $\dfrac{405}{450}$ (The sample space is reduced to the number of persons who were not cell phone users.)
Exercise $1$
Table shows the number of athletes who stretch before exercising and how many had injuries within the past year.
Injury in last year No injury in last year Total
Stretches 55 295 350
Does not stretch 231 219 450
Total 286 514 800
1. What is $P(\text{athlete stretches before exercising})$?
2. What is $P(\text{athlete stretches before exercising|no injury in the last year})$?
Answer
1. $P(\text{athlete stretches before exercising}) = \dfrac{350}{800} = 0.4375$
2. $P(\text{athlete stretches before exercising|no injury in the last year}) = \dfrac{295}{514} = 0.5739$
Example $2$
Table shows a random sample of 100 hikers and the areas of hiking they prefer.
Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 ___ 45
Male ___ ___ 14 55
Total ___ 41 ___ ___
1. Complete the table.
2. Are the events "being female" and "preferring the coastline" independent events? Let F = being female and let C = preferring the coastline.
1. Find P(F AND C).
2. Find P(F)P(C)
3. Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.
3. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let $\text{M} =$ being male, and let $\text{L} =$ prefers hiking near lakes and streams.
1. What word tells you this is a conditional?
2. Fill in the blanks and calculate the probability: $P$(___|___) = ___.
3. Is the sample space for this problem all 100 hikers? If not, what is it?
4. Find the probability that a person is female or prefers hiking on mountain peaks. Let $\text{F} =$ being female, and let $\text{P} =$ prefers mountain peaks.
1. Find $P(\text{F})$.
2. Find $P(\text{P})$.
3. Find $P(\text{F AND P})$.
4. Find $P(\text{F OR P})$.
Answers
a.
Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 11 45
Male 16 25 14 55
Total 34
41
25 100
b.
$P(\text{F AND C}) = \dfrac{18}{100} = 0.18$
$P(\text{F})P(\text{C}) = \left(\dfrac{45}{100}\right) \left(\dfrac{34}{100}\right) = (0.45)(0.34) = 0.153$
$P(\text{F AND C}) \neq P(\text{F})P(\text{C})$, so the events $\text{F}$ and $\text{C}$ are not independent.
c.
1. The word 'given' tells you that this is a conditional.
2. $P(\text{M|L}) = \dfrac{25}{41}$
3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.
d.
1. Find $P(\text{F})$.
2. Find $P(\text{P})$.
3. Find $P(\text{F AND P})$.
4. Find $P(\text{F OR P})$.
d.
1. $P(\text{F}) = \dfrac{45}{100}$
2. $P(\text{P}) = \dfrac{25}{100}$
3. $P(\text{F AND P}) = \dfrac{11}{100}$
4. $P(\text{F OR P}) = \dfrac{45}{100} + \dfrac{25}{100} - \dfrac{11}{100} = \dfrac{59}{100}$
Exercise $2$
Table shows a random sample of 200 cyclists and the routes they prefer. Let $\text{M} =$ males and $\text{H} =$ hilly path.
Gender Lake Path Hilly Path Wooded Path Total
Female 45 38 27 110
Male 26 52 12 90
Total 71 90 39 200
1. Out of the males, what is the probability that the cyclist prefers a hilly path?
2. Are the events “being male” and “preferring the hilly path” independent events?
Answer
1. P(H|M) = $\dfrac{52}{90}$ = 0.5778
2. For M and H to be independent, show P(H|M) = P(H)
P(H|M) = 0.5778, P(H) = $\dfrac{90}{200}$ = 0.45
P(H|M) does not equal P(H) so M and H are NOT independent.
Example $3$
Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is $\dfrac{1}{5}$ and the probability he is not caught is $\dfrac{4}{5}$. If he goes out the second door, the probability he gets caught by Alissa is $\dfrac{1}{4}$ and the probability he is not caught is $\dfrac{3}{4}$. The probability that Alissa catches Muddy coming out of the third door is $\dfrac{1}{2}$ and the probability she does not catch Muddy is $\dfrac{1}{2}$. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is $\dfrac{1}{3}$.
Door Choice
Caught or Not Door One Door Two Door Three Total
Caught $\dfrac{1}{15}$ $\dfrac{1}{12}$ $\dfrac{1}{6}$ ____
Not Caught $\dfrac{4}{15}$ $\dfrac{3}{12}$ $\dfrac{1}{6}$ ____
Total ____ ____ ____ 1
• The first entry $\dfrac{1}{15} = \left(\dfrac{1}{5}\right) \left(\dfrac{1}{3}\right)$ is $P(\text{Door One AND Caught})$
• The entry $\dfrac{4}{15} = \left(\dfrac{4}{5}\right) \left(\dfrac{1}{3}\right)$ is $P(\text{Door One AND Not Caught})$
Verify the remaining entries.
1. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
2. What is the probability that Alissa does not catch Muddy?
3. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?
Solution
Door Choice
Caught or Not Door One Door Two Door Three Total
Caught $\dfrac{1}{15}$ $\dfrac{1}{12}$ $\dfrac{1}{6}$ $\dfrac{19}{60}$
Not Caught $\dfrac{4}{15}$ $\dfrac{3}{12}$ $\dfrac{1}{6}$ $\dfrac{41}{60}$
Total $\dfrac{5}{15}$ $\dfrac{4}{12}$ $\dfrac{2}{6}$ 1
b. $\dfrac{41}{60}$
c. $\dfrac{9}{19}$
Example $4$
Table contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.
United States Crime Index Rates Per 100,000 Inhabitants 2008–2011
Year Robbery Burglary Rape Vehicle Total
2008 145.7 732.1 29.7 314.7
2009 133.1 717.7 29.1 259.2
2010 119.3 701 27.7 239.1
2011 113.7 702.2 26.8 229.6
Total
TOTAL each column and each row. Total data = 4,520.7
1. Find $P(\text{2009 AND Robbery})$.
2. Find $P(\text{2010 AND Burglary})$.
3. Find $P(\text{2010 OR Burglary})$.
4. Find $P(\text{2011|Rape})$.
5. Find $P(\text{Vehicle|2008})$.
Answer
a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575
Exercise $3$
Table relates the weights and heights of a group of individuals participating in an observational study.
Weight/Height Tall Medium Short Totals
Obese 18 28 14
Normal 20 51 28
Underweight 12 25 9
Totals
1. Find the total for each row and column
2. Find the probability that a randomly chosen individual from this group is Tall.
3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
4. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
7. Are the events Obese and Tall independent?
Answer
Weight/Height Tall Medium Short Totals
Obese 18 28 14 60
Normal 20 51 28 99
Underweight 12 25 9 46
Totals 50 104 51 205
1. Row Totals: 60, 99, 46. Column totals: 50, 104, 51.
2. $P(\text{Tall}) = \dfrac{50}{205} = 0.244$
3. $P(\text{Obese AND Tall}) = \dfrac{18}{205} = 0.088$
4. $P(\text{Tall|Obese}) = \dfrac{18}{60} = 0.3$
5. $P(\text{Obese|Tall}) = \dfrac{18}{50} = 0.36$
6. $P(\text{Tall AND Underweight}) = \dfrac{12}{205} = 0.0585$
7. No. $P(\text{Tall})$ does not equal $P(\text{Tall|Obese})$.
Review
There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilities that have multiple dependent variables.
Use the following information to answer the next four exercises. Table shows a random sample of musicians and how they learned to play their instruments.
Gender Self-taught Studied in School Private Instruction Total
Female 12 38 22 72
Male 19 24 15 58
Total 31 62 37 130
Exercise 3.5.4
Find P(musician is a female).
Exercise 3.5.5
Find $P(\text{musician is a male AND had private instruction})$.
Answer
$P(\text{musician is a male AND had private instruction}) = \dfrac{15}{130} = \dfrac{3}{26} = 0.12$
Exercise 3.5.6
Find P(musician is a female OR is self taught).
Exercise 3.5.7
Are the events “being a female musician” and “learning music in school” mutually exclusive events?
Answer
The events are not mutually exclusive. It is possible to be a female musician who learned music in school.
Bringing it Together
Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites.
Exercise 3.5.8
Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.
Smoking Levels by Ethnicity
Smoking Level African American Native Hawaiian Latino Japanese Americans White TOTALS
1–10
11–20
21–30
31+
TOTALS
Exercise 3.5.9
Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.
Answer
$\dfrac{35,065}{100,450}$
Exercise 3.5.10
Find the probability that the person was Latino.
Exercise 3.5.11
In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability.
Answer
To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is $\dfrac{4,715}{100,450}$.
Exercise 3.5.12
In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability.
Exercise 3.5.13
In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability.
Answer
To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is $\dfrac{4,715}{15,273}$.
Exercise 3.5.14
Prove that smoking level/day and ethnicity are dependent events.
Glossary
contingency table
the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.05%3A_Contingency_Tables.txt |
Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.
Tree Diagrams
A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
Example $1$: Probabilities from Sampling with replacement
In an urn, there are 11 balls. Three balls are red ($\text{R}$) and eight balls are blue ($\text{B}$). Draw two balls, one at a time, with replacement (remember that "with replacement" means that you put the first ball back in the urn before you select the second ball). The tree diagram using frequencies that show all the possible outcomes follows.
The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:
R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3
The other outcomes are similar.
There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.
Exercise $1$
In a standard deck, there are 52 cards. 12 cards are face cards (event $\text{F}$) and 40 cards are not face cards (event $\text{N}$). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate $P(\text{FF})$.
Answer
Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704.
$P(\text{FF}) = \frac{144}{144+480+480+1,600} = \frac{144}{2,704} = \frac{9}{169}$
1. a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
2. b. Using the tree diagram, calculate P(RR).
3. c. Using the tree diagram, calculate $P(\text{RB OR BR})$.
4. d. Using the tree diagram, calculate $P(\text{R on 1st draw AND B on 2nd draw})$.
5. e. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).
6. Using the tree diagram, calculate $P(\text{BB})$.
7. g. Using the tree diagram, calculate $P(\text{B on the 2nd draw given R on the first draw})$.
Solution
1. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3 B7R1; B7R2; B7R3; B8R1; B8R2; B8R3
2. $P(\text{RR}) = \left(\frac{3}{11}\right) \left(\frac{3}{11}\right) = \frac{9}{121}$
3. $P(\text{RB OR BR}) = \left(\frac{3}{11}\right) \left(\frac{8}{11}\right) + \left(\frac{8}{11}\right) \left(\frac{3}{11}\right) = \frac{48}{121}$
4. $P(\text{R on 1st draw AND B on 2nd draw}) = P(\text{RB}) = \left(\frac{3}{11}\right) \left(\frac{8}{11}\right) = \frac{24}{121}$
5. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = $\frac{24}{88}$ = $\frac{3}{11}$ This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. $\frac{24}{88}$ = $\frac{3}{11}$
6. $P(\text{BB}) = \frac{64}{121}$
7. $P(\text{B on 2nd draw|R on 1st draw}) = \frac{8}{11}$. There are 9 + 24 outcomes that have $\text{R}$ on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have $\text{B}$ on the second draw. The probability is then $\frac{24}{33}$.
Example $2$: Probabilities from Sampling without replacement
An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. (remember that "without replacement" means that you do not put the first ball back before you select the second marble). Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, $\left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}$.
If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.
Calculate the following probabilities using the tree diagram.
1. $P(\text{RR}) =$ ________
2. Fill in the blanks: $P(\text{RB OR BR}) = \left(\frac{3}{11}\right) \left(\frac{8}{10}\right) +$ (___)(___) $= \frac{48}{110}$
3. $P(\text{R on 2nd|B on 1st}) =$
4. Fill in the blanks: $P(\text{R on 1st AND B on 2nd}) = P(\text{RB}) =$ (___)(___) $= \frac{24}{110}$
5. Find $P(\text{BB})$.
6. Find $P(\text{B on 2nd|R on 1st})$.
Answers
1. $P(\text{RR}) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}$
2. $P(\text{RB OR BR}) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{48}{110}$
3. $P(\text{R on 2nd|B on 1st}) = \frac{3}{10}$
4. $P(\text{R on 1st AND B on 2nd}) = P(\text{RB}) = \left(\frac{3}{11}\right) \left(\frac{8}{10}\right) = \frac{24}{110}$
5. $P(\text{BB}) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right)$
6. Using the tree diagram, $P(\text{B on 2nd|R on 1st}) = P(\text{R|B}) = \frac{8}{10}$.
If we are using probabilities, we can label the tree in the following general way.
• $P(\text{R|R})$ here means $P(\text{R on 2nd|R on 1st})$
• $P(\text{B|R})$ here means $P(\text{B on 2nd|R on 1st})$
• $P(\text{R|B})$ here means $P(\text{R on 2nd|B on 1st})$
• $P(\text{B|B})$ here means $P(\text{B on 2nd|B on 1st})$
Exercise $2$
In a standard deck, there are 52 cards. Twelve cards are face cards ($\text{F}$) and 40 cards are not face cards ($\text{N}$). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.
1. Find $P(\text{FN OR NF})$.
2. Find $P(\text{N|F})$.
3. Find $P(\text{at most one face card})$.
Hint: "At most one face card" means zero or one face card.
4. Find $P(\text{at least on face card})$.
Hint: "At least one face card" means one or two face cards.
Answer
1. $P(\text{FN OR NF}) = \frac{480}{2,652} + \frac{480}{2,652} = \frac{960}{2,652} = \frac{80}{221}$
2. $P(\text{N|F}) = \frac{40}{51}$
3. $P(\text{at most one face card}) = \frac{(480+480+1,560)}{2,652} = \frac{2,520}{2,652}$
4. $P(\text{at least one face card}) = \frac{(132+480+480)}{2,652} = \frac{1,092}{2,652}$
Example $3$
A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.
1. What is the probability that both kittens are tabby?
a. $\left(\frac{1}{2}\right) \left(\frac{1}{2}\right)$ b. $\left(\frac{4}{9}\right) \left(\frac{4}{9}\right)$ c. $\left(\frac{4}{9}\right) \left(\frac{3}{8}\right)$ d. $\left(\frac{4}{9}\right) \left(\frac{5}{9}\right)$
2. What is the probability that one kitten of each coloring is selected?
a. $\left(\frac{4}{9}\right) \left(\frac{5}{9}\right)$ b. $\left(\frac{4}{9}\right) \left(\frac{5}{8}\right)$ c. $\left(\frac{4}{9}\right) \left(\frac{5}{9}\right)$ + $\left(\frac{5}{9}\right) \left(\frac{4}{9}\right)$ d. $\left(\frac{4}{9}\right) \left(\frac{5}{8}\right)$ + $\left(\frac{5}{9}\right) \left(\frac{4}{8}\right)$
3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
4. What is the probability of choosing two kittens of the same color?
Answer
a. c, b. d, c. $\frac{4}{8}$, d. $\frac{32}{72}$
Exercise $3$
Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?
Answer
$\left(\frac{4}{7}\right) \left(\frac{3}{6}\right)$ + $\left(\frac{3}{7}\right) \left(\frac{4}{6}\right)$
Venn Diagram
A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.
Example $4$
Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event $\text{A} =$ {1, 2, 3, 4, 5, 6} and event $\text{B} =$ {6, 7, 8, 9}. Then $\text{A AND B} =$ {6} and $\text{A OR B} =$ {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:
Exercise $4$
Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event $\text{C} =$ {green, blue, purple} and event $\text{P} =$ {red, yellow, blue}. Then $\text{C AND P} =$ {blue} and $\text{C OR P} =$ {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.
Answer
Example $5$
Flip two fair coins. Let $\text{A} =$ tails on the first coin. Let $\text{B} =$ tails on the second coin. Then $\text{A} =$ {TT,TH} and $\text{B} =$ {TT, HT}. Therefore, $\text{A AND B} =$ {TT}. $\text{A OR B} =$ {TH, TT, HT}.
The sample space when you flip two fair coins is $X =$ {HH, HT, TH, TT}. The outcome HH is in $\text{NEITHER A NOR B}$. The Venn diagram is as follows:
Exercise $5$
Roll a fair, six-sided die. Let $\text{A} =$ a prime number of dots is rolled. Let $\text{B} =$ an odd number of dots is rolled. Then $\text{A} =$ {2, 3, 5} and $\text{B} =$ {1, 3, 5}. Therefore, $\text{A AND B} =$ {3, 5}. $\text{A OR B} =$ {1, 2, 3, 5}. The sample space for rolling a fair die is $\text{S} =$ {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.
Answer
Example $6$: Probability and Venn Diagrams
Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let $\text{C} =$ student belongs to a club and $\text{PT} =$ student works part time.
If a student is selected at random, find
• the probability that the student belongs to a club. $P(\text{C}) = 0.40$
• the probability that the student works part time. $P(\text{PT}) = 0.50$
• the probability that the student belongs to a club AND works part time. $P(\text{C AND PT}) = 0.05$
• the probability that the student belongs to a club given that the student works part time. $P(\text{C|PT}) = \frac{P(\text{C AND PT})}{P(\text{PT})} = \frac{0.05}{0.50} = 0.1$
• the probability that the student belongs to a club OR works part time. $P(\text{C OR PT}) = P(\text{C}) + P(\text{PT}) - P(\text{C AND PT}) = 0.40 + 0.50 - 0.05 = 0.85$
Exercise $6$
Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let $\text{W} =$ works a second job and $\text{S} =$ spouse also works.
Answer
Example $7$
A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.
The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.
We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let $\text{O} =$ African American with Type O blood and $\text{R} =$ African American with Rh- factor.
1. $P(\text{O}) =$ ___________
2. $P(\text{R}) =$ ___________
3. $P(\text{O AND R}) =$ ___________
4. $P(\text{O OR R}) =$ ____________
5. In the Venn Diagram, describe the overlapping area using a complete sentence.
6. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.
Answer
a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.
Exercise $7$
In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.
1. Draw a Venn diagram representing the situation.
2. Find the probability that the customer buys either a novel or anon-fiction book.
3. In the Venn diagram, describe the overlapping area using a complete sentence.
4. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.
Answer
a. and d. In the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks.
b. $P(\text{novel or non-fiction}) = P(\text{Blue OR Red}) = P(\text{Blue}) + P(\text{Red}) - P(\text{Blue AND Red}) = 0.6 + 0.4 - 0.2 = 0.8$.
c. The overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book.
Review
A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.
Glossary
Tree Diagram
the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)
Venn Diagram
the visual representation of a sample space and events in the form of circles or ovals showing their intersections | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.06%3A_Tree_and_Venn_Diagrams.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will use theoretical and empirical methods to estimate probabilities.
• The student will appraise the differences between the two estimates.
• The student will demonstrate an understanding of long-term relative frequencies.
Do the Experiment
Count out 40 mixed-color M&Ms® which is approximately one small bag’s worth. Record the number of each color in Table. Use the information from this table to complete Table. Next, put the M&Ms in a cup. The experiment is to pick two M&Ms, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of Table. Do this 24 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of Table. After you record the pick, put both M&Ms back. Do this a total of 24 times, also. Use the data from Table to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions.
Population
Color Quantity
Yellow (Y)
Green (G)
Blue (BL)
Brown (B)
Orange (O)
Red (R)
Theoretical Probabilities
With Replacement Without Replacement
P(2 reds)
P(R1B2 OR B1R2)
P(R1 AND G2)
P(G2|R1)
P(no yellows)
P(doubles)
P(no doubles)
G2 = green on second pick; R1 = red on first pick; B1 = brown on first pick; B2 = brown on second pick; doubles = both picks are the same colour.
Empirical Results
With Replacement Without Replacement
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
Empirical Probabilities
With Replacement Without Replacement
P(2 reds)
P(R1B2 OR B1R2)
P(R1 AND G2)
P(G2|R1)
P(no yellows)
P(doubles)
P(no doubles)
Discussion Questions
1. Why are the “With Replacement” and “Without Replacement” probabilities different?
2. Convert P(no yellows) to decimal format for both Theoretical “With Replacement” and for Empirical “With Replacement”. Round to four decimal places.
1. Theoretical “With Replacement”: P(no yellows) = _______
2. Empirical “With Replacement”: P(no yellows) = _______
3. Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why?
3. If you increased the number of times you picked two M&Ms to 240 times, why would empirical probability values change?
4. Would this change (see part 3) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know?
5. Explain the differences in what P(G1 AND R2) and P(R1|G2) represent. Hint: Think about the sample space for each probability. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.07%3A_Probability_Topics_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
3.2: Terminology
Q 3.2.1
The graph in Figure 3.2.1 displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045.
1. Define three events in the graph.
2. Describe in words what the entry 40 means.
3. Describe in words the complement of the entry in question 2.
4. Describe in words what the entry 30 means.
5. Out of the males and females, what percent are males?
6. Out of the females, what percent disapprove of Mayor Ford?
7. Out of all the age groups, what percent approve of Mayor Ford?
8. Find $P(\text{Approve|Male})$.
9. Out of the age groups, what percent are more than 44 years old?
10. Find $P(\text{Approve|Age} < 35)$.
Q 3.2.2
Explain what is wrong with the following statements. Use complete sentences.
1. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend.
2. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit.
S 3.2.2
1. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%
2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.
3.3: Independent and Mutually Exclusive Events
Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews done by Gallup that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score.
Q 3.3.1
Find the probability that an Emotional Health Index Score is 82.7.
Q 3.3.2
Find the probability that an Emotional Health Index Score is 81.0.
0
Q 3.3.3
Find the probability that an Emotional Health Index Score is more than 81?
Q 3.3.4
Find the probability that an Emotional Health Index Score is between 80.5 and 82?
0.3571
Q 3.3.5
If we know an Emotional Health Index Score is 81.5 or more, what is the probability that it is 82.7?
Q 3.3.6
What is the probability that an Emotional Health Index Score is 80.7 or 82.7?
0.2142
Q 3.3.7
What is the probability that an Emotional Health Index Score is less than 80.2 given that it is already less than 81.
Q 3.3.8
What occupation has the highest emotional index score?
Physician (83.7)
Q 3.3.9
What occupation has the lowest emotional index score?
Q 3.3.10
What is the range of the data?
S 3.3.10
83.7 − 79.6 = 4.1
Q 3.3.11
Compute the average EHIS.
Q 3.3.12
If all occupations are equally likely for a certain individual, what is the probability that he or she will have an occupation with lower than average EHIS?
S 3.3.12
$P(\text{Occupation} < 81.3) = 0.5$
3.4: Two Basic Rules of Probability
Q 3.4.1
On February 28, 2013, a Field Poll Survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (California registered voters), the approval rating was 78%. Six in ten California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of California’s Proposition 8 was either very or somewhat important to them. Out of those CA registered voters who support same-sex marriage, 75% say the ruling is important to them.
In this problem, let:
• $\text{C} =$ California registered voters who support same-sex marriage.
• $\text{B} =$ California registered voters who say the Supreme Court’s ruling about the constitutionality of California’s Proposition 8 is very or somewhat important to them
• $\text{A} =$ California registered voters who are 18 to 39 years old.
1. Find $P(\text{C})$.
2. Find $P(\text{B})$.
3. Find $P(\text{C|A})$.
4. Find $P(\text{B|C})$.
5. In words, what is $\text{C|A}$?
6. In words, what is $\text{B|C}$?
7. Find $P(\text{C AND B})$.
8. In words, what is $\text{C AND B}$?
9. Find $P(\text{C OR B})$.
10. Are $\text{C}$ and $\text{B}$ mutually exclusive events? Show why or why not.
Q 3.4.2
After Rob Ford, the mayor of Toronto, announced his plans to cut budget costs in late 2011, the Forum Research polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results their poll produced:
• In early 2011, 60 percent of the population approved of Mayor Ford’s actions in office.
• In mid-2011, 57 percent of the population approved of his actions.
• In late 2011, the percentage of popular approval was measured at 42 percent.
1. What is the sample size for this study?
2. What proportion in the poll disapproved of Mayor Ford, according to the results from late 2011?
3. How many people polled responded that they approved of Mayor Ford in late 2011?
4. What is the probability that a person supported Mayor Ford, based on the data collected in mid-2011?
5. What is the probability that a person supported Mayor Ford, based on the data collected in early 2011?
S 3.4.2
1. The Forum Research surveyed 1,046 Torontonians.
2. 58%
3. 42% of 1,046 = 439 (rounding to the nearest integer)
4. 0.57
5. 0.60.
Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each number is assigned to a color and a range.
Q 3.4.3
1. List the sample space of the 38 possible outcomes in roulette.
2. You bet on red. Find $P(\text{red})$.
3. You bet on -1st 12- (1st Dozen). Find $P(\text{-1st 12-})$.
4. You bet on an even number. Find $P(\text{even number})$.
5. Is getting an odd number the complement of getting an even number? Why?
6. Find two mutually exclusive events.
7. Are the events Even and 1st Dozen independent?
Q 3.4.4
Compute the probability of winning the following types of bets:
1. Betting on two lines that touch each other on the table as in 1-2-3-4-5-6
2. Betting on three numbers in a line, as in 1-2-3
3. Betting on one number
4. Betting on four numbers that touch each other to form a square, as in 10-11-13-14
5. Betting on two numbers that touch each other on the table, as in 10-11 or 10-13
6. Betting on 0-00-1-2-3
7. Betting on 0-1-2; or 0-00-2; or 00-2-3
S 3.4.4
1. $P(\text{Betting on two line that touch each other on the table}) = \frac{6}{38}$
2. $P(\text{Betting on three numbers in a line}) = \frac{3}{38}$
3. $P(\text{Betting on one number}) = \frac{1}{38}$
4. $P(\text{Betting on four number that touch each other to form a square}) = \frac{4}{38}$
5. $P(\text{Betting on two number that touch each other on the table}) = \frac{2}{38}$
6. $P(\text{Betting on 0-00-1-2-3)} = \frac{5}{38}$
7. $P(\text{Betting on 0-1-2; or 0-00-2; or 00-2-3}) = \frac{3}{38}$
Q 3.4.5
Compute the probability of winning the following types of bets:
1. Betting on a color
2. Betting on one of the dozen groups
3. Betting on the range of numbers from 1 to 18
4. Betting on the range of numbers 19–36
5. Betting on one of the columns
6. Betting on an even or odd number (excluding zero)
Q 3.4.6
Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
• $\text{G} =$ card drawn is green
• $\text{E} =$ card drawn is even-numbered
1. List the sample space.
2. $P(\text{G}) =$ _____
3. $P(\text{G|E}) =$ _____
4. $P(\text{G AND E}) =$ _____
5. $P(\text{G OR E}) =$ _____
6. Are $\text{G}$ and $\text{E}$ mutually exclusive? Justify your answer numerically.
S 3.4.6
1. $\{G1, G2, G3, G4, G5, Y1, Y2, Y3\}$
2. $\frac{5}{8}$
3. $\frac{2}{3}$
4. $\frac{2}{8}$
5. $\frac{}6{8}$
6. No, because $P(\text{G AND E})$ does not equal 0.
Q 3.4.7
Roll two fair dice. Each die has six faces.
1. List the sample space.
2. Let $\text{A}$ be the event that either a three or four is rolled first, followed by an even number. Find $P(\text{A})$.
3. Let B be the event that the sum of the two rolls is at most seven. Find $P(\text{B})$.
4. In words, explain what “$P(\text{A|B})$” represents. Find $P(\text{A|B})$.
5. Are $\text{A}$ and $\text{B}$ mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification.
6. Are $\text{A}$ and $\text{B}$ independent events? Explain your answer in one to three complete sentences, including numerical justification.
Q 3.4.8
A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin.
1. List the sample space.
2. Let $\text{A}$ be the event that a blue card is picked first, followed by landing a head on the coin toss. Find $P(\text{A})$.
3. Let $\text{B}$ be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events $\text{A}$ and $\text{B}$ mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.
4. Let $\text{C}$ be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events $\text{A}$ and $\text{C}$ mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.
S 3.4.9
The coin toss is independent of the card picked first.
1. $\{(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)\}$
2. $P(\text{A}) = P(\text{blue})P(\text{head}) = \left(\frac{3}{10}\right)\left(\frac{1}{2}\right) = \frac{3}{20}$
3. Yes, $\text{A}$ and $\text{B}$ are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). $P(\text{A AND B}) = 0$
4. No, $\text{A}$ and $\text{C}$ are not mutually exclusive because they can occur at the same time. In fact, $\text{C}$ includes all of the outcomes of $\text{A}$; if the card chosen is blue it is also (red or blue). $P(\text{A AND C}) = P(\text{A}) = 320$
Q 3.4.10
An experiment consists of first rolling a die and then tossing a coin.
1. List the sample space.
2. Let $\text{A}$ be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find $P(\text{A})$.
3. Let $\text{B}$ be the event that the first and second tosses land on heads. Are the events $\text{A}$ and $\text{B}$ mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.
Q 3.4.11
An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on.
1. List the sample space.
2. Let $\text{A}$ be the event that there are at least two tails. Find $P(\text{A})$.
3. Let $\text{B}$ be the event that the first and second tosses land on heads. Are the events $\text{A}$ and $\text{B}$ mutually exclusive? Explain your answer in one to three complete sentences, including justification.
S 3.4.12
1. $S = {\text{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}}$
2. $\frac{4}{8}$
3. Yes, because if $\text{A}$ has occurred, it is impossible to obtain two tails. In other words, $P(\text{A AND B}) = 0$.
Q 3.4.13
Consider the following scenario:
Let $P(\text{C}) = 0.4$.
Let $P(\text{D}) = 0.5$.
Let $P(\text{C|D}) = 0.6$.
1. Find $P(C AND D)$.
2. Are $\text{C}$ and $\text{D}$ mutually exclusive? Why or why not?
3. Are $\text{C}$ and $\text{D}$ independent events? Why or why not?
4. Find $P(\text{C OR D})$.
5. Find $P(\text{D|C})$.
Q 3.4.14
$\text{Y}$ and $\text{Z}$ are independent events.
1. Rewrite the basic Addition Rule $P(\text{Y OR Z}) = P(\text{Y}) + P(\text{Z}) - P(\text{Y AND Z})$ using the information that $\text{Y}$ and $\text{Z}$ are independent events.
2. Use the rewritten rule to find $P(\text{Z})$ if $P(\text{Y OR Z}) = 0.71$ and $P(\text{Y}) = 0.42$.
S 3.4.14
1. If $\text{Y}$ and $\text{Z}$ are independent, then $P(\text{Y AND Z}) = P(\text{Y})P(\text{Z})$, so $P(\text{Y OR Z}) = P(\text{Y}) + P(\text{Z}) -P(\text{Y})P(\text{Z})$.
2. 0.5
Q 3.4.15
$\text{G}$ and $\text{H}$ are mutually exclusive events. $P(\text{G}) = 0.5 P(\text{H}) = 0.3$
1. Explain why the following statement MUST be false: $P(\text{H|G}) = 0.4$.
2. Find $P(\text{H OR G})$.
3. Are $\text{G}$ and $\text{H}$ independent or dependent events? Explain in a complete sentence.
Q 3.4.16
Approximately 281,000,000 people over age five live in the United States. Of these people, 55,000,000 speak a language other than English at home. Of those who speak another language at home, 62.3% speak Spanish.
Let: $\text{E} =$ speaks English at home; $\text{E′} =$ speaks another language at home; $\text{S} =$ speaks Spanish;
Finish each probability statement by matching the correct answer.
Probability Statements Answers
a. $P(\text{E′}) =$ i. 0.8043
b. $P(\text{E}) =$ ii. 0.623
c. $P(\text{S and E′}) =$ iii. 0.1957
d. $P(\text{S|E′}) =$ iv. 0.1219
1. iii
2. i
3. iv
4. ii
Q 3.4.17
1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let $G =$ won green card.
1. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement.
2. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Write your answer as a conditional probability statement. Let $\text{F} =$ was a finalist.
3. Are $\text{G}$ and $\text{F}$ independent or dependent events? Justify your answer numerically and also explain why.
4. Are $\text{G}$ and $\text{F}$ mutually exclusive events? Justify your answer numerically and explain why.
Q 3.4.18
Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with \$10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned.
Let: $\text{R} =$ money returned; $\text{E} =$ economics classes; $\text{O} =$ other classes
1. Write a probability statement for the overall percent of money returned.
2. Write a probability statement for the percent of money returned out of the economics classes.
3. Write a probability statement for the percent of money returned out of the other classes.
4. Is money being returned independent of the class? Justify your answer numerically and explain it.
5. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer.
S 3.4.18
1. $P(\text{R}) = 0.44$
2. $P(\text{R|E}) = 0.56$
3. $P(\text{R|O}) = 0.31$
4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; $P(\text{R|E}) \neq P(\text{R})$.
5. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; $P(\text{R|E}) > P(\text{R})$.
Q 3.4.19
The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected.
Name Single Double Triple Home Run Total Hits
Babe Ruth 1,517 506 136 714 2,873
Jackie Robinson 1,054 273 54 137 1,518
Ty Cobb 3,603 174 295 114 4,189
Hank Aaron 2,294 624 98 755 3,771
Total 8,471 1,577 583 1,720 12,351
Are "the hit being made by Hank Aaron" and "the hit being a double" independent events?
1. Yes, because $P(\text{hit by Hank Aaron|hit is a double}) = P(\text{hit by Hank Aaron})$
2. No, because $P(\text{hit by Hank Aaron|hit is a double}) \neq P(\text{hit is a double})$
3. No, because $P(\text{hit is by Hank Aaron|hit is a double}) \neq P(\text{hit by Hank Aaron})$
4. Yes, because $P(\text{hit is by Hank Aaron|hit is a double}) = P(\text{hit is a double})$
Q 3.4.29
United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Their data show that 43% of people have type O blood and 15% of people have Rh- factor; 52% of people have type O or Rh- factor.
1. Find the probability that a person has both type O blood and the Rh- factor.
2. Find the probability that a person does NOT have both type O blood and the Rh- factor.
S 3.4.30
1. $P(\text{type O OR Rh-}) = P(\text{type O}) + P(\text{Rh-}) - P(\text{type O AND Rh-})$
$0.52 = 0.43 + 0.15 - P(\text{type O AND Rh-})$; solve to find $P(\text{type O AND Rh-}) = 0.06$
6% of people have type O, Rh- blood
2. $P(\text{NOT(type O AND Rh-)}) = 1 - P(\text{type O AND Rh-}) = 1 - 0.06 = 0.94$
94% of people do not have type O, Rh- blood
Q 3.4.31
At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.
1. Find the probability that a course has a final exam or a research project.
2. Find the probability that a course has NEITHER of these two requirements.
Q 3.4.32
In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.
1. Find the probability that a cookie contains chocolate or nuts (he can't eat it).
2. Find the probability that a cookie does not contain chocolate or nuts (he can eat it).
S 3.4.32
1. Let $C =$ be the event that the cookie contains chocolate. Let $N =$ the event that the cookie contains nuts.
2. $P(\text{C OR N}) = P(\text{C}) + P(\text{N}) - P(\text{C AND N}) = 0.36 + 0.12 - 0.08 = 0.40$
3. $P(\text{NEITHER chocolate NOR nuts}) = 1 - P(\text{C OR N}) = 1 - 0.40 = 0.60$
Q 3.4.33
A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let $\text{D} =$ event that a student takes a distance learning class and $\text{E} =$ event that a student is a part time student
1. Find $P(\text{D AND E})$.
2. Find $P(\text{E|D})$.
3. Find $P(\text{D OR E})$.
4. Using an appropriate test, show whether $\text{D}$ and $\text{E}$ are independent.
5. Using an appropriate test, show whether $\text{D}$ and $\text{E}$ are mutually exclusive.
3.5: Contingency Tables
Use the information in the Table to answer the next eight exercises. The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection.
Up for reelection: Democratic Party Republican Party Other Total
November 2014 20 13 0
November 2016 10 24 0
Total
Q 3.5.1
What is the probability that a randomly selected senator has an “Other” affiliation?
0
Q 3.5.2
What is the probability that a randomly selected senator is up for reelection in November 2016?
Q 3.5.3
What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016?
S 3.5.3
$\frac{10}{67}$
Q 3.5.4
What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014?
Q 3.5.5
Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat?
S 3.5.5
$\frac{10}{34}$
Q 3.5.6
Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican?
Q 3.5.7
The events “Republican” and “Up for reelection in 2016” are ________
1. mutually exclusive.
2. independent.
3. both mutually exclusive and independent.
4. neither mutually exclusive nor independent.
d
Q 3.5.8
The events “Other” and “Up for reelection in November 2016” are ________
1. mutually exclusive.
2. independent.
3. both mutually exclusive and independent.
4. neither mutually exclusive nor independent.
Q 3.5.9
This table gives the number of participants in the recent National Health Interview Survey who had been treated for cancer in the previous 12 months. The results are sorted by age, race (black or white), and sex. We are interested in possible relationships between age, race, and sex.
Race and Sex 15-24 25-40 41-65 over 65 TOTALS
white, male 1,165 2,036 3,703 8,395
white, female 1,076 2,242 4,060 9,129
black, male 142 194 384 824
black, female 131 290 486 1,061
all others
TOTALS 2,792 5,279 9,354 21,081
Do not include "all others" for parts f and g.
1. Fill in the column for cancer treatment for individuals over age 65.
2. Fill in the row for all other races.
3. Find the probability that a randomly selected individual was a white male.
4. Find the probability that a randomly selected individual was a black female.
5. Find the probability that a randomly selected individual was black
6. Find the probability that a randomly selected individual was a black or white male.
7. Out of the individuals over age 65, find the probability that a randomly selected individual was a black or white male.
S 3.5.9
1. Race and Sex 1–14 15–24 25–64 over 64 TOTALS
white, male 210 3,360 13,610 4,870 22,050
white, female 80 580 3,380 890 4,930
black, male 10 460 1,060 140 1,670
black, female 0 40 270 20 330
all others 100
TOTALS 310 4,650 18,780 6,020 29,760
2. Race and Sex 1–14 15–24 25–64 over 64 TOTALS
white, male 210 3,360 13,610 4,870 22,050
white, female 80 580 3,380 890 4,930
black, male 10 460 1,060 140 1,670
black, female 0 40 270 20 330
all others 10 210 460 100 780
TOTALS 310 4,650 18,780 6,020 29,760
3. $\frac{22,050}{29,760}$
4. $\frac{330}{29,760}$
5. $\frac{2,000}{29,760}$
6. $\frac{23,720}{29,760}$
7. $\frac{5,010}{6,020}$
Use the following information to answer the next two exercises. The table of data obtained fromwww.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected.
NAME Single Double Triple Home Run TOTAL HITS
Babe Ruth 1,517 506 136 714 2,873
Jackie Robinson 1,054 273 54 137 1,518
Ty Cobb 3,603 174 295 114 4,189
Hank Aaron 2,294 624 98 755 3,771
TOTAL 8,471 1,577 583 1,720 12,351
Q 3.5.10
Find $P(\text{hit was made by Babe Ruth})$.
1. $\frac{1518}{2873}$
2. $\frac{2873}{12351}$
3. $\frac{583}{12351}$
4. $\frac{4189}{12351}$
Q 3.5.11
Find $P(\text{hit was made by Ty Cobb|The hit was a Home Run})$.
1. $\frac{4189}{12351}$
2. $\frac{114}{1720}$
3. $\frac{1720}{4189}$
4. $\frac{114}{12351}$
b
Q 3.5.12
Table identifies a group of children by one of four hair colors, and by type of hair.
Hair Type Brown Blond Black Red Totals
Wavy 20 15 3 43
Straight 80 15 12
Totals 20 215
1. Complete the table.
2. What is the probability that a randomly selected child will have wavy hair?
3. What is the probability that a randomly selected child will have either brown or blond hair?
4. What is the probability that a randomly selected child will have wavy brown hair?
5. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?
6. If $\text{B}$ is the event of a child having brown hair, find the probability of the complement of $\text{B}$.
7. In words, what does the complement of $\text{B}$ represent?
Q 3.5.13
In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.
Shirt# ≤ 210 211–250 251–290 > 290
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5
For the following, suppose that you randomly select one player from the 49ers or Cowboys.
1. Find the probability that his shirt number is from 1 to 33.
2. Find the probability that he weighs at most 210 pounds.
3. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.
4. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.
5. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.
S 3.5.13
1. $\frac{26}{106}$
2. $\frac{33}{106}$
3. $\frac{21}{106}$
4. $\left(\frac{26}{106}\right) + \left(\frac{33}{106}\right) - \left(\frac{21}{106}\right) = \left(\frac{38}{106}\right)$
5. $\frac{21}{33}$
3.6: Tree and Venn Diagrams
Exercise 3.6.8
The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: $\text{C} =$ a man develops cancer in his lifetime; $\text{P} =$ man has at least one false positive. Construct a tree diagram of the situation.
Answer
Bring It Together
Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.
Exercise 3.6.9
Suppose that you randomly draw two cards, one at a time, with replacement.
Let $\text{G}_{1} =$ first card is green
Let $\text{G}_{2} =$ second card is green
1. Draw a tree diagram of the situation.
2. Find $P(\text{G}_{1} \text{AND} \text{G}_{2})$.
3. Find $P(\text{at least one green})$.
4. Find $P(\text{G}_{2}|\text{G}_{1})$.
5. Are $\text{G}_{1}$ and $\text{G}_{2}$ independent events? Explain why or why not.
Answer
1. $P(\text{GG}) = \left(\frac{5}{8}\right)\left(\frac{5}{8}\right) = \frac{25}{64}$
2. $P(\text{at least one green}) = P(\text{GG}) + P(\text{GY}) + P(\text{YG}) = \frac{25}{64} + \frac{15}{64} + \frac{15}{64} = \frac{55}{64}$
3. $P(\text{G|G}) = \frac{5}{8}$
4. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.
Exercise 3.6.10
Suppose that you randomly draw two cards, one at a time, without replacement.
$\text{G}_{1} =$ first card is green
$\text{G}_{2} =$ second card is green
1. Draw a tree diagram of the situation.
2. Find $P(\text{G}_{1} \text{AND G}_{2})$.
3. Find $P(\text{at least one green})$.
4. Find $P(\text{G}_{2}|\text{G}_{1})$.
5. Are $\text{G}_{2}$ and $\text{G}_{1}$ independent events? Explain why or why not.
Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over.
Exercise 3.6.11
Complete the following.
1. Construct a table or a tree diagram of the situation.
2. Find $P(\text{driver is female})$.
3. Find $P(\text{driver is age 65 or over|driver is female})$.
4. Find $P(\text{driver is age 65 or over AND female})$.
5. In words, explain the difference between the probabilities in part c and part d.
6. Find $P(\text{driver is age 65 or over})$.
7. Are being age 65 or over and being female mutually exclusive events? How do you know?
Answer
1. <20 20–64 >64 Totals
Female 0.0244 0.3954 0.0661 0.486
Male 0.0259 0.4186 0.0695 0.514
Totals 0.0503 0.8140 0.1356 1
2. $P(\text{F}) = 0.486$
3. $P(\text{>64|F}) = 0.1361$
4. $P(\text{>64 and F}) = P(\text{F}) P(\text{>64|F}) = (0.486)(0.1361) = 0.0661$
5. $P(\text{>64|F})$ is the percentage of female drivers who are 65 or older and $P(\text{>64 and F})$ is the percentage of drivers who are female and 65 or older.
6. $P(\text{>64}) = P(\text{>64 and F}) + P(\text{>64 and M}) = 0.1356$
7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time $P(\text{>64 and F}) = 0.0661$.
Exercise 3.6.12
Suppose that 10,000 U.S. licensed drivers are randomly selected.
1. How many would you expect to be male?
2. Using the table or tree diagram, construct a contingency table of gender versus age group.
3. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.
Exercise 3.6.13
Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation.
1. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.
2. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?
3. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work?
4. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool?
Answer
1. Car, Truck or Van Walk Public Transportation Other Totals
Alone 0.7318
Not Alone 0.1332
Totals 0.8650 0.0390 0.0530 0.0430 1
2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: $P(\text{Alone}) = 0.7318 + 0.0390 = 0.7708$.
3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771
4. (0.1332)(1,000) = 133
Exercise 3.6.14
When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event $\text{H}$) while 110 showed a tail (event $\text{T}$). On that basis, they claimed that it is not a fair coin.
1. Based on the given data, find $P(\text{H})$ and $P(\text{T})$.
2. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.
3. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.
4. Use the tree to find the probability of obtaining at least one head.
Exercise 3.6.15
Use the following information to answer the next two exercises. The following are real data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059 documented cases of AIDS in the county. They were grouped into the following categories:
* includes homosexual/bisexual IV drug users
Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals
Female 0 70 136 49 ____
Male 2,146 463 60 135 ____
Totals ____ ____ ____ ____ ____
Suppose a person with AIDS in Santa Clara County is randomly selected.
1. Find $P(\text{Person is female})$.
2. Find $P(\text{Person has a risk factor heterosexual contact})$.
3. Find $P(\text{Person is female OR has a risk factor of IV drug user})$.
4. Find $P(\text{Person is female AND has a risk factor of homosexual/bisexual})$.
5. Find $P(\text{Person is male AND has a risk factor of IV drug user})$.
6. Find $P(\text{Person is female GIVEN person got the disease from heterosexual contact})$.
7. Construct a Venn diagram. Make one group females and the other group heterosexual contact.
Answer
The completed contingency table is as follows:
* includes homosexual/bisexual IV drug users
Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals
Female 0 70 136 49 255
Male 2,146 463 60 135 2,804
Totals 2,146 533 196 184 3,059
1. $\frac{255}{2059}$
2. $\frac{196}{3059}$
3. $\frac{718}{3059}$
4. 0
5. $\frac{463}{3059}$
6. $\frac{136}{196}$
Exercise 3.6.16
Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact.
1. Find $P(\text{Person is female})$.
2. Find $P(\text{Person obtained the disease through heterosexual contact})$.
3. Find $P(\text{Person is female GIVEN person got the disease from heterosexual contact})$
4. Construct a Venn diagram representing this situation. Make one group females and the other group heterosexual contact. Fill in all values as probabilities.
Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red $(\text{R})$, four yellow ($\text{Y}$) and five blue ($\text{B}$) beads. For the coin, $P(\text{H}) = \frac{2}{3}$ and $P(\text{T}) = \frac{1}{3}$ where $\text{H}$ is heads and $\text{T}$ is tails.
Q 3.6.1
Find $P(\text{tossing a Head on the coin AND a Red bead})$
1. $\frac{2}{3}$
2. $\frac{5}{15}$
3. $\frac{6}{36}$
4. $\frac{5}{36}$
Q 3.6.2
Find $P(\text{Blue bead})$.
1. $\frac{15}{36}$
2. $\frac{10}{36}$
3. $\frac{10}{12}$
4. $\frac{6}{36}$
a
Q 3.6.3
A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?)
1. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
2. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain.
3. For each complete path through the tree, write the event it represents and find the probabilities.
4. Let $\text{S}$ be the event that both cookies selected were the same flavor. Find $P(\text{S})$.
5. Let $\text{T}$ be the event that the cookies selected were different flavors. Find $P(\text{T})$ by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.
6. Let $\text{U}$ be the event that the second cookie selected is a butter cookie. Find $P(\text{U})$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/03%3A_Probability_Topics/3.E%3A_Probability_Topics_%28Exericses%29.txt |
• 4.1: Prelude to Discrete Random Variables
Random Variable (RV) a characteristic of interest in a population being studied
• 4.2: Probability Distribution Function (PDF) for a Discrete Random Variable
A discrete probability distribution function has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one.
• 4.3: Mean or Expected Value and Standard Deviation
The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μμ . In other words, after conducting many trials of an experiment, you would expect this average value.
• 4.4: Binomial Distribution
A statistical experiment can be classified as a binomial experiment if the following conditions are met: (1) There are a fixed number of trials. (2)There are only two possible outcomes: "success" or "failure" for each trial. (3) The trials are independent and are repeated using identical conditions. The outcomes of a binomial experiment fit a binomial probability distribution.
• 4.5: Geometric Distribution
There are three characteristics of a geometric experiment: (1) There are one or more Bernoulli trials with all failures except the last one, which is a success. (2) In theory, the number of trials could go on forever. There must be at least one trial. (3) The probability, p, of a success and the probability, q, of a failure are the same for each trial. In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success.
• 4.6: Hypergeometric Distribution
A hypergeometric experiment is a statistical experiment with the following properties: You take samples from two groups. You are concerned with a group of interest, called the first group. You sample without replacement from the combined groups. Each pick is not independent, since sampling is without replacement. You are not dealing with Bernoulli Trials. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution.
• 4.7: Poisson Distribution
A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is "small" (less than or equal to 0.05) and the number of trials is "large" (greater than or equal to 20).
• 4.8: Discrete Distribution (Playing Card Experiment)
A statistics Worksheet: The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution. The student will demonstrate an understanding of long-term probabilities.
• 4.9: Discrete Distribution (Lucky Dice Experiment)
A statistics Worksheet: The student will compare empirical data and a theoretical distribution to determine if a Tet gambling game fits a discrete distribution. The student will demonstrate an understanding of long-term probabilities.
• 4.E: Discrete Random Variables (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
04: Discrete Random Variables
CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Recognize and understand discrete probability distribution functions, in general.
• Calculate and interpret expected values.
• Recognize the binomial probability distribution and apply it appropriately.
• Recognize the Poisson probability distribution and apply it appropriately.
• Recognize the geometric probability distribution and apply it appropriately.
• Recognize the hypergeometric probability distribution and apply it appropriately.
• Classify discrete word problems by their distributions.
• A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, he or she could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%?
• Small companies might be interested in the number of long-distance phone calls their employees make during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees make more than 20 long-distance phone calls during the peak time?
These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment.
Random Variable Notation
Upper case letters such as \(X\) or \(Y\) denote a random variable. Lower case letters like \(x\) or \(y\) denote the value of a random variable. If \(X\) is a random variable, then \(X\) is written in words, and x is given as a number.
For example, let \(X =\) the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT; THH; HTH; HHT; HTT; THT; TTH; HHH. Then, \(x =\) 0, 1, 2, 3. \(X\) is in words and x is a number. Notice that for this example, the \(x\) values are countable outcomes. Because you can count the possible values that \(X\) can take on and the outcomes are random (the x values 0, 1, 2, 3), \(X\) is a discrete random variable.
Collaborative Exercise
Toss a coin ten times and record the number of heads. After all members of the class have completed the experiment (tossed a coin ten times and counted the number of heads), fill in Table. Let \(X =\) the number of heads in ten tosses of the coin.
\(x\) Frequency of \(x\) Relative Frequency of \(x\)
1. Which value(s) of \(x\) occurred most frequently?
2. If you tossed the coin 1,000 times, what values could \(x\) take on? Which value(s) of \(x\) do you think would occur most frequently?
3. What does the relative frequency column sum to?
Glossary
Random Variable (RV)
a characteristic of interest in a population being studied; common notation for variables are upper case Latin letters \(X, Y, Z\),...; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters \(x\), \(y\), and \(z\). For example, if \(X\) is the number of children in a family, then \(x\) represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in the two following ways.
• The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if \(X =\) hair color then the domain is {black, blond, gray, green, orange}.
• We can tell what specific value \(x\) the random variable \(X\) takes only after performing the experiment. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.01%3A_Prelude_to_Discrete_Random_Variables.txt |
A discrete probability distribution function has two characteristics:
1. Each probability is between zero and one, inclusive.
2. The sum of the probabilities is one.
Example \(1\)
A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let \(X =\) the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, \(x = 0, 1, 2, 3, 4, 5\).
\(P(x) =\) probability that \(X\) takes on a value \(x\).
\(x\) \(P(x)\)
0 \(P(x = 0) = \dfrac{2}{50}\)
1 \(P(x = 1) = \dfrac{11}{50}\)
2 \(P(x = 2) = \dfrac{23}{50}\)
3 \(P(x = 3) = \dfrac{9}{50}\)
4 \(P(x = 4) = \dfrac{4}{50}\)
5 \(P(x = 5) = \dfrac{1}{50}\)
\(X\) takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because:
1. Each \(P(x)\) is between zero and one, inclusive.
2. The sum of the probabilities is one, that is,
\[\dfrac{2}{50} + \dfrac{11}{50} + \dfrac{23}{50} + \dfrac{9}{50} + \dfrac{4}{50} + \dfrac{1}{50} = 1\]
Exercise \(1\)
A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let \(X =\) the number of times a patient rings the nurse during a 12-hour shift. For this exercise, \(x = 0, 1, 2, 3, 4, 5\). \(P(x) =\) the probability that \(X\) takes on value \(x\). Why is this a discrete probability distribution function (two reasons)?
\(X\) \(P(x)\)
0 \(P(x = 0) = \dfrac{4}{50}\)
1 \(P(x = 1) = \dfrac{8}{50}\)
2 \(P(x = 2) = \dfrac{16}{50}\)
3 \(P(x = 3) = \dfrac{14}{50}\)
4 \(P(x = 4) = \dfrac{6}{50}\)
5 \(P(x = 5) = \dfrac{2}{50}\)
Answer
Each \(P(x)\) is between 0 and 1, inclusive, and the sum of the probabilities is 1, that is:
\[\dfrac{4}{50} + \dfrac{8}{50} +\dfrac{16}{50} +\dfrac{14}{50} +\dfrac{6}{50} + \dfrac{2}{50} = 1\]
Example \(2\)
Suppose Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected.
1. Let \(X\) = the number of days Nancy ____________________.
2. \(X\) takes on what values?
3. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example. The table should have two columns labeled \(x\) and \(P(x)\). What does the \(P(x)\) column sum to?
Solutions
a. Let \(X\) = the number of days Nancy attends class per week.
b. 0, 1, 2, and 3
c
\(x\) \(P(x)\)
0 0.01
1 0.04
2 0.15
3 0.80
Exercise \(2\)
Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on?
Answer
\(X\) is the number of days Jeremiah attends basketball practice per week. X takes on the values 0, 1, and 2.
Review
The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows:
1. Each probability is between zero and one, inclusive (inclusive means to include zero and one).
2. The sum of the probabilities is one.
Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution.
Let \(X =\) the number of years a new hire will stay with the company.
Let \(P(x) =\) the probability that a new hire will stay with the company x years.
Exercise 4.2.3
Complete Table using the data provided.
\(x\) \(P(x)\)
0 0.12
1 0.18
2 0.30
3 0.15
4
5 0.10
6 0.05
Answer
\(x\) \(P(x)\)
0 0.12
1 0.18
2 0.30
3 0.15
4 0.10
5 0.10
6 0.05
Exercise 4.2.4
\(P(x = 4) =\) _______
Exercise 4.2.5
\(P(x \geq 5) =\) _______
Answer
0.10 + 0.05 = 0.15
Exercise 4.2.6
On average, how long would you expect a new hire to stay with the company?
Exercise 4.2.7
What does the column “P(x)” sum to?
Answer
1
Use the following information to answer the next six exercises: A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution.
\(x\) \(P(x)\)
1 0.15
2 0.35
3 0.40
4 0.10
Exercise 4.2.8
Define the random variable \(X\).
Exercise 4.2.9
What is the probability the baker will sell more than one batch? \(P(x > 1) =\) _______
Answer
0.35 + 0.40 + 0.10 = 0.85
Exercise 4.2.10
What is the probability the baker will sell exactly one batch? \(P(x = 1) =\) _______
Exercise 4.2.11
On average, how many batches should the baker make?
Answer
1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45
Use the following information to answer the next four exercises: Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random.
Exercise 4.2.12
Define the random variable \(X\).
Exercise 4.2.13
Construct a probability distribution table for the data.
Answer
\(x\) \(P(x)\)
0 0.03
1 0.04
2 0.08
3 0.85
Exercise 4.2.14
We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic?
Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time.
Exercise 4.2.15
Define the random variable \(X\).
Answer
Let \(X =\) the number of events Javier volunteers for each month.
Exercise 4.2.16
What values does \(x\) take on?
Exercise 4.2.17
Construct a PDF table.
Answer
\(x\) \(P(x)\)
0 0.05
1 0.05
2 0.10
3 0.20
4 0.25
5 0.35
Exercise 4.2.18
Find the probability that Javier volunteers for less than three events each month. \(P(x < 3) =\) _______
Exercise 4.2.19
Find the probability that Javier volunteers for at least one event each month. \(P(x > 0) =\) _______
Answer
1 – 0.05 = 0.95
Glossary
Probability Distribution Function (PDF)
a mathematical description of a discrete random variable (RV), given either in the form of an equation (formula) or in the form of a table listing all the possible outcomes of an experiment and the probability associated with each outcome. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.02%3A_Probability_Distribution_Function_%28PDF%29_for_a_Discrete_Random_Variable.txt |
The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.
You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. As you learned in Chapter 3, probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers.
The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter $\mu$. In other words, after conducting many trials of an experiment, you would expect this average value.
To find the expected value or long term average, $\mu$, simply multiply each value of the random variable by its probability and add the products.
Example $1$
A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, $\mu$, of the number of days per week the men's soccer team plays soccer.
Solution
To do the problem, first let the random variable $X =$ the number of days the men's soccer team plays soccer per week. $X$ takes on the values 0, 1, 2. Construct a PDF table adding a column $x*P(x)$. In this column, you will multiply each $x$ value by its probability.
Expected Value Table This table is called an expected value table. The table helps you calculate the expected value or long-term average.
$x$ $P(x)$ $x*P(x)$
0 0.2 (0)(0.2) = 0
1 0.5 (1)(0.5) = 0.5
2 0.3 (2)(0.3) = 0.6
Add the last column $x*P(x)$ to find the long term average or expected value:
$(0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1. \nonumber$
The expected value is 1.1. The men's soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week. We say $\mu = 1.1$.
Example $2$
Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times per week a newborn baby's crying wakes its mother after midnight. Calculate the standard deviation of the variable as well.
You expect a newborn to wake its mother after midnight 2.1 times per week, on the average.
$x$ $P(x)$ $x*P(x)$ (x – $\mu)^{2} ⋅ P(x)$
0 $P(x = 0) = \dfrac{2}{50}$ $(0)\left(\dfrac{2}{50}\right) = 0$ (0 – 2.1)2 ⋅ 0.04 = 0.1764
1 $P(x = 1) = \dfrac{11}{50}$ $(1)\left(\dfrac{11}{50}\right) = \dfrac{11}{50}$ (1 – 2.1)2 ⋅ 0.22 = 0.2662
2 $P(x = 2) = \dfrac{23}{50}$ $(2)\left(\dfrac{23}{50}\right) = \dfrac{46}{50}$ (2 – 2.1)2 ⋅ 0.46 = 0.0046
3 $P(x = 3) = \dfrac{9}{50}$ $(3)\left(\dfrac{9}{50}\right) = \dfrac{27}{50}$ (3 – 2.1)2 ⋅ 0.18 = 0.1458
4 $P(x = 4) = \dfrac{4}{50}$ $(4)\left(\dfrac{4}{50}\right) = \dfrac{16}{50}$ (4 – 2.1)2 ⋅ 0.08 = 0.2888
5 $P(x = 5) = \dfrac{1}{50}$ $(5)\left(\dfrac{1}{50}\right) = \dfrac{5}{50}$ (5 – 2.1)2 ⋅ 0.02 = 0.1682
Add the values in the third column of the table to find the expected value of $X$:
$\mu = \text{Expected Value} = \dfrac{105}{50} = 2.1 \nonumber$
Use $\mu$ to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value $x$, multiply the square of its deviation by its probability. (Each deviation has the format $x – \mu$.
Add the values in the fourth column of the table:
$0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05 \nonumber$
The standard deviation of $X$ is the square root of this sum: $\sigma = \sqrt{1.05} \approx 1.0247$
The mean, μ, of a discrete probability function is the expected value.
$μ=∑(x∙P(x))\nonumber$
The standard deviation, Σ, of the PDF is the square root of the variance.
$σ=\sqrt{∑[(x – μ)2 ∙ P(x)]}\nonumber$
When all outcomes in the probability distribution are equally likely, these formulas coincide with the mean and standard deviation of the set of possible outcomes.
Exercise $2$
A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?
$x$ $P(x)$
0 $P(x = 0) = \dfrac{4}{50}$
1 $P(x = 1) = \dfrac{8}{50}$
2 $P(x = 2) = \dfrac{16}{50}$
3 $P(x = 3) = \dfrac{14}{50}$
4 $P(x = 4) = \dfrac{6}{50}$
5 $P(x = 5) = \dfrac{2}{50}$
Answer
The expected value is 2.24
$(0)\dfrac{4}{50} + (1)\dfrac{8}{50} + (2)\dfrac{16}{50} + (3)\dfrac{14}{50} + (4)\dfrac{6}{50} + (5)\dfrac{2}{50} = 0 + \dfrac{8}{50} + \dfrac{32}{50} + \dfrac{42}{50} + \dfrac{24}{50} + \dfrac{10}{50} = \dfrac{116}{50} = 2.32$
Example $2$
Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit$100,000 if you match all five numbers in order (you get your $2 back plus$100,000). Over the long term, what is your expected profit of playing the game?
To do this problem, set up an expected value table for the amount of money you can profit.
Let $X =$ the amount of money you profit. The values of $x$ are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of $x$ are 100,000 dollars and −2 dollars.
To win, you must get all five numbers correct, in order. The probability of choosing one correct number is $\dfrac{1}{10}$ because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is
\begin{align*} \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) &= (1)(10^{-5}) \[5pt] &= 0.00001. \end{align*}
Therefore, the probability of winning is 0.00001 and the probability of losing is
$1−0.00001=0.99999.1−0.00001 = 0.99999.\nonumber$
The expected value table is as follows:
Αdd the last column. –1.99998 + 1 = –0.99998
$x$ $P(x)$ $xP(x)$
Loss –2 0.99999 (–2)(0.99999) = –1.99998
Profit 100,000 0.00001 (100000)(0.00001) = 1
Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose$2 or profit $100,000. The$1 is the average or expected LOSS per game after playing this game over and over.
Exercise $3$
You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and$256. What is your expected profit of playing the game over the long term?
Answer
Let $X =$ the amount of money you profit. The $x$-values are –$1 and$256.
The probability of guessing the right suit each time is $\left(\dfrac{1}{4}\right) \left(\dfrac{1}{4}\right) \left(\dfrac{1}{4}\right) \left(\dfrac{1}{4}\right) = \dfrac{1}{256} = 0.0039$
The probability of losing is $1 – \dfrac{1}{256} = \dfrac{255}{256} = 0.9961$
$(0.0039)256 + (0.9961)(–1) = 0.9984 + (–0.9961) = 0.0023$ or $0.23$ cents.
Example $4$
Suppose you play a game with a biased coin. You play each game by tossing the coin once. $P(\text{heads}) = \dfrac{2}{3}$ and $P(\text{tails}) = \dfrac{1}{3}$. If you toss a head, you pay $6. If you toss a tail, you win$10. If you play this game many times, will you come out ahead?
1. Define a random variable $X$.
2. Complete the following expected value table.
3. What is the expected value, $\mu$? Do you come out ahead?
Solutions
a.
$X$ = amount of profit
$x$ ____ ____
WIN 10 $\dfrac{1}{3}$ ____
LOSE ____ ____ $\dfrac{-12}{3}$
b.
$x$ $P(x)$ $xP(x)$
WIN 10 $\dfrac{1}{3}$ $\dfrac{10}{3}$
LOSE –6 $\dfrac{2}{3}$ $\dfrac{-12}{3}$
c.
Add the last column of the table. The expected value $\mu = \dfrac{-2}{3}$. You lose, on average, about 67 cents each time you play the game so you do not come out ahead.
Exercise $4$
Suppose you play a game with a spinner. You play each game by spinning the spinner once. $P(\text{red}) = \dfrac{2}{5}$, $P(\text{blue}) = \dfrac{2}{5}$, and $P(\text{green}) = \dfrac{1}{5}$. If you land on red, you pay $10. If you land on blue, you don't pay or win anything. If you land on green, you win$10. Complete the following expected value table.
$x$ $P(x)$
Red $-\dfrac{20}{5}$
Blue $\dfrac{2}{5}$
Green 10
Answer
$x$ $P(x)$ $x*P(x)$
Red –10 $\dfrac{2}{5}$ $-\dfrac{20}{5}$
Blue 0 $\dfrac{2}{5}$ $\dfrac{0}{5}$
Green 10 $\dfrac{1}{5}$ $\dfrac{1}{5}$
Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled $(x) – \mu^{2}P(x)$ and take the square root.
$x$ $P(x)$ $x*P(x)$ $(x – \mu)^{2}P(x)$
0 0.2 (0)(0.2) = 0 (0 – 1.1)2(0.2) = 0.242
1 0.5 (1)(0.5) = 0.5 (1 – 1.1)2(0.5) = 0.005
2 0.3 (2)(0.3) = 0.6 (2 – 1.1)2(0.3) = 0.243
Add the last column in the table. $0.242 + 0.005 + 0.243 = 0.490$. The standard deviation is the square root of 0.49, or $\sigma = \sqrt{0.49} = 0.7$
Generally for probability distributions, we use a calculator or a computer to calculate $\mu$ and $\sigma$ to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating $\mu$ and $\sigma$.
Example $5$
Toss a fair, six-sided die twice. Let $X$ = the number of faces that show an even number. Construct a table like Table and calculate the mean $\mu$ and standard deviation $\sigma$ of $X$.
Solution
Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Use the sample space to complete the following table:
Calculating $\mu$ and $\sigma$.
$x$ $P(x)$ $xP(x)$ $(x – \mu)^{2} ⋅ P(x)$
0 $\dfrac{9}{36}$ 0 $(0 – 1)^{2} ⋅ \dfrac{9}{36} = \dfrac{9}{36}$
1 $\dfrac{18}{36}$ $\dfrac{18}{36}$ $(1 – 1)^{2} ⋅ \dfrac{18}{36} = 0$
2 $\dfrac{9}{36}$ $\dfrac{18}{36}$ $(1 – 1)^{2} ⋅ \dfrac{9}{36} = \dfrac{9}{36}$
Add the values in the third column to find the expected value: $\mu$ = $\dfrac{36}{36}$ = 1. Use this value to complete the fourth column.
Add the values in the fourth column and take the square root of the sum:
$\sigma = \sqrt{\dfrac{18}{36}} \approx 0.7071.$
Example $6$
On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay$20. Let X = the amount of profit from a bet.
$P(\text{win}) = P(\text{one moderate earthquake will occur}) = 21.42%$
$P(\text{loss}) = P(\text{one moderate earthquake will not occur}) = 100% – 21.42%$
If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of $X$? Construct a table similar to Table and Table to help you answer these questions.
Answer
$x$ $P(x)$ $xP(x)$ $(x – \mu^{2})P(x)$
win 50 0.2142 10.71 [50 – (–5.006)]2(0.2142) = 648.0964
loss –20 0.7858 –15.716 [–20 – (–5.006)]2(0.7858) = 176.6636
Mean = Expected Value = 10.71 + (–15.716) = –5.006.
If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5.01 per bet. Standard Deviation $= \sqrt{648.0964+176.6636} \approx 28.7186$ Exercise $6$ On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about 1.08%. You bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win$100. If you lose the bet, you pay $10. Let $X$ = the amount of profit from a bet. Find the mean and standard deviation of $X$. Answer $x$ $P(x)$ $x ⋅ P(x)$ $(x - \mu^{2}) ⋅ P(x)$ win 100 0.0108 1.08 [100 – (–8.812)]2 ⋅ 0.0108 = 127.8726 loss –10 0.9892 –9.892 [–10 – (–8.812)]2 ⋅ 0.9892 = 1.3961 Mean = Expected Value $= \mu = 1.08 + (–9.892) = –8.812$ If you make this bet many times under the same conditions, your long term outcome will be an average loss of$8.81 per bet.
Standard Deviation $= \sqrt{127.7826+1.3961} \approx 11.3696$
Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions.
A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions.
Summary
The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes.
Formula Review
1. Mean or Expected Value: $\mu = \sum_{x \in X}xP(x)$
2. Standard Deviation: $\sigma = \sqrt{\sum_{x \in X}(x - \mu)^{2}P(x)}$
Glossary
Expected Value
expected arithmetic average when an experiment is repeated many times; also called the mean. Notations: $\mu$. For a discrete random variable (RV) with probability distribution function $P(x)$,the definition can also be written in the form $\mu = \sum{xP(x)}$.
Mean
a number that measures the central tendency; a common name for mean is ‘average.’ The term ‘mean’ is a shortened form of ‘arithmetic mean.’ By definition, the mean for a sample (detonated by $\bar{x}$) is $\bar{x} = \dfrac{\text{Sum of all values in the sample}}{\text{Number of values in the sample}}$ and the mean for a population (denoted by $\mu$) is $\mu = \dfrac{\text{Sum of all values in the population}}{\text{Number of values in the population}}$.
Mean of a Probability Distribution
the long-term average of many trials of a statistical experiment
Standard Deviation of a Probability Distribution
a number that measures how far the outcomes of a statistical experiment are from the mean of the distribution
The Law of Large Numbers
As the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.03%3A_Mean_or_Expected_Value_and_Standard_Deviation.txt |
The binomial distribution is frequently used to model the number of successes in a sample of size $n$ drawn with replacement from a population of size $N$.
Three characteristics of a binomial experiment
1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter $n$ denotes the number of trials.
2. There are only two possible outcomes, called "success" and "failure," for each trial. The letter $p$ denotes the probability of a success on one trial, and $q$ denotes the probability of a failure on one trial. $p + q = 1$.
3. The $n$ trials are independent and are repeated using identical conditions. Because the $n$ trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, $p$, of a success and probability, $q$, of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability $p = 0.6$. Then, $q = 0.4$. This means that for every true-false statistics question Joe answers, his probability of success ($p = 0.6$) and his probability of failure ($q = 0.4$) remain the same.
The outcomes of a binomial experiment fit a binomial probability distribution. The random variable $X =$ the number of successes obtained in the $n$ independent trials. The mean, $\mu$, and variance, $\sigma^{2}$, for the binomial probability distribution are
$\mu = np$
and
$\sigma^{2} = npq.$
The standard deviation, $\sigma$, is then
$\sigma = \sqrt{npq}.$
Any experiment that has characteristics two and three and where $n = 1$ is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.
Example $1$
At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could be defined as an individual who withdrew. The random variable $X =$ the number of students who withdraw from the randomly selected elementary physics class.
Exercise $1$
The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case?
Answer
a school that offers fruit in their lunch every day
Example $2$
Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define $X$ as the number of wins, then $X$ takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is $p = 0.55$. The probability of a failure is $q = 0.45$. The number of trials is $n = 20$. The probability question can be stated mathematically as $P(x = 15)$.
Exercise $2$
A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.
Answer
$P(x = 12)$
Example $3$
A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than ten heads? Let $X =$ the number of heads in 15 flips of the fair coin. $X$ takes on the values 0, 1, 2, 3, ..., 15. Since the coin is fair, $p = 0.5$ and $q = 0.5$. The number of trials is $n = 15$. State the probability question mathematically.
Solution
$P(x > 10)$
Exercise $4$
A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.
Answer
$P(x > 3)$
Example $5$
Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.
1. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.
2. If we are interested in the number of students who do their homework on time, then how do we define $X$?
3. What values does $x$ take on?
4. What is a "failure," in words?
5. If $p + q = 1$, then what is $q$?
6. The words "at least" translate as what kind of inequality for the probability question $P(x$ ____ $40$).
Solution
1. failure
2. $X$ = the number of statistics students who do their homework on time
3. 0, 1, 2, …, 50
4. Failure is defined as a student who does not complete his or her homework on time. The probability of a success is $p = 0.70$. The number of trials is $n = 50$.
5. $q = 0.30$
6. greater than or equal to ($\geq$). The probability question is $P(x \geq 40)$.
Exercise $5$
Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem.
Answer
This is a binomial problem because there is only a success or a failure, and there are a definite number of trials. The probability of a success stays the same for each trial.
Notation for the Binomial: $B =$ Binomial Probability Distribution Function
$X \sim B(n, p)$
Read this as "$X$ is a random variable with a binomial distribution." The parameters are $n$ and $p$; $n =$ number of trials, $p =$ probability of a success on each trial.
Example $6$
It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?
Let $X$ = the number of workers who have a high school diploma but do not pursue any further education.
$X$ takes on the values 0, 1, 2, ..., 20 where $n = 20, p = 0.41$, and $q = 1 – 0.41 = 0.59$. $X \sim B(20, 0.41)$
Find $P(x \leq 12)$. $P(x \leq 12) = 0.9738$. (calculator or computer)
Go into 2nd DISTR. The syntax for the instructions are as follows:
To calculate ($x = \text{value}): \text{binompdf}(n, p, \text{number}$) if "number" is left out, the result is the binomial probability table.
To calculate $P(x \leq \text{value}): \text{binomcdf}(n, p, \text{number})$ if "number" is left out, the result is the cumulative binomial probability table.
For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is $P(x \leq 12) = 0.9738$.
If you want to find $P(x = 12)$, use the pdf (binompdf). If you want to find $P(x > 12)$, use $1 - \text{binomcdf}(20,0.41,12)$.
The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.
The graph of $X \sim B(20, 0.41)$ is as follows:
The y-axis contains the probability of $x$, where $X =$ the number of workers who have only a high school diploma.
The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, $\mu = np = (20)(0.41) = 8.2$.
The formula for the variance is $\sigma^{2} = npq$. The standard deviation is $\sigma = \sqrt{npq}$.
$\sigma = \sqrt{(20)(0.41)(0.59)} = 2.20.$
Exercise 4.4.5
About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.
Answer
$P(x \leq 14) = 0.9695$
Example $7$
In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let $X =$ the number of pages that feature signature artists.
1. What values does $x$ take on?
2. What is the probability distribution? Find the following probabilities:
1. the probability that two pages feature signature artists
2. the probability that at most six pages feature signature artists
3. the probability that more than three pages feature signature artists.
3. Using the formulas, calculate the (i) mean and (ii) standard deviation.
Answer
1. $x = 0, 1, 2, 3, 4, 5, 6, 7, 8$
2. $X \sim B(100,8560)(100,8560)$
1. $P(x = 2) = \text{binompdf}\left(100,\dfrac{8}{560},2\right) = 0.2466$
2. $P(x \leq 6) = \text{binomcdf}\left(100,\dfrac{8}{560},6\right) = 0.9994$
3. $P(x > 3) = 1 – P(x \leq 3) = 1 – \text{binomcdf}\left(100,\dfrac{8}{560},3\right) = 1 – 0.9443 = 0.0557$
1. Mean $= np = (100)\left(\dfrac{8}{560}\right) = \dfrac{800}{560} \approx 1.4286$
2. Standard Deviation $= \sqrt{npq} = \sqrt{(100)\left(\dfrac{8}{560}\right)\left(\dfrac{552}{560}\right)} \approx 1.1867$
Exercise $7$
According to a Gallup poll, 60% of American adults prefer saving over spending. Let $X$ = the number of American adults out of a random sample of 50 who prefer saving to spending.
1. What is the probability distribution for $X$?
2. Use your calculator to find the following probabilities:
1. the probability that 25 adults in the sample prefer saving over spending
2. the probability that at most 20 adults prefer saving
3. the probability that more than 30 adults prefer saving
3. Using the formulas, calculate the (i) mean and (ii) standard deviation of $X$.
Answer
1. $X \sim B(50, 0.6)$
2. Using the TI-83, 83+, 84 calculator with instructions as provided in Example:
1. $P(x = 25) = \text{binompdf}(50, 0.6, 25) = 0.0405$
2. $P(x \leq 20) = \text{binomcdf}(50, 0.6, 20) = 0.0034$
3. $(x > 30) = 1 - \text{binomcdf}(50, 0.6, 30) = 1 – 0.5535 = 0.4465$
1. Mean $= np = 50(0.6) = 30$
2. Standard Deviation $= \sqrt{npq} = \sqrt{50(0.6)(0.4)} \approx 3.4641$
Example $8$
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let $X$ = the number of people who will develop pancreatic cancer.
1. What is the probability distribution for $X$?
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of $X$.
3. Use your calculator to find the probability that at most eight people develop pancreatic cancer
4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.
Answer
1. $X \sim B(200, 0.0128)$
1. Mean $= np = 200(0.0128) = 2.56$
2. Standard Deviation $= \sqrt{npq} = \sqrt{(200)(0.0128)(0.9872)} \approx 1.5897$
2. Using the TI-83, 83+, 84 calculator with instructions as provided in Example:
$P(x \leq 8) = \text{binomcdf}(200, 0.0128, 8) = 0.9988$
3. $P(x = 5) = \text{binompdf}(200, 0.0128, 5) = 0.0707$
$P(x = 6) = \text{binompdf}(200, 0.0128, 6) = 0.0298$
So $P(x = 5) > P(x = 6)$; it is more likely that five people will develop cancer than six.
Exercise $8$
During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let $X =$ the number of shots that scored points.
1. What is the probability distribution for $X$?
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of $X$.
3. Use your calculator to find the probability that DeAndre scored with 60 of these shots.
4. Find the probability that DeAndre scored with more than 50 of these shots.
Answer
1. $X \sim B(80, 0.613)$
1. Mean $= np = 80(0.613) = 49.04$
2. Standard Deviation $= \sqrt{npq} = \sqrt{80(0.613)(0.387)} \approx 4.3564$
2. Using the TI-83, 83+, 84 calculator with instructions as provided in Example:
$P(x = 60) = \text{binompdf}(80, 0.613, 60) = 0.0036$
3. $P(x > 50) = 1 – P(x \leq 50) = 1 – \text{binomcdf}(80, 0.613, 50) = 1 – 0.6282 = 0.3718$
Example $9$
The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is $\dfrac{6}{16}$. The probability of a student on the second draw is $\dfrac{5}{15}$, when the first draw selects a student. The probability is $\dfrac{6}{15}$, when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.
Exercise $9$
A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why.
Answer
This is not binomial because the names are not replaced, which means the probability changes for each time a name is drawn. This violates the condition of independence.
Review
A statistical experiment can be classified as a binomial experiment if the following conditions are met:
There are a fixed number of trials, $n$.
There are only two possible outcomes, called "success" and, "failure" for each trial. The letter $p$ denotes the probability of a success on one trial and $q$ denotes the probability of a failure on one trial.
The $n$ trials are independent and are repeated using identical conditions.
The outcomes of a binomial experiment fit a binomial probability distribution. The random variable $X =$ the number of successes obtained in the $n$ independent trials. The mean of $X$ can be calculated using the formula $\mu = np$, and the standard deviation is given by the formula $\sigma = \sqrt{npq}$.
Formula Review
• $X \sim B(n, p)$ means that the discrete random variable $X$ has a binomial probability distribution with $n$ trials and probability of success $p$.
• $X =$ the number of successes in $n$ independent trials
• $n =$ the number of independent trials
• $X$ takes on the values $x = 0, 1, 2, 3, \dotsc, n$
• $p =$ the probability of a success for any trial
• $q =$ the probability of a failure for any trial
• $p + q = 1$
• $q = 1 – p$
The mean of $X$ is $\mu = np$. The standard deviation of $X$ is $\sigma = \sqrt{npq}$.
Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status.
Exercise 4.4.9
In words, define the random variable $X$.
Answer
$X =$ the number that reply “yes”
Exercise 4.4.10
$X \sim$ _____(_____,_____)
Exercise 4.4.11
What values does the random variable $X$ take on?
Answer
0, 1, 2, 3, 4, 5, 6, 7, 8
Exercise 4.4.12
Construct the probability distribution function (PDF).
$x$ $P(x)$
Exercise 4.4.13
On average ($\mu$), how many would you expect to answer yes?
Answer
5.7
Exercise 4.4.14
What is the standard deviation ($\sigma$)?
Exercise 4.4.15
What is the probability that at most five of the freshmen reply “yes”?
Answer
0.4151
Exercise 4.4.16
What is the probability that at least two of the freshmen reply “yes”?
Glossary
Binomial Experiment
a statistical experiment that satisfies the following three conditions:
1. There are a fixed number of trials, $n$.
2. There are only two possible outcomes, called "success" and, "failure," for each trial. The letter $p$ denotes the probability of a success on one trial, and $q$ denotes the probability of a failure on one trial.
3. The $n$ trials are independent and are repeated using identical conditions.
Bernoulli Trials
an experiment with the following characteristics:
1. There are only two possible outcomes called “success” and “failure” for each trial.
2. The probability $p$ of a success is the same for any trial (so the probability $q = 1 − p$ of a failure is the same for any trial).
Binomial Probability Distribution
a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, $n$, of independent trials. “Independent” means that the result of any trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV $X$ is defined as the number of successes in $n$ trials. The notation is: $X ~ B(n, p)$. The mean is $\mu = np$ and the standard deviation is $\sigma = \sqrt{npq}$. The probability of exactly $x$ successes in $n$ trials is
$P(X = x) = {n \choose x}p^{x}q^{n-x}$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.04%3A_Binomial_Distribution.txt |
There are three main characteristics of a geometric experiment.
1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, $p$, of a success and the probability, $q$, of a failure is the same for each trial. $p + q = 1$ and $q = 1 − p$. For example, the probability of rolling a three when you throw one fair die is $\dfrac{1}{6}$. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is $q = \dfrac{5}{6}$, the probability of a failure. The probability of getting a three on the fifth roll is $\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\left(\dfrac{1}{6}\right) = 0.0804$.
$X =$ the number of independent trials until the first success.
You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is $p = 0.57$. What is the probability that it takes five games until you lose? Let $X =$ the number of games you play until you lose (includes the losing game). Then $X$ takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is $P(x = 5)$.
Example $1$
You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is $p = 0.57$. What is the probability that it takes five games until you lose? Let $X =$ the number of games you play until you lose (includes the losing game). Then $X$ takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is $P(x = 5)$.
Exercise $1$
You throw darts at a board until you hit the center area. Your probability of hitting the center area is $p = 0.17$. You want to find the probability that it takes eight throws until you hit the center. What values does $X$ take on?
Answer
$1, 2, 3, 4, \dotsc, n$. It can go on indefinitely.
Example $2$
A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions?
Let $X$ = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. $X$ takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find $P(x \geq 3)$. ("At least" translates to a "greater than or equal to" symbol).
Exercise $2$
An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically?
Answer
$P(x \leq 10)$
Example $3$
Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?
This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).
1. Let $X =$ the number of ____________ you must ask ____________ one says yes.
2. What values does $X$ take on?
3. What are $p$ and $q$?
4. The probability question is $P($_______$)$.
Solution
1. Let $X =$ the number of students you must ask until one says yes.
2. 1, 2, 3, …, (total number of students)
3. $p = 0.55; q = 0.45$
4. $P(x = 4)$
Exercise $3$
You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are $p$ and $q$?
Answer
$p = 0.1$
$q = 0.9$
Notation for the Geometric: $G =$ Geometric Probability Distribution Function
$X \sim G(p)$
Read this as "$X$ is a random variable with a geometric distribution." The parameter is $p$; $p =$ the probability of a success for each trial.
Example $4$
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?
Let $X$ = the number of computer components tested until the first defect is found.
$X$ takes on the values 1, 2, 3, ... where $p = 0.02$. $X \sim G(0.02)$
Find $P(x = 7)$. $P(x = 7) = 0.0177$.
To find the probability that $x = 7$,
• Enter 2nd, DISTR
• Scroll down and select geometpdf(
• Press ENTER
• Enter 0.02, 7); press ENTER to see the result: $P(x = 7) = 0.0177$
To find the probability that $x \leq 7$, follow the same instructions EXCEPT select E: geometcdf as the distribution function.
The probability that the seventh component is the first defect is 0.0177.
The graph of $X \sim G(0.02)$ is:
The y-axis contains the probability of $x$, where $X =$ the number of computer components tested.
The number of components that you would expect to test until you find the first defective one is the mean, $\mu = 50$.
The formula for the mean is
$\mu = \dfrac{1}{\text{p}} = \dfrac{1}{0.02} = 50$
The formula for the variance is
$\sigma^{2} = \left(\dfrac{1}{p}\right)\left(\dfrac{1}{p} - 1 \right) = \left(\dfrac{1}{0.02}\right)\left(\dfrac{1}{0.02} - 1 \right) = 2,450$
The standard deviation is
$\sigma = \sqrt{\left(\dfrac{1}{p}\right)\left(\dfrac{1}{p} - 1\right)} = \sqrt{\left(\dfrac{1}{0.02}\right)\left(\dfrac{1}{0.02} - 1\right)} = 49.5$
Exercise $4$
The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.
Answer
$P(x = 9) = 0.0092$
Example $5$
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let $X =$ the number of people you ask until one says he or she has pancreatic cancer. Then $X$ is a discrete random variable with a geometric distribution: $X \sim G\left(\dfrac{1}{78}\right)$ or $X \sim G(0.0128)$.
1. What is the probability of that you ask ten people before one says he or she has pancreatic cancer?
2. What is the probability that you must ask 20 people?
3. Find the (i) mean and (ii) standard deviation of $X$.
Answer
1. $P(x = 10) = \text{geometpdf}(0.0128, 10) = 0.0114$
2. $P(x = 20) = \text{geometpdf}(0.0128, 20) = 0.01$
1. Mean $= \mu = \dfrac{1}{p} = \dfrac{1}{0.0128} = 78$
2. Standard Deviation $= \sigma = \sqrt{\dfrac{1-p}{p^{2}}} = \sqrt{\dfrac{1-0.0128}{0.0128^{2}}} \approx 77.6234$
Exercise $5$
The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let $X =$ the number of Afghani women you ask until one says that she is literate.
1. What is the probability distribution of $X$?
2. What is the probability that you ask five women before one says she is literate?
3. What is the probability that you must ask ten women?
4. Find the (i) mean and (ii) standard deviation of $X$.
Answer
1. $X \sim G(0.12)$
2. $P(x = 5) = \text{geometpdf}(0.12, 5) = 0.0720$
3. $P(x = 10) = \text{geometpdf}(0.12, 10) = 0.0380$
1. Mean $= \mu = \dfrac{1}{p} = \dfrac{1}{0.12} \approx 3333$
2. Standard Deviation $= \sigma = \dfrac{1-p}{p^{2}} = \dfrac{1-0.12}{0.12^{2}} \approx 7.8174$
Review
There are three characteristics of a geometric experiment:
• There are one or more Bernoulli trials with all failures except the last one, which is a success.
• In theory, the number of trials could go on forever. There must be at least one trial.
• The probability, $p$, of a success and the probability, $q$, of a failure are the same for each trial.
In a geometric experiment, define the discrete random variable $X$ as the number of independent trials until the first success. We say that $X$ has a geometric distribution and write $X \sim G(p)$ where $p$ is the probability of success in a single trial. The mean of the geometric distribution $X \sim G(p)$ is $\mu = \dfrac{1-p}{p^{2}} = \sqrt{\dfrac{1}{p}\left(\dfrac{1}{p} - 1\right)}$.
Formula Review
$X \sim G(p)$ means that the discrete random variable $X$ has a geometric probability distribution with probability of success in a single trial $p$.
$X =$ the number of independent trials until the first success
$X$ takes on the values $x = 1, 2, 3, \dotsc$
$p =$ the probability of a success for any trial
$q =$ the probability of a failure for any trial $p + q = 1$
$q = 1 – p$
The mean is $\mu = \dfrac{1}{p}$.
The standard deviation is $\sigma = \dfrac{1-p}{p^{2}} = \sqrt{\dfrac{1}{p}\left(\dfrac{1}{p} - 1\right)}$.
Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask.
Exercise 4.5.6
In words, define the random variable $X$.
Answer
$X =$ the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.
Exercise 4.5.7
$X \sim$ _____(_____,_____)
Exercise 4.5.8
What values does the random variable $X$ take on?
Answer
1,2,...
Exercise 4.5.9
Construct the probability distribution function (PDF). Stop at $x = 6$.
$x$ $P(x)$
1
2
3
4
5
6
Exercise 4.5.10
On average ($\mu$), how many freshmen would you expect to have to ask until you found one who replies "yes?"
Answer
1.4
Exercise 4.5.11
What is the probability that you will need to ask fewer than three freshmen?
Footnotes
1”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/...pi_data_value- last&sort=desc (accessed May 15, 2013).
Glossary
Geometric Distribution
a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated until the first success. The geometric variable $X$ is defined as the number of trials until the first success. Notation: $X \sim G(p)$. The mean is $\mu = \dfrac{1}{p}$ and the standard deviation is $\sigma =$
$\sqrt{\dfrac{1}{p}\left(\dfrac{1}{p} - 1\right)}$
. The probability of exactly $x$ failures before the first success is given by the formula: $P(X = x) = p(1 –p)^{x-1}$.
Geometric Experiment
a statistical experiment with the following properties:
1. There are one or more Bernoulli trials with all failures except the last one, which is a success.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, $p$, of a success and the probability, $q$, of a failure do not change from trial to trial. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.05%3A_Geometric_Distribution.txt |
The hypergeometric distribution arises when one samples from a finite population, thus making the trials dependent on each other. There are five characteristics of a hypergeometric experiment.
Characteristics of a hypergeometric experiment
1. You take samples from two groups.
2. You are concerned with a group of interest, called the first group.
3. You sample without replacement from the combined groups. For example, you want to choose a softball team from a combined group of 11 men and 13 women. The team consists of ten players.
4. Each pick is not independent, since sampling is without replacement. In the softball example, the probability of picking a woman first is $\frac{13}{24}$. The probability of picking a man second is $\frac{11}{23}$ if a woman was picked first. It is $\frac{10}{23}$ if a man was picked first. The probability of the second pick depends on what happened in the first pick.
5. You are not dealing with Bernoulli Trials.
The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable $X$ = the number of items from the group of interest.
Example $1$
A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let $X =$ the number of gumdrops in the sample of 50. $X$ takes on the values $x = 0, 1, 2, ..., 50$. What is the probability statement written mathematically?
Answer
$P(x = 35)$
Exercise $1$
A bag contains letter tiles. Forty-four of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample?
Answer
The group of interest is the vowel letter tiles. The size of the group of interest is 44. The size of the sample is seven.
Example $2$
Suppose a shipment of 100 DVD players is known to have ten defective players. An inspector randomly chooses 12 for inspection. He is interested in determining the probability that, among the 12 players, at most two are defective. The two groups are the 90 non-defective DVD players and the 10 defective DVD players. The group of interest (first group) is the defective group because the probability question asks for the probability of at most two defective DVD players. The size of the sample is 12 DVD players. (They may be non-defective or defective.) Let $X =$ the number of defective DVD players in the sample of 12. $X$ takes on the values $0, 1, 2, \dotsc, 10$. $X$ may not take on the values 11 or 12. The sample size is 12, but there are only 10 defective DVD players. Write the probability statement mathematically.
Answer
$P(x \leq 2)$
Exercise $2$
A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is $X$, and what values does it take on?
Answer
Let $X =$ the number of cracked eggs in the sample of 15. $X$ takes on the values $0, 1, 2, \dotsc, 12$.
Example $3$
You are president of an on-campus special events organization. You need a committee of seven students to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. If the members of the committee are randomly selected, what is the probability that your committee has more than four men?
This is a hypergeometric problem because you are choosing your committee from two groups (men and women).
1. Are you choosing with or without replacement?
2. What is the group of interest?
3. How many are in the group of interest?
4. How many are in the other group?
5. Let $X =$ _________ on the committee. What values does $X$ take on?
6. The probability question is $P($_______$)$.
Solution
1. without
2. the men
3. 15 men
4. 18 women
5. Let $X =$ the number of men on the committee. $x = 0, 1, 2, \dotsc, 7$.
6. $P(x > 4)$
Exercise $3$
A palette has 200 milk cartons. Of the 200 cartons, it is known that ten of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18, no more than two are leaking. Give five reasons why this is a hypergeometric problem.
Answer
• There are two groups.
• You are concerned with a group of interest.
• You sample without replacement.
• Each pick is not independent.
• You are not dealing with Bernoulli Trials.
Notation for the Hypergeometric: $H =$ Hypergeometric Probability Distribution Function
$X \sim H(r, b, n)$
Read this as "$X$ is a random variable with a hypergeometric distribution." The parameters are $r, b$, and $n$; $r =$ the size of the group of interest (first group), $b =$ the size of the second group, $n =$ the size of the chosen sample.
Example $4$
A school site committee is to be chosen randomly from six men and five women. If the committee consists of four members chosen randomly, what is the probability that two of them are men? How many men do you expect to be on the committee?
Let $X$ = the number of men on the committee of four. The men are the group of interest (first group).
$X$ takes on the values $0, 1, 2, 3, 4$, where $r = 6, b = 5$, and $n = 4$. $X \sim H(6, 5, 4)$
Find $P(x = 2)$. $P(x = 2) = 0.4545$ (calculator or computer)
Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft Excel, that do.
The probability that there are two men on the committee is about 0.45.
The graph of $X \sim H(6, 5, 4)$ is:
The y-axis contains the probability of $X$, where $X =$ the number of men on the committee.
You would expect $m = 2.18$ (about two) men on the committee.
The formula for the mean is
$\mu = \frac{nr}{r+b} \frac{(4)(6)}{6+5} = 2.18$
Exercise $4$
An intramural basketball team is to be chosen randomly from 15 boys and 12 girls. The team has ten slots. You want to know the probability that eight of the players will be boys. What is the group of interest and the sample?
Answer
The group of interest is the 15 boys. The sample consists of the ten slots on the intramural basketball team.
Summary
A hypergeometric experiment is a statistical experiment with the following properties:
• You take samples from two groups.
• You are concerned with a group of interest, called the first group.
• You sample without replacement from the combined groups.
• Each pick is not independent, since sampling is without replacement.
• You are not dealing with Bernoulli Trials.
The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable $X$ = the number of items from the group of interest. The distribution of $X$ is denoted $X \sim H(r, b, n)$, where $r =$ the size of the group of interest (first group), $b =$ the size of the second group, and $n =$ the size of the chosen sample. It follows that $n \leq r + b$. The mean of $X$ is $\mu = \frac{nr}{r+b}$ and the standard deviation is $\sigma = \sqrt{\frac{rbn(r+b-n)}{(r+b)^{2}(r+b-1)}}$.
Formula Review
$X \sim H(r, b, n)$ means that the discrete random variable $X$ has a hypergeometric probability distribution with $r =$ the size of the group of interest (first group), $b =$ the size of the second group, and $n =$ the size of the chosen sample.
$X$ = the number of items from the group of interest that are in the chosen sample, and $X$ may take on the values $x = 0, 1, \dotsc,$ up to the size of the group of interest. (The minimum value for $X$ may be larger than zero in some instances.)
$n \leq r + b$
The mean of $X$ is given by the formula $\mu = \frac{nr}{r+b}$ and the standard deviation is $= \sqrt{\frac{rbn(r+b-n)}{(r+b)^{2}(r+b-1)}}$.
Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample.
Exercise $5$
In words, define the random variable $X$.
Answer
$X =$ the number of business majors in the sample.
Exercise $6$
$X \sim$ _____(_____,_____)
Exercise $7$
What values does $X$ take on?
Answer
$2, 3, 4, 5, 6, 7, 8, 9$
Exercise $8$
Find the standard deviation.
Exercise $9$
On average ($\mu$), how many would you expect to be business majors?
Answer
6.26
Glossary
Hypergeometric Experiment
a statistical experiment with the following properties:
1. You take samples from two groups.
2. You are concerned with a group of interest, called the first group.
3. You sample without replacement from the combined groups.
4. Each pick is not independent, since sampling is without replacement.
5. You are not dealing with Bernoulli Trials.
Hypergeometric Probability
a discrete random variable (RV) that is characterized by:
1. A fixed number of trials.
2. The probability of success is not the same from trial to trial.
We sample from two groups of items when we are interested in only one group. $X$ is defined as the number of successes out of the total number of items chosen. Notation: $X \sim H(r, b, n)$, where $r =$ the number of items in the group of interest, $b =$ the number of items in the group not of interest, and $n =$ the number of items chosen. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.06%3A_Hypergeometric_Distribution.txt |
The Poisson distribution is popular for modelling the number of times an event occurs in an interval of time or space. It is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.
two main characteristics of a Poisson experiment
1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.
2. The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000). You will verify the relationship in the homework exercises. $n$ is the number of trials, and $p$ is the probability of a "success."
The random variable $X =$ the number of occurrences in the interval of interest.
Example $1$
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?
Solution
Let $X =$ the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is $\left(\frac{5}{30}\right)(12) = 2$ loaves of bread.
The probability question asks you to find $P(x = 3)$.
Exercise $1$
The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes?
Answer
$\left(\frac{15}{60}\right)(8) = 2$ fish
Example $2$
A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let $X =$ the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.
Answer
$P(x < 5)$
Exercise $2$
An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically.
Answer
$P(x < 8)$
Example $3$
You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast. This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast.
1. What is the interval of interest?
2. What is the average number of times the news reporter says "uh" during one broadcast?
3. Let $X =$ ____________. What values does X take on?
4. The probability question is $P($______$)$.
Solutions
1. one broadcast
2. 2
3. Let $X =$ the number of times the news reporter says "uh" during one broadcast. $x = 0, 1, 2, 3, \dotsc$
4. $P(x > 2)$
Exercise $3$
An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution.
Answer
This problem wants to find the probability of events occurring in a fixed interval of time with a known average rate. The events are independent.
Notation for the Poisson: $P =$ Poisson Probability Distribution Function
$X \sim P(\mu)$
Read this as "$X$ is a random variable with a Poisson distribution." The parameter is $\mu$ (or $\lambda$); $\mu$ (or $\lambda) =$ the mean for the interval of interest.
Example $4$
Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
Solution
Let $X$ = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $\frac{1}{4}$ hour.)
$x = 0, 1, 2, 3, \dotsc$
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives
$\left(\frac{1}{8}\right)(6) = 0.75$ calls in 15 minutes, on average. So, $\mu = 0.75$ for this problem.
$X \sim P(0.75)$
Find $P(x > 1)$. $P(x > 1) = 0.1734$ (calculator or computer)
• Press 1 – and then press 2nd DISTR.
• Arrow down to poissoncdf. Press ENTER.
• Enter (.75,1).
• The result is $P(x > 1) = 0.1734$.
The TI calculators use $\lambda$ (lambda) for the mean.
The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:
$P(x > 1) = 1 − \text{poissoncdf}(0.75, 1)$.
The graph of $X \sim P(0.75)$ is:
The y-axis contains the probability of $x$ where $X =$ the number of calls in 15 minutes.
Exercise $4$
A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer.
Answer
$P(x > 4) = 0.0527$
Example $5$
According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let $X =$ the number of emails an email user receives per day. The discrete random variable $X$ takes on the values $x = 0, 1, 2 \dotsc$. The random variable $X$ has a Poisson distribution: $X \sim P(147)$. The mean is 147 emails.
1. What is the probability that an email user receives exactly 160 emails per day?
2. What is the probability that an email user receives at most 160 emails per day?
3. What is the standard deviation?
Solutions
1. $P(x = 160) = \text{poissonpdf}(147, 160) \approx 0.0180$
2. $P(x \leq 160) = \text{poissoncdf}(147, 160) \approx 0.8666$
3. Standard Deviation $= \sigma = \sqrt{\mu} = \sqrt{147} \approx 12.1244$
Exercise $5$
According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let $X =$ the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable $X$ takes on the values $x = 0, 1, 2 \dotsc$. The random variable $X$ has a Poisson distribution: $X \sim P(187)$. The mean is 187 text messages.
1. What is the probability that a teen girl sends exactly 175 texts per day?
2. What is the probability that a teen girl sends at most 150 texts per day?
3. What is the standard deviation?
Answer
1. $P(x = 175) = \text{poissonpdf}(187, 175) \approx 0.0203$
2. $P(x \leq 150) = \text{poissoncdf}(187, 150) \approx 0.0030$
3. Standard Deviation $= \sigma = \sqrt{\mu} = \sqrt{187} \approx 13.6748$
Example $6$
Text message users receive or send an average of 41.5 text messages per day.
1. How many text messages does a text message user receive or send per hour?
2. What is the probability that a text message user receives or sends two messages per hour?
3. What is the probability that a text message user receives or sends more than two messages per hour?
Solutions
1. Let $X =$ the number of texts that a user sends or receives in one hour. The average number of texts received per hour is $\frac{41.5}{24} \approx 1.7292$.
2. $X \sim P(1.7292)$, so $P(x = 2) = \text{poissonpdf}(1.7292, 2) \approx 0.2653$
3. $P(x > 2) = 1 – P(x \leq 2) = 1 – \text{poissoncdf}(1.7292, 2) \approx 1 – 0.7495 = 0.2505$
Exercise $6$
Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day.
1. How many airplanes arrive and depart the airport per hour?
2. What is the probability that there are exactly 100 arrivals and departures in one hour?
3. What is the probability that there are at most 100 arrivals and departures in one hour?
Answer
1. Let $X =$ the number of airplanes arriving and departing from Hartsfield-Jackson in one hour. The average number of arrivals and departures per hour is $\frac{2,500}{24} \approx 104.1667$.
2. $X \sim P(104.1667)$, so $P(x = 100) = \text{poissonpdf}(104.1667, 100) \approx 0.0366$.
3. $P(x \leq 100) = \text{poissoncdf}(104.1667, 100) \approx 0.3651$.
The Poisson distribution can be used to approximate probabilities for a binomial distribution. This next example demonstrates the relationship between the Poisson and the binomial distributions. Let $n$ represent the number of binomial trials and let $p$ represent the probability of a success for each trial. If $n$ is large enough and $p$ is small enough then the Poisson approximates the binomial very well. In general, $n$ is considered “large enough” if it is greater than or equal to 20. The probability $p$ from the binomial distribution should be less than or equal to 0.05. When the Poisson is used to approximate the binomial, we use the binomial mean $\mu = np$. The variance of $X$ is $\sigma^{2} = \sqrt{\mu}$ and the standard deviation is $\sigma = \sqrt{\mu}$. The Poisson approximation to a binomial distribution was commonly used in the days before technology made both values very easy to calculate.
Example $7$
On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
Answer
Let $X$ = the number of days with low seismic activity.
Using the binomial distribution:
• $P(x = 10) = \text{binompdf}(200, .0102, 10) \approx\ 0.000039$
Using the Poisson distribution:
• Calculate $\mu = np = 200(0.0102) \approx 2.04$
• $P(x = 10) = \text{poissonpdf}(2.04, 10) \approx 0.000045$
We expect the approximation to be good because $n$ is large (greater than 20) and $p$ is small (less than 0.05). The results are close—both probabilities reported are almost 0.
Exercise $7$
On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
Answer
Let $X =$ the number of days with moderate seismic activity.
Using the binomial distribution: $P(x = 5) = \text{binompdf}(100, 0.0143, 5) \approx 0.0115$
Using the Poisson distribution:
• Calculate $\mu = np = 100(0.0143) = 1.43$
• $P(x = 5) = \text{poissonpdf}(1.43, 5) = 0.0119$
We expect the approximation to be good because $n$ is large (greater than 20) and $p$ is small (less than 0.05). The results are close—the difference between the values is 0.0004.
Review
A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is "small" (less than or equal to 0.05) and the number of trials is "large" (greater than or equal to 20).
Formula Review
$X \sim P(\mu)$ means that $X$ has a Poisson probability distribution where $X =$ the number of occurrences in the interval of interest.
$X$ takes on the values $x = 0, 1, 2, 3, \dotsc$
The mean $\mu$ is typically given.
The variance is $\sigma = \mu$, and the standard deviation is
$\sigma = \sqrt{\mu}$.
When $P(\mu)$ is used to approximate a binomial distribution, $\mu = np$ where $n$ represents the number of independent trials and $p$ represents the probability of success in a single trial.
Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day.
Exercise $8$
Assume the event occurs independently in any given day. Define the random variable $X$.
Exercise $9$
What values does $X$ take on?
Answer
0, 1, 2, 3, 4, …
Exercise $10$
What is the probability of getting 150 customers in one day?
Exercise $11$
What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.
Answer
0.0485
Exercise $12$
What is the probability that the store will have more than 12 customers in the first hour?
Exercise $13$
What is the probability that the store will have fewer than 12 customers in the first two hours?
Answer
0.0214
Exercise $14$
Which type of distribution can the Poisson model be used to approximate? When would you do this?
Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age.
Exercise $15$
Assume the event occurs independently in any given day. In words, define the random variable $X$.
Answer
$X =$ the number of U.S. teens who die from motor vehicle injuries per day.
Exercise $16$
$X \sim$ _____(_____,_____)
Exercise $17$
What values does $X$ take on?
Answer
$0, 1, 2, 3, 4, \dotsc$
Exercise $18$
For the given values of the random variable $X$, fill in the corresponding probabilities.
Exercise $19$
Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically.
Answer
No
Exercise $20$
Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.
Glossary
Poisson Probability Distribution
a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable:
• The probability that the event occurs in a given interval is the same for all intervals.
• The events occur with a known mean and independently of the time since the last event.
The distribution is defined by the mean $\mu$ of the event in the interval. Notation: $X \sim P(\mu)$. The mean is $\mu = np$. The standard deviation is $\sigma = \sqrt{\mu}$. The probability of having exactly $x$ successes in $r$ trials is $P(X = x) =$
$\left(e^{-\mu}\right)\frac{\mu^{x}}{x!}$
. The Poisson distribution is often used to approximate the binomial distribution, when $n$ is “large” and $p$ is “small” (a general rule is that $n$ should be greater than or equal to 20 and $p$ should be less than or equal to 0.05). | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.07%3A_Poisson_Distribution.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution.
• The student will demonstrate an understanding of long-term probabilities.
Supplies
• One full deck of playing cards
Procedure
The experimental procedure is to pick one card from a deck of shuffled cards.
1. The theoretical probability of picking a diamond from a deck is _________.
2. Shuffle a deck of cards.
3. Pick one card from it.
4. Record whether it was a diamond or not a diamond.
5. Put the card back and reshuffle.
6. Do this a total of ten times.
7. Record the number of diamonds picked.
8. Let $X$ = number of diamonds. Theoretically, $X$ ~ B(_____,_____)
Organize the Data
1. Record the number of diamonds picked for your class in Table. Then calculate the relative frequency.
$x$ Frequency Relative Frequency
0 __________ __________
1 __________ __________
2 __________ __________
3 __________ __________
4 __________ __________
5 __________ __________
6 __________ __________
7 __________ __________
8 __________ __________
9 __________ __________
10 __________ __________
2. Calculate the following:
1. $\bar{x}$ = ________
2. s = ________
3. Construct a histogram of the empirical data.
Theoretical Distribution
1. Build the theoretical PDF chart based on the distribution in the Procedure section.
$x$ P($x$)
0
1
2
3
4
5
6
7
8
9
10
2. Calculate the following:
1. $\mu$ = ____________
2. $\sigma$ = ____________
3. Construct a histogram of the theoretical distribution.
Using the Data
Note 4.8.1
RF = relative frequency
Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.
• P($x$ = 3) = _______________________
• P(1 < $x$ < 4) = _______________________
• P($x \geq$ 8) = _______________________
Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.
• RF($x$ = 3) = _______________________
• RF(1 < $x$ < 4) = _______________________
• RF($x \geq$ 8) = _______________________
Discussion Questions
For questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations.
1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences.
2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions.
3. Using your answers from questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not.
4. Suppose that the experiment had been repeated 500 times. Would you expect Table orTable to change, and how would it change? Why? Why wouldn’t the other table change?
4.09: Discrete Distribution (Lucky Dice Experiment)
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will compare empirical data and a theoretical distribution to determine if a Tet gambling game fits a discrete distribution.
• The student will demonstrate an understanding of long-term probabilities.
Supplies
• one “Lucky Dice” game or three regular dice
Procedure
Round answers to relative frequency and probability problems to four decimal places.
1. The experimental procedure is to bet on one object. Then, roll three Lucky Dice and count the number of matches. The number of matches will decide your profit.
2. What is the theoretical probability of one die matching the object?
3. Choose one object to place a bet on. Roll the three Lucky Dice. Count the number of matches.
4. Let $X$ = number of matches. Theoretically, $X$ ~ B(______,______)
5. Let $Y$ = profit per game.
Organize the Data
In Table, fill in the $y$ value that corresponds to each x value. Next, record the number of matches picked for your class. Then, calculate the relative frequency.
1. Complete the table.
$x$ $y$ Frequency Relative Frequency
0
1
2
3
2. Calculate the following:
1. $\bar{x}$ = _______
2. $s_{x}$ = ________
3. $\bar{y}$ = _______
4. $s_{y}$ = _______
3. Explain what $\bar{x}$ represents.
4. Explain what $\bar{y}$ represents.
5. Based upon the experiment:
1. What was the average profit per game?
2. Did this represent an average win or loss per game?
3. How do you know? Answer in complete sentences.
6. Construct a histogram of the empirical data.
Theoretical Distribution
Build the theoretical PDF chart for x and y based on the distribution from the Procedure section.
1. $x$ $y$ P($x$) = P($y$)
0
1
2
3
2. Calculate the following:
1. $\mu_{x}$ = _______
2. $\sigma_{x}$ = _______
3. $\mu_{x}$ = _______
3. Explain what μx represents.
4. Explain what μy represents.
5. Based upon theory:
1. What was the expected profit per game?
2. Did the expected profit represent an average win or loss per game?
3. How do you know? Answer in complete sentences.
6. Construct a histogram of the theoretical distribution.
Use the Data
Note 4.9.1
RF = relative frequency
Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.
1. P(x = 3) = _________________
2. P(0 < x < 3) = _________________
3. P(x ≥ 2) = _________________
Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.
1. RF(x = 3) = _________________
2. RF(0 < x < 3) = _________________
3. RF(x ≥ 2) = _________________
Discussion Question
For questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations.
1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences.
2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions.
3. Thinking about your answers to questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not.
4. Suppose that the experiment had been repeated 500 times. Would you expect Table orTable to change, and how would it change? Why? Why wouldn’t the other table change? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.08%3A_Discrete_Distribution_%28Playing_Card_Experiment%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
4.2: Probability Distribution Function (PDF) for a Discrete Random Variable
Q 4.2.1
Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given in Table.
$x$ $P(x)$
3 0.05
4 0.40
5 0.30
6 0.15
7 0.10
1. In words, define the random variable $X$.
2. What does it mean that the values zero, one, and two are not included for $x$ in the PDF?
Exercise 4.3.5
Complete the expected value table.
$x$ $P(x)$ $x*P(x)$
0 0.2
1 0.2
2 0.4
3 0.2
Exercise 4.3.6
Find the expected value from the expected value table.
$x$ $P(x)$ $x*P(x)$
2 0.1
4 0.3 4(0.3) = 1.2
6 0.4 6(0.4) = 2.4
8 0.2 8(0.2) = 1.6
Answer
$0.2 + 1.2 + 2.4 + 1.6 = 5.4$
Exercise 4.3.7
Find the standard deviation.
$x$ $P(x)$ $x*P(x)$ $(x – \mu)^{2}P(x)$
2 0.1 2(0.1) = 0.2 (2–5.4)2(0.1) = 1.156
4 0.3 4(0.3) = 1.2 (4–5.4)2(0.3) = 0.588
6 0.4 6(0.4) = 2.4 (6–5.4)2(0.4) = 0.144
8 0.2 8(0.2) = 1.6 (8–5.4)2(0.2) = 1.352
Exercise 4.3.8
Identify the mistake in the probability distribution table.
$x$ $P(x)$ $x*P(x)$
1 0.15 0.15
2 0.25 0.50
3 0.30 0.90
4 0.20 0.80
5 0.15 0.75
Answer
The values of $P(x)$ do not sum to one.
Exercise 4.3.9
Identify the mistake in the probability distribution table.
$x$ $P(x)$ $x*P(x)$
1 0.15 0.15
2 0.25 0.40
3 0.25 0.65
4 0.20 0.85
5 0.15 1
Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He has the following probability distribution.
$x$ $P(x)$ $x*P(x)$
1 0.35
2 0.20
3 0.15
4
5 0.10
6 0.05
Exercise 4.3.10
Define the random variable $X$.
Answer
Let $X =$ the number of years a physics major will spend doing post-graduate research.
Exercise 4.3.11
Define $P(x)$, or the probability of $x$.
Exercise 4.3.12
Find the probability that a physics major will do post-graduate research for four years. $P(x = 4) =$ _______
Answer
$1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15$
Exercise 4.3.13
Find the probability that a physics major will do post-graduate research for at most three years. $P(x \leq 3) =$ _______
Exercise 4.3.14
On average, how many years would you expect a physics major to spend doing post-graduate research?
Answer
$1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6$ years
Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution.
• Let $X =$ the number of years a student will study ballet with the teacher.
• Let $P(x) =$ the probability that a student will study ballet $x$ years.
Exercise 4.3.15
Complete Table using the data provided.
$x$ $P(x)$ $x*P(x)$
1 0.10
2 0.05
3 0.10
4
5 0.30
6 0.20
7 0.10
Exercise 4.3.16
In words, define the random variable $X$.
Answer
$X$ is the number of years a student studies ballet with the teacher.
Exercise 4.3.17
$P(x = 4) =$ _______
Exercise 4.3.18
$P(x < 4) =$ _______
Answer
$0.10 + 0.05 + 0.10 = 0.25$
Exercise 4.3.19
On average, how many years would you expect a child to study ballet with this teacher?
Exercise 4.3.20
What does the column "$P(x)$" sum to and why?
Answer
The sum of the probabilities sum to one because it is a probability distribution.
Exercise 4.3.21
What does the column "$x*P(x)$" sum to and why?
Exercise 4.3.22
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay$2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?
Answer
$−2\left(\dfrac{40}{52}\right)+30\left(\dfrac{12}{52}\right) = −1.54 + 6.92 = 5.38$
Exercise 4.3.23
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay$2. There are 12 face cards in a deck of 52 cards. Should you play the game?
4.3: Mean or Expected Value and Standard Deviation
Q 4.3.1
A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of$150.
1. What are you interested in here?
2. In words, define the random variable $X$.
3. List the values that $X$ may take on.
4. Construct a PDF.
5. If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?
Q 4.3.2
A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails.
• If the card is a face card, and the coin lands on Heads, you win $6 • If the card is a face card, and the coin lands on Tails, you win$2
• If the card is not a face card, you lose $2, no matter what the coin shows. 1. Find the expected value for this game (expected net gain or loss). 2. Explain what your calculations indicate about your long-term average profits and losses on this game. 3. Should you play this game to win money? S 4.3.2 The variable of interest is $X$, or the gain or loss, in dollars. The face cards jack, queen, and king. There are $(3)(4) = 12$ face cards and $52 – 12 = 40$ cards that are not face cards. We first need to construct the probability distribution for $X$. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of $X$ to determine the expected value. Card Event $X$ net gain/loss $P(X)$ Face Card and Heads 6 $\left(\frac{12}{52}\right) \left(\frac{1}{2}\right) = \left(\frac{6}{52}\right)$ Face Card and Tails 2 $\left(\frac{12}{52}\right) \left(\frac{1}{2}\right) = \left(\frac{6}{52}\right)$ (Not Face Card) and (H or T) –2 $\left(\frac{40}{52}\right) (1) = \left(\frac{40}{52}\right)$ • $\text{Expected value} = (6)\left(\frac{6}{52}\right) + (2)\left(\frac{6}{52}\right) + (-2)\left(\frac{40}{52}\right) = -\frac{32}{52}$ • $\text{Expected value} = –0.62$, rounded to the nearest cent • If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average. • You should not play this game to win money because the expected value indicates an expected average loss. Q 4.3.3 You buy a lottery ticket to a lottery that costs$10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $500 prize, two$100 prizes, and four $25 prizes. Find your expected gain or loss. Q 4.3.4 Complete the PDF and answer the questions. $x$ $P(x)$ $xP(x)$ 0 0.3 1 0.2 2 3 0.4 1. Find the probability that $x = 2$. 2. Find the expected value. S 4.3.4 1. 0.1 2. 1.6 Q 4.3.5 Suppose that you are offered the following “deal.” You roll a die. If you roll a six, you win$10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay$6.
1. What are you ultimately interested in here (the value of the roll or the money you win)?
2. In words, define the Random Variable $X$.
3. List the values that $X$ may take on.
4. Construct a PDF.
5. Over the long run of playing this game, what are your expected average winnings per game?
6. Based on numerical values, should you take the deal? Explain your decision in complete sentences.
Q 4.3.6
A venture capitalist, willing to invest $1,000,000, has three investments to choose from. The first investment, a software company, has a 10% chance of returning$5,000,000 profit, a 30% chance of returning $1,000,000 profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning$3,000,000 profit, a 40% chance of returning $1,000,000 profit, and a 40% chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning$6,000,000 profit, a 70% of no profit or loss, and a 20% chance of losing the million dollars.
1. Construct a PDF for each investment.
2. Find the expected value for each investment.
3. Which is the safest investment? Why do you think so?
4. Which is the riskiest investment? Why do you think so?
5. Which investment has the highest expected return, on average?
S 4.3.6
1. Software Company
$x$ $P(x)$
5,000,000 0.10
1,000,000 0.30
–1,000,000 0.60
Hardware Company
$x$ $P(x)$
3,000,000 0.20
1,000,000 0.40
–1,000,00 0.40
Biotech Firm
$x$ $P(x)$
6,00,000 0.10
0 0.70
–1,000,000 0.20
2. $200,000;$600,000; $400,000 3. third investment because it has the lowest probability of loss 4. first investment because it has the highest probability of loss 5. second investment Q 4.3.7 Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let $X =$ the number of children married people have. $x$ $P(x)$ $xP(x)$ 0 0.10 1 0.20 2 0.30 3 4 0.10 5 0.05 6 (or more) 0.05 1. Find the probability that a married adult has three children. 2. In words, what does the expected value in this example represent? 3. Find the expected value. 4. Is it more likely that a married adult will have two to three children or four to six children? How do you know? Q 4.3.8 Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as in the Table. $x$ $P(x)$ 3 0.05 4 0.40 5 0.30 6 0.15 7 0.10 On average, how many years do you expect it to take for an individual to earn a B.S.? S 4.3.8 4.85 years Q 4.3.9 People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs. $x$ $P(x)$ 0 0.03 1 0.50 2 0.24 3 4 0.70 5 0.04 1. Describe the random variable $X$ in words. 2. Find the probability that a customer rents three DVDs. 3. Find the probability that a customer rents at least four DVDs. 4. Find the probability that a customer rents at most two DVDs. Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer. $x$ $P(x)$ 0 0.35 1 0.25 2 0.20 3 0.10 4 0.05 5 0.05 5. At which store is the expected number of DVDs rented per customer higher? 6. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form. 7. If Video to Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain. 8. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that? Q 4.3.10 A “friend” offers you the following “deal.” For a$10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift.
• Ten of the coupons are for a free gift worth $6. • Eighty of the coupons are for a free gift worth$8.
• Six of the coupons are for a free gift worth $12. • Four of the coupons are for a free gift worth$40.
Based upon the financial gain or loss over the long run, should you play the game?
1. Yes, I expect to come out ahead in money.
2. No, I expect to come out behind in money.
3. It doesn’t matter. I expect to break even.
b
Q 4.3.11
Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.
1. What is the average class size assuming each class is filled to capacity?
2. Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable $X$ equal the size of the student’s class. Define the PDF for $X$.
3. Find the mean of $X$.
4. Find the standard deviation of $X$.
In a lottery, there are 250 prizes of $5, 50 prizes of$25, and ten prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even? S 4.3.12 Let $X =$ the amount of money to be won on a ticket. The following table shows the PDF for $X$. $x$ $P(x)$ 0 0.969 5 $\frac{250}{10,000} = 0.025$ 25 $\frac{50}{10,000} = 0.005$ 100 $\frac{10}{10,000} = 0.001$ Calculate the expected value of $X$. $0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35$ A fair price for a ticket is$0.35. Any price over $0.35 will enable the lottery to raise money. 4.4: Binomial Distribution Q 4.4.1 According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Q 4.4.2 Define the random variable and list its possible values. S 4.4.2 $X =$ the number of patients calling in claiming to have the flu, who actually have the flu. $X = 0, 1, 2, ...25$ Q 4.4.3 State the distribution of $X$. Q 4.4.4 Find the probability that at least four of the 25 patients actually have the flu. S 4.4.4 0.0165 Q 4.4.5 On average, for every 25 patients calling in, how many do you expect to have the flu? Q 4.4.6 People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given Table. There is five-video limit per customer at this store, so nobody ever rents more than five DVDs. $x$ $P(x)$ 0 0.03 1 0.50 2 0.24 3 4 0.07 5 0.04 1. Describe the random variable $X$ in words. 2. Find the probability that a customer rents three DVDs. 3. Find the probability that a customer rents at least four DVDs. 4. Find the probability that a customer rents at most two DVDs. S 4.4.6 1. $X =$ the number of DVDs a Video to Go customer rents 2. 0.12 3. 0.11 4. 0.77 Q 4.4.7 A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. 1. In words, define the random variable $X$. 2. List the values that $X$ may take on. 3. Give the distribution of $X$. $X \sim$ _____(_____,_____) 4. How many of the 12 students do we expect to attend the festivities? 5. Find the probability that at most four students will attend. 6. Find the probability that more than two students will attend. Use the following information to answer the next three exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. Q 4.4.8 The expected number of wins for that upcoming month is: 1. 1.67 2. 12 3. $\frac{382}{1043}$ 4. 4.43 S 4.4.8 d. 4.43 Let $X =$ the number of games won in that upcoming month. Q 4.4.9 What is the probability that the San Jose Sharks win six games in that upcoming month? 1. 0.1476 2. 0.2336 3. 0.7664 4. 0.8903 Q 4.4.10 What is the probability that the San Jose Sharks win at least five games in that upcoming month? 1. 0.3694 2. 0.5266 3. 0.4734 4. 0.2305 S 4.4.10 c Q 4.4.11 A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct. Q 4.4.12 A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly. S 4.4.13 • $X =$ number of questions answered correctly • $X \sim B(32, \frac{1}{3})$ • We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find $P(x > 24)$. The event "more than 24" is the complement of "less than or equal to 24." • Using your calculator's distribution menu: $1 – \text{binomcdf}(32, \frac{1}{3}, 24)$ • $P(x > 24) = 0$ • The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero. Q 4.4.14 Six different colored dice are rolled. Of interest is the number of dice that show a one. 1. In words, define the random variable $X$. 2. List the values that $X$ may take on. 3. Give the distribution of $X$. $X \sim$ _____(_____,_____) 4. On average, how many dice would you expect to show a one? 5. Find the probability that all six dice show a one. 6. Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically. Q 4.4.15 More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses. 1. In words, define the random variable $X$. 2. List the values that $X$ may take on. 3. Give the distribution of $X$. $X \sim$ _____(_____,_____) 4. On average, how many schools would you expect to offer such courses? 5. Find the probability that at most ten offer such courses. 6. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence. S 4.4.15 1. $X =$ the number of college and universities that offer online offerings. 2. 0, 1, 2, …, 13 3. $X \sim B(13, 0.96)$ 4. 12.48 5. 0.0135 6. $P(x = 12) = 0.3186 P(x = 13) = 0.5882$ More likely to get 13. Q 4.4.16 Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen. 1. In words, define the random variable $X$. 2. List the values that $X$ may take on. 3. Give the distribution of $X$. $X \sim$ _____(_____,_____) 4. How many are expected to attend their graduation? 5. Find the probability that 17 or 18 attend. 6. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically. Q 4.4.17 At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey 25 fencers at The Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon. 1. In words, define the random variable $X$. 2. List the values that $X$ may take on. 3. Give the distribution of $X$. $X \sim$ _____(_____,_____) 4. How many are expected to not to use the foil as their main weapon? 5. Find the probability that six do not use the foil as their main weapon. 6. Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically. S 4.4.17 1. $X =$ the number of fencers who do not use the foil as their main weapon 2. 0, 1, 2, 3,... 25 3. $X \sim B(25,0.40)$ 4. 10 5. 0.0442 6. The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising. Q 4.4.18 Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school. 1. In words, define the random variable $X$. 2. List the values that $X$ may take on. 3. Give the distribution of $X$. $X \sim$ _____(_____,_____) 4. How many seniors are expected to have participated in after-school sports all four years of high school? 5. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. 6. Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. Q 4.4.19 The chance of an IRS audit for a tax return with over$25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many audits are expected in a 20-year period?
5. Find the probability that a person is not audited at all.
6. Find the probability that a person is audited more than twice.
S 4.4.19
1. $X =$ the number of audits in a 20-year period
2. 0, 1, 2, …, 20
3. $X \sim B(20, 0.02)$
4. 0.4
5. 0.6676
6. 0.0071
Q 4.4.20
It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. What is the probability that at least eight have adequate earthquake supplies?
5. Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why?
6. How many residents do you expect will have adequate earthquake supplies?
Q 4.4.21
There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The “house” rolls three dice. If none of the dice show the number or object that was bet, the house keeps the$1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back his or her $1 bet, plus$1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back his or her $1 bet, plus$2 profit. If all three dice show the number or object bet, the player gets back his or her $1 bet, plus$3 profit. Let $X =$ number of matches and $Y =$ profit per game.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. List the values that $Y$ may take on. Then, construct one PDF table that includes both $X$ and $Y$ and their probabilities.
5. Calculate the average expected matches over the long run of playing this game for the player.
6. Calculate the average expected earnings over the long run of playing this game for the player.
7. Determine who has the advantage, the player or the house.
S 4.4.21
1. $X =$ the number of matches
2. 0, 1, 2, 3
3. $X \sim B(3,16)(3,16)$
4. In dollars: −1, 1, 2, 3
5. $\frac{1}{2}$
6. Multiply each $Y$ value by the corresponding $X$ probability from the PDF table. The answer is −0.0787. You lose about eight cents, on average, per game.
7. The house has the advantage.
Q 4.4.22
According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let $X =$ the number of people who have access to electricity.
1. What is the probability distribution for $X$?
2. Using the formulas, calculate the mean and standard deviation of $X$.
3. Use your calculator to find the probability that 15 people in the sample have access to electricity.
4. Find the probability that at most ten people in the sample have access to electricity.
5. Find the probability that more than 25 people in the sample have access to electricity.
Q 4.4.23
The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in Afghanistan is 28.1%. Suppose you choose 15 people in Afghanistan at random. Let $X =$ the number of people who are literate.
1. Sketch a graph of the probability distribution of $X$.
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of $X$.
3. Find the probability that more than five people in the sample are literate. Is it is more likely that three people or four people are literate.
S 4.4.23
1. $X \sim B(15, 0.281)$
1. Mean $= \mu = np = 15(0.281) = 4.215$
2. Standard Deviation $= \sigma = \sqrt{npq} = \sqrt{15(0.281)(0.719)} = 1.7409$
2. $P(x > 5) = 1 – P(x ≤ 5) = 1 – \text{binomcdf}(15, 0.281, 5) = 1 – 0.7754 = 0.2246$
$P(x = 3) = \text{binompdf}(15, 0.281, 3) = 0.1927$
$P(x = 4) = \text{binompdf}(15, 0.281, 4) = 0.2259$
It is more likely that four people are literate that three people are.
4.5: Geometric Distribution
Q 4.5.1
A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. On average, how many dealerships would we expect her to have to call until she finds one that has the car?
5. Find the probability that she must call at most four dealerships.
6. Find the probability that she must call three or four dealerships.
Q 4.5.2
Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many adults in America do you expect to survey until you find one who will watch the Super Bowl?
5. Find the probability that you must ask seven people.
6. Find the probability that you must ask three or four people.
S 4.5.2
1. $X =$ the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
2. $X \sim G(0.40)$
3. 2.5
4. 0.0187
5. 0.2304
Q 4.5.3
It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies?
5. What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies?
6. How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies?
7. How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies?
Q 4.5.4
In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many pages do you expect to advertise footwear on them?
5. Is it probable that all twenty will advertise footwear on them? Why or why not?
6. What is the probability that fewer than ten will advertise footwear on them?
7. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable $X$ and give its distribution.
8. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it?
9. How many pages do you expect to need to survey in order to find one that advertises footwear?
S 4.5.4
1. $X =$ the number of pages that advertise footwear
2. $X$ takes on the values 0, 1, 2, ..., 20
3. $X \sim B(20, \frac{29}{192})$
4. 3.02
5. No
6. 0.9997
7. $X =$ the number of pages we must survey until we find one that advertises footwear. $X \sim G(\frac{29}{192})$
8. 0.3881
9. 6.6207 pages
Q 4.5.5
Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let $\text{F}$ be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome.
• $p =$ probability of success (event $\text{F}$ occurs)
• $q =$ probability of failure (event $\text{F}$ does not occur)
1. Write the description of the random variable $X$.
2. What are the values that $X$ can take on?
3. Find the values of $p$ and $q$.
4. Find the probability that the first occurrence of event $\text{F}$ (rolling a four or five) is on the second trial.
Q 4.5.5
Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does $X$ take on?
0, 1, 2, and 3
Q 4.5.6
The World Bank records the prevalence of HIV in countries around the world. According to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who are infected with HIV.”1 In South Africa, the prevalence of HIV is 17.3%. Let $X =$ the number of people you test until you find a person infected with HIV.
1. Sketch a graph of the distribution of the discrete random variable $X$.
2. What is the probability that you must test 30 people to find one with HIV?
3. What is the probability that you must ask ten people?
4. Find the (i) mean and (ii) standard deviation of the distribution of $X$.
Q 4.5.7
According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let $X =$ the number of millenials you ask until you find a person without a profile on a social networking site.
1. Describe the distribution of $X$.
2. Find the (i) mean and (ii) standard deviation of $X$.
3. What is the probability that you must ask ten people to find one person without a social networking site?
4. What is the probability that you must ask 20 people to find one person without a social networking site?
5. What is the probability that you must ask at most five people?
S 4.5.7
1. $X \sim \text{G}(0.25)$
1. Mean $= \mu = \frac{1}{p} = \frac{1}{0.25} = 4$
2. Standard Deviation $= \sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.25}{0.25^{2}}} \approx 3.4641$
2. $P(x = 10) = \text{geometpdf}(0.25, 10) = 0.0188$
3. $P(x = 20) = \text{geometpdf}(0.25, 20) = 0.0011$
4. $P(x \leq 5) = \text{geometcdf}(0.25, 5) = 0.7627$
Footnotes
1. ”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/...last&sort=desc (accessed May 15, 2013).
4.6: Hypergeometric Distribution
Q 4.6.1
A group of Martial Arts students is planning on participating in an upcoming demonstration. Six are students of Tae Kwon Do; seven are students of Shotokan Karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of Shotokan Karate students in that first demonstration.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many Shotokan Karate students do we expect to be in that first demonstration?
Q 4.6.2
In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many pages do you expect to advertise footwear on them?
5. Calculate the standard deviation.
S 4.6.2
1. $X =$ the number of pages that advertise footwear
2. 0, 1, 2, 3, ..., 20
3. $X \sim \text{H}(29, 163, 20); r = 29, b = 163, n = 20$
4. 3.03
5. 1.5197
Q 4.6.3
Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that ten people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many instructors do you expect on the committee who are not technically proficient?
5. Find the probability that at least five on the committee are not technically proficient.
6. Find the probability that at most three on the committee are not technically proficient.
Q 4.6.4
Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the Boston Celtics and four volunteers from the New England Patriots. We are interested in the number of Patriots picked.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. Are you choosing the nine athletes with or without replacement?
S 4.6.4
1. $X =$ the number of Patriots picked
2. 0, 1, 2, 3, 4
3. $X \sim H(4, 8, 9)$
4. Without replacement
Q 4.6.5
A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart?
1. What is the group of interest?
2. How many are in the group of interest?
3. How many are in the other group?
4. Let $X =$ _________. What values does $X$ take on?
5. The probability question is $P$(_______).
6. Find the probability in question.
7. Find the (i) mean and (ii) standard deviation of $X$.
4.7: Poisson Distribution
Q 4.7.1
The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let $X =$ the number of calls received at noon.
1. Find the mean and standard deviation of $X$.
2. What is the probability that the office receives at most six calls at noon on Monday?
3. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon?
4. What is the probability that the office receives more than eight calls at noon?
S 4.7.1
1. $X \sim P(5.5); \mu= 5.5; \sigma = \sqrt{5.5} \approx 2.3452$
2. $P(x \leq 6) = \text{poissoncdf}(5.5, 6) \approx 0.6860$
3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
4. $P(x > 8) = 1 – P(x \leq 8) = 1 – \text{poissoncdf}(5.5, 8) \approx 1 – 0.8944 = 0.1056$
Q 4.7.2
The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let $X =$ the number of births in an hour.
1. Find the mean and standard deviation of $X$.
2. Sketch a graph of the probability distribution of $X$.
3. What is the probability that the maternity ward will deliver three babies in one hour?
4. What is the probability that the maternity ward will deliver at most three babies in one hour?
5. What is the probability that the maternity ward will deliver more than five babies in one hour?
Q 4.7.3
A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.
S 4.7.3
Let $X =$ the number of defective bulbs in a string.
Using the Poisson distribution:
• $\mu = np = 100(0.03) = 3$
• $X \sim P(3)$
• $P(x \leq 4) = \text{poissoncdf}(3, 4) \approx 0.8153$
Using the binomial distribution:
• $X \sim \text{B}(100, 0.03)$
• $P(x \leq 4) = \text{binomcdf}(100, 0.03, 4) \approx 0.8179$
The Poisson approximation is very good—the difference between the probabilities is only 0.0026.
Q 4.7.4
The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. Find the probability that she has no children.
5. Find the probability that she has fewer children than the Japanese average.
6. Find the probability that she has more children than the Japanese average.
Q 4.7.5
The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen.
1. In words, define the Random Variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. Find the probability that she has no children.
5. Find the probability that she has fewer children than the Spanish average.
6. Find the probability that she has more children than the Spanish average .
S 4.7.5
1. $X =$ the number of children for a Spanish woman
2. 0, 1, 2, 3,...
3. $X \sim P(1.47)$
4. 0.2299
5. 0.5679
6. 0.4321
Q 4.7.6
Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces:
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _______
4. Find the probability that she has no litters in one year.
5. Find the probability that she has at least two litters in one year.
6. Find the probability that she has exactly three litters in one year.
Q 4.7.7
he chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many cookies do we expect to have an extra fortune?
5. Find the probability that none of the cookies have an extra fortune.
6. Find the probability that more than three have an extra fortune.
7. As $n$ increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.
S 4.7.7
1. $X =$ the number of fortune cookies that have an extra fortune
2. 0, 1, 2, 3,... 144
3. $X \sim B(144, 0.03)$ or $P(4.32)$
4. 4.32
5. 0.0124 or 0.0133
6. 0.6300 or 0.6264
7. As $n$ gets larger, the probabilities get closer together.
Q 4.7.8
According to the South Carolina Department of Mental Health web site, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many are expected to suffer from anorexia?
5. Find the probability that no one suffers from anorexia.
6. Find the probability that more than four suffer from anorexia.
Q 4.7.9
The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. Suppose that 100 people with tax returns over$25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to answer the following questions.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many are expected to be audited?
5. Find the probability that no one was audited.
6. Find the probability that at least three were audited.
S 4.7.9
1. $X =$ the number of people audited in one year
2. 0, 1, 2, ..., 100
3. $X \sim P(2)$
4. 2
5. 0.1353
6. 0.3233
Q 4.7.10
Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. How many seniors are expected to have participated in after-school sports all four years of high school?
5. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
6. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
Q 4.7.11
On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes.
1. In words, define the random variable $X$.
2. List the values that $X$ may take on.
3. Give the distribution of $X$. $X \sim$ _____(_____,_____)
4. On average, how many pieces of egg shell do you expect to be in the cake?
5. What is the probability that there will not be any pieces of egg shell in the cake?
6. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes?
7. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?
S 4.7.11
1. $X =$ the number of shell pieces in one cake
2. 0, 1, 2, 3,...
3. $X \sim P(1.5)$
4. 1.5
5. 0.2231
6. 0.0001
7. Yes
Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week.
Q 4.7.12
In words, the random variable $X =$ _________________
1. the number of times Mrs. Plum’s cats wake her up each week.
2. the number of times Mrs. Plum’s cats wake her up each hour.
3. the number of times Mrs. Plum’s cats wake her up each night.
4. the number of times Mrs. Plum’s cats wake her up.
Q 4.7.13
Find the probability that her cats will wake her up no more than five times next week.
1. 0.5000
2. 0.9329
3. 0.0378
4. 0.0671
d | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.E%3A_Discrete_Random_Variables_%28Exercises%29.txt |
Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables.
• 5.1: Introduction
The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve. The curve is called the probability density function (abbreviated as pdf).
• 5.2: Continuous Probability Functions
The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P(a<x<b)P(a<x<b) . The cumulative distribution function (cdf) gives the probability as an area.
• 5.3: The Uniform Distribution
The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive.
• 5.4: The Exponential Distribution
The exponential distribution is often concerned with the amount of time until some specific event occurs. Values for an exponential random variable occur in the following way. There are fewer large values and more small values. The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.
• 5.5: Continuous Distribution (Worksheet)
A statistics Worksheet: The student will compare and contrast empirical data from a random number generator with the uniform distribution.
• 5.E: Continuous Random Variables (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
• 5.E: Exercises
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
05: Continuous Random Variables
CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Recognize and understand continuous probability density functions in general.
• Recognize the uniform probability distribution and apply it appropriately.
• Recognize the exponential probability distribution and apply it appropriately.
Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables.
The values of discrete and continuous random variables can be ambiguous. For example, if $X$ is equal to the number of miles (to the nearest mile) you drive to work, then $X$ is a discrete random variable. You count the miles. If $X$ is the distance you drive to work, then you measure values of $X$ and $X$ is a continuous random variable. For a second example, if $X$ is equal to the number of books in a backpack, then $X$ is a discrete random variable. If $X$ is the weight of a book, then $X$ is a continuous random variable because weights are measured. How the random variable is defined is very important.
Properties of Continuous Probability Distributions
The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve. The curve is called the probability density function (abbreviated as pdf). We use the symbol $f(x)$ to represent the curve. $f(x)$ is the function that corresponds to the graph; we use the density function $f(x)$ to draw the graph of the probability distribution. Area under the curve is given by a different function called the cumulative distribution function(abbreviated as cdf). The cumulative distribution function is used to evaluate probability as area.
• The outcomes are measured, not counted.
• The entire area under the curve and above the x-axis is equal to one.
• Probability is found for intervals of $x$ values rather than for individual $x$ values.
• $P(c < x < d)$ is the probability that the random variable $X$ is in the interval between the values $c$ and $d$. $P(c < x < d)$ is the area under the curve, above the x-axis, to the right of $c$ and the left of $d$.
• $P(x = c) = 0$ The probability that $x$ takes on any single individual value is zero. The area below the curve, above the x-axis, and between $x = c$ and $x = c$ has no width, and therefore no area (area = 0). Since the probability is equal to the area, the probability is also zero.
• $P(c < x < d)$ is the same as $P(c \leq x \leq d)$ because probability is equal to area.
We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, the formulas were found by using the techniques of integral calculus. However, because most students taking this course have not studied calculus, we will not be using calculus in this textbook. There are many continuous probability distributions. When using a continuous probability distribution to model probability, the distribution used is selected to model and fit the particular situation in the best way.
In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions.
Glossary
Uniform Distribution
a continuous random variable (RV) that has equally likely outcomes over the domain, $a < x < b$; it is often referred as the rectangular distribution because the graph of the pdf has the form of a rectangle. Notation: $X \sim U(a,b)$. The mean is $\mu = \frac{a+b}{2}$ and the standard deviation is $\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$. The probability density function is $f(x) = \frac{1}{b-a}$ for $a < x < b$ or $a \leq x \leq b$. The cumulative distribution is $P(X \leq x) = \frac{x-a}{b-a}$.
Exponential Distribution
a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital; the notation is $X \sim \text{Exp}(m)$. The mean is $\mu = \frac{1}{m}$ and the standard deviation is $\sigma = \frac{1}{m}$. The probability density function is $f(x) = me^{-mx}$, $x \geq 0$ and the cumulative distribution function is $P(X \leq x) = 1 − e^{mx}$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/05%3A_Continuous_Random_Variables/5.01%3A_Introduction.txt |
We begin by defining a continuous probability density function. We use the function notation $f(x)$. Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function $f(x)$ so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA.
Example $1$
Consider the function $f(x) = \frac{1}{20}$ for $0 \leq x \leq 20$. $x =$ a real number. The graph of $f(x) = \frac{1}{20}$ is a horizontal line. However, since $0 \leq x \leq 20$, $f(x)$ is restricted to the portion between $x = 0$ and $x = 20$, inclusive.
$f(x) = \frac{1}{20} \text{ for } 0 \leq x \leq 20.$
The graph of $f(x) = \frac{1}{20}$ is a horizontal line segment when $0 \leq x \leq 20$.
The area between $f(x) = \frac{1}{20}$ where $0 \leq x \leq 20$ and the x-axis is the area of a rectangle with base = 20 and height = $\frac{1}{20}$.
$AREA = 20 \left(\frac{1}{20} \right) = 1$
Suppose we want to find the area between $f(x) = \frac{1}{20}$ and the x-axis where $0 < x < 2$.
$AREA = (2 - 0) \left(\dfrac{1}{20} \right) = 0.1$
$(2 - 0) = 2 = \text{base of a rectangle}$
REMINDER: area of a rectangle = (base)(height).
The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written mathematically as $P(0 < x < 2) = P(x < 2) = 0.1$.
Suppose we want to find the area between $f(x) = \frac{1}{20}$ and the x-axis where $4 < x < 15$.
$\text{AREA} = (15 – 4)(\frac{1}{20}) = 0.55$
$\text{AREA} = (15 – 4)(\frac{1}{20}) = 0.55$
$(15 – 4) = 11 = \text{the base of a rectangle}(15 – 4) = 11 = \text{the base of a rectangle}$
The area corresponds to the probability $P(4 < x < 15) = 0.55$.
Suppose we want to find $P(x = 15)$. On an x-y graph, $x = 15$ is a vertical line. A vertical line has no width (or zero width). Therefore, $P(x = 15) = (\text{base})(\text{height}) = (0)\left(\frac{1}{20}\right) = 0$
$P(X \leq x)$ (can be written as $P(X < x)$ for continuous distributions) is called the cumulative distribution function or CDF. Notice the "less than or equal to" symbol. We can use the CDF to calculate $P(X > x)$. The CDF gives "area to the left" and $P(X > x)$ gives "area to the right." We calculate $P(X > x)$ for continuous distributions as follows: $P(X > x) = 1 – P(X < x)$.
Label the graph with $f(x)$ and $x$. Scale the $x$ and $y$ axes with the maximum $x$ and $y$ values. $f(x) = \frac{1}{20}$, $0 \leq x \leq 20$.
To calculate the probability that $x$ is between two values, look at the following graph. Shade the region between $x = 2.3$ and $x = 12.7$. Then calculate the shaded area of a rectangle.
$P(2.3 < x < 12.7) = (\text{base})(\text{height}) = (12.7−2.3)\left(\dfrac{1}{20}\right) = 0.52$
Exercise $1$
Consider the function $f(x) = \frac{1}{8}$ for $0 \leq x \leq 8$. Draw the graph of $f(x)$ and find $P(2.5 < x < 7.5)$.
Answer
$P (2.5 < x < 7.5) = 0.625$
Summary
The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to $P(a < x < b)$. The cumulative distribution function (cdf) gives the probability as an area. If $X$ is a continuous random variable, the probability density function (pdf), $f(x)$, is used to draw the graph of the probability distribution. The total area under the graph of $f(x)$ is one. The area under the graph of $f(x)$ and between values a and b gives the probability $P(a < x < b)$.
The cumulative distribution function (cdf) of $X$ is defined by $P(X \leq x)$. It is a function of $x$ that gives the probability that the random variable is less than or equal to $x$.
Formula Review
Probability density function (pdf) $f(x)$:
• $f(x) \geq 0$
• The total area under the curve $f(x)$ is one.
Cumulative distribution function (cdf): $P(X \leq x)$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/05%3A_Continuous_Random_Variables/5.02%3A_Continuous_Probability_Functions.txt |
The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive.
Example 5.3.1
The data in Table $1$ are 55 smiling times, in seconds, of an eight-week-old baby.
Table $1$
10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9
12.8 14.8 22.8 20.0 15.9 16.3 13.4 17.1 14.5 19.0 22.8
1.3 0.7 8.9 11.9 10.9 7.3 5.9 3.7 17.9 19.2 9.8
5.8 6.9 2.6 5.8 21.7 11.8 3.4 2.1 4.5 6.3 10.7
8.9 9.4 9.4 7.6 10.0 3.3 6.7 7.8 11.6 13.8 18.6
The sample mean = 11.49 and the sample standard deviation = 6.23.
We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.
Let $X =$ length, in seconds, of an eight-week-old baby's smile.
The notation for the uniform distribution is
$X \sim U(a, b)$ where $a =$ the lowest value of $x$ and $b =$ the highest value of $x$.
The probability density function is $f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$.
For this example, $X \sim U(0, 23)$ and $f(x) = \frac{1}{23-0}$ for $0 \leq X \leq 23$.
Formulas for the theoretical mean and standard deviation are
$\mu = \frac{a+b}{2} \nonumber$
and
$\sigma = \sqrt{\frac{(b-a)^{2}}{12}} \nonumber$
For this problem, the theoretical mean and standard deviation are
$\mu = \frac{0+23}{2} = 11.50 \, seconds \nonumber$
and
$\sigma = \frac{(23-0)^{2}}{12} = 6.64\, seconds. \nonumber$
Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.
Exercise $1$
The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and $b$. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.
Table $2$
1 12 4 10 4 14 11
7 11 4 13 2 4 6
3 10 0 12 6 9 10
5 13 4 10 14 12 11
6 10 11 0 11 13 2
Answer
$a$ is zero; $b$ is $14$; $X \sim U (0, 14)$; $\mu = 7$ passengers; $\sigma = 4.04$ passengers
Example 5.3.2A
a. Refer to Example 5.3.1. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?
Answer
a. Find $P(2 < x < 18)$.
$P(2 < x < 18) = (\text{base})(\text{height}) = (18 – 2)\left(\frac{1}{23}\right) = \left(\frac{16}{23}\right)$.
Exercise $2$B
b. Find the 90th percentile for an eight-week-old baby's smiling time.
Answer
b. Ninety percent of the smiling times fall below the 90th percentile, $k$, so $P(x < k) = 0.90$
$P(x < k)= 0.90$
$(\text{base})(\text{height}) = 0.90$
$(k−0)\left(\frac{1}{23}\right) = 0.90$
$k = (23)(0.90) = 20.7$
Exercise $3$C
c. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN EIGHT SECONDS.
Answer
c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds.
Find $P(x > 12 | x > 8)$ There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds.
Write a new $f(x): f(x) = \frac{1}{23-8} = \frac{1}{15}$
for $8 < x < 23$
$P(x > 12 | x > 8) = (23 − 12)\left(\frac{1}{15}\right) = \left(\frac{11}{15}\right)$
For the second way, use the conditional formula from Probability Topics with the original distribution $X \sim U(0, 23)$:
$P(\text{A|B}) = \frac{P(\text{A AND B})}{P(\text{B})}$
For this problem, $\text{A}$ is ($x > 12$) and $\text{B}$ is ($x > 8$).
So, $P(x > 12|x > 8) = \frac{(x > 12 \text{ AND } x > 8)}{P(x > 8)} = \frac{P(x > 12)}{P(x > 8)} = \frac{\frac{11}{23}}{\frac{15}{23}} = \frac{11}{15}$
Exercise $2$
A distribution is given as $X \sim U(0, 20)$. What is $P(2 < x < 18)$? Find the 90thpercentile.
Answer
$P(2 < x < 18) = 0.8$; 90th percentile $= 18$
Example 5.3.3
The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.
Exercise $3$.1
a. What is the probability that a person waits fewer than 12.5 minutes?
Answer
a. Let $X =$ the number of minutes a person must wait for a bus. $a = 0$ and $b = 15$. $X \sim U(0, 15)$. Write the probability density function. $f(x) = \frac{1}{15-0} = \frac{1}{15}$ for $0 \leq x \leq 15$.
Find $P(x < 12.5)$. Draw a graph.
$P(x < k) = (\text{base})(\text{height}) = (12.5−0)\left(\frac{1}{15}\right) = 0.8333$
The probability a person waits less than 12.5 minutes is 0.8333.
Exercise $3$.2
b. On the average, how long must a person wait? Find the mean, $\mu$, and the standard deviation, $\sigma$.
Answer
b. $\mu = \frac{a+b}{2} = \frac{15+0}{2} = 7.5$. On the average, a person must wait 7.5 minutes.
$\sigma = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(12-0)^{2}}{12}} = 4.3$. The Standard deviation is 4.3 minutes.
Exercise $3$.3
c. Ninety percent of the time, the time a person must wait falls below what value?
Note 5.3.3.3.1
This asks for the 90th percentile.
Answer
c. Find the 90th percentile. Draw a graph. Let $k =$ the 90th percentile.
$P(x < k) = (\text{base})(\text{height}) = (k−0)\left(\frac{1}{15}\right)$
$0.90 = (k)\left(\frac{1}{15}\right)$
$k = (0.90)(15) = 13.5$
$k$ is sometimes called a critical value.
The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.
Exercise $4$
The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.
1. Find $a$ and $b$ and describe what they represent.
2. Write the distribution.
3. Find the mean and the standard deviation.
4. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?
5. What is the 65th percentile for the duration of games for a team for the 2011 season?
Answer
1. $a$ is $447$, and $b$ is $521$. a is the minimum duration of games for a team for the 2011 season, and $b$ is the maximum duration of games for a team for the 2011 season.
2. $X \sim U(447, 521)$.
3. $\mu = 484$, and $\sigma = 21.36$
4. $P(480 < x < 500) = 0.2703$
5. 65th percentile is 495.1 hours.
Example 5.3.4
Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let $X =$ the time, in minutes, it takes a nine-year old child to eat a donut. Then $X \sim U(0.5, 4)$.
a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______.
Solution
a. 0.5714
Exercise $4$.1
b. Find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.
The second question has a conditional probability. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example). You must reduce the sample space. First way: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes.
Write a new $f(x)$:
$f(x) = \frac{1}{4-1.5} = \frac{2}{5}$ for $1.5 \leq x \leq 4$.
Find $P(x > 2|x > 1.5)$. Draw a graph.
$P(x > 2|x > 1.5) = (\text{base})(\text{new height}) = (4 − 2)(25)\left(\frac{2}{5}\right) =$ ?
Answer
b. $\frac{4}{5}$
The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is $\frac{4}{5}$.
Second way: Draw the original graph for $X \sim U(0.5, 4)$. Use the conditional formula
$P(x > 2 | x > 1.5) = \frac{P(x > 2 \text{AND} x > 1.5)}{P(x > 1.5)} = \frac{P(x>2)}{P(x>1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}$
Exercise $5$
Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let $X =$ the time, in minutes, it takes a student to finish a quiz. Then $X \sim U(6, 15)$.
Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes.
Answer
$P(x > 8) = 0.7778$
$P(x > 8 | x > 7) = 0.875$
Example 5.3.5
Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let $x =$ the time needed to fix a furnace. Then $x \sim U(1.5, 4)$.
1. Find the probability that a randomly selected furnace repair requires more than two hours.
2. Find the probability that a randomly selected furnace repair requires less than three hours.
3. Find the 30th percentile of furnace repair times.
4. The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent?
5. Find the mean and standard deviation
Solution
a. To find $f(x): f(x) = \frac{1}{4-1.5} = \frac{1}{2.5}$ so $f(x) = 0.4$
$P(x > 2) = (\text{base})(\text{height}) = (4 – 2)(0.4) = 0.8$
b. $P(x < 3) = (\text{base})(\text{height}) = (3 – 1.5)(0.4) = 0.6$
The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between $x = 1.5$ and $x = 3$. Note that the shaded area starts at $x = 1.5$ rather than at $x = 0$; since $X \sim U(1.5, 4)$, $x$ can not be less than 1.5.
c.
$P(x < k) = 0.30$
$P(x < k) = (\text{base})(\text{height}) = (k – 1.5)(0.4)$
$0.3 = (k – 1.5) (0.4)$; Solve to find $k$:
$0.75 = k – 1.5$, obtained by dividing both sides by 0.4
$k = 2.25$ , obtained by adding 1.5 to both sides
The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.25 hours or less.
d.
$P(x > k) = 0.25$
$P(x > k) = (\text{base})(\text{height}) = (4 – k)(0.4)$
$0.25 = (4 – k)(0.4)$; Solve for $k$:
$0.625 = 4 − k$,
obtained by dividing both sides by 0.4
$−3.375 = −k$,
obtained by subtracting four from both sides: $k = 3.375$
The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).
Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times.
e. $\mu = \frac{a+b}{2}$ and $\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$
$\mu = \frac{1.5+4}{2} = 2.75$ hours and $\sigma = \sqrt{\frac{(4-1.5)^{2}}{12}} = 0.7217$ hours
Exercise $6$
The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let $X =$ the time needed to change the oil on a car.
1. Write the random variable $X$ in words. $X =$ __________________.
2. Write the distribution.
3. Graph the distribution.
4. Find $P(x > 19)$.
5. Find the 50th percentile.
Answer
1. Let $X =$ the time needed to change the oil in a car.
2. $X \sim U(11, 21)$.
3. $P(x > 19) = 0.2$
4. the 50th percentile is 16 minutes.
Review
If $X$ has a uniform distribution where $a < x < b$ or $a \leq x \leq b$, then $X$ takes on values between $a$ and $b$ (may include $a$ and $b$). All values $x$ are equally likely. We write $X \sim U(a, b)$. The mean of $X$ is $\mu = \frac{a+b}{2}$. The standard deviation of $X$ is $\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$. The probability density function of $X$ is $f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$. The cumulative distribution function of $X$ is $P(X \leq x) = \frac{x-a}{b-a}$. $X$ is continuous.
The probability $P(c < X < d)$ may be found by computing the area under $f(x)$, between $c$ and $d$. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.
Formula Review
$X =$ a real number between $a$ and $b$ (in some instances, $X$ can take on the values $a$ and $b$). $a =$ smallest $X$; $b =$ largest $X$
$X \sim U(a, b)$
The mean is $\mu = \frac{a+b}{2}$
The standard deviation is $\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$
Probability density function: $f(x) = \frac{1}{b-a} \text{for} a \leq X \leq b$
Area to the Left of $x$: $P(X < x) = (x – a)\left(\frac{1}{b-a}\right)$
Area to the Right of $x$: P($X$ > $x$) = (b – x)$\left(\frac{1}{b-a}\right)$
Area Between $c$ and $d$: $P(c < x < d) = (\text{base})(\text{height}) = (d – c)\left(\frac{1}{b-a}\right)$
Uniform: $X \sim U(a, b)$ where $a < x < b$
• pdf: $f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$
• cdf: $P(X \leq x) = \frac{x-a}{b-a}$
• mean $\mu = \frac{a+b}{2}$
• standard deviation $\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$
• $P(c < X < d) = (d – c)\left(\frac{1}{b-a}\right)$
References
McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.
Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.
1.5 2.4 3.6 2.6 1.6 2.4 2.0
3.5 2.5 1.8 2.4 2.5 3.5 4.0
2.6 1.6 2.2 1.8 3.8 2.5 1.5
2.8 1.8 4.5 1.9 1.9 3.1 1.6
The sample mean = 2.50 and the sample standard deviation = 0.8302.
The distribution can be written as $X \sim U(1.5, 4.5)$.
Exercise $7$
What type of distribution is this?
Exercise $8$
In this distribution, outcomes are equally likely. What does this mean?
Answer
It means that the value of x is just as likely to be any number between 1.5 and 4.5.
Exercise $9$
What is the height of $f(x)$ for the continuous probability distribution?
Exercise $10$
What are the constraints for the values of $x$?
Answer
$1.5 \leq x \leq 4.5$
Exercise $11$
Graph $P(2 < x < 3)$.
Exercise $12$
What is $P(2 < x < 3)$?
Answer
0.3333
Exercise $13$
What is $P(x < 3.5 | x < 4)$?
Exercise $14$
What is $P(x = 1.5)$?
Answer
zero
Exercise $15$
What is the 90th percentile of square footage for homes?
Exercise $16$
Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet.
Answer
0.6
Exercise $17$
What is $a$? What does it represent?
Exercise $18$
What is $b$? What does it represent?
Answer
$b$ is $12$, and it represents the highest value of $x$.
Exercise $19$
What is the probability density function?
Exercise $20$
What is the theoretical mean?
Answer
six
Exercise $21$
What is the theoretical standard deviation?
Exercise $22$
Draw the graph of the distribution for $P(x > 9)$.
Answer
Exercise $23$
Find $P(x > 9)$.
Exercise $24$
Find the 40th percentile.
Answer
4.8
Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.
Exercise $25$
What is being measured here?
Exercise $26$
In words, define the random variable $X$.
Answer
$X$ = The age (in years) of cars in the staff parking lot
Exercise $27$
Are the data discrete or continuous?
Exercise $28$
The interval of values for $x$ is ______.
Answer
0.5 to 9.5
Exercise $29$
The distribution for $X$ is ______.
Exercise $30$
Write the probability density function.
Answer
$f(x) = \frac{1}{9}$ where $x$ is between 0.5 and 9.5, inclusive.
Exercise $31$
Graph the probability distribution.
1. Sketch the graph of the probability distribution.
Figure $10$.
2. Identify the following values:
1. Lowest value for $\bar{x}$: _______
2. Highest value for $\bar{x}$: _______
3. Height of the rectangle: _______
4. Label for x-axis (words): _______
5. Label for y-axis (words): _______
Exercise $32$
Find the average age of the cars in the lot.
Answer
$\mu$ = 5
Exercise $33$
Find the probability that a randomly chosen car in the lot was less than four years old.
1. Sketch the graph, and shade the area of interest.
Figure $11$.
2. Find the probability. $P(x < 4) =$ _______
Exercise $34$
Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old.
1. Sketch the graph, shade the area of interest.
Figure $12$.
2. Find the probability. $P(x < 4 | x < 7.5) =$ _______
Answer
1. Check student’s solution.
2. $\frac{3.5}{7}$
Exercise $35$
What has changed in the previous two problems that made the solutions different
Exercise $36$
Find the third quartile of ages of cars in the lot. This means you will have to find the value such that $\frac{3}{4}$, or 75%, of the cars are at most (less than or equal to) that age.
1. Sketch the graph, and shade the area of interest.
Figure $13$.
2. Find the value $k$ such that $P(x < k) = 0.75$.
3. The third quartile is _______
Answer
1. Check student's solution.
2. $k = 7.25$
3. $7.25$
Glossary
Conditional Probability
the likelihood that an event will occur given that another event has already occurred | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/05%3A_Continuous_Random_Variables/5.03%3A_The_Uniform_Distribution.txt |
The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.
Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.
The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.
Example $1$
Let $X$ = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.
$X$ is a continuous random variable since time is measured. It is given that $\mu = 4$ minutes. To do any calculations, you must know $m$, the decay parameter.
$m = \dfrac{1}{\mu}$. Therefore, $m = \dfrac{1}{4} = 0.25$.
The standard deviation, $\sigma$, is the same as the mean. $\mu = \sigma$
The distribution notation is $X \sim Exp(m)$. Therefore, $X \sim Exp(0.25)$.
The probability density function is $f(x) = me^{-mx}$. The number $e = 2.71828182846$... It is a number that is used often in mathematics. Scientific calculators have the key "$e^{x}$." If you enter one for $x$, the calculator will display the value $e$.
The curve is:
$f(x) = 0.25e^{-0.25x}$ where $x$ is at least zero and $m = 0.25$.
For example, $f(5) = 0.25e^{-(0.25)(5)} = 0.072$. The value 0.072 is the height of the curve when x = 5. In Example $2$ below, you will learn how to find probabilities using the decay parameter.
The graph is as follows:
Notice the graph is a declining curve. When $x = 0$,
$f(x) = 0.25e^{(-0.25)(0)} = (0.25)(1) = 0.25 = m$. The maximum value on the y-axis is m.
Exercise $1$
The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.
Answer
$X \sim Exp(0.125)$;
$f(x) = 0.125e^{-0.125x}$;
Example $2$
1. Using the information in Exercise, find the probability that a clerk spends four to five minutes with a randomly selected customer.
2. Half of all customers are finished within how long? (Find the 50th percentile)
3. Which is larger, the mean or the median?
Answer
a. Find $P(4 < x < 5)$.
The cumulative distribution function (CDF) gives the area to the left.
$P(x < x) = 1 – e^{-mx}$
$P(x < 5) = 1 – e(–0.25)(5) = 0.7135$
and
$(P(x < 4) = 1 – e^{(-0.25)(4)} = 0.6321$
You can do these calculations easily on a calculator.
The probability that a postal clerk spends four to five minutes with a randomly selected customer is
$P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814.$
On the home screen, enter (1 – e^(–0.25*5)) – (1 – e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5).
b. Find the 50th percentile.
$P(x < k) = 0.50$, $k = 2.8$ minutes (calculator or computer)
Half of all customers are finished within 2.8 minutes.
You can also do the calculation as follows:
$P(x < k) = 0.50$
and
$P(x < k) = 1 – e^{-0.25k}$
Therefore,
$0.50 = 1 − e^{-0.25k}$
and
$e^{-0.25k} = 1 − 0.50 = 0.5$
Take natural logs:
$\ln(e^{-0.25k}) = \ln(0.50).$
So,
$-0.25k = ln(0.50).$
Solve for $k: k = \dfrac{ln(0.50)}{-0.25} = 0.28$ minutes. The calculator simplifies the calculation for percentile k. See the following two notes.
A formula for the percentile $k$ is $k = ln(1 − \text{Area To The Left}) - mk = ln(1 - \text{Area To The Left}) - m$ where $ln$ is the natural log.
c. From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.
Exercise $2$
The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?
Answer
$P(x < 10) = 0.4866$
50th percentile = 10.40
Collaborative Exercise
Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.
Let $X =$ the amount of money a student in your class has in his or her pocket or purse.
The distribution for $X$ is approximately exponential with mean, $\mu =$ _______ and $m =$ _______. The standard deviation, $\sigma =$ ________.
Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than \$.40 in his or her pocket or purse. (Shade $P(x < 0.40)$).
Example $3$
On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.
1. What is the probability that a computer part lasts more than 7 years?
2. On the average, how long would five computer parts last if they are used one after another?
3. Eighty percent of computer parts last at most how long?
4. What is the probability that a computer part lasts between nine and 11 years?
Answer
a. Let $x =$ the amount of time (in years) a computer part lasts.
$\mu = 10$
so
$m = \dfrac{1}{\mu} = \dfrac{1}{10} = 0.1$
Find $P(x > 7)$. Draw the graph.
$P(x > 7) = 1 – P(x < 7).$
Since $P(X < x) = 1 – e^{-mx}$ then
$P(X > x) = 1 –(1 –e^{-mx}) = e^{-mx}$
$P(x > 7) = e^{(–0.1)(7)} = 0.4966.$
The probability that a computer part lasts more than seven years is 0.4966.
On the home screen, enter e^(-.1*7).
b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.
c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile.
Solve for $k: k = \dfrac{ln(1-0.80)}{-0.1} = 16.1$ years
Eighty percent of the computer parts last at most 16.1 years.
On the home screen, enter $\dfrac{ln(1-0.80)}{-0.1}$
d. Find $P(9 < x < 11)$. Draw the graph.
$P(9 < x < 11) = P(x < 11) - P(x < 9) = (1 - e^{(–0.1)(11)}) - (1 - e^{(–0.1)(9)}) = 0.6671 - 0.5934 = 0.0737.$
The probability that a computer part lasts between nine and 11 years is 0.0737.
On the home screen, enter e^(–0.1*9) – e^(–0.1*11).
Exercise $3$
On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?
Answer
$P(x > 15) = 0.4346$
Six pairs of running shoes would last 108 months on average.
80th percentile = 28.97 months
Example $4$
Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = $\dfrac{1}{12}$. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let $X$ = the length of a phone call, in minutes.
What is $m$, $\mu$, and $\sigma$? The probability that you must wait more than five minutes is _______ .
Answer
• $m = \dfrac{1}{12}$
• $\mu = 12$
• $\sigma = 12$
$P(x > 5) = 0.6592$
Exercise $4$
Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter $\dfrac{1}{20}$. Let $S =$ the distance people are willing to commute in miles. What is $m$, $\mu$, and $\sigma$? What is the probability that a person is willing to commute more than 25 miles?
Answer
$m = \dfrac{1}{20}$; $\mu = 20$; $\sigma = 20$; $P(x > 25) = 0.2865$
Example $5$
The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.
1. On average, how many minutes elapse between two successive arrivals?
2. When the store first opens, how long on average does it take for three customers to arrive?
3. After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
4. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
5. Seventy percent of the customers arrive within how many minutes of the previous customer?
6. Is an exponential distribution reasonable for this situation?
Answer
1. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
2. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
3. Let $X =$ the time between arrivals, in minutes. By part a, $\mu = 2$, so $m = \dfrac{1}{2} = 0.5$.
Therefore, $X \sim Exp(0.5)$.
The cumulative distribution function is $P(X < x) = 1 – e(–0.5x)^{e}$.
Therefore $P(X < 1) = 1 - e^{(–0.5)(1)} \approx 0.3935$.
$1 - e^(–0.5) \approx 0.3935$
4. $P(X > 5) = 1 - P(X < 5) = 1 - (1 - e^{(-5)(0.5)}) = e^{-2.5} \approx 0.0821$.
Figure $9$.
$1 - (1 - e^{( – 5*0.5)}$ or $e^{(-5*0.5)}$
5. We want to solve $0.70 = P(X < x)$ for $x$.
Substituting in the cumulative distribution function gives $0.70 = 1 – e^{–0.5x}$, so that $e^{–0.5x} = 0.30$. Converting this to logarithmic form gives $-0.5x = ln(0.30)$, or $x = \dfrac{ln(0.30)}{-0.5} \approx 2.41$ minutes.
Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer.
You are finding the 70th percentile $k$ so you can use the formula $k = \dfrac{ln(1-\text{Area To The left Of k})}{-m}$
$k = \dfrac{ln(1-0.70)}{(-0.5)} \approx 2.41$ minutes
Figure $10$.
This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.
Exercise $5$
Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.
1. On average, how many seconds elapse between two successive cars?
2. After a car passes by, how long on average will it take for another seven cars to pass by?
3. Find the probability that after a car passes by, the next car will pass within the next 20 seconds.
4. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds.
Answer
1. At a rate of five cars per minute, we expect $\dfrac{60}{5} = 12$ seconds to pass between successive cars on average.
2. Using the answer from part a, we see that it takes $(12)(7) = 84$ seconds for the next seven cars to pass by.
3. Let $T =$ the time (in seconds) between successive cars.
The mean of $T$ is 12 seconds, so the decay parameter is $\dfrac{1}{12}$ and $T \sim Exp\dfrac{1}{12}$. The cumulative distribution function of $T$ is $P(T < t) = 1 – e^{−\dfrac{t}{12}}$. Then $P(T < 20) = 1 –e^{−\dfrac{20}{12}} \approx 0.8111$.
Figure $11$.
$P(T > 15) = 1 – P(T < 15) = 1 – (1 – e^{−\dfrac{15}{12}}) = e^{−\dfrac{15}{12}} \approx 0.2865$.
Memorylessness of the Exponential Distribution
In Example recall that the amount of time between customers is exponentially distributed with a mean of two minutes ($X \sim Exp (0.5)$). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says that
$P(X > r + t | X > r) = P(X > t)$
for all $r \geq 0$ and $t \geq 0$.
For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using $r = 5$ and $t = 1$ in the foregoing equation.
$P(X > 5 + 1 | X > 5) = P(X > 1) = e(–0.5)(1) e(–0.5)(1) \approx 0.6065$.
This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical or mechanical device. In Example, the lifetime of a certain computer part has the exponential distribution with a mean of ten years ($X \sim Exp(0.1)$). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is $P(X > 17 | X > 10) = P(X > 7) = 0.4966$.
Example $6$
Refer to Example where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?
The decay parameter of $X$ is $m = \dfrac{1}{4} = 0.25$, so $X \sim Exp(0.25)$.
The cumulative distribution function is $P(X < x) = 1 - e^{–0.25x}$.
We want to find $P(X > 7 | X > 4)$. The memoryless property says that $P(X > 7 | X > 4) = P(X > 3)$, so we just need to find the probability that a customer spends more than three minutes with a postal clerk.
This is $P(X > 3) = 1 - P(X < 3) = 1 - (1 - e^{-0.25 \cdot 3}) = e^{–0.75} \approx 0.4724$.
$1 - (1 - e^(-0.25*2)) = e^(-0.25*2)$.
Exercise $6$
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.
Answer
Let $T =$ the lifetime of the light bulb. Then $T \sim Exp\left(\dfrac{1}{8}\right)$.
The cumulative distribution function is $P(T < t) = 1 − e^{-\dfrac{t}{8}}$
We need to find $P(T > 19 | T = 12)$. By the memoryless property,
$P(T > 19 | T = 12) = P(T > 7) = 1 - P(T < 7) = 1 - (1 - e^{-7/8}) = e^{-7/8} \approx 0.4169$.
1 - (1 – e^(–7/8)) = e^(–7/8).
Relationship between the Poisson and the Exponential Distribution
There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of $\mu$ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean $\lambda = \dfrac{1}{\mu}$. Recall from the chapter on Discrete Random Variables that if $X$ has the Poisson distribution with mean $\lambda$, then $P(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}$. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. $(k! = k*(k-1*)(k - 2)*(k - 3) \dotsc 3*2*1)$
Suppose $X$ has the Poisson distribution with mean $\lambda$. Compute $P(X = k)$ by entering 2nd, VARS(DISTR), C: poissonpdf$(\lambda, k$). To compute $P(X \leq k$), enter 2nd, VARS (DISTR), D:poissoncdf($\lambda, k$).
Example $7$
At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.
1. Find the average time between two successive calls.
2. Find the probability that after a call is received, the next call occurs in less than ten seconds.
3. Find the probability that exactly five calls occur within a minute.
4. Find the probability that less than five calls occur within a minute.
5. Find the probability that more than 40 calls occur in an eight-minute period.
Answer
1. On average there are four calls occur per minute, so 15 seconds, or $\dfrac{15}{60} = 0.25$ minutes occur between successive calls on average.
2. Let $T =$ time elapsed between calls. From part a, $\mu = 0.25$, so $m = \dfrac{1}{0.25} = 4$. Thus, $T \sim Exp(4)$.
The cumulative distribution function is $P(T < t) = 1 - e^{–4t}$.
The probability that the next call occurs in less than ten seconds (ten seconds $= \dfrac{1}{6}$ minute) is $P(T < \dfrac{1}{6}) = 1 - e^{-4 (\dfrac{1}{6})} \approx 0.4866)$.
Figure $13$
3. Let $X =$ the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.
Therefore, $X \sim Poisson(4)$, and so $P(X = 5) = \dfrac{4^{5}e^{-4}}{5!} \approx 0.1563$. ($5! = (5)(4)(3)(2)(1)$)
$\text{poissonpdf}(4, 5) = 0.1563$.
4. Keep in mind that $X$ must be a whole number, so $P(X < 5) = P(X \leq 4)$.
To compute this, we could take $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$.
Using technology, we see that $P(X \approx 4) = 0.6288$.
$\text{poisssoncdf}(4, 4) = 0.6288$
5. Let $Y =$ the number of calls that occur during an eight minute period.
Since there is an average of four calls per minute, there is an average of $(8)(4) = 32$ calls during each eight minute period.
Hence, $Y \sim Poisson(32)$. Therefore, $P(Y > 40) = 1 - P(Y \leq 40) = 1 - 0.9294 = 0.0707$.
$1 - \text{poissoncdf}(32, 40). = 0.0707$
Exercise $7$
In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.
1. Calculate the probability that there are at most 2 accidents occur in any given week.
2. What is the probability that there is at least two weeks between any 2 accidents?
Answer
1. Let $X =$ the number of accidents per week, so that $X \sim Poisson(3)$. We need to find $P(X \leq 2) \approx 0.4232$
$\text{poissoncdf}(3, 2)$
2. Let $T =$ the time (in weeks) between successive accidents.
Since the number of accidents occurs with a Poisson distribution, the time between accidents follows the exponential distribution.
If there are an average of three per week, then on average there is $\mu = \dfrac{1}{3}$ of a week between accidents, and the decay parameter is $m = \dfrac{1}{\left(\dfrac{1}{3}\right)} = 3$.
To find the probability that there are at least two weeks between two accidents, $P(T > 2) = 1 - P(T < 2) = 1 – (1 – e(–3)(2)) = e^{–6} \approx 0.0025$.
e^(-3*2).
Review
If $X$ has an exponential distribution with mean $\mu$, then the decay parameter is $m = \dfrac{1}{\mu}$, and we write $X \sim Exp(m)$ where $x \geq 0$ and $m > 0$. The probability density function of $X$ is $f(x) = me^{-mx}$ (or equivalently $f(x) = \dfrac{1}{\mu}e^{-\dfrac{x}{\mu}}$). The cumulative distribution function of $X$ is $P(X \leq X) = 1 - e^{-mx}$.
The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that $P(X > x + k | X > x) = P(X > k)$.
If $T$ represents the waiting time between events, and if $T \sim Exp(\lambda)$, then the number of events $X$ per unit time follows the Poisson distribution with mean $\lambda$. The probability density function of $PX$ is $(X = k) = \dfrac{\lambda^{k}e^{-k}}{k!}$. This may be computed using a TI-83, 83+, 84, 84+ calculator with the command $\text{poissonpdf}(\lambda, k)$. The cumulative distribution function $P(X \leq k)$ may be computed using the TI-83, 83+,84, 84+ calculator with the command $\text{poissoncdf}(\lambda, k)$.
Formula Review
Exponential: $X \sim Exp(m)$ where $m =$ the decay parameter
• pdf: $f(x) = me^{(–mx)}$ where $x \geq 0$ and $m > 0$
• cdf: $P(X \leq x) = 1 - e^{(–mx)}$
• mean $\mu = \dfrac{1}{m}$
• standard deviation $\sigma = \mu$
• percentile $k: k = \dfrac{ln(1 - \text{Area To The Left Of k})}{-m}$
• Additionally
• $P(X > x) = e^{(–mx)}$
• $P(a < X < b) = e^{(–ma)} - e^{(–mb)}$
• Memoryless Property: $P(X > x + k | X > x) = P(X > k)$
• Poisson probability: $P(X = k) = \dfrac{\lambda^{k}e^{k}}{k!}$ with mean $\lambda$
• $k! = k*(k - 1)*(k - 2)*(k - 3) \dotsc 3*2*1$
Glossary
decay parameter
The decay parameter describes the rate at which probabilities decay to zero for increasing values of $x$. It is the value $m$ in the probability density function $f(x) = me^{(-mx)}$ of an exponential random variable. It is also equal to $m = \dfrac{1}{\mu}$, where $\mu$ is the mean of the random variable.
memoryless property
For an exponential random variable $X$, the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that $X$ exceeds $x + k$, given that it has exceeded $x$, is the same as the probability that $X$ would exceed $k$ if we had no knowledge about it. In symbols we say that $P(X > x + k | X > x) = P(X > k)$
Poisson distribution
If there is a known average of $\lambda$ events occurring per unit time, and these events are independent of each other, then the number of events $X$ occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to $P(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/05%3A_Continuous_Random_Variables/5.04%3A_The_Exponential_Distribution.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will compare and contrast empirical data from a random number generator with the uniform distribution.
Collect the Data
Use a random number generator to generate 50 values between zero and one (inclusive). List them in Table. Round the numbers to four decimal places or set the calculator MODE to four places.
1. Complete the table.
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
2. Calculate the following:
1. $\bar{x}$= _______
2. s = _______
3. first quartile = _______
4. third quartile = _______
5. median = _______
Organize the Data
1. Construct a histogram of the empirical data. Make eight bars.
2. Construct a histogram of the empirical data. Make five bars.
Describe the Data
1. In two to three complete sentences, describe the shape of each graph. (Keep it simple. Does the graph go straight across, does it have a V shape, does it have a hump in the middle or at either end, and so on. One way to help you determine a shape is to draw a smooth curve roughly through the top of the bars.)
2. Describe how changing the number of bars might change the shape.
Theoretical Distribution
1. In words, $X$ = _____________________________________.
2. The theoretical distribution of $X$ is $X$ ~ U(0,1).
3. In theory, based upon the distribution $X$ ~ U(0,1), complete the following.
1. $\mu$ = ______
2. $\sigma$ = ______
3. first quartile = ______
4. third quartile = ______
5. median = __________
4. Are the empirical values (the data) in the section titled Collect the Data close to the corresponding theoretical values? Why or why not?
Plot the Data
1. Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges.
2. Do you notice any potential outliers? If so, which values are they? Either way, justify your answer numerically. (Recall that any DATA that are less than Q1 – 1.5(IQR) or more than Q3 + 1.5(IQR) are potential outliers. IQR means interquartile range.)
Compare the Data
1. For each of the following parts, use a complete sentence to comment on how the value obtained from the data compares to the theoretical value you expected from the distribution in the section titled Theoretical Distribution.
1. minimum value: _______
2. first quartile: _______
3. median: _______
4. third quartile: _______
5. maximum value: _______
6. width of IQR: _______
7. overall shape: _______
2. Based on your comments in the section titled Collect the Data, how does the box plot fit or not fit what you would expect of the distribution in the section titledTheoretical Distribution?
Discussion Question
1. Suppose that the number of values generated was 500, not 50. How would that affect what you would expect the empirical data to be and the shape of its graph to look like? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/05%3A_Continuous_Random_Variables/5.05%3A_Continuous_Distribution_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
Continuous Probability Functions
For each probability and percentile problem, draw the picture.
Q 5.2.1
Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor.
1. What part of the experiment will yield discrete data?
2. What part of the experiment will yield continuous data?
Q 5.2.2
When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why?
S 5.2.2
Age is a measurement, regardless of the accuracy used.
Exercise 5.2.2
Which type of distribution does the graph illustrate?
Answer
Uniform Distribution
Exercise 5.2.3
Which type of distribution does the graph illustrate?
Exercise 5.2.4
Which type of distribution does the graph illustrate?
Answer
Normal Distribution
Exercise 5.2.5
What does the shaded area represent? $P($___$< x <$ ___$)$
Exercise 5.2.6
What does the shaded area represent? $P($___$< x <$ ___$)$
Answer
$P(6 < x < 7)$
Exercise 5.2.7
For a continuous probability distribution, $0 \leq x \leq 15$. What is $P(x > 15)$?
Exercise 5.2.8
What is the area under $f(x)$ if the function is a continuous probability density function?
Answer
one
Exercise 5.2.9
For a continuous probability distribution, $0 \leq x \leq 10$. What is $P(x = 7)$?
Exercise 5.2.11
A continuous probability function is restricted to the portion between $x = 0$ and $7$. What is $P(x = 10)$?
Answer
zero
Exercise 5.2.12
$f(x)$ for a continuous probability function is $\frac{1}{5}$, and the function is restricted to $0 \leq x \leq 5$. What is $P(x < 0)$?
Exercise 5.2.13
$f(x)$, a continuous probability function, is equal to $\frac{1}{12}$, and the function is restricted to $0 \leq x \leq 12$. What is $P(0 < x) < 12)$?
Answer
one
Exercise 5.2.14
Find the probability that $x$ falls in the shaded area.
Exercise 5.2.15
Find the probability that $x$ falls in the shaded area.
Answer
0.625
Exercise 5.2.16
Find the probability that $x$ falls in the shaded area.
Exercise 5.2.17
$f(x)$, a continuous probability function, is equal to $\frac{1}{3}$ and the function is restricted to $1 \leq x \leq 4$. Describe $P(x > \frac{3}{2})$.
Answer
The probability is equal to the area from $x = \frac{3}{2}$ to $x = 4$ above the x-axis and up to $f(x) = \frac{1}{3}$.
The Uniform Distribution
For each probability and percentile problem, draw the picture.
Q 5.3.1
Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks).
1. $X \sim$ _________
2. Graph the probability distribution.
3. $f(x) =$ _________
4. $\mu =$ _________
5. $\sigma =$ _________
6. Find the probability that a person is born at the exact moment week 19 starts. That is, find $P(x = 19) =$ _________
7. $P(2 < x < 31) =$ _________
8. Find the probability that a person is born after week 40.
9. $P(12 < x|x < 28) =$ _________
10. Find the 70th percentile.
11. Find the minimum for the upper quarter.
Q 5.3.2
A random number generator picks a number from one to nine in a uniform manner.
1. $X \sim$ _________
2. Graph the probability distribution.
3. $f(x) =$ _________
4. $\mu =$ _________
5. $\mu =$ _________
6. $P(3.5 < x < 7.25) =$ _________
7. $P(x > 5.67) =$ _________
8. $P(x > 5|x > 3) =$ _________
9. Find the 90th percentile.
S 5.3.2
1. $X \sim U(1, 9)$
2. Check student’s solution.
3. $f(x) = 18$ where $1 \leq x \leq 9$
4. five
5. 2.3
6. $\frac{15}{32}$
7. $\frac{333}{800}$
8. $\frac{2}{3}$
9. 8.2
Q 5.3.3
According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month.
1. Define the random variable. $X =$ _________
2. $X \sim$ _________
3. Graph the probability distribution.
4. $f(x) =$ _________
5. $\mu =$ _________
6. $\sigma =$ _________
7. Find the probability that the individual lost more than ten pounds in a month.
8. Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 12 pounds in the month.
9. $P(7 < x < 13|x > 9) =$ __________. State this in a probability question, similarly to parts g and h, draw the picture, and find the probability.
Q 5.3.4
A subway train on the Red Line arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution.
1. Define the random variable. $X =$ _______
2. $X \sim$ _______
3. Graph the probability distribution.
4. $f(x) =$ _______
5. $\mu =$ _______
6. $\sigma =$ _______
7. Find the probability that the commuter waits less than one minute.
8. Find the probability that the commuter waits between three and four minutes.
9. Sixty percent of commuters wait more than how long for the train? State this in a probability question, similarly to parts g and h, draw the picture, and find the probability.
S 5.3.5
1. $X$ represents the length of time a commuter must wait for a train to arrive on the Red Line.
2. $X \sim U(0, 8)$
3. $f(x) = \frac{1}{8}$ where $leq x leq 8$
4. four
5. 2.31
6. $\frac{1}{8}$
7. $\frac{1}{8}$
8. 3.2
Q 5.3.6
The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class.
1. Define the random variable. $X =$ _________
2. $X \sim$ _________
3. Graph the probability distribution.
4. $f(x) =$ _________
5. $\mu =$ _________
6. $\sigma =$ _________
7. Find the probability that she is over 6.5 years old.
8. Find the probability that she is between four and six years old.
9. Find the 70th percentile for the age of first graders on September 1 at Garden Elementary School.
Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.
Q 5.3.7
What is the average waiting time (in minutes)?
1. zero
2. two
3. three
4. four
d
Q 5.3.8
Find the 30th percentile for the waiting times (in minutes).
1. two
2. 2.4
3. 2.75
4. three
Q 5.3.9
The probability of waiting more than seven minutes given a person has waited more than four minutes is?
1. 0.125
2. 0.25
3. 0.5
4. 0.75
b
Q 5.3.10
The time (in minutes) until the next bus departs a major bus depot follows a distribution with $f(x) = \frac{1}{20}$ where $x$ goes from 25 to 45 minutes.
1. Define the random variable. $X =$ ________
2. $X \sim$ ________
3. Graph the probability distribution.
4. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous).
5. $\mu =$ ________
6. $\sigma =$ ________
7. Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement.
8. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement.
9. $P(25 < x < 55) =$ _________. State this in a probability statement, similarly to parts g and h, draw the picture, and find the probability.
10. Find the 90th percentile. This means that 90% of the time, the time is less than _____ minutes.
11. Find the 75th percentile. In a complete sentence, state what this means. (See part j.)
12. Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes.
Q 5.3.11
Suppose that the value of a stock varies each day from $16 to$25 with a uniform distribution.
1. Find the probability that the value of the stock is more than $19. 2. Find the probability that the value of the stock is between$19 and $22. 3. Find the upper quartile - 25% of all days the stock is above what value? Draw the graph. 4. Given that the stock is greater than$18, find the probability that the stock is more than $21. S 5.3.11 1. The probability density function of $X$ is $\frac{1}{25-16} = \frac{1}{9}$. $P(X > 19) = (25 - 19)\left(\frac{1}{9}\right) = \frac{6}{9} = \frac{2}{3}$. 2. The area must be 0.25, and $0.25 = (\text{width})\left(\frac{1}{9}\right)$, so $\text{width} = (0.25)(9) = 2.25$. Thus, the value is $25 – 2.25 = 22.75$. 3. This is a conditional probability question. $P(x > 21| x > 18)$. You can do this two ways: • Draw the graph where a is now 18 and b is still 25. The height is $\frac{1}{25 - 18} = \frac{1}{7}$ So, $P(x > 21|x > 18) = (25 – 21)\left(\frac{1}{7}\right) = \frac{4}{7}$. • Use the formula: $P(x > 21|x > 18) = \frac{P(x > 21 \text{AND} x > 18)}{P(x > 18)} = \frac{P(x > 21)}{P(x > 18)} = \frac{(25 - 21)}{25 - 18} = \frac{4}{7}$. Q 5.3.12 A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution. 1. Find the average time between fireworks. 2. Find probability that the time between fireworks is greater than four seconds. Q 5.3.13 The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. 1. Find the probability that the truck driver goes more than 650 miles in a day. 2. Find the probability that the truck drivers goes between 400 and 650 miles in a day. 3. At least how many miles does the truck driver travel on the furthest 10% of days? S 5.3.13 1. $P(X > 650) = \frac{700-650}{700-300} = \frac{500}{400} = \frac{1}{8} = 0.125$. 2. $P(400 < X < 650) = \frac{700-650}{700-300} = \frac{250}{400} = 0.625$ 3. $0.10 = \frac{\text{width}}{700-300}$, so $\text{width} = 400(0.10) = 40$. Since $700 – 40 = 660$, the drivers travel at least 660 miles on the furthest 10% of days. The Exponential Distribution Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: $X \sim Exp(0.2)$ Exercise 5.4.8 What type of distribution is this? Exercise 5.4.9 Are outcomes equally likely in this distribution? Why or why not? Answer No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time. Exercise 5.4.10 What is $m$? What does it represent? Exercise 5.4.11 What is the mean? Answer five Exercise 5.4.12 What is the standard deviation? Exercise 5.4.13 State the probability density function. Answer $f(x) = 0.2e^{-0.2x}$ Exercise 5.4.14 Graph the distribution. Exercise 5.4.15 Find $P(2 < x < 10)$. Answer 0.5350 Exercise 5.4.16 Find $P(x > 6)$. Exercise 5.4.17 Find the 70th percentile. Answer 6.02 Use the following information to answer the next seven exercises. A distribution is given as $X \sim Exp(0.75)$. Exercise 5.4.18 What is $m$? Exercise 5.4.19 What is the probability density function? Answer $f(x) = 0.75e^{-0.75x}$ Exercise 5.4.20 What is the cumulative distribution function? Exercise 5.4.21 Draw the distribution. Answer Exercise 5.4.22 Find $P(x < 4)$. Exercise 5.4.23 Find the 30th percentile. Answer 0.4756 Exercise 5.4.24 Find the median. Exercise 5.4.25 Which is larger, the mean or the median? Answer The mean is larger. The mean is $\frac{1}{m} = \frac{1}{0.75} \approx 1.33$, which is greater than $0.9242$. Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14. Exercise 5.4.26 What is being measured here? Exercise 5.4.27 Are the data discrete or continuous? Answer continuous Exercise 5.4.28 In words, define the random variable $X$. Exercise 5.4.29 What is the decay rate ($m$)? Answer $m = 0.000121$ Exercise 5.4.30 The distribution for $X$ is ______. Exercise 5.4.31 Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find $P(x < 5,730)$. 1. Sketch the graph, and shade the area of interest. 2. Find the probability. $P(x < 5,730) =$_ __________ Answer 1. Check student's solution 2. $P(x < 5,730) = 0.5001$ Exercise 5.4.32 Find the percentage of carbon-14 lasting longer than 10,000 years. 1. Sketch the graph, and shade the area of interest. 2. Find the probability. P($x$ > 10,000) = ________ Exercise 5.2.33 Thirty percent (30%) of carbon-14 will decay within how many years? 1. Sketch the graph, and shade the area of interest. 2. Find the value $k$ such that $P(x < k) = 0.30$. Answer 1. Check student's solution. 2. $k = 2947.73$ Q 5.4.1 Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. 1. Define the random variable. $X =$ ________________. 2. Is $X$ continuous or discrete? 3. $X \sim$ ________ 4. $\mu =$ ________ 5. $\sigma =$ ________ 6. Draw a graph of the probability distribution. Label the axes. 7. Find the probability that a phone call lasts less than nine minutes. 8. Find the probability that a phone call lasts more than nine minutes. 9. Find the probability that a phone call lasts between seven and nine minutes. 10. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why? Q 5.4.2 Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery. 1. Define the random variable. $X =$ _________________________________. 2. Is $X$ continuous or discrete? 3. $X \sim$ ________ 4. On average, how long would you expect one car battery to last? 5. On average, how long would you expect nine car batteries to last, if they are used one after another? 6. Find the probability that a car battery lasts more than 36 months. 7. Seventy percent of the batteries last at least how long? S 5.4.2 1. $X =$ the useful life of a particular car battery, measured in months. 2. $X$ is continuous. 3. $X ~\sim \text{Exp}(0.025)$ 4. 40 months 5. 360 months 6. 0.4066 7. 14.27 Q 5.4.3 The percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state. 1. Define the random variable. $X =$ _________________________________. 2. Is $X$ continuous or discrete? 3. $X ~\sim$ ________ 4. $\mu =$ ________ 5. $\sigma =$ ________ 6. Draw a graph of the probability distribution. Label the axes. 7. Find the probability that the percent is less than 12. 8. Find the probability that the percent is between eight and 14. 9. The percent of all individuals living in the United States who speak a language at home other than English is 13.8. 1. Why is this number different from 9.848%? 2. What would make this number higher than 9.848%? Q 5.4.4 The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement. 1. Define the random variable. $X =$ _________________________________. 2. Is $X$ continuous or discrete? 3. $X \sim$ ________ 4. $\mu =$ ________ 5. $\sigma =$ ________ 6. Draw a graph of the probability distribution. Label the axes. 7. Find the probability that the person retired after age 70. 8. Do more people retire before age 65 or after age 65? 9. In a room of 1,000 people over age 80, how many do you expect will NOT have retired yet? S 5.4.4 1. $X =$ the time (in years) after reaching age 60 that it takes an individual to retire 2. $X$ is continuous. 3. $X ~ \text{Exp}\left(\frac{1}{5}\right)$ 4. five 5. five 6. Check student’s solution. 7. 0.1353 8. before 9. 18.3 Q 5.4.5 The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of$150.
1. Define the random variable. $X =$ _________________________________.
2. $X \sim$ ________
3. $\mu =$ ________
4. $\sigma =$ ________
5. Draw a graph of the probability distribution. Label the axes.
6. Find the probability that a car required over \$300 for maintenance during its first year.
Use the following information to answer the next three exercises. The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution.
Q 5.4.6
The decay rate is:
1. 0.3333
2. 0.5000
3. 2
4. 3
a
Q 5.4.7
What is the probability that a phone will fail within two years of the date of purchase?
1. 0.8647
2. 0.4866
3. 0.2212
4. 0.9997
Q 5.4.8
What is the median lifetime of these phones (in years)?
1. 0.1941
2. 1.3863
3. 2.0794
4. 5.5452
c
Q 5.4.9
Let $X \sim \text{Exp}(0.1)$.
1. decay rate = ________
2. $\mu =$ ________
3. Graph the probability distribution function.
4. On the graph, shade the area corresponding to $P(x < 6)$ and find the probability.
5. Sketch a new graph, shade the area corresponding to $P(3 < x < 6)$ and find the probability.
6. Sketch a new graph, shade the area corresponding to $P(x < 7)$ and find the probability.
7. Sketch a new graph, shade the area corresponding to the 40th percentile and find the value.
8. Find the average value of $x$.
Q 5.4.10
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years.
1. Find the probability that a light bulb lasts less than one year.
2. Find the probability that a light bulb lasts between six and ten years.
3. Seventy percent of all light bulbs last at least how long?
4. A company decides to offer a warranty to give refunds to light bulbs whose lifetime is among the lowest two percent of all bulbs. To the nearest month, what should be the cutoff lifetime for the warranty to take place?
5. If a light bulb has lasted seven years, what is the probability that it fails within the 8thyear.
S 5.4.10
Let $T =$ the life time of a light bulb.
The decay parameter is $m = \frac{1}{8}$, and $T \sim \text{Exp}(\frac{1}{8})$. The cumulative distribution function is $P(T < t) = 1 - e^{-\frac{t}{s}}$
1. Therefore, $P(T < t) = 1 - ^{-\frac{1}{s}} \approx 0.1175$.
2. We want to find $P(6 < t < 10)$.
To do this, $P(6 < t < 10) – P(t < 6) = \left(1 - e^{-\frac{1}{s} * 10}\right) - \left(1 - e^{\frac{1}{s}*6}\right) \approx 0.7135 - 0.5276 = 0.1859$
3. We want to find $0.70 = P(T < t) = 1 - \left(1 - e^{-\frac{t}{s}}\right) = e^{-\frac{t}{s}}$.
Solving for $t$, $e^{-\frac{t}{s}} = 0.70$, so $-\frac{t}{8} = \text{ln}(0.70)$ and $t = -8 \cdot \text{ln}(0.70) \approx 2.85$ years.
Or use $t = \frac{\text{ln(area_to_the_right)}}{(-m)} = \frac{\text{ln}(0.70)}{-\frac{1}{8}} \approx 2.85$ years
4. We want to find $0.02 = P(T < t) = 1 – e^{-\frac{t}{8}}$
Solving for $t, e^{-\frac{t}{8}} = 0.98$, so $\frac{t}{8} = \text{ln}(0.98)$, and $t = –8 \cdot \text{ln}(0.98) \approx 0.1616$ years, or roughly two months.
The warranty should cover light bulbs that last less than 2 months.
Or use $\frac{\text{ln(area_to_the_right)}}{(-m)} = \frac{\text{ln}(1-0.2)}{-\frac{1}{8}} = 0.1616$.
5. We must find $P(T < 8|T > 7)$.
Notice that by the rule of complement events, $P(T < 8|T > 7) = 1 – P(T > 8|T > 7)$.
By the memoryless property $(P(X > r + t|X > r) = P(X > t))$.
So $P(T > 8|T > 7) = P(T > 1) = 1 - \left(1 - e^{-\frac{1}{8}}\right) = e^{-\frac{1}{8}} \approx 0.8825$
Therefore, $P(T < 8|T > 7) = 1 – 0.8825 = 0.1175$.
Q 5.4.11
At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution.
1. On average, how much time occurs between five consecutive calls?
2. Find the probability that after a call is received, it takes more than three minutes for the next call to occur.
3. Ninety-percent of all calls occur within how many minutes of the previous call?
4. Suppose that two minutes have elapsed since the last call. Find the probability that the next call will occur within the next minute.
5. Find the probability that less than 20 calls occur within an hour.
Q 5.4.12
In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential.
1. What is the probability that an entire season elapses with a single no-hitter?
2. If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season?
3. What is the probability that there are more than 3 no-hitters in a single season?
S 5.4.12
Let $X =$ the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean $\lambda = 3$.
Therefore, $(X = 0) = \frac{3^{0}e^{3}}{0!} = e^{-3} \approx 0.0498$
NOTE
You could let $T =$ duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is $\frac{1}{3}$season. For the exponential, $\mu = \frac{1}{3}$.
Therefore, $m = \frac{1}{\mu} = 3$ and $T \sim \text{Exp}(3)$.
1. The desired probability is $P(T > 1) = 1 – P(T < 1) = 1 – (1 – e^{-3}) = e^{-3} \approx 0.0498$.
2. Let $T =$ duration of time between no-hitters. We find $P(T > 2|T > 1)$, and by the memoryless property this is simply $P(T > 1)$, which we found to be 0.0498 in part a.
3. Let $X =$ the number of no-hitters is a season. Assume that $X$ is Poisson with mean $\lambda = 3$. Then $P(X > 3) = 1 – P(X \leq 3) = 0.3528$.
Q 5.4.13
During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential.
1. What is the probability that the next earthquake occurs within the next three months?
2. Given that six months has passed without an earthquake in Papua New Guinea, what is the probability that the next three months will be free of earthquakes?
3. What is the probability of zero earthquakes occurring in 2014?
4. What is the probability that at least two earthquakes will occur in 2014?
Q 5.4.14
According to the American Red Cross, about one out of nine people in the U.S. have Type B blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B blood types that arrive roughly follows the Poisson distribution.
1. If 100 people arrive, how many on average would be expected to have Type B blood?
2. What is the probability that over 10 people out of these 100 have type B blood?
3. What is the probability that more than 20 people arrive before a person with type B blood is found?
S 5.4.14
1. $\frac{100}{9} = 11.11$
2. $P(X > 10) = 1 – P(X \leq 10) = 1 – \text{Poissoncdf}(11.11, 10) \approx 0.5532$.
3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people $X$ who arrive between successive Type B arrivals is roughly exponential with mean $μ = 9$ and $m = \frac{1}{9}$. The cumulative distribution function of $X$ is $P(X < x) = 1 - e^{-\frac{x}{9}}$. Thus, $P(X > 20) = 1 - P(X \leq 20) = 1 − (1 − e^{-\frac{20}{9}}) \approx 0.1084.$.
NOTE
We could also deduce that each person arriving has a 8/9 chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is $\left(\frac{8}{9}\right)^{20} \approx 0.0948$. (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)
Q 5.4.15
A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution.
1. Find the probability that the duration between two successive visits to the web site is more than ten minutes.
2. The top 25% of durations between visits are at least how long?
3. Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes?
4. Find the probability that less than 7 visits occur within a one-hour period.
Q 5.4.16
At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.
1. Find the probability that the time between two successive visits to the urgent care facility is less than 2 minutes.
2. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes.
3. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes?
4. Find the probability that more than eight patients arrive during a half-hour period.
S 5.4.17
Let $T =$ duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, $\mu = 7$ and the decay constant is $m = \frac{1}{7}$. The cdf is $P(T < t) = 1 - e^{\frac{t}{\tau}}$
1. $P(T < 2) = 1 - 1 - e^{-\frac{2}{7}} \approx 0.2485$.
2. $P(T > 15) = 1 - P(T < 15) = 1 - \left(1 - e^{-\frac{15}{7}}\right) \approx e^{-\frac{15}{7}} \approx 0.1173$.
3. $P(T > 15|T > 10) = P(T > 5) = 1 - \left(1 - e^{ \frac{5}{7}}\right) = e^{-\frac{5}{7}} \approx 0.4895$.
4. Let $X =$ # of patients arriving during a half-hour period. Then $X$ has the Poisson distribution with a mean of $\frac{30}{7}$, $X \sim \text{Poisson}\left(\frac{30}{7}\right)$. Find $P(X > 8) = 1 – P(X \leq 8) \approx 0.0311$.
5.E: Exercises
5.2: Continuous Probability Functions
Which type of distribution does the graph illustrate?
Which type of distribution does the graph illustrate?
Which type of distribution does the graph illustrate?
What does the shaded area represent? P(___< x < ___)
What does the shaded area represent? P(___< x < ___)
For a continuous probability distribution, 0 ≤ x ≤ 15. What is P(x > 15)?
What is the area under f(x) if the function is a continuous probability density function?
For a continuous probability distribution, 0 ≤ x ≤ 10. What is P(x = 7)?
A continuous probability function is restricted to the portion between x = 0 and 7. What is P(x = 10)?
f(x) for a continuous probability function is 15
, and the function is restricted to 0 ≤ x ≤ 5. What is P(x < 0)?
f(x), a continuous probability function, is equal to 112
, and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)?
one
Find the probability that x falls in the shaded area.
Find the probability that x falls in the shaded area.
Find the probability that x falls in the shaded area.
f(x), a continuous probability function, is equal to 13
and the function is restricted to 1 ≤ x ≤ 4. Describe P(x>32).
The probability is equal to the area from x = 32to x = 4 above the x-axis and up to f(x) = 13
Homework
For each probability and percentile problem, draw the picture.
Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor.
1. What part of the experiment will yield discrete data?
2. What part of the experiment will yield continuous data?
When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why?
Age is a measurement, regardless of the accuracy used. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/05%3A_Continuous_Random_Variables/5.E%3A_Continuous_Random_Variables_%28Exercises%29.txt |
In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them. The normal distribution has two parameters (two numerical descriptive measures), the mean (μ) and the standard deviation (σ).
• 6.1: Prelude to The Normal Distribution
The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them. The normal distribution has two parameters (two numerical descriptive measures), the mean (μ) and the standard deviation (σ).
• 6.2: The Standard Normal Distribution
A z-score is a standardized value. Its distribution is the standard normal, Z∼N(0,1). The mean of the z-scores is zero and the standard deviation is one. If y is the z-score for a value x from the normal distribution N(μ,σ) then z tells you how many standard deviations x is above (greater than) or below (less than) μ.
• 6.3: Using the Normal Distribution
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean μ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.
• 6.4: Normal Distribution - Lap Times (Worksheet)
A statistics Worksheet: The student will compare and contrast empirical data and a theoretical distribution to determine if Terry Vogel's lap times fit a continuous distribution.
• 6.5: Normal Distribution - Pinkie Length (Worksheet)
A statistics Worksheet: The student will compare empirical data and a theoretical distribution to determine if data from the experiment follow a continuous distribution.
• 6.E: The Normal Distribution (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
06: The Normal Distribution
Learning Objectives
By the end of this chapter, the student should be able to:
• Recognize the normal probability distribution and apply it appropriately.
• Recognize the standard normal probability distribution and apply it appropriately.
• Compare normal probabilities by converting to the standard normal distribution.
The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines. Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world.
In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them. The normal distribution has two parameters (two numerical descriptive measures), the mean ($\mu$) and the standard deviation ($\sigma$). If $X$ is a quantity to be measured that has a normal distribution with mean ($\mu$) and standard deviation ($\sigma$), we designate this by writing
$f(x) = \dfrac{1}{\sigma \cdot \sqrt{2 \cdot \pi}} \cdot e^{\left(-\dfrac{1}{2}\right) \cdot \left(\dfrac{x-\mu}{\sigma}\right)^{2}}$
The probability density function is a rather complicated function. Do not memorize it. It is not necessary.
The cumulative distribution function is $P(X < x)$. It is calculated either by a calculator or a computer, or it is looked up in a table. Technology has made the tables virtually obsolete. For that reason, as well as the fact that there are various table formats, we are not including table instructions.
The curve is symmetrical about a vertical line drawn through the mean, $\mu$. In theory, the mean is the same as the median, because the graph is symmetric about $\mu$. As the notation indicates, the normal distribution depends only on the mean and the standard deviation. Since the area under the curve must equal one, a change in the standard deviation, $\sigma$, causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on $\sigma$. A change in $\mu$ causes the graph to shift to the left or right. This means there are an infinite number of normal probability distributions. One of special interest is called the standard normal distribution.
COLLABORATIVE CLASSROOM ACTIVITY
Your instructor will record the heights of both men and women in your class, separately. Draw histograms of your data. Then draw a smooth curve through each histogram. Is each curve somewhat bell-shaped? Do you think that if you had recorded 200 data values for men and 200 for women that the curves would look bell-shaped? Calculate the mean for each data set. Write the means on the x-axis of the appropriate graph below the peak. Shade the approximate area that represents the probability that one randomly chosen male is taller than 72 inches. Shade the approximate area that represents the probability that one randomly chosen female is shorter than 60 inches. If the total area under each curve is one, does either probability appear to be more than 0.5?
Formula Review
• $X \sim N(\mu, \sigma)$
• $\mu =$ the mean $\sigma =$ the standard deviation
Glossary
Normal Distribution
a continuous random variable (RV) with pdf $f(x) = \dfrac{1}{\sigma \sqrt{2 \pi}}e^{\dfrac{(x \cdot \mu)}{2 \sigma^{2}}^{2}}$, where $\mu$ is the mean of the distribution and $\sigma$ is the standard deviation; notation: $X \sim N(\mu, \sigma)$. If $\mu = 0$ and $\sigma = 1$, the RV is called the standard normal distribution. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/06%3A_The_Normal_Distribution/6.01%3A_Prelude_to_The_Normal_Distribution.txt |
Z-Scores
The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation.
Definition: Z-Score
If $X$ is a normally distributed random variable and $X \sim N(\mu, \sigma)$, then the z-score is:
$z = \dfrac{x - \mu}{\sigma} \label{zscore}$
The z-score tells you how many standard deviations the value $x$ is above (to the right of) or below (to the left of) the mean, $\mu$. Values of $x$ that are larger than the mean have positive $z$-scores, and values of $x$ that are smaller than the mean have negative $z$-scores. If $x$ equals the mean, then $x$ has a $z$-score of zero. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:
\begin{align*} x &= \mu + (z)(\sigma) \[5pt] &= 5 + (3)(2) = 11 \end{align*}
The z-score is three.
Since the mean for the standard normal distribution is zero and the standard deviation is one, then the transformation in Equation \ref{zscore} produces the distribution $Z \sim N(0, 1)$. The value $x$ comes from a normal distribution with mean $\mu$ and standard deviation $\sigma$.
A z-score is measured in units of the standard deviation.
Example $1$
Suppose $X \sim N(5, 6)$. This says that $x$ is a normally distributed random variable with mean $\mu = 5$ and standard deviation $\sigma = 6$. Suppose $x = 17$. Then (via Equation \ref{zscore}):
$z = \dfrac{x-\mu}{\sigma} = \dfrac{17-5}{6} = 2 \nonumber$
This means that $x = 17$ is two standard deviations (2$\sigma$) above or to the right of the mean $\mu = 5$. The standard deviation is $\sigma = 6$.
Notice that: $5 + (2)(6) = 17$ (The pattern is $\mu + z \sigma = x$)
Now suppose $x = 1$. Then:
$z = \dfrac{x-\mu}{\sigma} = \dfrac{1-5}{6} = -0.67 \nonumber$
(rounded to two decimal places)
This means that $x = 1$ is $0.67$ standard deviations ($–0.67\sigma$) below or to the left of the mean $\mu = 5$. Notice that: $5 + (–0.67)(6)$ is approximately equal to one (This has the pattern $\mu + (–0.67)\sigma = 1$)
Summarizing, when $z$ is positive, $x$ is above or to the right of $\mu$ and when $z$ is negative, $x$ is to the left of or below $\mu$. Or, when $z$ is positive, $x$ is greater than $\mu$, and when $z$ is negative $x$ is less than $\mu$.
Exercise $1$
What is the $z$-score of $x$, when $x = 1$ and $X \sim N(12, 3)$?
Answer
$z = \dfrac{1-12}{3} \approx -3.67$
Example $2$
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let $X =$ the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. $X \sim N(5, 2)$. Fill in the blanks.
1. Suppose a person lost ten pounds in a month. The $z$-score when $x = 10$ pounds is $x = 2.5$ (verify). This $z$-score tells you that $x = 10$ is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose a person gained three pounds (a negative weight loss). Then $z =$ __________. This $z$-score tells you that $x = -3$ is ________ standard deviations to the __________ (right or left) of the mean.
Answers
a. This $z$-score tells you that $x = 10$ is 2.5 standard deviations to the right of the mean five.
b. Suppose the random variables $X$ and $Y$ have the following normal distributions: $X \sim N(5, 6)$ and $Y \sim N(2, 1)$. If $x = 17$, then $z = 2$. (This was previously shown.) If $y = 4$, what is $z$?
$z = \dfrac{y-\mu}{\sigma} = \dfrac{4-2}{1} = 2 \nonumber$
where $\mu = 2$ and $\sigma = 1$.
The $z$-score for $y = 4$ is $z = 2$. This means that four is $z = 2$ standard deviations to the right of the mean. Therefore, $x = 17$ and $y = 4$ are both two (of their own) standard deviations to the right of their respective means.
The z-score allows us to compare data that are scaled differently. To understand the concept, suppose $X \sim N(5, 6)$ represents weight gains for one group of people who are trying to gain weight in a six week period and $Y \sim N(2, 1)$ measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since $x = 17$ and $y = 4$ are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.
Exercise $2$
Fill in the blanks.
Jerome averages 16 points a game with a standard deviation of four points. $X \sim N(16, 4)$. Suppose Jerome scores ten points in a game. The $z$–score when $x = 10$ is $-1.5$. This score tells you that $x = 10$ is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).
Answer
1.5, left, 16
The Empirical Rule
If $X$ is a random variable and has a normal distribution with mean $\mu$ and standard deviation $\sigma$, then the Empirical Rule says the following:
• About 68% of the $x$ values lie between –1$\sigma$ and +1$\sigma$ of the mean $\mu$ (within one standard deviation of the mean).
• About 95% of the $x$ values lie between –2$\sigma$ and +2$\sigma$ of the mean $\mu$ (within two standard deviations of the mean).
• About 99.7% of the $x$ values lie between –3$\sigma$ and +3$\sigma$ of the mean $\mu$ (within three standard deviations of the mean). Notice that almost all the $x$ values lie within three standard deviations of the mean.
• The $z$-scores for +1$\sigma$ and –1$\sigma$ are +1 and –1, respectively.
• The $z$-scores for +2$\sigma$ and –2$\sigma$ are +2 and –2, respectively.
• The $z$-scores for +3$\sigma$ and –3$\sigma$ are +3 and –3 respectively.
The empirical rule is also known as the 68-95-99.7 rule.
Example $3$
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let $X =$ the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then $X \sim N(170, 6.28)$.
1. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The $z$-score when $x = 168$ cm is $z =$ _______. This $z$-score tells you that $x = 168$ is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a $z$-score of $z = 1.27$. What is the male’s height? The $z$-score ($z = 1.27$) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
Answers
1. –0.32, 0.32, left, 170
2. 177.98, 1.27, right
Exercise $3$
Use the information in Example $3$ to answer the following questions.
1. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The $z$-score when $x = 176$ cm is $z =$ _______. This $z$-score tells you that $x = 176$ cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a $z$-score of $z = –2$. What is the male’s height? The $z$-score ($z = –2$) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
Answer
Solve the equation $z = \dfrac{x-\mu}{\sigma}$ for $z$. $x = \mu+ (z)(\sigma)$
$z = \dfrac{176-170}{6.28}$, This z-score tells you that $x = 176$ cm is 0.96 standard deviations to the right of the mean 170 cm.
Answer
Solve the equation $z = \dfrac{x-\mu}{\sigma}$ for $z$. $x = \mu+ (z)(\sigma)$
$X = 157.44$ cm, The $z$-score($z = –2$) tells you that the male’s height is two standard deviations to the left of the mean.
Example $4$
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let $Y =$ the height of 15 to 18-year-old males from 1984 to 1985. Then $Y \sim N(172.36, 6.34)$.
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let $X =$ the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then $X \sim N(170, 6.28)$.
Find the z-scores for $x = 160.58$ cm and $y = 162.85$ cm. Interpret each $z$-score. What can you say about $x = 160.58$ cm and $y = 162.85$ cm?
Answer
• The $z$-score (Equation \ref{zscore}) for $x = 160.58$ is $z = –1.5$.
• The $z$-score for $y = 162.85$ is $z = –1.5$.
Both $x = 160.58$ and $y = 162.85$ deviate the same number of standard deviations from their respective means and in the same direction.
Exercise $4$
In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean $\mu = 496$ and a standard deviation $\sigma = 114$. Let $X =$ a SAT exam verbal section score in 2012. Then $X \sim N(496, 114)$.
Find the $z$-scores for $x_{1} = 325$ and $x_{2} = 366.21$. Interpret each $z$-score. What can you say about $x_{1} = 325$ and $x_{2} = 366.21$?
Answer
The z-score (Equation \ref{zscore}) for $x_{1} = 325$ is $z_{1} = –1.15$.
The z-score (Equation \ref{zscore}) for $x_{2} = 366.21$ is $z_{2} = –1.14$.
Student 2 scored closer to the mean than Student 1 and, since they both had negative $z$-scores, Student 2 had the better score.
Example $5$
Suppose x has a normal distribution with mean 50 and standard deviation 6.
• About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively.
• About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively.
• About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 99.7% of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 from the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively.
Exercise $5$
Suppose $X$ has a normal distribution with mean 25 and standard deviation five. Between what values of $x$ do 68% of the values lie?
Answer
between 20 and 30.
Example $6$
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let $Y =$ the height of 15 to 18-year-old males in 1984 to 1985. Then $Y \sim N(172.36, 6.34)$.
1. About 68% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively.
2. About 95% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________ respectively.
3. About 99.7% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively.
Answer
1. About 68% of the values lie between 166.02 and 178.7. The $z$-scores are –1 and 1.
2. About 95% of the values lie between 159.68 and 185.04. The $z$-scores are –2 and 2.
3. About 99.7% of the values lie between 153.34 and 191.38. The $z$-scores are –3 and 3.
Exercise $6$
The scores on a college entrance exam have an approximate normal distribution with mean, $\mu = 52$ points and a standard deviation, $\sigma = 11$ points.
1. About 68% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively.
2. About 95% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively.
3. About 99.7% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively.
Answer a
About 68% of the values lie between the values 41 and 63. The $z$-scores are –1 and 1, respectively.
Answer b
About 95% of the values lie between the values 30 and 74. The $z$-scores are –2 and 2, respectively.
Answer c
About 99.7% of the values lie between the values 19 and 85. The $z$-scores are –3 and 3, respectively.
Summary
A $z$-score is a standardized value. Its distribution is the standard normal, $Z \sim N(0,1)$. The mean of the $z$-scores is zero and the standard deviation is one. If $y$ is the z-score for a value $x$ from the normal distribution $N(\mu, \sigma)$ then $z$ tells you how many standard deviations $x$ is above (greater than) or below (less than) $\mu$.
Formula Review
$Z \sim N(0, 1)$
$z = a$ standardized value ($z$-score)
mean = 0; standard deviation = 1
To find the $K$th percentile of $X$ when the $z$-scores is known:
$k = \mu + (z)\sigma$
$z$-score: $z = \dfrac{x-\mu}{\sigma}$
$Z =$ the random variable for z-scores
$Z \sim N(0, 1)$
Glossary
Standard Normal Distribution
a continuous random variable (RV) $X \sim N(0, 1)$; when $X$ follows the standard normal distribution, it is often noted as $Z \sim N(0, 1)\. \(z$-score
the linear transformation of the form $z = \dfrac{x-\mu}{\sigma}$; if this transformation is applied to any normal distribution $X \sim N(\mu, \sigma$ the result is the standard normal distribution $Z \sim N(0,1)$. If this transformation is applied to any specific value $x$ of the RV with mean $\mu$ and standard deviation $\sigma$, the result is called the $z$-score of $x$. The $z$-score allows us to compare data that are normally distributed but scaled differently.
6.02: The Standard Normal Distribution
Exercise $7$
A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable $X$ in words. $X =$ ____________.
Answer
ounces of water in a bottle
Exercise $8$
A normal distribution has a mean of 61 and a standard deviation of 15. What is the median?
Exercise $9$
$X \sim N(1, 2)$
$\sigma =$ _______
Answer
2
Exercise $10$
A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable $X$ in words. $X =$ ______________.
Exercise $11$
$X \sim N(-4, 1)$
What is the median?
Answer
–4
Exercise $12$
$X \sim N(3, 5)$
$\sigma =$ _______
Exercise $13$
$X \sim N(-2, 1)$
$\mu =$ _______
Answer
–2
Exercise $14$
What does a $z$-score measure?
Exercise $15$
What does standardizing a normal distribution do to the mean?
Answer
The mean becomes zero.
Exercise $16$
Is $X \sim N(0, 1)$ a standardized normal distribution? Why or why not?
Exercise $17$
What is the $z$-score of $x = 12$, if it is two standard deviations to the right of the mean?
Answer
$z = 2$
Exercise $18$
What is the $z$-score of $x = 9$, if it is 1.5 standard deviations to the left of the mean?
Exercise $19$
What is the $z$-score of $x = -2$, if it is 2.78 standard deviations to the right of the mean?
Answer
$z = 2.78$
Exercise $20$
What is the $z$-score of $x = 7$, if it is 0.133 standard deviations to the left of the mean?
Exercise $21$
Suppose $X \sim N(2, 6)$. What value of x has a z-score of three?
Answer
$x = 20$
Exercise $22$
Suppose $X \sim N(8, 1)$. What value of $x$ has a $z$-score of –2.25?
Exercise $23$
Suppose $X \sim N(9, 5)$. What value of $x$ has a $z$-score of –0.5?
Answer
$x = 6.5$
Exercise $24$
Suppose $X \sim N(2, 3)$. What value of $x$ has a $z$-score of –0.67?
Exercise $25$
Suppose $X \sim N(4, 2)$. What value of $x$ is 1.5 standard deviations to the left of the mean?
Answer
$x = 1$
Exercise $26$
Suppose $X \sim N(4, 2)$. What value of $x$ is two standard deviations to the right of the mean?
Exercise $27$
Suppose $X \sim N(8, 9)$. What value of $x$ is 0.67 standard deviations to the left of the mean?
Answer
$x = 1.97$
Exercise $28$
Suppose $X \sim N(-1, 12)$. What is the $z$-score of $x = 2$?
Exercise $29$
Suppose $X \sim N(12, 6)$. What is the $z$-score of $x = 2$?
Answer
$z = –1.67$
Exercise $30$
Suppose $X \sim N(9, 3)$. What is the $z$-score of $x = 9$?
Exercise $31$
Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the $z$-score of $x = 5.5$?
Answer
$z \approx –0.33$
Exercise $32$
In a normal distribution, $x = 5$ and $z = –1.25$. This tells you that $x = 5$ is ____ standard deviations to the ____ (right or left) of the mean.
Exercise $33$
In a normal distribution, $x = 3$ and $z = 0.67$. This tells you that $x = 3$ is ____ standard deviations to the ____ (right or left) of the mean.
Answer
0.67, right
Exercise $34$
In a normal distribution, $x = –2$ and $z = 6$. This tells you that $z = –2$ is ____ standard deviations to the ____ (right or left) of the mean.
Exercise $35$
In a normal distribution, $x = –5$ and $z = –3.14$. This tells you that $x = –5$ is ____ standard deviations to the ____ (right or left) of the mean.
Answer
3.14, left
Exercise $36$
In a normal distribution, $x = 6$ and $z = –1.7$. This tells you that $x = 6$ is ____ standard deviations to the ____ (right or left) of the mean.
Exercise $37$
About what percent of $x$ values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution?
Answer
about 68%
Exercise $38$
About what percent of the $x$ values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution?
Exercise $39$
About what percent of $x$ values lie between the second and third standard deviations (both sides)?
Answer
about 4%
Exercise $40$
Suppose $X \sim N(15, 3)$. Between what $x$ values does 68.27% of the data lie? The range of $x$ values is centered at the mean of the distribution (i.e., 15).
Exercise $41$
Suppose $X \sim N(-3, 1)$. Between what $x$ values does 95.45% of the data lie? The range of $x$ values is centered at the mean of the distribution (i.e., –3).
Answer
between –5 and –1
Exercise $42$
Suppose $X \sim N(-3, 1)$. Between what $x$ values does 34.14% of the data lie?
Exercise $43$
About what percent of $x$ values lie between the mean and three standard deviations?
Answer
about 50%
Exercise $44$
About what percent of $x$ values lie between the mean and one standard deviation?
Exercise $45$
About what percent of $x$ values lie between the first and second standard deviations from the mean (both sides)?
Answer
about 27%
Exercise $46$
About what percent of $x$ values lie between the first and third standard deviations(both sides)?
Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.
Exercise $47$
Define the random variable $X$ in words. $X =$ _______________.
Answer
The lifetime of a Sunshine CD player measured in years.
Exercise $48$
$X \sim$ _____(_____,_____) | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/06%3A_The_Normal_Distribution/6.02%3A_The_Standard_Normal_Distribution/6.2E%3A_The_Standard_Normal_Distribution_%28Exercises%29.txt |
The shaded area in the following graph indicates the area to the left of $x$. This area is represented by the probability $P(X < x)$. Normal tables, computers, and calculators provide or calculate the probability $P(X < x)$.
The area to the right is then $P(X > x) = 1 – P(X < x)$. Remember, $P(X < x) =$ Area to the left of the vertical line through $x$. $P(X > x) = 1 – P(X < x) =$ Area to the right of the vertical line through $x$. $P(X < x)$ is the same as $P(X \leq x)$ and $P(X > x)$ is the same as $P(X \geq x)$ for continuous distributions.
Calculations of Probabilities
Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.To calculate the probability, use the probability tables provided in [link] without the use of technology. The tables include instructions for how to use them.
Example $1$
If the area to the left is 0.0228, then the area to the right is $1 - 0.0228 = 0.9772$.
Exercise $1$
If the area to the left of $x$ is $0.012$, then what is the area to the right?
Answer
$1 - 0.012 = 0.988$
Example $2$
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.
1. Find the probability that a randomly selected student scored more than 65 on the exam.
2. Find the probability that a randomly selected student scored less than 85.
3. Find the 90th percentile (that is, find the score $k$ that has 90% of the scores below k and 10% of the scores above $k$).
4. Find the 70th percentile (that is, find the score $k$ such that 70% of scores are below $k$ and 30% of the scores are above $k$).
Answer
a. Let $X$ = a score on the final exam. $X \sim N(63, 5)$, where $\mu = 63$ and $\sigma = 5$
Draw a graph.
Then, find $P(x > 65)$.
$P(x > 65) = 0.3446\nonumber$
The probability that any student selected at random scores more than 65 is 0.3446.
USING THE TI-83, 83+, 84, 84+ CALCULATOR
Go into 2nd DISTR.
After pressing 2nd DISTR, press 2:normalcdf.
The syntax for the instructions are as follows:
normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter 10^99instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve.
Historical Note
The TI probability program calculates a $z$-score and then the probability from the $z$-score. Before technology, the $z$-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the $z$-score was used. You calculate the $z$-score and look up the area to the left. The probability is the area to the right.
$z = 65 – 63565 – 635 = 0.4\nonumber$
Area to the left is 0.6554.
$P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446\nonumber$
USING THE TI-83, 83+, 84, 84+ CALCULATOR
Find the percentile for a student scoring 65:
*Press 2nd Distr
*Press 2:normalcdf(
*Enter lower bound, upper bound, mean, standard deviation followed by )
*Press ENTER.
For this Example, the steps are
2nd Distr
2:normalcdf(65,1,2nd EE,99,63,5) ENTER
The probability that a selected student scored more than 65 is 0.3446.
To find the probability that a selected student scored more than 65, subtract the percentile from 1.
Answer
b. Draw a graph.
Then find $P(x < 85)$, and shade the graph.
Using a computer or calculator, find $P(x < 85) = 1$.
$\text{normalcdf}(0,85,63,5) = 1$ (rounds to one)
The probability that one student scores less than 85 is approximately one (or 100%).
Answer
c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the $x$-axis. Shade the area that corresponds to the 90th percentile.
Let $k =$ the 90th percentile. The variable $k$ is located on the $x$-axis. $P(x < k)$ is the area to the left of $k$. The 90th percentile $k$ separates the exam scores into those that are the same or lower than $k$ and those that are the same or higher. Ninety percent of the test scores are the same or lower than $k$, and ten percent are the same or higher. The variable $k$ is often called a critical value.
$k = 69.4$
The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:
invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)
For this problem, $\text{invNorm}(0.90,63,5) = 69.4$
Answer
d. Find the 70th percentile.
Draw a new graph and label it appropriately. $k = 65.6$
The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.6 and 30% fall at or above.
$\text{invNorm}(0.70,63,5) = 65.6$
Exercise $2$
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65.
Answer
$\text{normalcdf}(10^{99},65,68,3) = 0.1587$
Example $3$
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.
1. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
2. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.
Answer
a. Let $X =$ the amount of time (in hours) a household personal computer is used for entertainment. $X \sim N(2, 0.5)$ where $\mu = 2$ and $\sigma = 0.5$.
Find $P(1.8 < x < 2.75)$.
The probability for which you are looking is the area between $x = 1.8$ and $x = 2.75$. $P(1.8 < x < 2.75) = 0.5886$
$\text{normalcdf}(1.8,2.75,2,0.5) = 0.5886\nonumber$
The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
b.
To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, $k$, where $P(x < k) = 0.25$.
$\text{invNorm}(0.25,2,0.5) = 1.66\nonumber$
The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.
Exercise $3$
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.
Answer
$\text{normalcdf}(66,70,68,3) = 0.4950$
Example $4$
There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
1. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
2. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
3. Find the 80th percentile of this distribution, and interpret it in a complete sentence.
Answer
1. $\text{normalcdf}(23,64.7,36.9,13.9) = 0.8186$
2. $\text{normalcdf}(-10^{99},50.8,36.9,13.9) = 0.8413$
3. $\text{invNorm}(0.80,36.9,13.9) = 48.6$
The 80th percentile is 48.6 years.
80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.
Use the information in Example to answer the following questions.
Exercise $4$
1. Find the 30th percentile, and interpret it in a complete sentence.
2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old.
70.
Answer
Let $X =$ a smart phone user whose age is 13 to 55+. $X \sim N(36.9, 13.9)$
To find the 30th percentile, find $k$ such that $P(x < k) = 0.30$.
$\text{invNorm}(0.30, 36.9, 13.9) = 29.6$ years
Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years. Find $P(x < 27)$
(Note that $\text{normalcdf}(-10^{99},27,36.9,13.9) = 0.2382$. The two answers differ only by 0.0040.)
$\text{normalcdf}(0,27,36.9,13.9) = 0.2342\nonumber$
Example $5$
In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).
1. Calculate the interquartile range ($IQR$).
2. Forty percent of the ages that range from 13 to 55+ are at least what age?
Answer
a.
$IQR = Q_{3} – Q_{1}\nonumber$
Calculate $Q_{3} =$ 75th percentile and $Q_{1} =$ 25th percentile.
\begin{align*} \text{invNorm}(0.75,36.9,13.9) &= Q_{3} = 46.2754 \[4pt] \text{invNorm}(0.25,36.9,13.9) &= Q_{1} = 27.5246 \[4pt] IQR &= Q_{3} - Q_{1} = 18.7508 \end{align*}
b.
Find $k$ where $P(x > k) = 0.40$ ("At least" translates to "greater than or equal to.")
$0.40 =$ the area to the right.
Area to the left $= 1 – 0.40 = 0.60$.
The area to the left of $k = 0.60$.
$\text{invNorm}(0.60,36.9,13.9) = 40.4215$.
$k = 40.42$.
Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years.
Exercise $5$
Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean $\mu = 81$ points and standard deviation $\sigma = 15$ points.
1. Calculate the first- and third-quartile scores for this exam.
2. The middle 50% of the exam scores are between what two values?
Answer
1. $Q_{1} =$ 25th percentile $= \text{invNorm}(0.25,81,15) = 70.9$
$Q_{3} =$ 75th percentile $= \text{invNorm}(0.75,81,15) = 91.9$
2. The middle 50% of the scores are between 70.9 and 91.1.
Example $6$
A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.
1. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.
2. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.
3. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.
Answer
a. $\text{normalcdf}(6,10^{99},5.85,0.24) = 0.2660$
Answer
b.
$1 – 0.20 = 0.80$
The tails of the graph of the normal distribution each have an area of 0.40.
Find $k1$, the 40th percentile, and $k2$, the 60th percentile ($0.40 + 0.20 = 0.60$).
$k1 = \text{invNorm}(0.40,5.85,0.24) = 5.79$ cm
$k2 = \text{invNorm}(0.60,5.85,0.24) = 5.91$ cm
Answer
c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.
Exercise $6$
Using the information from Example, answer the following:
1. The middle 45% of mandarin oranges from this farm are between ______ and ______.
2. Find the 16th percentile and interpret it in a complete sentence.
Answer a
The middle area $= 0.40$, so each tail has an area of 0.30.
$– 0.40 = 0.60$
The tails of the graph of the normal distribution each have an area of 0.30.
Find $k1$, the 30th percentile and $k2$, the 70th percentile ($0.40 + 0.30 = 0.70$).
$k1 = \text{invNorm}(0.30,5.85,0.24) = 5.72$ cm
$k2 = \text{invNorm}(0.70,5.85,0.24) = 5.98$ cm
Answer b
$\text{normalcdf}(5,10^{99},5.85,0.24) = 0.9998$
Review
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean $\mu$ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.
Formula Review
• Normal Distribution: $X \sim N(\mu, \sigma)$ where $\mu$ is the mean and σ is the standard deviation.
• Standard Normal Distribution: $Z \sim N(0, 1)$.
• Calculator function for probability: normalcdf (lower $x$ value of the area, upper $x$ value of the area, mean, standard deviation)
• Calculator function for the $k$th percentile: $k = \text{invNorm}$ (area to the left of $k$, mean, standard deviation)
Exercise $7$
How would you represent the area to the left of one in a probability statement?
Answer
$P(x < 1)$
Exercise $8$
Is $P(x < 1)$ equal to $P(x \leq 1)$? Why?
Answer
Yes, because they are the same in a continuous distribution: $P(x = 1) = 0$
Exercise $9$
How would you represent the area to the left of three in a probability statement?
Exercise $10$
What is the area to the right of three?
Answer
$1 – P(x < 3)$ or $P(x > 3)$
Exercise $11$
If the area to the left of $x$ in a normal distribution is 0.123, what is the area to the right of $x$?
Exercise $12$
If the area to the right of $x$ in a normal distribution is 0.543, what is the area to the left of $x$?
Answer
$1 - 0.543 = 0.457$
Use the following information to answer the next four exercises:
$X \sim N(54, 8)$
Exercise $13$
Find the probability that $x > 56$.
Exercise $14$
Find the probability that $x < 30$.
Answer
0.0013
Exercise $15$
Find the 80th percentile.
Exercise $16$
Find the 60th percentile.
Answer
56.03
Exercise $17$
$X \sim N(6, 2)$
Find the probability that $x$ is between three and nine.
Exercise $18$
$X \sim N(–3, 4)$
Find the probability that $x$ is between one and four.
Answer
0.1186
Exercise $19$
$X \sim N(4, 5)$
Find the maximum of $x$ in the bottom quartile.
Exercise $20$
Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
Figure $12$.
$P(0 < x <$ ____________$) =$ ___________ (Use zero for the minimum value of $x$.)
Answer
1. Check student’s solution.
2. 3, 0.1979
Exercise $21$
Find the probability that a CD player will last between 2.8 and six years.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
Figure $13$.
$P($__________ $< x <$ __________$)$ = __________
Exercise $22$
Find the 70th percentile of the distribution for the time a CD player lasts.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%.
Figure $14$.
$P(x < k) =$ __________ Therefore, $k =$ _________
Answer
1. Check student’s solution.
2. 0.70, 4.78 years | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/06%3A_The_Normal_Distribution/6.03%3A_Using_the_Normal_Distribution.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcome
• The student will compare and contrast empirical data and a theoretical distribution to determine if Terry Vogel's lap times fit a continuous distribution.
Directions
Round the relative frequencies and probabilities to four decimal places. Carry all other decimal answers to two places.
Collect the Data
1. Use the data from Appendix C. Use a stratified sampling method by lap (races 1 to 20) and a random number generator to pick six lap times from each stratum. Record the lap times below for laps two to seven.
_______ _______ _______ _______ _______ _______
_______ _______ _______ _______ _______ _______
_______ _______ _______ _______ _______ _______
_______ _______ _______ _______ _______ _______
_______ _______ _______ _______ _______ _______
_______ _______ _______ _______ _______ _______
2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes.
3. Calculate the following:
1. $\bar{x}$ = _______
2. $s$ = _______
4. Draw a smooth curve through the tops of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a v-shape, does it have a hump in the middle or at either end, and so on?)
Analyze the Distribution
Using your sample mean, sample standard deviation, and histogram to help, what is the approximate theoretical distribution of the data?
• $X \sim$ _____(_____,_____)
• How does the histogram help you arrive at the approximate distribution?
Describe the Data
Use the data you collected to complete the following statements.
• The IQR goes from __________ to __________.
• IQR = __________. (IQR = Q3Q1)
• The 15th percentile is _______.
• The 85th percentile is _______.
• The median is _______.
• The empirical probability that a randomly chosen lap time is more than 130 seconds is _______.
• Explain the meaning of the 85th percentile of this data.
Theoretical Distribution
Using the theoretical distribution, complete the following statements. You should use a normal approximation based on your sample data.
• The IQR goes from __________ to __________.
• IQR = _______.
• The 15th percentile is _______.
• The 85th percentile is _______.
• The median is _______.
• The probability that a randomly chosen lap time is more than 130 seconds is _______.
• Explain the meaning of the 85th percentile of this distribution.
Discussion Questions
Do the data from the section titled Collect the Data give a close approximation to the theoretical distribution in the section titled Analyze the Distribution? In complete sentences and comparing the result in the sections titled Describe the Data and Theoretical Distribution, explain why or why not.
6.05: Normal Distribution - Pinkie Length (Worksheet)
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will compare empirical data and a theoretical distribution to determine if data from the experiment follow a continuous distribution.
Collect the Data
Measure the length of your pinky finger (in centimeters).
1. Randomly survey 30 adults for their pinky finger lengths. Round the lengths to the nearest 0.5 cm.
_______ _______ _______ _______ _______
_______ _______ _______ _______ _______
_______ _______ _______ _______ _______
_______ _______ _______ _______ _______
_______ _______ _______ _______ _______
_______ _______ _______ _______ _______
2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes.
3. Calculate the following.
1. $\bar{x}$ = _______
2. $s$ = _______
4. Draw a smooth curve through the top of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a v-shape, does it have a hump in the middle or at either end, and so on?)
Analyze the Distribution
Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution of the data you collected?
• $X \sim$ _____(_____,_____)
• How does the histogram help you arrive at the approximate distribution?
Describe the Data
Using the data you collected complete the following statements. (Hint: order the data)
REMEMBER
(IQR = Q3Q1)
• IQR = _______
• The 15th percentile is _______.
• The 85th percentile is _______.
• Median is _______.
• What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm?
• Explain the meaning of the 85th percentile of this data.
Theoretical Distribution
Using the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation.
• IQR = _______
• The 15th percentile is _______.
• The 85th percentile is _______.
• Median is _______.
• What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm?
• Explain the meaning of the 85th percentile of this data.
Discussion Questions
Do the data you collected give a close approximation to the theoretical distribution? In complete sentences and comparing the results in the sections titled Describe the Data and Theoretical Distribution, explain why or why not. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/06%3A_The_Normal_Distribution/6.04%3A_Normal_Distribution_-_Lap_Times_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
6.2: The Standard Normal Distribution
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
Q 6.2.1
What is the median recovery time?
1. 2.7
2. 5.3
3. 7.4
4. 2.1
Q 6.2.2
What is the z-score for a patient who takes ten days to recover?
1. 1.5
2. 0.2
3. 2.2
4. 7.3
c
Q 6.2.3
The length of time to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true?
1. The data cannot follow the uniform distribution.
2. The data cannot follow the exponential distribution..
3. The data cannot follow the normal distribution.
1. I only
2. II only
3. III only
4. I, II, and III
Q 6.2.4
The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences.
1. 77 inches
2. 85 inches
3. If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer.
S 6.2.4
1. Use the $z$-score formula. $z = –0.5141$. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.
2. Use the $z$-score formula. $z = 1.5424$. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.
3. Height $= 79 + 3.5(3.89) = 90.67$ inches, which is over 7.7 feet tall. There are very few NBA players this tall so the answer is no, not likely.
Q 6.2.5
The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean $\mu = 125$ and standard deviation $\sigma = 14$. Systolic blood pressure for males follows a normal distribution.
1. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters.
2. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him?
Q 6.2.6
Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean $\mu = 125$ and standard deviation $\sigma = 14$. If $X =$ a systolic blood pressure score then $X \sim N(125, 14)$.
1. Which answer(s) is/are correct?
1. Kyle’s systolic blood pressure is 175.
2. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age.
3. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age.
4. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.
2. Calculate Kyle’s blood pressure.
S 6.2.6
1. iv
2. Kyle’s blood pressure is equal to $125 + (1.75)(14) = 149.5$.
Q 6.2.7
Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean $\mu = 10.2$ kg and standard deviation $\sigma = 0.8$ kg. Weights are normally distributed. $X \sim N(10.2, 0.8)$. Calculate the z-scores that correspond to the following weights and interpret them.
1. 11 kg
2. 7.9 kg
3. 12.2 kg
Q 6.2.8
In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean $\mu = 520$ and standard deviation $\sigma = 115$.
1. Calculate the $z$-score for an SAT score of 720. Interpret it using a complete sentence.
2. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score?
3. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?
S 6.2.8
Let $X =$ an SAT math score and $Y =$ an ACT math score.
1. $X = 720 \frac{720-520}{15} = 1.74$ The exam score of 720 is 1.74 standard deviations above the mean of 520.
2. $z = 1.5$
The math SAT score is $520 + 1.5(115) \approx 692.5$. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.
3. $\frac{X-\mu}{\sigma} = \frac{700-514}{117} \approx 1.59$, the z-score for the SAT. $\frac{Y-\mu}{\sigma} = \frac{30-21}{5.3} \approx 1.70$, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).
6.3: Using the Normal Distribution
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
Q 6.3.1
What is the probability of spending more than two days in recovery?
1. 0.0580
2. 0.8447
3. 0.0553
4. 0.9420
Q 6.3.2
The 90th percentile for recovery times is?
1. 8.89
2. 7.07
3. 7.99
4. 4.32
S 6.3.2
c
Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.
Q 6.3.3
Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space?
1. Yes
2. No
3. Unable to determine
Q 6.3.4
Find the probability that it takes at least eight minutes to find a parking space.
1. 0.0001
2. 0.9270
3. 0.1862
4. 0.0668
d
Q 6.3.5
Seventy percent of the time, it takes more than how many minutes to find a parking space?
1. 1.24
2. 2.41
3. 3.95
4. 6.05
Q 6.3.6
According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let $X =$ height of the individual.
1. $X \sim$ _____(_____,_____)
2. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement.
3. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.
4. The middle 40% of heights fall between what two values? Sketch the graph, and write the probability statement.
S 6.3.6
1. $X \sim N(66, 2.5)$
2. 0.5404
3. No, the probability that an Asian male is over 72 inches tall is 0.0082
Q 6.3.7
IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let $X =$ IQ of an individual.
1. $X \sim$ _____(_____,_____)
2. Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement.
3. MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement.
4. The middle 50% of IQs fall between what two values? Sketch the graph and write the probability statement.
Q 6.3.8
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let $X =$ percent of fat calories.
1. $X \sim$ _____(_____,_____)
2. Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined.
3. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.
S 6.3.8
1. $X \sim N(36, 10)$
2. The probability that a person consumes more than 40% of their calories as fat is 0.3446.
3. Approximately 25% of people consume less than 29.26% of their calories as fat.
Q 6.3.9
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.
1. If $X =$ distance in feet for a fly ball, then $X \sim$ _____(_____,_____)
2. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.
3. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.
Q 6.3.10
In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
1. In words, define the random variable $X$.
2. $X \sim$ _____(_____,_____)
3. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.
4. What percent of the children spend over ten hours per day unsupervised?
5. Seventy percent of the children spend at least how long per day unsupervised?
S 6.3.10
1. $X =$ number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
2. $X ~ N(3, 1.5)$
3. The probability that the child spends less than one hour a day unsupervised is 0.0918.
4. The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
5. 2.21 hours
Q 6.3.11
In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let $X =$ number of votes for President Clinton for an election district.
1. State the approximate distribution of $X$.
2. Is 1,956.8 a population mean or a sample mean? How do you know?
3. Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement.
4. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.
5. Find the third quartile for votes for President Clinton.
Q 6.3.12
Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days.
1. In words, define the random variable $X$.
2. $X \sim$ _____(_____,_____)
3. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement.
4. Sixty percent of all trials of this type are completed within how many days?
S 6.3.12
1. $X =$ the distribution of the number of days a particular type of criminal trial will take
2. $X \sim N(21, 7)$
3. The probability that a randomly selected trial will last more than 24 days is 0.3336.
4. 22.77
Q 6.3.13
Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps.
1. In words, define the random variable $X$.
2. $X \sim$ _____(_____,_____)
3. Find the percent of her laps that are completed in less than 130 seconds.
4. The fastest 3% of her laps are under _____.
5. The middle 80% of her laps are from _______ seconds to _______ seconds.
Q 6.3.14
Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let $X =$ time in line. Table displays the ordered real data (in minutes):
0.50 4.25 5 6 7.25
1.75 4.25 5.25 6 7.25
2 4.25 5.25 6.25 7.25
2.25 4.25 5.5 6.25 7.75
2.25 4.5 5.5 6.5 8
2.5 4.75 5.5 6.5 8.25
2.75 4.75 5.75 6.5 9.5
3.25 4.75 5.75 6.75 9.5
3.75 5 6 6.75 9.75
3.75 5 6 6.75 10.75
1. Calculate the sample mean and the sample standard deviation.
2. Construct a histogram.
3. Draw a smooth curve through the midpoints of the tops of the bars.
4. In words, describe the shape of your histogram and smooth curve.
5. Let the sample mean approximate $\mu$ and the sample standard deviation approximate $\sigma$. The distribution of $X$ can then be approximated by $X \sim$ _____(_____,_____)
6. Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes.
7. Determine the cumulative relative frequency for waiting less than 6.1 minutes.
8. Why aren’t the answers to part f and part g exactly the same?
9. Why are the answers to part f and part g as close as they are?
10. If only ten customers has been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion.
S 6.3.14
1. $\text{mean} = 5.51$, $s = 2.15$
2. Check student's solution.
3. Check student's solution.
4. Check student's solution.
5. $X \sim N(5.51, 2.15)$
6. 0.6029
7. The cumulative frequency for less than 6.1 minutes is 0.64.
8. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.
9. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
10. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.
Q 6.3.15
Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false.
1. Ricardo’s actual GPA is lower than Anita’s actual GPA.
2. Ricardo is not passing because his z-score is zero.
3. Anita is in the 70th percentile of students at her college.
Q 6.3.16
Table shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums.
40,000 40,000 45,050 45,500 46,249 48,134
49,133 50,071 50,096 50,466 50,832 51,100
51,500 51,900 52,000 52,132 52,200 52,530
52,692 53,864 54,000 55,000 55,000 55,000
55,000 55,000 55,000 55,082 57,000 58,008
59,680 60,000 60,000 60,492 60,580 62,380
62,872 64,035 65,000 65,050 65,647 66,000
66,161 67,428 68,349 68,976 69,372 70,107
70,585 71,594 72,000 72,922 73,379 74,500
75,025 76,212 78,000 80,000 80,000 82,300
1. Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums (the data).
2. Construct a histogram.
3. Draw a smooth curve through the midpoints of the tops of the bars of the histogram.
4. In words, describe the shape of your histogram and smooth curve.
5. Let the sample mean approximate $\mu$ and the sample standard deviation approximate $\sigma$. The distribution of $X$ can then be approximated by $X \sim$ _____(_____,_____).
6. Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators.
7. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample.
8. Why aren’t the answers to part f and part g exactly the same?
S 6.3.16
1. $\text{mean} = 60,136$, $s = 10,468$
2. Answers will vary.
3. Answers will vary.
4. Answers will vary.
5. $X \sim N(60136, 10468)$
6. 0.7440
7. The cumulative relative frequency is $\frac{43}{60} = 0.717$.
8. The answers for part f and part g are not the same, because the normal distribution is only an approximation.
Q 6.3.17
An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability.
Q 6.3.18
A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.
S 6.3.18
• $n = 100; p = 0.1; q = 0.9$
• $\mu = np = (100)(0.10) = 10$
• $\sigma = \sqrt{npq} = \sqrt{(100)(0.1)(0.9)} = 3$
1. $z = \pm: x_{1} = \mu + z\sigma = 10 + 1(3) = 13$ and $x_{2} = \mu = z\sigma = 10 - 1(3) = 7.68%$ of the defective cars will fall between seven and 13.
2. $z = \pm: x_{1} = \mu + z\sigma = 10 + 2(3) = 16$ and $x_{2} = \mu = z\sigma = 10 - 2(3) = 4.95%$ of the defective cars will fall between four and 16
3. $z = \pm: x_{1} = \mu + z\sigma = 10 + 3(3) = 19$ and $x_{2} = \mu = z\sigma = 10 - 3(3) = 1.997%$ of the defective cars will fall between one and 19.
Q 6.3.19
We flip a coin 100 times ($n = 100$) and note that it only comes up heads 20% ($p = 0.20$) of the time. The mean and standard deviation for the number of times the coin lands on heads is $\mu = 20$ and $\sigma = 4$ (verify the mean and standard deviation). Solve the following:
1. There is about a 68% chance that the number of heads will be somewhere between ___ and ___.
2. There is about a ____chance that the number of heads will be somewhere between 12 and 28.
3. There is about a ____ chance that the number of heads will be somewhere between eight and 32.
Q 6.3.20
A \$1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of $n = 190$ lotto tickets, find the probability for the lotto tickets that there are
1. somewhere between 34 and 54 prizes.
2. somewhere between 54 and 64 prizes.
3. more than 64 prizes.
S 6.3.21
• $n = 190; p = 1515 = 0.2; q = 0.8$
• $\mu = np = (190)(0.2) = 38$
• $\sigma = \sqrt{npq} = \sqrt{(190)(0.2)(0.8)} = 5.5136$
1. For this problem: $P(34 < x < 54) = \text{normalcdf}(34,54,48,5.5136) = 0.7641$
2. For this problem: $P(54 < x < 64) = \text{normalcdf}(54,64,48,5.5136) = 0.0018$
3. For this problem: $P(x > 64) = \text{normalcdf}(64,10^{99},48,5.5136) = 0.0000012$ (approximately 0)
Q 6.3.22
Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
1. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
2. Find the 95th percentile, and express it in a sentence. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/06%3A_The_Normal_Distribution/6.E%3A_The_Normal_Distribution_%28Exercises%29.txt |
In this chapter, you will study means and the central limit theorem, which is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, $\mu$, and a known standard deviation, $\sigma$. The first alternative says that if we collect samples of size $n$ with a "large enough $n$," calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size $n$ that are "large enough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape.
• 7.1: Prelude to the Central Limit Theorem
The central limit theorem states that, given certain conditions, the arithmetic mean of a sufficiently large number of iterates of independent random variables, each with a well-defined expected value and well-defined variance, will be approximately normally distributed.
• 7.2: The Central Limit Theorem for Sample Means (Averages)
In a population whose distribution may be known or unknown, if the size (n) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n).
• 7.3: The Central Limit Theorem for Sums
The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed.
• 7.4: Using the Central Limit Theorem
The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean <x> gets to μ . The central limit theorem illustrates the law of large numbers.
• 7.5: Central Limit Theorem - Pocket Change (Worksheet)
A statistics Worksheet: The student will demonstrate and compare properties of the central limit theorem.
• 7.6: Central Limit Theorem - Cookie Recipes (Worksheet)
A statistics Worksheet: The student will demonstrate and compare properties of the central limit theorem.
• 7.E: The Central Limit Theorem (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected].
07: The Central Limit Theorem
Learning Objectives
By the end of this chapter, the student should be able to:
• Recognize central limit theorem problems.
• Classify continuous word problems by their distributions.
• Apply and interpret the central limit theorem for means.
• Apply and interpret the central limit theorem for sums.
Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem. The central limit theorem (clt for short) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, $\mu$, and a known standard deviation, $\sigma$. The first alternative says that if we collect samples of size $n$ with a "large enough $n$," calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size $n$ that are "large enough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape.
In either case, it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distribution of sample means and the sums tend to follow the normal distribution.
The size of the sample, $n$, that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement.
COLLABORATIVE CLASSROOM ACTIVITY
Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times.
Each time a person rolls more than one die, he or she calculates the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, 6 on one roll.
The mean is $\frac{2+2+3+4+6}{5} = 3.4$. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means.
Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5.
Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Since the "mean" when you roll one die is just the face on the die, what distribution do these means appear to be representing?
• Draw the graph for the means using two dice. Do the sample means show any kind of pattern?
• Draw the graph for the means using five dice. Do you see any pattern emerging?
• Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice?
As the number of dice rolled increases from one to two to five to ten, the following is happening:
1. The mean of the sample means remains approximately the same.
2. The spread of the sample means (the standard deviation of the sample means) gets smaller.
3. The graph appears steeper and thinner.
You have just demonstrated the central limit theorem (clt). The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution).
Glossary
Sampling Distribution
Given simple random samples of size $n$ from a given population with a measured characteristic such as mean, proportion, or standard deviation for each sample, the probability distribution of all the measured characteristics is called a sampling distribution. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.01%3A_Prelude_to_the_Central_Limit_Theorem.txt |
Suppose $X$ is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:
1. $\mu_{x} =$ the mean of $X$
2. $\sigma_{x} =$ the standard deviation of $X$
If you draw random samples of size $n$, then as $n$ increases, the random variable $\bar{X}$ which consists of sample means, tends to be normally distributed and
$\bar{X} \sim N \left(\mu_{x}, \dfrac{\sigma_{x}}{\sqrt{n}}\right).$
The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by, the sample size. The variable $n$ is the number of values that are averaged together, not the number of times the experiment is done.
To put it more formally, if you draw random samples of size $n$, the distribution of the random variable $\bar{X}$, which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as $n$, the sample size, increases.
The random variable $\bar{X}$ has a different $z$-score associated with it from that of the random variable $X$. The mean $\bar{x}$ is the value of $\bar{X}$ in one sample.
$z = \dfrac{\bar{x}-\mu_{x}}{\left(\dfrac{\sigma_{x}}{\sqrt{n}}\right)}$
• $\mu_{x}$ is the average of both $X$ and $\bar{X}$.
• $\sigma \bar{x} = \dfrac{\sigma_{x}}{\sqrt{n}} =$ standard deviation of $\bar{X}$ and is called the standard error of the mean.
Howto: Find probabilities for means on the calculator
2nd DISTR
2:normalcdf
$\text{normalcdf} \left(\text{lower value of the area, upper value of the area, mean}, \dfrac{\text{standard deviation}}{\sqrt{\text{sample size}}}\right)$
where:
• mean is the mean of the original distribution
• standard deviation is the standard deviation of the original distribution
• sample size $= n$
Example $1$
An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size $n = 25$ are drawn randomly from the population.
1. Find the probability that the sample mean is between 85 and 92.
2. Find the value that is two standard deviations above the expected value, 90, of the sample mean.
Answer
a.
Let $X =$ one value from the original unknown population. The probability question asks you to find a probability for the sample mean.
Let $\bar{X} =$ the mean of a sample of size 25. Since $\mu_{x} = 90, \sigma_{x} = 15$, and $n = 25$,
$\bar{X} \sim N(90, \dfrac{15}{\sqrt{25}}). \nonumber$
Find $P(85 < x < 92)$. Draw a graph.
$P(85 < x < 92) = 0.6997 \nonumber$
The probability that the sample mean is between 85 and 92 is 0.6997.
normalcdf(lower value, upper value, mean, standard error of the mean)
The parameter list is abbreviated (lower value, upper value, $\mu$, $\dfrac{\sigma}{\sqrt{n}}$)
normalcdf$(85,92,90,\dfrac{15}{\sqrt{25}}) = 0.6997$
b.
To find the value that is two standard deviations above the expected value 90, use the formula:
\begin{align*} \text{value} &= \mu_{x} + (\#\text{ofTSDEVs})\left(\dfrac{\sigma_{x}}{\sqrt{n}}\right) \[5pt] &= 90 + 2 \left(\dfrac{15}{\sqrt{25}}\right) = 96 \end{align*}
The value that is two standard deviations above the expected value is 96.
The standard error of the mean is
$\dfrac{\sigma_{x}}{\sqrt{n}} = \dfrac{15}{\sqrt{25}} = 3. \nonumber$
Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size $n$.
Exercise $1$
An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size $n$ = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.
Answer
$P(42 < \bar{x} < 50) = \left(42, 50, 45, \dfrac{8}{\sqrt{30}}\right) = 0.9797$
Example $2$
The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours. A sample of size $n = 50$ is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours.
Answer
Let $X =$ the time, in hours, it takes to play one soccer match.
The probability question asks you to find a probability for the sample mean time, in hours, it takes to play one soccer match.
Let $\bar{X} =$ the mean time, in hours, it takes to play one soccer match.
If $\mu_{x} =$ _________, $\sigma_{x} =$ __________, and $n =$ ___________, then $X \sim N$(______, ______) by the central limit theorem for means.
$\mu_{x} = 2, \sigma_{x} = 0.5, n = 50$, and $X \sim N \left(2, \dfrac{0.5}{\sqrt{50}}\right)$
Find $P(1.8 < \bar{x} < 2.3)$. Draw a graph.
$P(1.8 < \bar{x} < 2.3) = 0.9977$
normalcdf$\left(1.8,2.3,2,\dfrac{.5}{\sqrt{50}}\right) = 0.9977$
The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.
Exercise $2$
The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of $n = 60$ is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.
Answer
$P(2 < \bar{x} < 3) = \text{normalcdf}\left(2, 3, 2.5, \dfrac{0.25}{\sqrt{60}}\right) = 1 \nonumber$
Calculator SKills
To find percentiles for means on the calculator, follow these steps.
• 2nd DIStR
• 3:invNorm
$k = \text{invNorm} \left(\text{area to the left of} k, \text{mean}, \dfrac{\text{standard deviation}}{\sqrt{sample size}}\right)$
where:
• $k$ = the $k$th percentile
• mean is the mean of the original distribution
• standard deviation is the standard deviation of the original distribution
• sample size = $n$
Example $3$
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size $n = 100$.
1. What are the mean and standard deviation for the sample mean ages of tablet users?
2. What does the distribution look like?
3. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).
4. Find the 95th percentile for the sample mean age (to one decimal place).
Answer
1. Since the sample mean tends to target the population mean, we have $\mu_{x} = \mu = 34$. The sample standard deviation is given by: $\sigma_{x} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{15}{\sqrt{100}} = \dfrac{15}{10} = 1.5 \nonumber$
2. The central limit theorem states that for large sample sizes ($n$), the sampling distribution will be approximately normal.
3. The probability that the sample mean age is more than 30 is given by: $P(Χ > 30) = \text{normalcdf}(30,E99,34,1.5) = 0.9962 \nonumber$
4. Let $k$ = the 95th percentile. $k = \text{invNorm}\left(0.95, 34, \dfrac{15}{\sqrt{100}}\right) = 36.5 \nonumber$
Exercise $3$
In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy?
Answer
You need to determine the probability for men whose mean age is between 29 and 35 years of age wanting to play a strategy game.
$P(29 < \bar{x} < 35) = \text{normalcdf} \left(29, 35, 28,\dfrac{4.8}{\sqrt{100}}\right) = 0.0186$
You can conclude there is approximately a 1.9% chance that your game will be played by men whose mean age is between 29 and 35.
Example $4$
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.
1. What are the mean and standard deviation for the sample mean number of app engagement by a tablet user?
2. What is the standard error of the mean?
3. Find the 90th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.
Answer
1. $\mu = \mu = 8.2 \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{1}{\sqrt{60}} = 0.13$
2. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
3. Let $k$ = the 90th percentile
$k = \text{invNorm}\left(0.90, 8.2, \dfrac{1}{\sqrt{60}}\right) = 8.37$. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
4. $P(8 < \bar{x} < 8.5) = \text{normalcdf}\left(8, 8.5, 8.2, \dfrac{1}{\sqrt{60}}\right) = 0.9293$
Exercise $4$
Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are $n = 34$, $\bar{x} = 16.01$ ounces. If the cans are filled so that $\mu = 16.00$ ounces (as labeled) and $\sigma = 0.143$ ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?
Answer
We have $P(\bar{x} > 16.01) = \text{normalcdf} \left(16.01,E99,16, \dfrac{0.143}{\sqrt{34}}\right) = 0.3417$. Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company’s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.
Summary
In a population whose distribution may be known or unknown, if the size ($n$) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ($n$).
Formula Review
• The Central Limit Theorem for Sample Means: $\bar{X} \sim N\left(\mu_{x}, \dfrac{\sigma_{x}}{\sqrt{n}}\right) \nonumber$
• The Mean $\bar{X}: \sigma_{x}$
• Central Limit Theorem for Sample Means z-score and standard error of the mean: $z = \dfrac{\bar{x}-\mu_{x}}{\left(\dfrac{\sigma_{x}}{\sqrt{n}}\right)} \nonumber$
• Standard Error of the Mean (Standard Deviation ($\bar{X}$)): $\dfrac{\sigma_{x}}{\sqrt{n}} \nonumber$
Glossary
Average
a number that describes the central tendency of the data; there are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean.
Central Limit Theorem
Given a random variable (RV) with known mean $\mu$ and known standard deviation, $\sigma$, we are sampling with size $n$, and we are interested in two new RVs: the sample mean, $\bar{X}$, and the sample sum, $\sum X$. If the size ($n$) of the sample is sufficiently large, then $\bar{X} \sim N\left(\mu, \dfrac{\sigma}{\sqrt{n}}\right)$ and $\sum X \sim N(n\mu, (\sqrt{n})(\sigma))$. If the size ($n$) of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distributions regardless of the shape of the population. The mean of the sample means will equal the population mean, and the mean of the sample sums will equal $n$ times the population mean. The standard deviation of the distribution of the sample means, $\dfrac{\sigma}{\sqrt{n}}$, is called the standard error of the mean.
Normal Distribution
a continuous random variable (RV) with pdf $f(x) = \dfrac{1}{\sigma \sqrt{2 \pi}}e^{\dfrac{-(x-\mu)^{2}}{2 \sigma^{2}}}$, where $\mu$ is the mean of the distribution and $\sigma$ is the standard deviation; notation: $X \sim N(\mu, \sigma)$. If $\mu = 0$ and $\sigma = 1$, the RV is called a standard normal distribution.
Standard Error of the Mean
the standard deviation of the distribution of the sample means, or $\dfrac{\sigma}{\sqrt{n}}$.
7.02: The Central Limit Theorem for Sample Means (Averages)
Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let $X$ be the random variable representing the time it takes her to complete one review. Assume $X$ is normally distributed. Let $\bar{X}$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
Example $1$
What is the mean, standard deviation, and sample size?
Answer
mean = 4 hours; standard deviation = 1.2 hours; sample size = 16
Exercise $2$
Complete the distributions.
1. $X \sim$ _____(_____,_____)
2. $\bar{X} \sim$ _____(_____,_____)
Example $3$
Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
Figure $2$.
2. $P$(________ $< x <$ ________) = _______
Answer
1. Check student's solution.
2. 3.5, 4.25, 0.2441
Exercise $4$
Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
Figure $3$.
2. $P$(________________) = _______
Example $5$
What causes the probabilities in Exercise and Exercise to be different?
Answer
The fact that the two distributions are different accounts for the different probabilities.
Exercise $6$
Find the 95th percentile for the mean time to complete one month's reviews. Sketch the graph.
Figure $4$.
1. The 95th Percentile =____________ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.02%3A_The_Central_Limit_Theorem_for_Sample_Means_%28Averages%29/7.2E%3A_The_Central_Limit_Theorem_for_Sample_Means_%28Exercises%.txt |
Suppose $X$ is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
• $\mu_{x}$ = the mean of $X$
• $\sigma_{x}$ = the standard deviation of $X$
If you draw random samples of size $n$, then as $n$ increases, the random variable $\sum X$ consisting of sums tends to be normally distributed and
$\sum X \sim N((n)(\mu_{x}), (\sqrt{n})(\sigma_{x})).$
The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.
The random variable $\sum X$ has the following z-score associated with it:
1. $\sum x$ is one sum.
2. $z = \frac{\sum x - (n)(\mu_{x})}{(\sqrt{n})(\sigma_{x})}$
1. $(n)(\mu_{x})$= the mean of $\sum X$
2. $(\sqrt{n})(\sigma_{x})$= standard deviation of $\sum X$
Calculator
To find probabilities for sums on the calculator, follow these steps.
2nd DISTR
2:normalcdf
normalcdf(lower value of the area, upper value of the area, ($n$)(mean), ($\sqrt{n}$)(standard deviation))
where:
• mean is the mean of the original distribution
• standard deviation is the standard deviation of the original distribution
• sample size $= n$
Example $1$
An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.
1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
2. Find the sum that is 1.5 standard deviations above the mean of the sums.
Answer
Let $X =$ one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.
$\sum X =$ the sum or total of 80 values. Since $\mu_{x} = 90$, $\sigma_{x} = 15$, and $n = 80$, $\sum X \sim N((80)(90),(\sqrt{80})(15))$
• mean of the sums $= (n)(\mu_{x}) = (80)(90) = 7,200$
• standard deviation of the sums $= (\sqrt{n})(\sigma_{x}) = (\sqrt{80})(15) = (80)(15)$
• sum of 80 values $= \sum X = 7,500$
a. Find $P(\sum X > 7,500)$
$P(\sum X > 7,500) = 0.0127$
normalcdf(lower value, upper value, mean of sums, stdev of sums)
The parameter list is abbreviated $\left(lower, upper, (n)(\mu_{x}, (\sqrt{n}(\sigma_{x})\right)$
normalcdf $\left(7500,1E99,(80)(90),(\sqrt{80})(15)\right) = 0.0127$
REMINDER
1E99 = 1099.
Press the EE key for E.
b. Find $\sum x$ where $z = 1.5$.
$\sum x = (n)(\nu_{x}) + (z)(\sqrt{n})(\sigma_{x}) = (80)(90) + (1.5)(\sqrt{80})(15) = 7,401.2$
Exercise $1$
An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.
Answer
0.0040
Calculator
To find percentiles for sums on the calculator, follow these steps.
2nd DIStR
3:invNorm
$k = \text{invNorm} (\text{area to the left of} k, (n)(\text{mean}), (\sqrt{n})(\text{standard deviation}))$
where:
• $k$ is the $k$th percentile
• mean is the mean of the original distribution
• standard deviation is the standard deviation of the original distribution
• sample size $= n$
Example $2$
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.
1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
2. Find the probability that the sum of the ages is between 1,500 and 1,800 years.
3. Find the 80th percentile for the sum of the 50 ages.
Answer
1. $\mu_{x} - n\mu_{x} = 1,700$ and $\sigma_{\sum X} = \sqrt{n}\sigma_{X} = (\sqrt{50})(15) = 106.01$
The distribution is normal for sums by the central limit theorem.
2. $P(1500 < \sum X < 1800) = (1,500, 1,800, (50)(34), (\sqrt{50})(15)) = 0.7974$
3. Let $k$ = the 80th percentile.
$k = (0.80,(50)(34),(\sqrt{50})(15)) = 1,789.3$
Exercise $2$
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 35 years. Suppose the standard deviation is ten years. The sample size is 39.
1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
2. Find the probability that the sum of the ages is between 1,400 and 1,500 years.
3. Find the 90th percentile for the sum of the 39 ages.
Answer
1. $\mu_{\sum X} = n\mu_{X} = 1,365$ and $\sigma_{\sum X} = \sqrt{n}\sigma_{x} = 62.4$
The distribution is normal for sums by the central limit theorem.
2. $P(1400 < \sum_{X} < 1500) = \text{normalcdf} (1400,1500,(39)(35),(\sqrt{39})(10)) = 0.2723$
3. Let $k$ = the 90th percentile.
$k = \text{invNorm} (0.90,(39)(35),(\sqrt{39}) (10)) = 1445.0$
Example $3$
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70.
1. What are the mean and standard deviation for the sums?
2. Find the 95th percentile for the sum of the sample. Interpret this value in a complete sentence.
3. Find the probability that the sum of the sample is at least ten hours.
Answer
1. $\mu_{\sum X} = n\mu_{X}= 70(8.2) = 574$ minutes and $\sigma_{\sum X} (\sqrt{n})(\sigma_{x}) = (\sqrt{70})(1) = 8.37$ minutes
2. Let $k$ = the 95th percentile.
$k = \text{invNorm} (0.95,(70)(8.2),(\sqrt{70})(1)) = 587.76$ minutes
Ninety five percent of the app engagement times are at most 587.76 minutes.
3. ten hours = 600 minutes
$P(\sum X \geq 600) = \text{normalcdf}(600,E99,(70)(8.2),(\sqrt{70})(1)) = 0.0009$
Exercise $3$
The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.
1. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?
2. Find the 84th and 16th percentiles for the sum of the sample. Interpret these values in context.
Answer
1. 7 hours = 420 minutes
10 hours = 600 minutes
$\text{normalcdf} P(420 \leq \sum X \leq 600) = \text{normalcdf}(420,600,(70)(8.2),\sqrt{70}(1)) = 0.9991$
This means that for this sample sums there is a 99.9% chance that the sums of usage minutes will be between 420 minutes and 600 minutes.
2. $\text{invNorm}(0.84,(70)(8.2)$,$\sqrt{70}(1)) = 582.32$
$\text{invNorm}(0.16,(70)(8.2),(\sqrt{70}(1)) = 565.68$
Since 84% of the app engagement times are at most 582.32 minutes and 16% of the app engagement times are at most 565.68 minutes, we may state that 68% of the app engagement times are between 565.68 minutes and 582.32 minutes.
Review
The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of $\mu_{x}$ and a standard deviation of $\sigma_{x}$, the mean of the sums is $n$$\mu_{x}$ and the standard deviation is ($\sqrt{n}$)($\sigma_{x}$) where $n$ is the sample size.
Formula Review
• The Central Limit Theorem for Sums: $\sum X ~ N[(n)(\mu_{x}, (\sqrt{n})(\sigma_{x}))]$
• Mean for Sums $(\sum X): (n)(\mu_{x})$
• The Central Limit Theorem for Sums $z$-score and standard deviation for sums: $z \text{ for the sample mean} = \frac{\sum x - (n)(\mu_{x})}{(\sqrt{n})(\sigma_{x})}$
• Standard deviation for Sums $(\sum X): (\sqrt{n})(\sigma_{x})$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.03%3A_The_Central_Limit_Theorem_for_Sums.txt |
It is important for you to understand when to use the central limit theorem (clt). If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.
If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.
Law of Large Numbers
The law of large numbers says that if you take samples of larger and larger size from any population, then the mean $\bar{x}$ of the sample tends to get closer and closer to $\mu$. From the central limit theorem, we know that as $n$ gets larger and larger, the sample means follow a normal distribution. The larger $n$ gets, the smaller the standard deviation gets. (Remember that the standard deviation for $\bar{X}$ is $\dfrac{\sigma}{\sqrt{n}}$.) This means that the sample mean $\bar{x}$ must be close to the population mean $\mu$. We can say that $\mu$ is the value that the sample means approach as $n$ gets larger. The central limit theorem illustrates the law of large numbers.
Example $1$
A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:
1. The probability that the mean stress score for the 75 students is less than two.
2. The 90th percentile for the mean stress score for the 75 students.
3. The probability that the total of the 75 stress scores is less than 200.
4. The 90th percentile for the total stress score for the 75 students.
Solutions
Let $X =$ one stress score.
Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d ask you to find a probability or a percentile for a total or sum. The sample size, $n$, is equal to 75.
Since the individual stress scores follow a uniform distribution, $X \sim U(1, 5)$ where $a = 1$ and $b = 5$.
$\mu_{x} = \dfrac{a+b}{2} = \dfrac{1+5}{2} = 3$
$\sigma_{x} = \sqrt{\dfrac{(b-a)^{2}}{12}} = \sqrt{\dfrac{(5-1)^{2}}{12}} = 1.15$
For problems 1. and 2., let $\bar{X} =$ the mean stress score for the 75 students. Then,
$\bar{X} \sim N\left(3, \dfrac{1,15}{\sqrt{75}}\right)$
where $n = 75$.
1. Find $P(\bar{x} < 2)$. Draw the graph.
2. Find the 90th percentile for the mean of 75 stress scores. Draw a graph.
3. Find $P(\sum x < 2000)$. Draw the graph.
4. Find the 90th percentile for the total of 75 stress scores. Draw a graph.
Answers
a. $P(\bar{x} < 2) = 0$
The probability that the mean stress score is less than two is about zero.
normalcdf $\left(1,2,3,\dfrac{1.15}{\sqrt{75}}\right) = 0$
REMINDER
The smallest stress score is one
b. Let $k =$ the 90th percentile.
Find $k$, where $P(\bar{x} < k) = 0.90$.
$k = 3.2$
The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.
invNorm$\left(0.90,3,1.\dfrac{1.15}{\sqrt{75}}\right) = 3.2$
For problems c and d, let $\sum X =$ the sum of the 75 stress scores. Then,
$\sum X \sim N((75)(3), (\sqrt{75})(1.15))$
c. The mean of the sum of 75 stress scores is $(75)(3) = 225$
The standard deviation of the sum of 75 stress scores is $(\sqrt{75})(1.15) = 9.96$
$P(\sum x < 200)$
The probability that the total of 75 scores is less than 200 is about zero.
normalcdf $75,200,(75)(3),(\sqrt{75})(1.15)$.
REMINDER
The smallest total of 75 stress scores is 75, because the smallest single score is one.
d. Let $k =$ the 90th percentile.
Find $k$ where $P(\sum x < k) = 0.90$.
$k = 237.8$
The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.
invNorm$\left(0.90, (75)(3), (\sqrt{75})(1.15)\right) = 237.8$
Exercise $1$
Use the information in Example $1$, but use a sample size of 55 to answer the following questions.
1. Find $P(\bar{x} < 7)$.
2. Find $P(\sum x < 7)$.
3. Find the 80th percentile for the mean of 55 scores.
4. Find the 85th percentile for the sum of 55 scores.
Answer
1. 0.0265
2. 0.2789
3. 3.13
4. 173.84
Example $2$
Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.
Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.
Let $X =$ the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.
$X \sim Exp\left(\dfrac{1}{22}\right)$. From previous chapters, we know that $\mu = 22$ and $\sigma = 22$.
Let $\bar{X}$ = the mean excess time used by a sample of $n = 80$ customers who exceed their contracted time allowance.
$\bar{X} \sim N\left(22,\dfrac{22}{\sqrt{80}}\right)$
by the central limit theorem for sample means
1. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find $P(\bar{x} > 20)$. Draw the graph.
2. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find $P(x > 20)$.
3. Explain why the probabilities in parts a and b are different.
4. Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph.
Answer
1. Find: $P(\bar{x} > 20)$
$P(\bar{x} > 20) = 0.79199$ using normalcdf$\left(20,1\text{E}99,22,\dfrac{22}{\sqrt{80}}\right)$
The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
Figure $5$.
REMINDER
1E99 = 1099 and –1E99 = –1099. Press the EE key for E. Or just use 1099 instead of 1E99.
2. Find $P(x > 20)$. Remember to use the exponential distribution for an individual: $X \sim Exp\left(\dfrac{1}{22}\right)$. $P(x > 20) = e^{(−\left(\dfrac{1}{22}\right)(20))}$ or $e^{(–0.04545(20))} = 0.4029$
1. $P(x > 20) = 0.4029$ but $P(\bar{x} > 20) = 0.7919$
2. The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.
3. When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean.
3. Let $k$ = the 95th percentile. Find $k$ where $P(\bar{x} < k) = 0.95$
$k = 26.0$ using invNorm $\left(0.95,22,\dfrac{22}{\sqrt{80}}\right) = 26.0$
Figure $6$.
The 95th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.
Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.
Exercise $2$
Use the information in Example $2$, but change the sample size to 144.
1. Find $P(20 < \bar{x} < 30)$.
2. Find $P(\sum x \text{ is at least } 3,000)$.
3. Find the 75th percentile for the sample mean excess time of 144 customers.
4. Find the 85th percentile for the sum of 144 excess times used by customers.
Answer
1. 0.8623
2. 0.7377
3. 23.2
4. 3,441.6
Example $3$
In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.
1. Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.
2. Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.
3. Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.
4. Find the value that is two standard deviations above the sample mean.
5. Find the IQR for the sum of the sample times.
Answer
1. We have, $\mu_{x} = \mu = 2$ and $\sigma_{x} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{10} = 0.05$. Therefore:
1. 50th percentile $= \mu_{x} = \mu = 2$
2. 25th percentile $= \text{invNorm}(0.25,2,0.05) = 1.97$
3. 75th percentile $= \text{invNorm}(0.75,2,0.05) = 2.03$
2. We have $\mu_{\sum X} = n(\mu_{x}) = 100(2)$ and $\sigma_{\mu X} = \sqrt{n}(\sigma_{x}) = 10(0.5) = 5$. Therefore
1. 50th percentile = $\mu_{\sum X} = n(\mu_{X}) = 100(2) = 200$
2. 25th percentile $= \text{invNorm}(0.25,200,5) = 196.63$
3. 75th percentile $= \text{invNorm}(0.75,200,5) = 203.37$
3. $P(1.75 < bar{x} < 1.85) =$ normalcdf$(1.75,1.85,2,0.05) = 0.0013$
4. Using the $z$-score equation, $z = \dfrac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}$, and solving for $x$, we have $x = 2(0.05) + 2 = 2.1$
5. The $IQR$ is 75th percentile – 25th percentile $= 203.37 – 196.63 = 6.74$
Exercise $3$
Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.
1. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.
2. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
3. If the sample were four women between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used?
Answer
1. $P(x > 120)$ = normalcdf$(120,99,114.8,13.1) = 0.0272$. There is about a 3%, that the randomly selected woman will have systolics blood pressure greater than 120.
2. $P(\bar{x} > 120) =$ normalcdf$\left(120,114.8,\dfrac{13.1}{\sqrt{40}}\right) = 0.006$. There is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.
3. The central limit theorem could not be used if the sample size were four and we did not know the original distribution was normal. The sample size would be too small.
Example $4$
A study was done about violence against prostitutes and the symptoms of the posttraumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.
1. In a sample of 25 prostitutes, what is the probability that the mean age of the prostitutes is less than 35?
2. Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.
3. In a sample of 49 prostitutes, what is the probability that the sum of the ages is no less than 1,600?
4. Is it likely that the sum of the ages of the 49 prostitutes is at most 1,595? Interpret the results.
5. Find the 95th percentile for the sample mean age of 65 prostitutes. Interpret the results.
6. Find the 90th percentile for the sum of the ages of 65 prostitutes. Interpret the results.
Answer
1. $P(\bar{x} < 35) =$ normalcdf$(-E99,35,30.9,1.8) = 0.9886$
2. $P(\bar{x} > 50) =$ normalcdf$(50, E99,30.9,1.8) \approx 0$. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.
3. $P(\sum x \geq 1,600) =$ normalcdf$(1600,E99,1514.10,63) = 0.0864$
4. $P(\sum x \leq 1,595) =$ normalcdf$(-E99,1595,1514.10,63) = 0.9005$. This means that there is a 90% chance that the sum of the ages for the sample group $n = 49$ is at most 1595.
5. The 95th percentile = invNorm$(0.95,30.9,1.1) = 32.7$. This indicates that 95% of the prostitutes in the sample of 65 are younger than 32.7 years, on average.
6. The 90th percentile = invNorm$(0.90,2008.5,72.56) = 2101.5$. This indicates that 90% of the prostitutes in the sample of 65 have a sum of ages less than 2,101.5 years.
Exercise $4$
According to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches.
1. What mean doorway height would allow 95% of men to enter the aircraft without bending?
2. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?
3. For engineers designing the 757, which result is more relevant: the height from part a or part b? Why?
Answer
1. We know that $\mu_{x} = \mu = 69$ and we have $\sigma_{x} = 2.8$. The height of the doorway is found to be invNorm$(0.95,69,2.8) = 73.61$
2. We know that $\mu_{x} = \mu = 69$ and we have $\sigma_{x} = 2.8$. So, invNorm$(0.95,69,0.28) = 69.49$
3. When designing the doorway heights, we need to incorporate as much variability as possible in order to accommodate as many passengers as possible. Therefore, we need to use the result based on part a.
Historical Note: Normal Approximation to the Binomial
Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for $n$(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of $n$, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution:
• there are a certain number $n$ of independent trials
• the outcomes of any trial are success or failure
• each trial has the same probability of a success $p$
Recall that if $X$ is the binomial random variable, then $X \sim B(n, p)$. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities $np$ and $nq$ must both be greater than five ($np > 5$ and $nq > 5$); the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean $\mu = np$ and standard deviation $\sigma = \sqrt{npq}$. Remember that $q = 1 - p$. In order to get the best approximation, add 0.5 to $x$ or subtract 0.5 from $x$ (use $x + 0.5$ or $x - 0.5$). The number 0.5 is called the continuity correction factor and is used in the following example.
Example $5$
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.
1. Find the probability that at least 150 favor a charter school.
2. Find the probability that at most 160 favor a charter school.
3. Find the probability that more than 155 favor a charter school.
4. Find the probability that fewer than 147 favor a charter school.
5. Find the probability that exactly 175 favor a charter school.
Let $X =$ the number that favor a charter school for grades K trough 5. $X \sim B(n, p)$ where $n = 300$ and $p = 0.53$. Since $np > 5$ and $nq > 5$, use the normal approximation to the binomial. The formulas for the mean and standard deviation are $\mu = np$ and $\sigma = \sqrt{npq}$. The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is $X$. $Y \sim N(159, 8.6447)$. See The Normal Distribution for help with calculator instructions.
For part a, you include 150 so $P(X \geq 150)$ has normal approximation $P(Y \geq 149.5) = 0.8641$.
normalcdf$(149.5,10^{99},159,8.6447) = 0.8641$.
For part b, you include 160 so $P(X \leq 160)$ has normal approximation $P(Y \leq 160.5) = 0.5689$.
normalcdf$(0,160.5,159,8.6447) = 0.5689$
For part c, you exclude 155 so $P(X > 155)$ has normal approximation $P(y > 155.5) = 0.6572$.
normalcdf$(155.5,10^{99},159,8.6447) = 0.6572$.
For part d, you exclude 147 so $P(X < 147)$ has normal approximation $P(Y < 146.5) = 0.0741$.
normalcdf$(0,146.5,159,8.6447) = 0.0741$
For part e, $P(X = 175)$ has normal approximation $P(174.5 < Y < 175.5) = 0.0083$.
normalcdf$(174.5,175.5,159,8.6447) = 0.0083$
Because of calculators and computer software that let you calculate binomial probabilities for large values of $n$ easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in "binomial probability distribution calculation" in an Internet browser, you can find at least one online calculator for the binomial.
For Example, the probabilities are calculated using the following binomial distribution: ($n = 300 and p = 0.53$). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial.
$P(X \geq 150)$ :1 - binomialcdf$(300,0.53,149) = 0.8641$
$P(X \leq 160)$ :binomialcdf$(300,0.53,160) = 0.5684$
$P(X > 155)$ :1 - binomialcdf$(300,0.53,155) = 0.6576$
$P(X < 147)$ :binomialcdf$(300,0.53,146) = 0.0742$
$P(X = 175)$ :(You use the binomial pdf.)binomialpdf$(300,0.53,175) = 0.0083$
Exercise $5$
In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor.
Answer
0.0401
Glossary
Exponential Distribution
a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital, notation: $X \sim Exp(m)$. The mean is $\mu = \dfrac{1}{m}$ and the standard deviation is $\sigma = \dfrac{1}{m}$. The probability density function is $f(x) = me^{-mx}$, $x \geq 0$ and the cumulative distribution function is $P(X \leq x) = 1 - e^{-mx}$.
Mean
a number that measures the central tendency; a common name for mean is "average." The term "mean" is a shortened form of "arithmetic mean." By definition, the mean for a sample (denoted by $\bar{x}$) is $\bar{x} = \dfrac{\text{Sum of all values in the sample}}{\text{Number of values in the sample}}$, and the mean for a population (denoted by $\mu$) is $\mu = \dfrac{\text{Sum of all values in the population}}{\text{Number of values in the population}}$.
Normal Distribution
a continuous random variable (RV) with pdf $f(x) = \dfrac{1}{\sigma \sqrt{2\pi}}e^{\dfrac{(x - \mu)^{2}}{2\sigma^{2}}}$, where $\mu$ is the mean of the distribution and $\sigma$ is the standard deviation.; notation: $X \sim N(\mu, \sigma)$. If $\mu = 0$ and $\sigma = 1$, the RV is called the standard normal distribution.
Uniform Distribution
a continuous random variable (RV) that has equally likely outcomes over the domain, $a < x < b$; often referred as the Rectangular Distribution because the graph of the pdf has the form of a rectangle. Notation: $X \sim U(a, b)$. The mean is $\mu = \dfrac{a+b}{2}$ and the standard deviation is $\sigma = \sqrt{\dfrac{(b-a)^{2}}{12}}$. The probability density function is $f(x) = \dfrac{a+b}{2}$ for $a < x < b$ or $a \leq x \leq b$. The cumulative distribution is $P(X \leq x) = \dfrac{x-a}{b-a}$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.04%3A_Using_the_Central_Limit_Theorem.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will demonstrate and compare properties of the central limit theorem.
NOTE
This lab works best when sampling from several classes and combining data.
Collect the Data
1. Count the change in your pocket. (Do not include bills.)
2. Randomly survey 30 classmates. Record the values of the change in Table.
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
3. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes.
4. Calculate the following ($n = 1$; surveying one person at a time):
1. $\bar{x}$= _______
2. $s$= _______
5. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Collecting Averages of Pairs
Repeat steps one through five of the section Collect the Data. with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs.
1. Randomly survey 30 pairs of classmates.
2. Record the values of the average of their change in Table.
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
3. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data. Sketch the graph using a ruler and a pencil.
4. Calculate the following ($n = 2$; surveying two people at a time):
1. $\bar{x}$= _______
2. $s$= _______
5. Draw a smooth curve through tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Collecting Averages of Groups of Five
Repeat steps one through five (of the section titled Collect the Data) with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of five.
1. Randomly survey 30 groups of five classmates.
2. Record the values of the average of their change.
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
3. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data. Sketch the graph using a ruler and a pencil.
4. Calculate the following ($n = 5$; surveying five people at a time):
1. $\bar{x}$= _______
2. $s$= _______
5. Draw a smooth curve through tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Discussion Questions
1. Why did the shape of the distribution of the data change, as n changed? Use one to two complete sentences to explain what happened.
2. In the section titled Collect the Data, what was the approximate distribution of the data? $X ~$_____(_____,_____)
3. In the section titled Collecting Averages of Groups of Five, what was the approximate distribution of the averages? $\bar{X} ~$ _____(_____,_____)
4. In one to two complete sentences, explain any differences in your answers to the previous two questions. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.05%3A_Central_Limit_Theorem_-_Pocket_Change_%28Worksheet%29.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will demonstrate and compare properties of the central limit theorem.
Q1
$X$= length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. (Assume that each of the different recipes makes the same quantity of cookies.)
Recipe # X Recipe # X Recipe # X Recipe # X
1 1 16 2 31 3 46 2
2 5 17 2 32 4 47 2
3 2 18 4 33 5 48 11
4 5 19 6 34 6 49 5
5 6 20 1 35 6 50 5
6 1 21 6 36 1 51 4
7 2 22 5 37 1 52 6
8 6 23 2 38 2 53 5
9 5 24 5 39 1 54 1
10 2 25 1 40 6 55 1
11 5 26 6 41 1 56 2
12 1 27 4 42 6 57 4
13 1 28 1 43 2 58 3
14 3 29 6 44 6 59 6
15 2 30 2 45 2 60 5
Calculate the following:
1. $\mu_{x}$= _______
2. $\sigma_{x}$= _______
Collect the Data
Use a random number generator to randomly select four samples of size $n = 5$ from the given population. Record your samples in Table. Then, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class.
Q2
Complete the table:
Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups:
Means: x¯x¯ = ____ x¯x¯ = ____ x¯x¯ = ____ x¯x¯ = ____
Q3
Calculate the following:
1. $\bar{x}$= _______
2. $s_{\bar{x}}$= _______
Q4
Again, use a random number generator to randomly select four samples from the population. This time, make the samples of size $n = 10$. Record the samples in Table. As before, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class.
Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups
Means: $\bar{x}$ = ____ $\bar{x}$ = ____ $\bar{x}$ = ____ $\bar{x}$ = ____
Calculate the following:
1. $\bar{x}$ = ______
2. $s_{\bar{x}}$ = ______
Q4
For the original population, construct a histogram. Make intervals with a bar width of one day. Sketch the graph using a ruler and pencil. Scale the axes.
Q5
Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Repeat the Procedure for $n = 5$
For the sample of $n = 5$ days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of $\frac{1}{2}$ a day. Sketch the graph using a ruler and pencil. Scale the axes.
Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Repeat the Procedure for n = 10
1. For the sample of $n = 10$ days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of $\frac{1}{2}$ a day. Sketch the graph using a ruler and pencil. Scale the axes.
2. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Discussion Questions
1. Compare the three histograms you have made, the one for the population and the two for the sample means. In three to five sentences, describe the similarities and differences.
2. State the theoretical (according to the clt) distributions for the sample means.
1. $n = 5$: $\bar{x} ~$ _____(_____,_____)
2. $n = 10$: $\bar{x} ~$ _____(_____,_____)
3. Are the sample means for $n = 5$ and $n = 10$ “close” to the theoretical mean, $\mu_{x}$? Explain why or why not.
4. Which of the two distributions of sample means has the smaller standard deviation? Why?
5. As $n$ changed, why did the shape of the distribution of the data change? Use one to two complete sentences to explain what happened. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.06%3A_Central_Limit_Theorem_-_Cookie_Recipes_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
7.2: The Central Limit Theorem for Sample Means (Averages)
Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. 1. In words, $Χ =$ ____________ 2. $Χ \sim$ _____(_____,_____) 3. In words, $\bar{X} =$ ____________ 4. $\bar{X} \sim$ ______ (______, ______) 5. Find the probability that an individual had between$0.80 and $1.00. Graph the situation, and shade in the area to be determined. 6. Find the probability that the average of the 25 students was between$0.80 and $1.00. Graph the situation, and shade in the area to be determined. 7. Explain why there is a difference in part e and part f. S 7.2.1 1. $Χ =$ amount of change students carry 2. $Χ \sim E(0.88, 0.88)$ 3. $\bar{X} =$ average amount of change carried by a sample of 25 students. 4. $\bar{X} \sim N(0.88, 0.176)$ 5. 0.0819 6. 0.1882 7. The distributions are different. Part a is exponential and part b is normal. Q 7.2.2 Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. 1. If $\bar{X} =$ average distance in feet for 49 fly balls, then $\bar{X} \sim$ _______(_______,_______) 2. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for $\bar{X}$. Shade the region corresponding to the probability. Find the probability. 3. Find the 80th percentile of the distribution of the average of 49 fly balls. Q 7.2.3 According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers. 1. In words, $Χ =$ _____________ 2. In words, $\bar{X} =$ _____________ 3. $\bar{X} \sim$ _____(_____,_____) 4. Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. 5. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why. S 7.2.3 1. length of time for an individual to complete IRS form 1040, in hours. 2. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours. 3. $N\left(10.53, \frac{1}{3}\right)$ 4. Yes. I would be surprised, because the probability is almost 0. 5. No. I would not be totally surprised because the probability is 0.2312 Q 7.2.4 Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let $\bar{X}$ the average of the 49 races. 1. $\bar{X} \sim$ _____(_____,_____) 2. Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. 3. Find the 80th percentile for the average of these 49 marathons. 4. Find the median of the average running times. Q 7.2.5 The length of songs in a collector’s iTunes album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums. 1. In words, $Χ =$ _________ 2. $\bar{X} \sim$ _____________ 3. In words, $\bar{X} =$ _____________ 4. $\bar{X} \sim$ _____(_____,_____) 5. Find the first quartile for the average song length. 6. The IQR(interquartile range) for the average song length is from _______–_______. S 7.2.6 1. the length of a song, in minutes, in the collection 2. $U(2, 3.5)$ 3. the average length, in minutes, of the songs from a sample of five albums from the collection 4. $N(2.75, 0.0220)$ 5. 2.74 minutes 6. 0.03 minutes Q 7.2.7 In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940. 1. In words, $Χ =$ _____________ 2. In words, $\bar{X} =$ _____________ 3. $\bar{X} \sim$ _____(_____,_____) 4. The IQR for $\bar{X}$ is from _______ acres to _______ acres. Q 7.2.8 Determine which of the following are true and which are false. Then, in complete sentences, justify your answers. 1. When the sample size is large, the mean of $\bar{X}$ is approximately equal to the mean of $Χ$. 2. When the sample size is large, $\bar{X}$ is approximately normally distributed. 3. When the sample size is large, the standard deviation of $\bar{X}$ is approximately the same as the standard deviation of $Χ$. S 7.2.8 1. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution. 2. True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal. 3. The standard deviation of the sampling distribution of the means will decrease making it approximately the same as the standard deviation of $Χ$ as the sample size increases. Q 7.2.9 The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let $\bar{X} =$ average percent of fat calories. 1. $\bar{X} \sim$ ______(______, ______) 2. For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined. 3. Find the first quartile for the average percent of fat calories. Q 7.2.10 The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be$2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country. 1. In words, $Χ =$ _____________ 2. In words, $\bar{X} =$ _____________ 3. $\bar{X} \sim$ _____(_____,_____) 4. How is it possible for the standard deviation to be greater than the average? 5. Why is it more likely that the average of the 1,000 residents will be from$2,000 to $2,100 than from$2,100 to $2,200? S 7.2.10 1. $Χ =$ the yearly income of someone in a third world country 2. the average salary from samples of 1,000 residents of a third world country 3. $\bar{X} \sim N\left(2000, \frac{8000}{\sqrt{1000}}\right)$ 4. Very wide differences in data values can have averages smaller than standard deviations. 5. The distribution of the sample mean will have higher probabilities closer to the population mean. $P(2000 < \bar{X} < 2100) = 0.1537$ $P(2100 < \bar{X} < 2200) = 0.1317$ Q 7.2.11 Which of the following is NOT TRUE about the distribution for averages? 1. The mean, median, and mode are equal. 2. The area under the curve is one. 3. The curve never touches the x-axis. 4. The curve is skewed to the right. Q 7.2.12 The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of$4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is: 1. $\bar{X} \sim N(4.59, 0.10)$ 2. $\bar{X} \sim N\left(4.59, \frac{0.10}{\sqrt{16}}\right)$ 3. $\bar{X} \sim N\left(4.59, \frac{16}{0.10}\right)$ 4. $\bar{X} \sim N\left(4.59, \frac{\sqrt{16}}{0.10}\right)$ S 7.2.12 b 7.3: The Central Limit Theorem for Sums Q 7.3.1 Which of the following is NOT TRUE about the theoretical distribution of sums? 1. The mean, median and mode are equal. 2. The area under the curve is one. 3. The curve never touches the x-axis. 4. The curve is skewed to the right. Q 7.3.2 Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials. 1. In words, $\sum X =$ ______________ 2. $\sum X \sim$ _____(_____,_____) 3. Find the probability that the total length of the nine trials is at least 225 days. 4. Ninety percent of the total of nine of these types of trials will last at least how long? S 7.3.2 1. the total length of time for nine criminal trials 2. $N(189, 21)$ 3. 0.0432 4. 162.09; ninety percent of the total nine trials of this type will last 162 days or more. Q 7.3.3 Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children. 1. In words, $X =$_____________ 2. The distribution is _______. 3. In words, $\sum X =$ _______________ 4. $\sum X \sim$ _____(_____,_____) 5. Find the probability that the total weight of open boxes is less than 250 pounds. 6. Find the 35th percentile for the total weight of open boxes of cereal. Q 7.3.4 Salaries for teachers in a particular elementary school district are normally distributed with a mean of$44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district. 1. In words, $X =$ ______________ 2. $X \sim$ _____(_____,_____) 3. In words, $\sum X =$ _____________ 4. $\sum X \sim$ _____(_____,_____) 5. Find the probability that the teachers earn a total of over$400,000.
6. Find the 90th percentile for an individual teacher's salary.
7. Find the 90th percentile for the sum of ten teachers' salary.
8. If we surveyed 70 teachers instead of ten, graphically, how would that change the distribution in part d?
9. If each of the 70 teachers received a $3,000 raise, graphically, how would that change the distribution in part b? S 7.3.4 1. $X =$ the salary of one elementary school teacher in the district 2. $X \sim N(44,000, 6,500)$ 3. $\sum X \sim$ sum of the salaries of ten elementary school teachers in the sample 4. $\sum X \sim N(44000, 20554.80)$ 5. 0.9742 6.$52,330.09
7. 466,342.04
8. Sampling 70 teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.
1. almost zero
2. 0.1587
3. 0.0943
4. unknown
a
2. \$46,634
Q 7.4.15
The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.
1. In words, $X =$ _____________
2. In words, $\bar{X} =$ ___________________
3. $\bar{X} \sim$ _____(_____,_____)
4. In words, $\sum X =$ _______________
5. $\sum X \sim$ _____(_____,_____)
6. Is it likely that an individual stayed more than five days in the hospital? Why or why not?
7. Is it likely that the average stay for the 80 women was more than five days? Why or why not?
8. Which is more likely:
1. An individual stayed more than five days.
2. the average stay of 80 women was more than five days.
9. If we were to sum up the women’s stays, is it likely that, collectively they spent more than a year in the hospital? Why or why not?
For each problem, wherever possible, provide graphs and use the calculator.
Q 7.4.16
NeverReady batteries has engineered a newer, longer lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? Is the company’s claim reasonable?
S 7.4.16
• We have $\mu = 17, \sigma = 0.8, \bar{x} = 16.7$, and $n = 30$. To calculate the probability, we usenormalcdf(lower, upper, $\mu$, $\frac{\sigma}{\sqrt{n}}$) = normalcdf$\left(E–99,16.7,17,\frac{0.8}{\sqrt{30}}\right) = 0.0200$.
• If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 lifetime hours is only 2%. Therefore, the class was justified to question the claim.
Q 7.4.17
Men have an average weight of 172 pounds with a standard deviation of 29 pounds.
1. Find the probability that 20 randomly selected men will have a sum weight greater than 3600 lbs.
2. If 20 men have a sum weight greater than 3500 lbs, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain.
Q 7.4.18
M&M candies large candy bags have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class.
Red Orange Yellow Brown Blue Green
0.751 0.735 0.883 0.696 0.881 0.925
0.841 0.895 0.769 0.876 0.863 0.914
0.856 0.865 0.859 0.855 0.775 0.881
0.799 0.864 0.784 0.806 0.854 0.865
0.966 0.852 0.824 0.840 0.810 0.865
0.859 0.866 0.858 0.868 0.858 1.015
0.857 0.859 0.848 0.859 0.818 0.876
0.942 0.838 0.851 0.982 0.868 0.809
0.873 0.863 0.803 0.865
0.809 0.888 0.932 0.848
0.890 0.925 0.842 0.940
0.878 0.793 0.832 0.833
0.905 0.977 0.807 0.845
0.850 0.841 0.852
0.830 0.932 0.778
0.856 0.833 0.814
0.842 0.881 0.791
0.778 0.818 0.810
0.786 0.864 0.881
0.853 0.825
0.864 0.855
0.873 0.942
0.880 0.825
0.882 0.869
0.931 0.912
0.887
The bag contained 465 candies and he listed weights in the table came from randomly selected candies. Count the weights.
1. Find the mean sample weight and the standard deviation of the sample weights of candies in the table.
2. Find the sum of the sample weights in the table and the standard deviation of the sum the of the weights.
3. If 465 M&Ms are randomly selected, find the probability that their weights sum to at least 396.9.
4. Is the Mars Company’s M&M labeling accurate?
S7.4.19
1. For the sample, we have $n = 100, \bar{x} = 0.862, s = 0.05$
2. $\sum \bar{x} = 85.65, \sum s = 5.18$
3. normalcdf$(396.9,E99,(465)(0.8565),(0.05)\left(\sqrt{465}\right)) \approx 1$
4. Since the probability of a sample of size 465 having at least a mean sum of 396.9 is approximately 1, we can conclude that Mars is correctly labeling their M&M packages.
Q7.4.20
The Screw Right Company claims their $\frac{3}{4}$ inch screws are within $\pm 0.23$ of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded.
0.757 0.723 0.754 0.737 0.757 0.741 0.722 0.741 0.743 0.742
0.740 0.758 0.724 0.739 0.736 0.735 0.760 0.750 0.759 0.754
0.744 0.758 0.765 0.756 0.738 0.742 0.758 0.757 0.724 0.757
0.744 0.738 0.763 0.756 0.760 0.768 0.761 0.742 0.734 0.754
0.758 0.735 0.740 0.743 0.737 0.737 0.725 0.761 0.758 0.756
The screws were randomly selected from the local home repair store.
1. Find the mean diameter and standard deviation for the sample
2. Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company’s diameter claim plausible?
Q 7.4.21
Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?
Q 7.4.22
Use normalcdf
$(E–99,1.1,1,\frac{1}{\sqrt{70}}) = 0.7986.$
This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours.
Q 7.4.23
A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points?
Q 7.4.24
Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine?
S 7.4.25
Since we have normalcdf$(5.111,5.291,5.201,\frac{0.065}{\sqrt{280}}) \approx 1$, we can conclude that practically all the coins are within the limits, therefore, there should be no rejected coins out of a well selected sample of size 280.
Chapter Review
The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean $\bar{x}$ gets to $\mu$.
Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.
Exercise 7.4.7
1. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deivation?
2. What is the distribution for the mean weight of 100 25-pound lifting weights?
3. Find the probability that the mean actual weight for the 100 weights is less than 24.9.
Answer
1. $U(24, 26), 25, 0.5774$
2. $N(25, 0.0577)$
3. 0.0416
Exercise 7.4.8
Draw the graph from Exercise
Exercise 7.4.9
Find the probability that the mean actual weight for the 100 weights is greater than 25.2.
Answer
0.0003
Exercise 7.4.10
Draw the graph from Exercise
Exercise 7.4.11
Find the 90th percentile for the mean weight for the 100 weights.
Answer
25.07
Exercise 7.4.12
Draw the graph from Exercise
Exercise 7.4.13
1. What is the distribution for the sum of the weights of 100 25-pound lifting weights?
2. Find $P(\sum x < 2,450)$.
Answer
1. $N(2,500, 5.7735)$
2. 0
Exercise 7.4.14
Draw the graph from Exercise
Exercise 7.4.15
Find the 90th percentile for the total weight of the 100 weights.
Answer
2,507.40
Exercise 7.4.16
Draw the graph from Exercise
Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.
Exercise 7.4.17
1. What is the standard deviation?
2. What is the parameter $m$?
Answer
1. 10
2. $\dfrac{1}{10}$
Exercise 7.4.18
What is the distribution for the length of time one battery lasts?
Exercise 7.4.19
What is the distribution for the mean length of time 64 batteries last?
Answer
$N\left(10, \dfrac{10}{8}\right)$
Exercise 7.4.20
What is the distribution for the total length of time 64 batteries last?
Exercise 7.4.21
Find the probability that the sample mean is between seven and 11.
Answer
0.7799
Exercise 7.4.22
Find the 80th percentile for the total length of time 64 batteries last.
Exercise 7.4.23
Find the $IQR$ for the mean amount of time 64 batteries last.
Answer
1.69
Exercise 7.4.24
Find the middle 80% for the total amount of time 64 batteries last.
Use the following information to answer the next eight exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.
Exercise 7.4.25
Find $P(\sum x > 420)$
Answer
0.0072
Exercise 7.4.26
Find the 90th percentile for the sums.
Exercise 7.4.27
Find the 15th percentile for the sums.
Answer
391.54
Exercise 7.4.28
Find the first quartile for the sums.
Exercise 7.4.29
Find the third quartile for the sums.
Answer
405.51
Exercise 7.4.30
Find the 80th percentile for the sums. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.E%3A_The_Central_Limit_Theorem_%28Exercises%29.txt |
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
• 8.1: Prelude to Confidence Intervals
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
• 8.2: A Single Population Mean using the Normal Distribution
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution.
• 8.3: A Single Population Mean using the Student t-Distribution
We rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation ss as an estimate for σσ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.
• 8.4: A Population Proportion
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different.
• 8.5: Confidence Interval - Home Costs (Worksheet)
A statistics Worksheet: The student will calculate the 90% confidence interval for the mean cost of a home in the area in which this school is located. The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval.
• 8.6: Confidence Interval -Place of Birth (Worksheet)
A statistics Worksheet: The student will calculate the 90% confidence interval the proportion of students in this school who were born in this state. The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval.
• 8.7: Confidence Interval -Women's Heights (Worksheet)
A statistics Worksheet: The student will calculate a 90% confidence interval using the given data. The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean.
• 8.E: Confidence Intervals (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
• 8.S: Confidence Intervals (Summary)
In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known.
08: Confidence Intervals
Learning Objectives
By the end of this chapter, the student should be able to:
• Calculate and interpret confidence intervals for estimating a population mean and a population proportion.
• Interpret the Student's t probability distribution as the sample size changes.
• Discriminate between problems applying the normal and the Student's t distributions.
• Calculate the sample size required to estimate a population mean and a population proportion given a desired confidence level and margin of error.
Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion.
We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics. The sample data help us to make an estimate of a population parameter. We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals.
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from iTunes. If so, you could conduct a survey and calculate the sample mean, $\bar{x}$, and the sample standard deviation, $s$. You would use $\bar{x}$ to estimate the population mean and $s$ to estimate the population standard deviation. The sample mean, $\bar{x}$, is the point estimate for the population mean, $\mu$. The sample standard deviation, $s$, is the point estimate for the population standard deviation, $\sigma$.
Each of $\bar{x}$ and $s$ is called a statistic.
A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter.
Suppose, for the iTunes example, we do not know the population mean $\mu$, but we do know that the population standard deviation is $\sigma = 1$ and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is
$\dfrac{\sigma}{\sqrt{n}} = \dfrac{1}{\sqrt{100}} = 0.1.$
The empirical rule, which applies to bell-shaped distributions, says that in approximately 95% of the samples, the sample mean, $\bar{x}$, will be within two standard deviations of the population mean $\mu$. For our iTunes example, two standard deviations is (2)(0.1) = 0.2. The sample mean $\bar{x}$ is likely to be within 0.2 units of $\mu$.
Because $\bar{x}$ is within 0.2 units of $\mu$, which is unknown, then $\mu$ is likely to be within 0.2 units of $\bar{x}$ in 95% of the samples. The population mean $\mu$ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, $\mu$ is between $\bar{x} - 0.2$ and $\bar{x} + 0.2$ in 95% of all the samples.
For the iTunes example, suppose that a sample produced a sample mean $\bar{x} = 2$. Then the unknown population mean $\mu$ is between
$\bar{x} - 0.2 = 2 - 0.2 = 1.8$
and
$\bar{x} + 0.2 = 2 + 0.2 = 2.2$
We say that we are 95% confident that the unknown population mean number of songs downloaded from iTunes per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2). This 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean $\mu$ or our sample produced an $\bar{x}$ that is not within 0.2 units of the true mean $\mu$. The second possibility happens for only 5% of all the samples (95–100%).
Remember that a confidence interval is created for an unknown population parameter like the population mean, $\bar{x}$. Confidence intervals for some parameters have the form:
(point estimate – margin of error, point estimate + margin of error)
The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean.
When you read newspapers and journals, some reports will use the phrase "margin of error." Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. These are two ways of expressing the same concept.
Although the text only covers symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation).
Collaborative Exercise
Have your instructor record the number of meals each student in your class eats out in a week. Assume that the standard deviation is known to be three meals. Construct an approximate 95% confidence interval for the true mean number of meals students eat out each week.
1. Calculate the sample mean.
2. Let $\sigma = 3$ and $n$ = the number of students surveyed.
3. Construct the interval $\left(\bar{x} - 2 \cdot \frac{\sigma}{\sqrt{n}}, \bar{x} + 2 \cdot \frac{\sigma}{\sqrt{n}}\right)$.
We say we are approximately 95% confident that the true mean number of meals that students eat out in a week is between __________ and ___________.
Glossary
Confidence Interval (CI)
an interval estimate for an unknown population parameter. This depends on:
• the desired confidence level,
• information that is known about the distribution (for example, known standard deviation),
• the sample and its size.
Inferential Statistics
also called statistical inference or inductive statistics; this facet of statistics deals with estimating a population parameter based on a sample statistic. For example, if four out of the 100 calculators sampled are defective we might infer that four percent of the production is defective.
Parameter
a numerical characteristic of a population
Point Estimate
a single number computed from a sample and used to estimate a population parameter | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.01%3A_Prelude_to_Confidence_Intervals.txt |
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of $\bar{x} = 10$ and we have constructed the 90% confidence interval (5, 15) where $EBM = 5$.
Calculating the Confidence Interval
To construct a confidence interval for a single unknown population mean $\mu$, where the population standard deviation is known, we need $\bar{x}$ as an estimate for $\mu$ and we need the margin of error. Here, the margin of error ($EBM$) is called the error bound for a population mean (abbreviated EBM). The sample mean $\bar{x}$ is the point estimate of the unknown population mean $\mu$.
The confidence interval estimate will have the form:
$(\text{point estimate} - \text{error bound}, \text{point estimate} + \text{error bound})\nonumber$
or, in symbols,
$(\bar{x} - EBM, \bar{x} + EBM)\nonumber$
The margin of error ($EBM$) depends on the confidence level (abbreviated $CL$). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions.
There is another probability called alpha $(\alpha)$. $\alpha$ is related to the confidence level, $CL$. $\alpha$ is the probability that the interval does not contain the unknown population parameter. Mathematically,
$\alpha + CL = 1.\nonumber$
Example $1$
Suppose we have collected data from a sample. We know the sample mean but we do not know the mean for the entire population. The sample mean is seven, and the error bound for the mean is 2.5: $\bar{x} = 7$ and $EBM = 2.5$
The confidence interval is (7 – 2.5, 7 + 2.5) and calculating the values gives (4.5, 9.5). If the confidence level ($CL$) is 95%, then we say that, "We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5."
Exercise $1$
Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean?
Answer
(11.8, 18.2)
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of $\bar{x} = 10$, and we have constructed the 90% confidence interval (5, 15) where $EBM = 5$. To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of $\alpha = 10%$ in both tails, or 5% in each tail, of the normal distribution.
To capture the central 90%, we must go out 1.645 "standard deviations" on either side of the calculated sample mean. The value 1.645 is the z-score from a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail.
It is important that the "standard deviation" used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is
$\dfrac{\sigma}{\sqrt{n}}\nonumber$
This fraction is commonly called the "standard error of the mean" to distinguish clearly the standard deviation for a mean from the population standard deviation $\sigma$.
In summary, as a result of the central limit theorem:
• $\bar{X}$ is normally distributed, that is, $\bar{X} \sim N(\mu_{x},\dfrac{\sigma}{\sqrt{n}})$.
• When the population standard deviation σ is known, we use a normal distribution to calculate the error bound.
Calculating the Confidence Interval
To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:
• Calculate the sample mean $\bar{x}$ from the sample data. Remember, in this section we already know the population standard deviation $\sigma$.
• Find the z-score that corresponds to the confidence level.
• Calculate the error bound $EBM$.
• Construct the confidence interval.
• Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)
We will first examine each step in more detail, and then illustrate the process with some examples.
Finding the $z$-score for the Stated Confidence Level
When we know the population standard deviation $\sigma$, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of $z$ that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution $Z \sim N(0, 1)$.
The confidence level, $CL$, is the area in the middle of the standard normal distribution. $CL = 1 - \alpha$, so $\alpha$ is the area that is split equally between the two tails. Each of the tails contains an area equal to $\dfrac{\alpha}{2}$.
The $z$-score that has an area to the right of $\dfrac{\alpha}{2}$ is denoted by $z_{\dfrac{\alpha}{2}}$.
For example, when $CL = 0.95, \alpha = 0.05$ and $\dfrac{\alpha}{2} = 0.025$; we write $z_{\dfrac{\alpha}{2}} = z_{0.025}$.
The area to the right of $z_{0.025}$ is $0.025$ and the area to the left of $z_{0.025}$ is $1 - 0.025 = 0.975$.
$z_{\dfrac{\alpha}{2}} = z_{0.025} = 1.96\nonumber$
using a calculator, computer or a standard normal probability table.
invNorm$(0.975, 0, 1) = 1.96$
Remember to use the area to the LEFT of $z_{\dfrac{\alpha}{2}}$; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution $Z \sim N(0, 1)$.
Calculating the Error Bound
The error bound formula for an unknown population mean $\mu$ when the population standard deviation $\sigma$ is known is
$EBM = z_{\alpha/2} \left(\dfrac{\sigma}{\sqrt{n}}\right)\nonumber$
Constructing the Confidence Interval
The confidence interval estimate has the format $(\bar{x} - EBM, \bar{x} + EBM)$.
The graph gives a picture of the entire situation.
$CL + \dfrac{\alpha}{2} + \dfrac{\alpha}{2} = CL + \alpha = 1.\nonumber$
Writing the Interpretation
The interpretation should clearly state the confidence level ($CL$), explain what population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints). "We estimate with ___% confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)."
Example $2$
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
Answer
• You can use technology to calculate the confidence interval directly.
• The first solution is shown step-by-step (Solution A).
• The second solution uses the TI-83, 83+, and 84+ calculators (Solution B).
Solution A
To find the confidence interval, you need the sample mean, $\bar{x}$, and the $EBM$.
$\bar{x} = 68\nonumber$
$EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)\nonumber$
$\sigma = 3; n = 36\nonumber$
The confidence level is 90% (CL = 0.90)
$CL = 0.90\nonumber$
so
$\alpha = 1 – CL = 1 – 0.90 = 0.10\nonumber$
$\dfrac{\alpha}{2} = 0.05 z_{\dfrac{\alpha}{2}} = z_{0.05}\nonumber$
The area to the right of $z_{0.05}$ is $0.05$ and the area to the left of $z_{0.05}$ is $1 - 0.05 = 0.95$.
$z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\nonumber$
using $\text{invNorm}(0.95, 0, 1)$ on the TI-83,83+, and 84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.
$EBM = (1.645)\left(\dfrac{3}{\sqrt{36}}\right) = 0.8225\nonumber$
$\bar{x} - EBM = 68 - 0.8225 = 67.1775\nonumber$
$\bar{x} + EBM = 68 + 0.8225 = 68.8225\nonumber$
The 90% confidence interval is (67.1775, 68.8225).
Solution B
Press STAT and arrow over to TESTS.
Arrow down to 7:ZInterval.
Press ENTER.
Arrow to Stats and press ENTER.
Arrow down and enter three for σ, 68 for $\bar{x}$, 36 for $n$, and .90 for C-level.
Arrow down to Calculate and press ENTER.
The confidence interval is (to three decimal places)(67.178, 68.822).
Interpretation
We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score.
Exercise $2$
Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time.
Answer
(34.1347, 37.8653)
Example $3$: Specific Absorption Rate
The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Table shows the highest SAR level for a random selection of cell phone models as measured by the FCC.
Phone Model SAR Phone Model SAR Phone Model SAR
Apple iPhone 4S 1.11 LG Ally 1.36 Pantech Laser 0.74
BlackBerry Pearl 8120 1.48 LG AX275 1.34 Samsung Character 0.5
BlackBerry Tour 9630 1.43 LG Cosmos 1.18 Samsung Epic 4G Touch 0.4
Cricket TXTM8 1.3 LG CU515 1.3 Samsung M240 0.867
HP/Palm Centro 1.09 LG Trax CU575 1.26 Samsung Messager III SCH-R750 0.68
HTC One V 0.455 Motorola Q9h 1.29 Samsung Nexus S 0.51
HTC Touch Pro 2 1.41 Motorola Razr2 V8 0.36 Samsung SGH-A227 1.13
Huawei M835 Ideos 0.82 Motorola Razr2 V9 0.52 SGH-a107 GoPhone 0.3
Kyocera DuraPlus 0.78 Motorola V195s 1.6 Sony W350a 1.48
Kyocera K127 Marbl 1.25 Nokia 1680 1.39 T-Mobile Concord 1.38
Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones. Assume that the population standard deviation is $\sigma = 0.337$.
Solution A
To find the confidence interval, start by finding the point estimate: the sample mean.
$\bar{x} = 1.024\nonumber$
Next, find the $EBM$. Because you are creating a 98% confidence interval, $CL = 0.98$.
You need to find $z_{0.01}$ having the property that the area under the normal density curve to the right of $z_{0.01}$ is $0.01$ and the area to the left is 0.99. Use your calculator, a computer, or a probability table for the standard normal distribution to find $z_{0.01} = 2.326$.
$EBM = (z_{0.01})\dfrac{\sigma}{\sqrt{n}} = (2.326)\dfrac{0.337}{\sqrt{30}} =0.1431$
To find the 98% confidence interval, find $\bar{x} \pm EBM$.
$\bar{x} - EBM = 1.024 – 0.1431 = 0.8809$
$\bar{x} - EBM = 1.024 – 0.1431 = 1.1671$
We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram.
Solution B
• Press STAT and arrow over to TESTS.
• Arrow down to 7:Z Interval.
• Press ENTER.
• Arrow to Stats and press ENTER.
• Arrow down and enter the following values:
• $\sigma: 0.337$
• $\bar{x}: 1024$
• $n: 30$
• C-level: 0.98
• Arrow down to Calculate and press ENTER.
• The confidence interval is (to three decimal places) (0.881, 1.167).
Exercise $3$
Table shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is $\sigma = 0.337$.
Phone Model SAR Phone Model SAR
Blackberry Pearl 8120 1.48 Nokia E71x 1.53
HTC Evo Design 4G 0.8 Nokia N75 0.68
HTC Freestyle 1.15 Nokia N79 1.4
LG Ally 1.36 Sagem Puma 1.24
LG Fathom 0.77 Samsung Fascinate 0.57
LG Optimus Vu 0.462 Samsung Infuse 4G 0.2
Motorola Cliq XT 1.36 Samsung Nexus S 0.51
Motorola Droid Pro 1.39 Samsung Replenish 0.3
Motorola Droid Razr M 1.3 Sony W518a Walkman 0.73
Nokia 7705 Twist 0.7 ZTE C79 0.869
Answer
$\bar{x} = 0.940\nonumber$
$\dfrac{\alpha}{2} = \dfrac{1 - CL}{2} = \dfrac{1 - 0.93}{2} = 0.035\nonumber$
$z_{0.035} = 1.812\nonumber$
$EBM = (z_{0.035})\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.812)\left(\dfrac{0.337}{\sqrt{20}}\right) = 0.1365\nonumber$
$\bar{x} - EBM = 0.940 - 0.1365 = 0.8035\nonumber$
$\bar{x} + EBM = 0.940 + 0.1365 = 1.0765\nonumber$
We estimate with 93% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8035 and 1.0765 watts per kilogram.
Notice the difference in the confidence intervals calculated in Example and the following Try It exercise. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter.
Changing the Confidence Level or Sample Size
Example $4$
Suppose we change the original problem in Example by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.
Answer
To find the confidence interval, you need the sample mean, $\bar{x}$, and the $EBM$.
$\bar{x} = 68$
$EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)$
$\sigma = 3; n = 36$; The confidence level is 95% (CL = 0.95).
$CL = 0.95$ so $\alpha = 1 – CL = 1 – 0.95 = 0.05$
$\dfrac{\alpha}{2} = 0.025 z_{\dfrac{\alpha}{2}} = z_{0.025}$
The area to the right of $z_{0.025}$ is 0.025 and the area to the left of $z_{0.025}$ is $1 – 0.025 = 0.975$.
$z_{\dfrac{\alpha}{2}} = z_{0.025} = 1.96$
when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.)
$EBM = (1.96)\left(\dfrac{3}{\sqrt{36}}\right) = 0.98$
$\bar{x} – EBM = 68 – 0.98 = 67.02$
$\bar{x} + EBM = 68 + 0.98 = 68.98$
Notice that the $EBM$ is larger for a 95% confidence level in the original problem.
Interpretation
We estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.
Explanation of 95% Confidence Level
Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score.
Comparing the results
The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider.
Summary: Effect of Changing the Confidence Level
• Increasing the confidence level increases the error bound, making the confidence interval wider.
• Decreasing the confidence level decreases the error bound, making the confidence interval narrower.
Exercise $4$
Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time.
Answer
(33.37, 38.63)
Example $5$
Suppose we change the original problem in Example to see what happens to the error bound if the sample size is changed.
Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use $n = 100$ instead of $n = 36$? What happens if we decrease the sample size to $n = 25$ instead of $n = 36$?
• $\bar{x} = 68$
• $EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)$
• $\sigma = 3$; The confidence level is 90% (CL=0.90); $z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645$.
Answer
Solution A
If we increase the sample size $n$ to 100, we decrease the error bound.
When $n = 100: EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.645)\left(\dfrac{3}{\sqrt{100}}\right) = 0.4935$.
Solution B
If we decrease the sample size $n$ to 25, we increase the error bound.
When $n = 25: EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.645)\left(\dfrac{3}{\sqrt{25}}\right) = 0.987$.
Summary: Effect of Changing the Sample Size
• Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
• Decreasing the sample size causes the error bound to increase, making the confidence interval wider.
Exercise $5$
Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.
Answer
(34.6041, 37.3958)
Working Backwards to Find the Error Bound or Sample Mean
When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.
Finding the Error Bound
• From the upper value for the interval, subtract the sample mean,
• OR, from the upper value for the interval, subtract the lower value. Then divide the difference by two.
Finding the Sample Mean
• Subtract the error bound from the upper value of the confidence interval,
• OR, average the upper and lower endpoints of the confidence interval.
Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.
Example $6$
Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean.
Calculate the Error Bound:
• If we know that the sample mean is $68: EBM = 68.82 – 68 = 0.82$.
• If we don't know the sample mean: $EBM = \dfrac{(68.82−67.18)}{2} = 0.82$.
Calculate the Sample Mean:
• If we know the error bound: $\bar{x} = 68.82 – 0.82 = 68$
• If we don't know the error bound: $\bar{x} = \dfrac{(67.18+68.82)}{2} = 68$.
Exercise $6$
Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean.
Answer
Sample mean is 45, error bound is 2.88
Calculating the Sample Size $n$
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population mean when the population standard deviation is known is
$EBM = \left(z_{\dfrac{a}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) \label{samplesize}\nonumber$
The formula for sample size is $n = \dfrac{z^{2}\sigma^{2}}{EBM^{2}}$, found by solving the error bound formula for $n$. In Equation \ref{samplesize}, $z$ is $z_{\dfrac{a}{2}}$, corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.
Example $7$
The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed?
Solution
• From the problem, we know that $\sigma = 15$ and $EBM = 2$.
• $z = z_{0.025} = 1.96$, because the confidence level is 95%.
• $n = \dfrac{z^{2}\sigma^{2}}{EBM^{2}} = \dfrac{(1.96)^{2}(15)^{2}}{2^{2}}$ using the sample size equation.
• Use $n = 217$: Always round the answer UP to the next higher integer to ensure that the sample size is large enough.
Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students.
Exercise $7$
The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed?
Answer
35 students
Glossary
Confidence Level ($CL$)
the percent expression for the probability that the confidence interval contains the true population parameter; for example, if the $CL = 90%$, then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
Error Bound for a Population Mean ($EBM$)
the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.02%3A_A_Single_Population_Mean_using_the_Normal_Distribution.txt |
In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation $s$ as an estimate for $\sigma$ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.
William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing $\sigma$ with $s$ did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name "Student."
Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and only used the Student's $t$-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever $s$ is used as an estimate for $\sigma$. If you draw a simple random sample of size $n$ from a population that has an approximately a normal distribution with mean $\mu$ and unknown population standard deviation $\sigma$ and calculate the $t$-score
$t = \dfrac{\bar{x} - \mu}{\left(\dfrac{s}{\sqrt{n}}\right)},$
then the $t$-scores follow a Student's t-distribution with $n – 1$ degrees of freedom. The $t$-score has the same interpretation as the z-score. It measures how far $\bar{x}$ is from its mean $\mu$. For each sample size $n$, there is a different Student's t-distribution.
The degrees of freedom, $n – 1$, come from the calculation of the sample standard deviation $s$. Previously, we used $n$ deviations ($x - \bar{x}$ values) to calculate $s$. Because the sum of the deviations is zero, we can find the last deviation once we know the other $n – 1$ deviations. The other $n – 1$ deviations can change or vary freely. We call the number $n – 1$ the degrees of freedom (df).
For each sample size $n$, there is a different Student's t-distribution.
Properties of the Student's $t$-Distribution
• The graph for the Student's $t$-distribution is similar to the standard normal curve.
• The mean for the Student's $t$-distribution is zero and the distribution is symmetric about zero.
• The Student's $t$-distribution has more probability in its tails than the standard normal distribution because the spread of the $t$-distribution is greater than the spread of the standard normal. So the graph of the Student's $t$-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution.
• The exact shape of the Student's $t$-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's $t$-distribution becomes more like the graph of the standard normal distribution.
• The underlying population of individual observations is assumed to be normally distributed with unknown population mean $\mu$ and unknown population standard deviation $\sigma$. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn't need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.
Calculators and computers can easily calculate any Student's $t$-probabilities. The TI-83,83+, and 84+ have a tcdf function to find the probability for given values of $t$. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.
For the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom) The output is the t-score that corresponds to the area we specified.
The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.)
A probability table for the Student's $t$-distribution can also be used. The table gives $t$-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's $t$-Distribution.) When using a $t$-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.
A Student's $t$-table gives $t$-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's $t$-probabilities.
The notation for the Student's t-distribution (using T as the random variable) is:
• $T \sim t_{df}$ where $df = n – 1$.
• For example, if we have a sample of size $n = 20$ items, then we calculate the degrees of freedom as $df = n - 1 = 20 - 1 = 19$ and we write the distribution as $T \sim t_{19}$.
If the population standard deviation is not known, the error bound for a population mean is:
• $EBM = \left(t_{\frac{\alpha}{2}}\right)\left(\frac{s}{\sqrt{n}}\right)$,
• $t_{\frac{\alpha}{2}}$ is the $t$-score with area to the right equal to $\frac{\alpha}{2}$,
• use $df = n – 1$ degrees of freedom, and
• $s =$ sample standard deviation.
The format for the confidence interval is:
$(\bar{x} - EBM, \bar{x} + EBM). \label{confint}$
To calculate the confidence interval directly:
Press STAT.
Arrow over to TESTS.
Arrow down to 8:TInterval and press ENTER (or just press 8).
Example $1$: Acupuncture
Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.
The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators.
8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9
Answer
• The first solution is step-by-step (Solution A).
• The second solution uses the TI-83+ and TI-84 calculators (Solution B).
Solution A
To find the confidence interval, you need the sample mean, $\bar{x}$, and the $EBM$.
$\bar{x} = 8.2267$
$s = 1.6722$ $n = 15$
$df = 15 – 1 = 14 CL so \alpha = 1 – CL = 1 – 0.95 = 0.05$
$\frac{\alpha}{2} = 0.025 t_{\frac{\alpha}{2}} = t_{0.025}$
The area to the right of $t_{0.025}$ is 0.025, and the area to the left of $t_{0.025}$ is 1 – 0.025 = 0.975
$t_{\frac{\alpha}{2}} = t_{0.025} = 2.14$ using invT(.975,14) on the TI-84+ calculator.
\begin{align*} EBM &= \left(t_{\frac{\alpha}{2}}\right)\left(\frac{s}{\sqrt{n}}\right) \[4pt] &= (2.14)\left(\frac{1.6722}{\sqrt{15}}\right) = 0.924 \end{align*}
Now it is just a direct application of Equation \ref{confint}:
\begin{align*} \bar{x} – EBM &= 8.2267 – 0.9240 = 7.3 \[4pt] \bar{x} + EBM &= 8.2267 + 0.9240 = 9.15 \end{align*}
The 95% confidence interval is (7.30, 9.15).
We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval and press ENTER (or you can just press 8).
Arrow to Data and press ENTER.
Arrow down to List and enter the list name where you put the data.
There should be a 1 after Freq.
Arrow down to C-level and enter 0.95
Arrow down to Calculate and press ENTER.
The 95% confidence interval is (7.3006, 9.1527)
When calculating the error bound, a probability table for the Student's t-distribution can also be used to find the value of $t$. The table gives $t$-scores that correspond to the confidence level (column) and degrees of freedom (row); the $t$-score is found where the row and column intersect in the table.
Exercise $1$
You do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.
8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5
Answer
(8.1634, 9.8032)
Example $2$: The Human Toxome Project
The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the "In utero/newborn" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table $1$ shows how many of the targeted chemicals were found in each infant’s cord blood.
Table $1$
79 145 147 160 116 100 159 151 156 126
137 83 156 94 121 144 123 114 139 99
Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood.
Solution A
From the sample, you can calculate $\bar{x} = 127.45$ and $s = 25.965$. There are 20 infants in the sample, so $n = 20$, and $df = 20 – 1 = 19$.
You are asked to calculate a 90% confidence interval: $CL = 0.90$, so
$\alpha = 1 – CL = 1 – 0.90 = 0.10 \frac{\alpha}{2} = 0.05, t_{\frac{\alpha}{2}} = t_{0.05}$
By definition, the area to the right of $t_{0.05}$ is 0.05 and so the area to the left of $t_{0.05}$ is $1 – 0.05 = 0.95$.
Use a table, calculator, or computer to find that $t_{0.05} = 1.729$.
$EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right) = 1.729\left(\frac{25.965}{\sqrt{20}}\right) \approx 10.038$
$\bar{x} – EBM = 127.45 – 10.038 = 117.412$
$\bar{x} + EBM = 127.45 + 10.038 = 137.488$
We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.
Solution B
Enter the data as a list.
Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER.
Arrow down to List and enter the list name where you put the data.
Arrow down to Freq and enter 1.
Arrow down to C-level and enter 0.90
Arrow down to Calculate and press ENTER.
The 90% confidence interval is (117.41, 137.49).
Example $3$
A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table $2$. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.
Table $2$
0 3 1 20 9
5 10 1 10 4
14 2 4 4 5
Solution A
• $\bar{x} = 6.133$,
• $s = 5.514$,
• $n = 15$, and
• $df = 15 – 1 = 14$.
$CL = 0.98$, so $\alpha = 1 - CL = 1 - 0.98 = 0.02$
$\frac{\alpha}{2} = 0.01 t_{\frac{\alpha}{2}} = t_{0.01} 2.624$
$EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right) = 2.624\left(\frac{5.514}{\sqrt{15}}\right) - 3.736$
$\bar{x} – EBM = 6.133 – 3.736 = 2.397$
$\bar{x} + EBM = 6.133 + 3.736 = 9.869$
We estimate with 98% confidence that the mean number of all hours that statistics students spend watching television in one week is between 2.397 and 9.869.
Solution B
Enter the data as a list.
Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval.
Press ENTER.
Arrow to Data and press ENTER.
Arrow down and enter the name of the list where the data is stored.
Enter Freq: 1
Enter C-Level: 0.98
Arrow down to Calculate and press Enter.
The 98% confidence interval is (2.3965, 9,8702).
Reference
1. “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013).
2. Data from Microsoft Bookshelf.
3. Data from http://www.businessweek.com/.
4. Data from http://www.forbes.com/.
5. “Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at www.fec.gov/data/index.jsp (accessed July 2,2013).
6. “Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at www.ewg.org/sites/humantoxome...tero%2Fnewborn (accessed July 2, 2013).
7. “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at www.fec.gov/finance/disclosur...pPacList.shtml (accessed July 2, 2013).
Glossary
Degrees of Freedom ($df$)
the number of objects in a sample that are free to vary
Normal Distribution
a continuous random variable (RV) with pdf $f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^{2}}$, where $\mu$ is the mean of the distribution and $\sigma$ is the standard deviation, notation: $X \sim N(\mu,\sigma)$. If $\mu = 0$ and $\sigma = 1$, the RV is called the standard normal distribution.
Standard Deviation
a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: $s$ for sample standard deviation and $\sigma$ for population standard deviation
Student's t-Distribution
investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student; the major characteristics of the random variable (RV) are:
• It is continuous and assumes any real values.
• The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution.
• It approaches the standard normal distribution as n get larger.
• There is a "family" of t–distributions: each representative of the family is completely defined by the number of degrees of freedom, which is one less than the number of data. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.03%3A_A_Single_Population_Mean_using_the_Student_t-Distribution.txt |
During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If $X$ is a binomial random variable, then
$X \sim B(n, p)\nonumber$
where $n$ is the number of trials and $p$ is the probability of a success.
To form a proportion, take $X$, the random variable for the number of successes and divide it by $n$, the number of trials (or the sample size). The random variable $P′$ (read "P prime") is that proportion,
$P' = \dfrac{X}{n}\nonumber$
(Sometimes the random variable is denoted as $\hat{P}$, read "P hat".)
When $n$ is large and $p$ is not close to zero or one, we can use the normal distribution to approximate the binomial.
$X \sim N(np, \sqrt{npq})\nonumber$
If we divide the random variable, the mean, and the standard deviation by $n$, we get a normal distribution of proportions with $P′$, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by $n$.)
$\dfrac{X}{n} = P' - N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber$
Using algebra to simplify:
$\dfrac{\sqrt{npq}}{n} = \sqrt{\dfrac{pq}{n}}\nonumber$
P′ follows a normal distribution for proportions:
$\dfrac{X}{n} = P' - N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber$
The confidence interval has the form
$(p′ – EBP, p′ + EBP).\nonumber$
where
• $EBP$ is error bound for the proportion.
• $p′ = \dfrac{x}{n}$
• $p′ =$ the estimated proportion of successes (p′ is a point estimate for p, the true proportion.)
• $x =$ the number of successes
• $n =$ the size of the sample
The error bound (EBP) for a proportion is
$EBP = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right)\nonumber$
where $q\ = 1 - p'$.
This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is $\dfrac{\sigma}{\sqrt{n}}$. For a proportion, the appropriate standard deviation is
$\sqrt{\dfrac{pq}{n}}.\nonumber$
However, in the error bound formula, we use
$\sqrt{\dfrac{p'q'}{n}}\nonumber$
as the standard deviation, instead of
$\sqrt{\dfrac{pq}{n}}.\nonumber$
In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q. The estimated proportions $p′$ and $q′$ are used because $p$ and $q$ are not known. The sample proportions $p′$ and $q′$ are calculated from the data: $p′$ is the estimated proportion of successes, and $q′$ is the estimated proportion of failures.
The confidence interval can be used only if the number of successes $np′$ and the number of failures $nq′$ are both greater than five.
Normal Distribution of Proportions
For the normal distribution of proportions, the $z$-score formula is as follows.
If
$P' - N\left(p, \sqrt{\dfrac{pq}{n}}\right)$
then the $z$-score formula is
$z = \dfrac{p'-p}{\sqrt{\dfrac{pq}{n}}}$
Example $1$
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
Solution A
• The first solution is step-by-step (Solution A).
• The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).
Let $X =$ the number of people in the sample who have cell phones. $X$ is binomial.
$X \sim B(500,\dfrac{421}{500}).\nonumber$
To calculate the confidence interval, you must find $p′$, $q′$, and $EBP$.
• $n = 500$
• $x =$ the number of successes $= 421$
$p′ = \dfrac{x}{n} = \dfrac{421}{500} = 0.842\nonumber$
• $p′ = 0.842$ is the sample proportion; this is the point estimate of the population proportion.
$q′ = 1 – p′ = 1 – 0.842 = 0.158\nonumber$
Since $CL = 0.95$, then
$\alpha = 1 – CL = 1 – 0.95 = 0.05\left(\dfrac{\alpha}{2}\right) = 0.025.\nonumber$
Then
$z_{\dfrac{\alpha}{2}} = z_{0.025 = 1.96}\nonumber$
Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find $z_{0.025}$. Remember that the area to the right of $z_{0.025}$ is $0.025$ and the area to the left of $z_{0.025}$ is $0.975$. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
$EBP = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.96)\sqrt{\dfrac{(0.842)(0.158)}{500}} = 0.032\nonumber$
$p' – EBP = 0.842 – 0.032 = 0.81\nonumber$
$p′ + EBP = 0.842 + 0.032 = 0.874\nonumber$
The confidence interval for the true binomial population proportion is $(p′ – EBP, p′ +EBP) = (0.810, 0.874)$.
Interpretation
We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
Explanation of 95% Confidence Level
Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to xx and enter 421.
Arrow down to nn and enter 500.
Arrow down to C-Level and enter .95.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.81003, 0.87397).
Exercise $1$
Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.
Answer
(0.3315, 0.4525)
Example $2$
For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.
Answer
• The first solution is step-by-step (Solution A).
• The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B).
Solution A
• $x = 300$ and
• $n = 500$
$p' = \dfrac{x}{n} = \dfrac{300}{500} = 0.600\nonumber$
$q′ = 1 − p′ = 1 − 0.600 = 0.400\nonumber$
Since $CL = 0.90$, then
$\alpha = 1 – CL = 1 – 0.90 = 0.10\left(\dfrac{\alpha}{2}\right) = 0.05$
$z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\nonumber$
Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find $z_{0.05}$. Remember that the area to the right of $z_{0.05}$ is 0.05 and the area to the left of $z_{0.05}$ is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.
$EBP = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.645)\sqrt{\dfrac{(0.60)(0.40)}{500}} = 0.036\nonumber$
$p′ – EBP = 0.60 − 0.036 = 0.564\nonumber$
$p′ + EBP = 0.60 + 0.036 = 0.636\nonumber$
The confidence interval for the true binomial population proportion is $(p′ – EBP, p′ +EBP) = (0.564,0.636)$.
Interpretation
• We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
• Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to xx and enter 300.
Arrow down to nn and enter 500.
Arrow down to C-Level and enter 0.90.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.564, 0.636).
Exercise $2$
A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.
1. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.
2. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.
Answer a
(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.
Answer b
Sixty-eight percent (68%) of students own an iPod and a smart phone.
$p′ = 0.68\nonumber$
$q′ = 1–p′ = 1 – 0.68 = 0.32\nonumber$
Since $CL = 0.97$, we know
$\alpha = 1 – 0.97 = 0.03\nonumber$
and
$\dfrac{\alpha}{2} = 0.015.\nonumber$
The area to the left of $z_{0.05}$ is 0.015, and the area to the right of $z_{0.05}$ is 1 – 0.015 = 0.985.
Using the TI 83, 83+, or 84+ calculator function InvNorm(0.985,0,1),
$z_{0.05} = 2.17\nonumber$
$EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = 2.17\sqrt{\dfrac{0.68(0.32)}{300}} \approx 0.0269\nonumber$
$p′ – EPB = 0.68 – 0.0269 = 0.6531\nonumber$
$p′ + EPB = 0.68 + 0.0269 = 0.7069\nonumber$
We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6531 and 0.7069.
Calculator
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to x and enter 300*0.68.
Arrow down to n and enter 300.
Arrow down to C-Level and enter 0.97.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.6531, 0.7069).
"Plus Four" Confidence Interval for $p$
There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.
Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is $n + 4$, and the new count of successes is $x + 2$. Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten.
Example $3$
A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke.
Solution A
Six students out of 25 reported smoking within the past week, so $x = 6$ and $n = 25$. Because we are using the “plus-four” method, we will use $x = 6 + 2 = 8$ and $n = 25 + 4 = 29$.
$p' = \dfrac{x}{n} = \dfrac{8}{29} \approx 0.276\nonumber$
$q′ = 1 – p′ = 1 – 0.276 = 0.724\nonumber$
Since $CL = 0.95$, we know $\alpha = 1 – 0.95 = 0.05$ and $\dfrac{\alpha}{2} = 0.025$.
$z_{0.025} = 1.96\nonumber$
$EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.96)\sqrt{\dfrac{0.276(0.724)}{29}} \approx 0.163$
$p′ – EPB = 0.276 – 0.163 = 0.113\nonumber$
$p′ + EPB = 0.276 + 0.163 = 0.439\nonumber$
We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
REMINDER
Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials.
Arrow down to $x$ and enter eight.
Arrow down to $n$ and enter 29.
Arrow down to C-Level and enter 0.95.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.113, 0.439).
Exercise $3$
Out of a random sample of 65 freshmen at State University, 31 students have declared a major. Use the “plus-four” method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.
Solution A
Using “plus four,” we have $x = 31 + 2 = 33$ and $n = 65 + 4 = 69$.
$p′ = 3369 \approx 0.478\nonumber$
$q′ = 1 – p′ = 1 – 0.478 = 0.522\nonumber$
Since $CL = 0.96$, we know $\alpha = 1 – 0.96 = 0.04$ and $\dfrac{\alpha}{2} = 0.02$.
$z_{0.02} = 2.054\nonumber$
$EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (2.054)\left(\sqrt{\dfrac{(0.478)(0.522)}{69}}\right) - 0.124\nonumber$
$p′ – EPB = 0.478 – 0.124 = 0.354\nonumber$
$p′ + EPB = 0.478 + 0.124 = 0.602\nonumber$
We are 96% confident that between 35.4% and 60.2% of all freshmen at State U have declared a major.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to $x$ and enter 33.
Arrow down to $n$ and enter 69.
Arrow down to C-Level and enter 0.96.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.355, 0.602).
Example $4$
The Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on Facebook. Use the “plus four” method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends.
Solution A
Using “plus-four,” we have $x = 13 + 2 = 15$ and $n = 50 + 4 = 54$.
$p′ = 1554 \approx 0.278\nonumber$
$q′ = 1 – p′ = 1 − 0.241 = 0.722\nonumber$
Since $CL = 0.90$, we know $\alpha = 1 – 0.90 = 0.10$ and $\dfrac{\alpha}{2} = 0.05$.
$z_{0.05} = 1.645\nonumber$
$EPB = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.278)(0.722)}{54}}\right) \approx 0.100\nonumber$
$p′ – EPB = 0.278 – 0.100 = 0.178\nonumber$
$p′ + EPB = 0.278 + 0.100 = 0.378\nonumber$
We are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to $x$ and enter 15.
Arrow down to $n$ and enter 54.
Arrow down to C-Level and enter 0.90.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.178, 0.378).
Exercise $4$
The Berkman Center Study referenced in Example talked to teens in smaller focus groups, but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their Facebook friends with 159 saying that they have more than 500 friends. Use the “plus-four” method to find a 90% confidence interval for the true proportion of teens that would report having more than 500 Facebook friends based on this larger sample. Compare the results to those in Example.
Answer
Solution A
Using “plus-four,” we have $x = 159 + 2 = 161$ and $n = 588 + 4 = 592$.
$p′ = 161592 \approx 0.272\nonumber$
$q′ = 1 – p′ = 1 – 0.272 = 0.728\nonumber$
Since CL = 0.90, we know $\alpha = 1 – 0.90 = 0.10$ and $\dfrac{\alpha}{2} = 0.05$
$EPB = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.272)(0.728)}{592}}\right) \approx 0.030\nonumber$
$p′ – EPB = 0.272 – 0.030 = 0.242\nonumber$
$p′ + EPB = 0.272 + 0.030 = 0.302\nonumber$
We are 90% confident that between 24.2% and 30.2% of all teens would report having more than 500 friends on Facebook.
Solution B
• Press STAT and arrow over to TESTS.
• Arrow down to A:1-PropZint. Press ENTER.
• Arrow down to $x$ and enter 161.
• Arrow down to $n$ and enter 592.
• Arrow down to C-Level and enter 0.90.
• Arrow down to Calculate and press ENTER.
• The confidence interval is (0.242, 0.302).
Conclusion: The confidence interval for the larger sample is narrower than the interval from Example. Larger samples will always yield more precise confidence intervals than smaller samples. The “plus four” method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13/50) to 0.278 (15/54). It has a smaller impact on the EPB, changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159/588) to 0.272 (161/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.
Calculating the Sample Size $n$
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population proportion is
$EBP = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right)\nonumber$
Solving for $n$ gives you an equation for the sample size.
$n = \dfrac{\left(z_{\frac{\alpha}{2}}\right)^{2}(p'q')}{EBP^{2}}\nonumber$
Example $5$
Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones.
Answer
From the problem, we know that $\bf{EBP = 0.03}$ (3%=0.03) and $z_{\dfrac{\alpha}{2}} z_{0.05} = 1.645$ because the confidence level is 90%.
However, in order to find $n$, we need to know the estimated (sample) proportion $p′$. Remember that $q′ = 1 – p′$. But, we do not know $p′$ yet. Since we multiply $p′$ and $q′$ together, we make them both equal to 0.5 because $p′q′ = (0.5)(0.5) = 0.25$ results in the largest possible product. (Try other products: $(0.6)(0.4) = 0.24$; $(0.3)(0.7) = 0.21$; $(0.2)(0.8) = 0.16$ and so on). The largest possible product gives us the largest $n$. This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size $n$, use the formula and make the substitutions.
$n = \dfrac{z^{2}p'q'}{EBP^{2}}\nonumber$
gives
$n = \dfrac{1.645^{2}(0.5)(0.5)}{0.03^{2}} = 751.7\nonumber$
Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.
Exercise $5$
Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?
Answer
271 customers should be surveyed. Check the Real Estate section in your local
Glossary
Binomial Distribution
a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, $n$, of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV $X$ is defined as the number of successes in $n$ trials. The notation is: $X \sim B(\mathbf{n},\mathbf{p})$. The mean is $\mu = np$ and the standard deviation is $\sigma = \sqrt{npq}$. The probability of exactly $x$ successes in $n$ trials is $P(X = x = \left(\binom{n}{x}\right))p^{x}q^{n-x}$.
Error Bound for a Population Proportion ($EBP$)
the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.04%3A_A_Population_Proportion.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will calculate the 90% confidence interval for the mean cost of a home in the area in which this school is located.
• The student will interpret confidence intervals.
• The student will determine the effects of changing conditions on the confidence interval.
Collect the Data
Check the Real Estate section in your local newspaper. Record the sale prices for 35 randomly selected homes recently listed in the county.
NOTE
Many newspapers list them only one day per week. Also, we will assume that homes come up for sale randomly.
1. Complete the table:
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
Describe the Data
1. Compute the following:
1. $\bar{x}$ = _____
2. $s_{x}$ = _____
3. $n$ = _____
2. In words, define the random variable $\bar{X}$.
3. State the estimated distribution to use. Use both words and symbols.
Find the Confidence Interval
1. Calculate the confidence interval and the error bound.
1. Confidence Interval: _____
2. Error Bound: _____
2. How much area is in both tails (combined)? $\alpha$ = _____
3. How much area is in each tail? $\frac{\alpha}{2}$ = _____
4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample mean.
5. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data on the first page and count how many of the data values lie within the confidence interval. What percent is this? Is this percent close to 90%? Explain why this percent should or should not be close to 90%.
Describe the Confidence Interval
1. In two to three complete sentences, explain what a confidence interval means (in general), as if you were talking to someone who has not taken statistics.
2. In one to two complete sentences, explain what this confidence interval means for this particular study.
Use the Data to Construct Confidence Intervals
1. Using the given information, construct a confidence interval for each confidence level given.
Confidence level EBM/Error Bound Confidence Interval
50%
80%
95%
99%
2. What happens to the $EBM$ as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens.
8.06: Confidence Interval -Place of Birth (Worksheet)
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will calculate the 90% confidence interval the proportion of students in this school who were born in this state.
• The student will interpret confidence intervals.
• The student will determine the effects of changing conditions on the confidence interval.
Collect the Data
1. Survey the students in your class, asking them if they were born in this state. Let $X =$ the number that were born in this state.
1. $n =$ ____________
2. $x =$ ____________
2. In words, define the random variable $P′$.
3. State the estimated distribution to use.
Find the Confidence Interval and Error Bound
1. Calculate the confidence interval and the error bound.
1. Confidence Interval: _____
2. Error Bound: _____
2. How much area is in both tails (combined)? $\alpha =$ _____
3. How much area is in each tail? $\frac{\alpha}{2} =$ _____
4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample proportion. <figure >
Describe the Confidence Interval
1. In two to three complete sentences, explain what a confidence interval means (in general), as though you were talking to someone who has not taken statistics.
2. In one to two complete sentences, explain what this confidence interval means for this particular study.
3. Construct a confidence interval for each confidence level given.
Confidence level EBP/Error Bound Confidence Interval
50%
80%
95%
99%
4. What happens to the $EBP$ as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.05%3A_Confidence_Interval_-_Home_Costs_%28Worksheet%29.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will calculate a 90% confidence interval using the given data.
• The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean.
Given:
Heights of 100 Women (in Inches)
59.4 71.6 69.3 65.0 62.9 66.5 61.7 55.2
67.5 67.2 63.8 62.9 63.0 63.9 68.7 65.5
61.9 69.6 58.7 63.4 61.8 60.6 69.8 60.0
64.9 66.1 66.8 60.6 65.6 63.8 61.3 59.2
64.1 59.3 64.9 62.4 63.5 60.9 63.3 66.3
61.5 64.3 62.9 60.6 63.8 58.8 64.9 65.7
62.5 70.9 62.9 63.1 62.2 58.7 64.7 66.0
60.5 64.7 65.4 60.2 65.0 64.1 61.1 65.3
64.6 59.2 61.4 62.0 63.5 61.4 65.5 62.3
65.5 64.7 58.8 66.1 64.9 66.9 57.9 69.8
58.5 63.4 69.2 65.9 62.2 60.0 58.1 62.5
62.4 59.1 66.4 61.2 60.4 58.7 66.7 67.5
63.2 56.6 67.7 62.5
1. Table lists the heights of 100 women. Use a random number generator to select ten data values randomly.
2. Calculate the sample mean and the sample standard deviation. Assume that the population standard deviation is known to be 3.3 inches. With these values, construct a 90% confidence interval for your sample of ten values. Write the confidence interval you obtained in the first space of Table.
3. Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into Table.
90% Confidence Intervals
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
Discussion Questions
1. The actual population mean for the 100 heights given Table is $\mu = 63.4$. Using the class listing of confidence intervals, count how many of them contain the population mean $\mu$; i.e., for how many intervals does the value of $\mu$ lie between the endpoints of the confidence interval?
2. Divide this number by the total number of confidence intervals generated by the class to determine the percent of confidence intervals that contains the mean $\mu$. Write this percent here: _____________.
3. Is the percent of confidence intervals that contain the population mean $\mu$ close to 90%?
4. Suppose we had generated 100 confidence intervals. What do you think would happen to the percent of confidence intervals that contained the population mean?
5. When we construct a 90% confidence interval, we say that we are 90% confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean by this phrase.
6. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data given (the heights of women) and count how many of the data values lie within the confidence interval that you generated based on that data. How many of the 100 data values lie within your confidence interval? What percent is this? Is this percent close to 90%?
7. Explain why it does not make sense to count data values that lie in a confidence interval. Think about the random variable that is being used in the problem.
8. Suppose you obtained the heights of ten women and calculated a confidence interval from this information. Without knowing the population mean $\mu$, would you have any way of knowing for certain if your interval actually contained the value of $\mu$? Explain. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.07%3A_Confidence_Interval_-Women%27s_Heights_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
8.2: A Single Population Mean using the Normal Distribution
Q 8.2.1
Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.
1. $\bar{x}$ =________
2. $\sigma$ =________
3. $n$ =________
1. In words, define the random variables $X$ and $\bar{X}$.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population mean height of male Swedes.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. What will happen to the error bound obtained if 1,000 male Swedes are surveyed instead of 48? Why?
S 8.2.1
1. 71
2. 3
3. 48
1. X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males.
2. Normal. We know the standard deviation for the population, and the sample size is greater than 30.
1. CI: (70.151, 71.849)
ii. EBM = 0.849
e. The error bound will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.
Q 8.2.2
Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal.
1. In words, define the random variables $X$ and $\bar{X}$.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population mean length of engineering conferences.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
Q 8.2.3
Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.
1. $\bar{x} =$ ________
2. $\sigma =$ ________
3. $n =$ ________
1. In words, define the random variables $X$ and $\bar{X}$.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population mean time to complete the tax forms.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?
5. If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?
6. Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why?
S 8.2.3
1. $\bar{x} = 23.6$
2. $\sigma = 7$
3. $n = 100$
1. $X$ is the time needed to complete an individual tax form. $\bar{X}$ is the mean time to complete tax forms from a sample of 100 customers.
2. $N\left(23.6, \frac{7}{\sqrt{100}}\right)$ because we know sigma.
1. (22.228, 24.972)
2. $EBM = 1.372$
3. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size.
4. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
5. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size.
Q 8.2.4
A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce.
1. $\bar{x} =$________
2. $\sigma =$________
3. $s_{x} =$________
1. In words, define the random variable $X$.
2. In words, define the random variable $\bar{X}$.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 90% confidence interval for the population mean weight of the candies.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
5. Construct a 98% confidence interval for the population mean weight of the candies.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
6. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e.
7. In complete sentences, give an interpretation of what the interval in part f means.
Q 8.2.5
A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8.
1. $\bar{x} =$________
2. $\sigma =$________
3. $x =$________
1. Define the random variables $X$ and $\bar{X}$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 90% confidence interval for the population mean number of letters campers send home.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?
S 8.2.5
1. 7.9
2. 2.5
3. 20
1. $X$ is the number of letters a single camper will send home. $\bar{X}$ is the mean number of letters sent home from a sample of 20 campers.
2. $N 7.9\left(\frac{2.5}{\sqrt{20}}\right)$
3.
1. CI: (6.98, 8.82)
1. $EBM: 0.92$
4. The error bound and confidence interval will decrease.
Q 8.2.6
What is meant by the term “90% confident” when constructing a confidence interval for a mean?
1. If we took repeated samples, approximately 90% of the samples would produce the same confidence interval.
2. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean.
3. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean.
4. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples.
Q 8.2.7
The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is $\sigma$ =$909,200.
$3,600$1,243,900 $10,900$385,200 $581,500$7,400 $2,900$400 $3,714,500$632,500
$391,000$467,400 $56,800$5,800 $405,200$733,200 $8,000$468,700 $75,200$41,000
$13,300$9,500 $953,800$1,113,500 $1,109,300$353,900 $986,100$88,600 $378,200$13,200
$3,800$745,100 $5,800$3,072,100 $1,626,700$512,900 $2,309,200$6,600 $202,400$15,800
1. Find the point estimate for the population mean.
2. Using 95% confidence, calculate the error bound.
3. Create a 95% confidence interval for the mean total individual contributions.
4. Interpret the confidence interval in the context of the problem.
S 8.2.7
1. $\bar{x} = 568,873$
2. $CL = 0.95 \alpha = 1 - 0.95 = 0.05 z_{\frac{\alpha}{2}} = 1.96$
$EBM = z_{0.025 } \frac{\sigma}{\sqrt{n}} = 1.96 \frac{909200}{\sqrt{40}} = 281,764$
3. $\bar{x}$ − EBM = 568,873 − 281,764 = 287,109
$\bar{x}$ + EBM = 568,873 + 281,764 = 850,637
Alternate solution:
1. Press STAT and arrow over to TESTS.
2. Arrow down to 7:ZInterval.
3. Press ENTER.
4. Arrow to Stats and press ENTER.
5. Arrow down and enter the following values:
• $\sigma: 909,200$
• $\bar{x}: 568,873$
• $n: 40$
• $CL: 0.95$
6. Arrow down to Calculate and press ENTER.
7. The confidence interval is ($287,114,$850,632).
8. Notice the small difference between the two solutions–these differences are simply due to rounding error in the hand calculations.
4. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and$850,637.
Q 8.2.8
The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The most recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and$69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income.
Q 8.2.9
The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure?
S 8.2.9
Use the formula for $EBM$, solved for $n$:
$n = \frac{z^{2}\sigma^{2}}{EBM^{2}}$
From the statement of the problem, you know that $\sigma$ = 2.5, and you need $EBM = 1$.
$z = z_{0.035} = 1.812$
(This is the value of $z$ for which the area under the density curve to the right of $z$ is 0.035.)
$n = \frac{z^{2}\sigma^{2}}{EBM^{2}} = \frac{1.812^{2}2.5^{2}}{1^{2}} \approx 20.52$
You need to measure at least 21 male students to achieve your goal.
8.3: A Single Population Mean using the Student t Distribution
Q 8.3.1
In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces.
1. Define the random variables $X$ and $P′$ in words.
2. Which distribution should you use for this problem? Explain your choice
3. Calculate $p′$.
4. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
5. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?
Q 8.3.2
A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal.
1. $\bar{x} =$ __________
2. $s_{x} =$ __________
3. $n =$ __________
4. $n – 1 =$ __________
1. Define the random variables $X$ and $\bar{X}$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why?
S 8.3.2
1. 8629
2. 6944
3. 35
4. 34
1. $t_{34}$
1. $CI: (6244, 11,014)$
2. $EB = 2385$
2. It will become smaller
Q 8.3.3
Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours.
1. $\bar{x} =$ __________
2. $s_{x} =$ __________
3. $n =$ __________
4. $n – 1 =$ __________
1. Define the random variables $X$ and $\bar{X}$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population mean time wasted.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. Explain in a complete sentence what the confidence interval means.
Q 8.3.4
A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.
1. $\bar{x} =$ __________
2. $s_{x} =$ __________
3. $n =$ __________
4. $n – 1 =$ __________
1. Define the random variable $X$ in words.
2. Define the random variable $\bar{X}$ in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95% confidence interval for the population mean length of time.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
5. What does it mean to be “95% confident” in this problem?
S 8.3.4
1. $\bar{x} = 2.51$
2. $s_{x} = 0.318$
3. $n = 9$
4. $n - 1 = 8$
1. the effective length of time for a tranquilizer
2. the mean effective length of time of tranquilizers from a sample of nine patients
3. We need to use a Student’s-t distribution, because we do not know the population standard deviation.
1. $CI: (2.27, 2.76)$
2. Check student's solution.
3. $EBM: 0.25$
4. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time.
Q 8.3.5
Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal.
1. $\bar{x} =$ __________
2. $s_{x} =$ __________
3. $n =$ __________
4. $n – 1 =$ __________
1. Define the random variable $X$ in words.
2. Define the random variable $\bar{X}$ in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 99% confidence interval for the population mean length of time using training wheels.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
5. Why would the error bound change if the confidence level were lowered to 90%?
Q 8.3.6
The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns.
The FEC has reported financial information for 556 Leadership PACs that operating during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs.
$46,500.00$0 $40,966.50$105,887.20 $5,175.00$29,050.00 $19,500.00$181,557.20 $31,500.00$149,970.80
$2,555,363.20$12,025.00 $409,000.00$60,521.70 $18,000.00$61,810.20 $76,530.80$119,459.20 $0$63,520.00
$6,500.00$502,578.00 $705,061.10$708,258.90 $135,810.00$2,000.00 $2,000.00$0 $1,287,933.80$219,148.30
$\bar{x} = 251,854.23$
$s = 521,130.41$
Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's $t$-distribution.
S 8.3.6
$\bar{x} = 251,854.23$
$s = 521,130.41$
Note that we are not given the population standard deviation, only the standard deviation of the sample.
There are 30 measures in the sample, so $n = 30$, and $df = 30 - 1 = 29$
$CL = 0.96$, so $\alpha = 1 - CL = 1 - 0.96 = 0.04$
$\frac{\alpha}{2} = 0.02 t_{0.02} = t_{0.02} = 2.150$
$EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right) = 2.150\left(\frac{521,130.41}{\sqrt{30}}\right) - 204,561.66$
$\bar{x} - EBM = 251,854.23 - 204,561.66 = 47,292.57$
$\bar{x} + EBM = 251,854.23+ 204,561.66 = 456,415.89$
We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and$456,415.89.
Alternate Solution
Enter the data as a list.
Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval.
Press ENTER.
Arrow to Data and press ENTER.
Arrow down and enter the name of the list where the data is stored.
Enter Freq: 1
Enter C-Level: 0.96
Arrow down to Calculate and press Enter.
The 96% confidence interval is ($47,262,$456,447).
The difference between solutions arises from rounding differences.
Forbes magazine published data on the best small firms in 2012. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between$5 million and $1 billion. The Table shows the ages of the corporate CEOs for a random sample of these firms. 48 58 51 61 56 59 74 63 53 50 59 60 60 57 46 55 63 57 47 55 57 43 61 62 49 67 67 55 55 49 Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution. Q 8.3.8 Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. 1. $\bar{x} =$ __________ 2. $s_{x} =$ __________ 3. $n =$ __________ 4. $n-1 =$ __________ 1. Define the random variables $X$ and $\bar{X}$ in words. 2. Which distribution should you use for this problem? Explain your choice. 3. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight. 1. State the confidence interval. 2. Sketch the graph. 3. Calculate the error bound. S 8.3.8 1. $\bar{x} =$ 2. $s_{x} =$ 3. $n =$ 4. $n-1 =$ 1. $X$ is the number of unoccupied seats on a single flight. $\bar{X}$ is the mean number of unoccupied seats from a sample of 225 flights. 2. We will use a Student’s $t$-distribution, because we do not know the population standard deviation. 1. $CI: (11.12 , 12.08)$ 2. Check student's solution. 3. $EBM$: 0.48 Q 8.3.9 In a recent sample of 84 used car sales costs, the sample mean was$6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. 1. Which distribution should you use for this problem? Explain your choice. 2. Define the random variable $\bar{X}$ in words. 3. Construct a 95% confidence interval for the population mean cost of a used car. 1. State the confidence interval. 2. Sketch the graph. 3. Calculate the error bound. 4. Explain what a “95% confidence interval” means for this study. Q 8.3.10 Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal. 1. Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets. 1. State the confidence interval. 2. Sketch the graph. 3. Calculate the error bound. 2. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done? 3. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies. 4. Calculate the mean. 5. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not? S 8.3.10 1. CI: (7.64 , 9.36) 2. $EBM: 0.86$ 1. The sample should have been increased. 2. Answers will vary. 3. Answers will vary. 4. Answers will vary. Q 8.3.11 A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢;$1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.
1. $\bar{x} =$ __________
2. $s_{x} =$ __________
3. $n =$ __________
4. $n-1 =$ __________
1. Define the random variables $X$ and $\bar{X}$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population mean worth of coupons.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why.
Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed.
Q 8.3.12
Find the 95% Confidence Interval for the true population mean for the amount of soda served.
1. (12.42, 14.18)
2. (12.32, 14.29)
3. (12.50, 14.10)
4. Impossible to determine
b
Q 8.3.13
What is the error bound?
1. 0.87
2. 1.98
3. 0.99
4. 1.74
8.4: A Population Proportion
Q 8.4.1
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.
1. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03?
2. If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why?
S 8.4.1
1. 1,068
2. The sample size would need to be increased since the critical value increases as the confidence level increases.
Q 8.4.2
Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.
1. $x =$ __________
2. $n =$ __________
3. $p′ =$ __________
1. Define the random variables $X$ and $P′$, in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population proportion who claim they always buckle up.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.
Q 8.4.3
According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job.
1. Define the random variables $X$ and $P′$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
S 8.4.3
1. $X =$ the number of people who feel that the president is doing an acceptable job;
$P′ =$ the proportion of people in a sample who feel that the president is doing an acceptable job.
2. $N\left(0.61, \sqrt{\frac{(0.61)(0.39)}{1200}}\right)$
1. $CI: (0.59, 0.63)$
2. Check student’s solution
3. $EBM: 0.02$
Q 8.4.4
An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
1. We are interested in finding the 95% confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables $X$ and $P′$, in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
Q 8.4.5
Refer to the information in Exercise.
1. Construct three 95% confidence intervals.
1. percent of all Asians who would welcome a white person into their families.
2. percent of all Asians who would welcome a Latino into their families.
3. percent of all Asians who would welcome a black person into their families.
2. Even though the three point estimates are different, do any of the confidence intervals overlap? Which?
3. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?
4. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?
S 8.4.5
1. (0.72, 0.82)
2. (0.65, 0.76)
3. (0.60, 0.72)
1. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
2. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
3. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.
Q 8.4.6
Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period.
1. Define the random variables $X$ and $P′$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. Explain what a “97% confidence interval” means for this study.
Q 8.4.7
A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem.
1. Define the random variables $X$ and $P′$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. Suppose we want to lower the sampling error. What is one way to accomplish that?
5. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is $\pm 3%$. In one to three complete sentences, explain what the ±3% represents.
S 8.4.7
1. $X =$ the number of adult Americans who feel that crime is the main problem; $P′ =$ the proportion of adult Americans who feel that crime is the main problem
2. Since we are estimating a proportion, given $P′ = 0.2$ and $n = 1000$, the distribution we should use is $N\left(0.61, \sqrt{\frac{(0.2)(0.8)}{1000}}\right)$.
1. $CI: (0.18, 0.22)$
2. Check student’s solution.
3. $EBM: 0.02$
3. One way to lower the sampling error is to increase the sample size.
4. The stated “$\pm 3%$” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%.
Q 8.4.8
Refer to Exercise. Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.
1. Define the random variables $X$ and $P′$ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
4. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is $\pm 3%$. In one to three complete sentences, explain what the ±3% represents.
Use the following information to answer the next three exercises: According to a Field Poll, 79% of California adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California.
Q 8.4.9
A point estimate for the true population proportion is:
1. 0.90
2. 1.27
3. 0.79
4. 400
c
Q 8.4.10
A 90% confidence interval for the population proportion is _______.
1. (0.761, 0.820)
2. (0.125, 0.188)
3. (0.755, 0.826)
4. (0.130, 0.183)
Q 8.4.11
The error bound is approximately _____.
1. 1.581
2. 0.791
3. 0.059
4. 0.030
S 8.4.11
d
Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not.
Q 8.4.12
Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.
1. (0.2975, 0.3796)
2. (0.6270, 0.6959)
3. (0.3041, 0.3730)
4. (0.6204, 0.7025)
Q 8.4.13
The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______.
1. 0.6614
2. 0.3386
3. 173
4. 338
a
Q 8.4.14
On May 23, 2013, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a $\pm 3%$ margin of error.
1. Determine the estimated proportion from the sample.
2. Determine the sample size.
3. Identify $CL$ and $\alpha$.
4. Calculate the error bound based on the information provided.
5. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the values.
6. Create a confidence interval for the results of this study.
7. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience?
Q 8.4.15
A national survey of 1,000 adults was conducted on May 13, 2013 by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education.
1. Find the point estimate and the error bound for this confidence interval.
2. Can we (with 95% confidence) conclude that more than half of all American adults believe this?
3. Use the point estimate from part a and $n = 1,000$ to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education.
4. Can we (with 75% confidence) conclude that at least half of all American adults believe this?
S 8.4.15
1. $p′ = \frac{(0.55+0.49)}{2} = 0.52; EBP = 0.55 - 0.52 = 0.03$
2. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
3. $CL = 0.75$, so $\alpha = 1 – 0.75 = 0.25$ and $\frac{\alpha}{2} = 0.125 z_{\frac{\alpha}{2}} = 1.150$. (The area to the right of this $z$ is 0.125, so the area to the left is $1 – 0.125 = 0.875$.)
$EBP = (1.150)\sqrt{\frac{0.52(0.48)}{1,000}} \approx 0.018$
$(p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)$
Alternate Solution
STAT TESTS A: 1-PropZinterval with $x = (0.52)(1,000), n = 1,000, CL = 0.75$.
Answer is (0.502, 0.538)
4. Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.
Q 8.4.16
Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music.
1. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music.
2. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error.
3. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%.
Q 8.4.17
You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview?
S 8.4.17
$CL = 0.95 \alpha = 1 - 0.95 = 0.05 \frac{\alpha}{2} = 0.025 z_{\frac{\alpha}{2}} = 1.96.$ Use $p′ = q′ = 0.5$.
$n = \frac{z_{\frac{\alpha}{2}}^{2}p'q'}{EPB^{2}} = \frac{1.96^{2}(0.5)(0.5)}{0.05^{2}} = 384.16$
You need to interview at least 385 students to estimate the proportion to within 5% at 95% confidence.
Q 8.4.18
In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as “likely” or “very likely.” Use the “plus four” method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.E%3A_Confidence_Intervals_%28Exercises%29.txt |
Review
In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean (EBM). A confidence interval has the general form:
$(\text{lower bound, upper bound}) = (\text{point estimate} – EBM, \text{point estimate} + EBM)$
The calculation of $EBM$ depends on the size of the sample and the level of confidence desired. The confidence level is the percent of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding $EBM$ increases as well. As the sample size increases, the $EBM$ decreases. By the central limit theorem,
$EBM = z\frac{\sigma}{\sqrt{n}}$
Given a confidence interval, you can work backwards to find the error bound ($EBM$) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean.
Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the $EBM$ formula for $n$ to discover the size of the sample that is needed to achieve this goal:
$n = \frac{z^{2}\sigma^{2}}{EBM^{2}}$
Formula Review
$\bar{X} - N \left(\mu_{x}, \frac{\sigma}{\sqrt{n}}\right)$ The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size.
The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by
$(\text{lower bound, upper bound}) = (\text{point estimate} – EBM, \text{point estimate} + EBM)$
$= \bar{x} - EBM, \bar{x} + EBM$
$= \left(\bar{x} - z\frac{\sigma}{\sqrt{n}}, \bar{x} + z\frac{\sigma}{\sqrt{n}}\right)$
$EBM = z\frac{\sigma}{\sqrt{n}} =$ the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known.
$CL =$ confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter
$\alpha = 1 – CL =$ the proportion of confidence intervals that will not contain the population parameter
$z_{\frac{\alpha}{2}}$ = the $z$-score with the property that the area to the right of the z-score is $\frac{\propto}{2}$ this is the $z$-score used in the calculation of "$EBM$ where $\alpha = 1 – CL$".
$n = \frac{z^{2}\sigma^{2}}{EBM^{2}}$ the formula used to determine the sample size ($n$) needed to achieve a desired margin of error at a given level of confidence
General form of a confidence interval
$(\text{lower value, upper value}) = (\text{point estimate} - \text{error bound, point estimate} + \text{error bound})$
To find the error bound when you know the confidence interval
$\text{error bound} = \text{upper value} - \text{point estimate}$ OR $\text{error bound} = \frac{\text{upper value - lower value}}{2}$
Single Population Mean, Known Standard Deviation, Normal Distribution
Use the Normal Distribution for Means, Population Standard Deviation is Known $EBM = z\frac{\alpha}{2} \cdot \frac{\sigma}{\sqrt{n}}$
The confidence interval has the format $(\bar{x} - EBM, \bar{x} + EBM)$.
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds.
Exercise 8.2.8
Identify the following:
1. $\bar{x} =$ _____
2. $\sigma =$ _____
3. $n =$ _____
Answer
1. 244
2. 15
3. 50
Exercise 8.2.9
In words, define the random variables $X$ and $\bar{X}$.
Exercise 8.2.10
Which distribution should you use for this problem?
Answer
$N\left(244, \frac{15}{\sqrt{50}}\right)$
Exercise 8.2.11
Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound.
Exercise 8.2.12
What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?
Answer
As the sample size increases, there will be less variability in the mean, so the interval size decreases.
Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.
Exercise 8.2.13
Identify the following:
1. $\bar{x} =$ _____
2. $\sigma =$ _____
3. $n =$ _____
Exercise 8.2.14
In words, define the random variables $X$ and $\bar{X}$.
Answer
$X$ is the time in minutes it takes to complete the U.S. Census short form. $\bar{X}$ is the mean time it took a sample of 200 people to complete the U.S. Census short form.
Exercise 8.2.15
Which distribution should you use for this problem?
Exercise 8.2.16
Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
$CI: (7.9441, 8.4559)$
$EBM = 0.26$
Exercise 8.2.17
If the Census wants to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?
Exercise 8.2.18
If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why?
Answer
The level of confidence would decrease because decreasing $n$ makes the confidence interval wider, so at the same error bound, the confidence level decreases.
Exercise 8.2.19
Suppose the Census needed to be 98% confident of the population mean length of time. Would the Census have to survey more people? Why or why not?
Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds.
Exercise 8.2.20
Identify the following:
1. $\bar{x} =$ _____
2. $\sigma =$ _____
3. $n =$ _____
Answer
1. $\bar{x} = 2.2$
2. $\sigma = 0.2$
3. $n = 20$
Exercise 8.2.21
In words, define the random variable $X$.
Exercise 8.2.22
In words, define the random variable $\bar{X}$.
Answer
$\bar{X}$ is the mean weight of a sample of 20 heads of lettuce.
Exercise 8.2.23
Which distribution should you use for this problem?
Exercise 8.2.24
Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
$EBM = 0.07$
$CI: (2.1264, 2.2736)$
Exercise 8.2.25
Construct a 95% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.
Exercise 8.2.26
In complete sentences, explain why the confidence interval in Exercise is larger than in Exercise.
Answer
The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.
Exercise 8.2.27
In complete sentences, give an interpretation of what the interval in Exercise means.
Exercise 8.2.28
What would happen if 40 heads of lettuce were sampled instead of 20, and the error bound remained the same?
Answer
The confidence level would increase.
Exercise 8.2.29
What would happen if 40 heads of lettuce were sampled instead of 20, and the confidence level remained the same?
Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let $X$ = the age of a Winter Foothill College student.
Exercise 8.2.30
$\bar{x} =$ _____
Answer
30.4
Exercise 8.2.31
$n =$ _____
Exercise 8.2.32
________ $= 15$
Answer
$\sigma$
Exercise 8.2.33
In words, define the random variable $\bar{X}$.
Exercise 8.2.34
What is $\bar{x}$ estimating?
Answer
$\mu$
Exercise 8.2.35
Is $\sigma_{x}$ known?
Exercise 8.2.36
As a result of your answer to Exercise, state the exact distribution to use when calculating the confidence interval.
Answer
normal
Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.
Exercise 8.2.37
How much area is in both tails (combined)? $\alpha =$ ________
Exercise 8.2.38
How much area is in each tail? $\frac{\alpha}{2} =$ ________
Answer
0.025
Exercise 8.2.39
Identify the following specifications:
1. lower limit
2. upper limit
3. error bound
Exercise 8.2.40
The 95% confidence interval is:__________________.
Answer
(24.52,36.28)
Exercise 8.2.41
Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean.
Exercise 8.2.42
In one complete sentence, explain what the interval means.
Answer
We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28.
Exercise 8.2.43
Using the same mean, standard deviation, and level of confidence, suppose that $n$ were 69 instead of 25. Would the error bound become larger or smaller? How do you know?
Exercise 8.2.44
Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why?
Answer
The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.
Review
In many cases, the researcher does not know the population standard deviation, $\sigma$, of the measure being studied. In these cases, it is common to use the sample standard deviation, $s$, as an estimate of $\sigma$. The normal distribution creates accurate confidence intervals when $\sigma$ is known, but it is not as accurate when $s$ is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula:
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$
The $t$-score follows the Student’s $t$-distribution with $n – 1$ degrees of freedom. The confidence interval under this distribution is calculated with $EBM = \left(t_{\frac{\alpha}{2}}\right)\frac{s}{\sqrt{n}}$ where $t_{\frac{\alpha}{2}}$ is the $t$-score with area to the right equal to $\frac{\alpha}{2}$, $s$ is the sample standard deviation, and $n$ is the sample size. Use a table, calculator, or computer to find $t_{\frac{\alpha}{2}}$ for a given $\alpha$.
Formula Review
$s =$ the standard deviation of sample values.
$t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ is the formula for the $t$-score which measures how far away a measure is from the population mean in the Student’s $t$-distribution
$df = n - 1$; the degrees of freedom for a Student’s $t$-distribution where n represents the size of the sample
$T \sim t_{df}$ the random variable, $T$, has a Student’s $t$-distribution with $df$ degrees of freedom
$EBM = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}} =$ the error bound for the population mean when the population standard deviation is unknown
$t_{\frac{\alpha}{2}}$ is the $t$-score in the Student’s $t$-distribution with area to the right equal to $\frac{\alpha}{2}$
The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound)
$= (\text{point estimate} – EBM, \text{point estimate} + EBM)$
$= \left(\bar{x} - \frac{ts}{\sqrt{n}}, \bar{x} + \frac{ts}{\sqrt{n}}\right)$
Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.
Exercise 8.3.3
Identify the following:
1. $\bar{x} =$_______
2. $s_{x} =$_______
3. $n =$_______
4. $n - 1=$_______
Exercise 8.3.4
Define the random variables $X$ and $\bar{X}$ in words.
Answer
$X$ is the number of hours a patient waits in the emergency room before being called back to be examined. $\bar{X}$ is the mean wait time of 70 patients in the emergency room.
Exercise 8.3.5
Which distribution should you use for this problem?
Exercise 8.3.6
Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
$CI: (1.3808, 1.6192)$
$EBM = 0.12$
Exercise 8.3.7
Explain in complete sentences what the confidence interval means.
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Exercise 8.3.8
1. $\bar{x} =$_______
2. $s_{x} =$_______
3. $n =$_______
4. $n - 1=$_______
Answer
1. $\bar{x} = 151$
2. $s_{x} = 32$
3. $n = 108$
4. $n - 1= 107$
Exercise 8.3.9
Define the random variable $X$ in words.
Exercise 8.3.10
Define the random variable $\bar{X}$ in words.
Answer
$\bar{X}$ is the mean number of hours spent watching television per month from a sample of 108 Americans.
Exercise 8.3.11
Which distribution should you use for this problem?
Exercise 8.3.12
Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound.
Answer
$CI: (142.92, 159.08)$
$EBM = 8.08$
Exercise 8.3.13
Why would the error bound change if the confidence level were lowered to 95%?
Use the following information to answer the next 13 exercises: The data in Table are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let $X =$ the number of colors on a national flag.
$X$ Freq.
1 1
2 7
3 18
4 7
5 6
Exercise 8.3.14
1. $\bar{x} =$_______
2. $s_{x} =$_______
3. $n =$_______
Answer
1. 3.26
2. 1.02
3. 39
Exercise 8.3.15
Define the random variable $\bar{X}$ in words.
Exercise 8.3.16
What is $\bar{x}$ estimating?
Answer
$\mu$
Exercise 8.3.17
Is $\sigma_{x}$ known?
Exercise 8.3.18
As a result of your answer to Exercise, state the exact distribution to use when calculating the confidence interval.
Answer
$t_{38}$
Construct a 95% confidence interval for the true mean number of colors on national flags.
Exercise 8.3.19
How much area is in both tails (combined)?
Exercise 8.3.20
How much area is in each tail?
Answer
0.025
Exercise 8.3.21
Calculate the following:
1. lower limit
2. upper limit
3. error bound
Exercise 8.3.22
The 95% confidence interval is_____.
Answer
(2.93, 3.59)
Exercise 8.3.23
Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean.
Exercise 8.3.24
In one complete sentence, explain what the interval means.
Answer
We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.
Exercise 8.3.25
Using the same $\bar{x}$, $s_{x}$, and level of confidence, suppose that $n$ were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
Answer
The error bound would become $EBM = 0.245$. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.
Exercise 8.3.26
Using the same $\bar{x}$, $s_{x}$, and $n = 39$, how would the error bound change if the confidence level were reduced to 90%? Why?
Review
Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning.
Let $p′$ represent the sample proportion, $\frac{x}{n}$, where $x$ represents the number of successes and $n$ represents the sample size. Let $q′ = 1 – p′$. Then the confidence interval for a population proportion is given by the following formula:
(lower bound, upper bound) = $\left(p' - EBP, p' + EBP\right) = \left(p' - z\sqrt{\frac{p'q'}{n}}, p' + z\sqrt{\frac{p'q'}{n}}\right)$
The “plus four” method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate $p′ = \frac{x+2}{n_4}$, and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples.
Formula Review
$p′ = \frac{x}{n}$ where $x$ represents the number of successes and $n$ represents the sample size. The variable $p'$ is the sample proportion and serves as the point estimate for the true population proportion.
$q′ = 1 – p′$
$p' - N\left(p, \sqrt{\frac{pq}{n}}\right)$ The variable $p′$ has a binomial distribution that can be approximated with the normal distribution shown here.
$EBP =$ the error bound for a proportion $= z_{\frac{\alpha}{2}}\sqrt{\frac{p'q'}{n}}$
Confidence interval for a proportion:
$(\text{lower bound, upper bound}) = (p′ – EBP, p′ + EBP) = \left(p' - z\sqrt{\frac{p'q'}{n}}\right), p' + z\sqrt{\frac{p'q'}{n}}$
$n = \frac{z_{\frac{\alpha}{2}}p'q'}{EBP^{2}}$ provides the number of participants needed to estimate the population proportion with confidence $1 - \alpha$ and margin of error $EBP$.
Use the normal distribution for a single population proportion $p' = \frac{x}{n}$
$EBP = \left(z_{\frac{\alpha}{2}}\right)\sqrt{\frac{p'q'}{n}}p' + q' = 1$
The confidence interval has the format $(p′ – EBP, p′ + EBP)$.
• $\bar{x}$ is a point estimate for $\mu$
• $p′$ is a point estimate for $\rho$
• $s$ is a point estimate for $\sigma$
Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions.
Exercise 8.4.6
When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05?
Exercise 8.4.7
If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why?
Answer
It would decrease, because the $z$-score would decrease, which reducing the numerator and lowering the number.
Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.
Exercise 8.4.8
Identify the following:
1. $x =$ ______
2. $n =$ ______
3. $p′ =$ ______
Exercise 8.4.9
Define the random variables $X$ and $P′$ in words.
Answer
$X$ is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. $P′$ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.
Exercise 8.4.10
Which distribution should you use for this problem?
Exercise 8.4.11
Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
$CI: (0.5321, 0.6679)$
$EBM: 0.0679$
Exercise 8.4.12
List two difficulties the company might have in obtaining random results, if this survey were done by email.
Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks.
Exercise 8.4.13
We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables $X$ and $P′$ in words.
Answer
$X$ is the number of “successes” where an executive prefers a truck. $P′$ is the percentage of executives sampled who prefer a truck.
Exercise 8.4.14
Which distribution should you use for this problem?
Exercise 8.4.15
Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
$CI: (0.19432, 0.33068)$
$EBM: 0.0707$
Exercise 8.4.16
Suppose we want to lower the sampling error. What is one way to accomplish that?
Exercise 8.4.17
The sampling error given in the survey is $\pm 2%$. Explain what the $\pm 2%$ means.
Answer
The sampling error means that the true mean can be 2% above or below the sample mean.
Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important.
Exercise 8.4.18
Define the random variable $X$ in words.
Exercise 8.4.19
Define the random variable $P′$ in words.
Answer
$P′$ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.
Exercise 8.4.20
Which distribution should you use for this problem?
Exercise 8.4.21
Construct a 90% confidence interval, and state the confidence interval and the error bound.
Answer
$CI: (0.62735, 0.67265)$
$EBM: 0.02265$
Exercise 8.4.22
What would happen to the confidence interval if the level of confidence were 95%?
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
Exercise 8.4.23
What is being counted?
Answer
The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class.
Exercise 8.4.24
In words, define the random variable $X$.
Exercise 8.4.25
Calculate the following:
1. $x =$ _______
2. $n =$ _______
3. $p′ =$ _______
Answer
1. $x = 64$
2. $n = 80$
3. $p′ = 0.8$
Exercise 8.4.26
State the estimated distribution of $X$. $X \sim$________
Exercise 8.4.27
Define a new random variable $P′$. What is $p′$ estimating?
Answer
$p$
Exercise 8.4.28
In words, define the random variable $P′$.
Exercise 8.4.29
State the estimated distribution of $P′$. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet.
Answer
$P\ - N\left(0.8, \sqrt{\frac{(0.8)(0.2)}{80}}\right)$. $(0.72171,0.87829)$.
Exercise 8.4.30
How much area is in both tails (combined)?
Exercise 8.4.31
How much area is in each tail?
Answer
0.04
Exercise 8.4.32
Calculate the following:
1. lower limit
2. upper limit
3. error bound
Exercise 8.4.33
The 92% confidence interval is _______.
Answer
(0.72; 0.88)
Exercise 8.4.34
Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion.
Exercise 8.4.35
In one complete sentence, explain what the interval means.
Answer
With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%.
Exercise 8.4.36
Using the same $p′$ and level of confidence, suppose that $n$ were increased to 100. Would the error bound become larger or smaller? How do you know?
Exercise 8.4.37
Using the same $p′$ and $n = 80$, how would the error bound change if the confidence level were increased to 98%? Why?
Answer
The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.
Exercise 8.4.38
If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/08%3A_Confidence_Intervals/8.S%3A_Confidence_Intervals_%28Summary%29.txt |
One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of \$60,000 per year.
09: Hypothesis Testing with One Sample
CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Differentiate between Type I and Type II Errors
• Describe hypothesis testing in general and in practice
• Conduct and interpret hypothesis tests for a single population mean, population standard deviation known.
• Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown.
• Conduct and interpret hypothesis tests for a single population proportion
One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of \$60,000 per year.
A statistician will make a decision about these claims. This process is called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analysis of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests.
Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will:
• Set up two contradictory hypotheses.
• Collect sample data (in homework problems, the data or summary statistics will be given to you).
• Determine the correct distribution to perform the hypothesis test.
• Analyze sample data by performing the calculations that ultimately will allow you to reject or decline to reject the null hypothesis.
• Make a decision and write a meaningful conclusion.
To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See Appendix E.
Glossary
Confidence Interval (CI)
an interval estimate for an unknown population parameter. This depends on:
• The desired confidence level.
• Information that is known about the distribution (for example, known standard deviation).
• The sample and its size.
Hypothesis Testing
Based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.01%3A_Prelude_to_Hypothesis_Testing.txt |
The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints.
$H_0$: The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.
$H_a$: The alternative hypothesis: It is a claim about the population that is contradictory to $H_0$ and what we conclude when we reject $H_0$. This is usually what the researcher is trying to prove.
Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.
After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are "reject $H_0$" if the sample information favors the alternative hypothesis or "do not reject $H_0$" or "decline to reject $H_0$" if the sample information is insufficient to reject the null hypothesis.
Table $1$: Mathematical Symbols Used in $H_{0}$ and $H_{a}$:
$H_{0}$ $H_{a}$
equal (=) not equal $(\neq)$ or greater than (>) or less than (<)
greater than or equal to $(\geq)$ less than (<)
less than or equal to $(\geq)$ more than (>)
$H_{0}$ always has a symbol with an equal in it. $H_{a}$ never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.
Example $1$
• $H_{0}$: No more than 30% of the registered voters in Santa Clara County voted in the primary election. $p \leq 30$
• $H_{a}$: More than 30% of the registered voters in Santa Clara County voted in the primary election. $p > 30$
Exercise $1$
A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.
Answer
• $H_{0}$: The drug reduces cholesterol by 25%. $p = 0.25$
• $H_{a}$: The drug does not reduce cholesterol by 25%. $p \neq 0.25$
Example $2$
We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:
• $H_{0}: \mu = 2.0$
• $H_{a}: \mu \neq 2.0$
Exercise $2$
We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol $(=, \neq, \geq, <, \leq, >)$ for the null and alternative hypotheses.
• $H_{0}: \mu \_ 66$
• $H_{a}: \mu \_ 66$
Answer
• $H_{0}: \mu = 66$
• $H_{a}: \mu \neq 66$
Example $3$
We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:
• $H_{0}: \mu \geq 5$
• $H_{a}: \mu < 5$
Exercise $3$
We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
1. $H_{0}: \mu \_ 45$
2. $H_{a}: \mu \_ 45$
Answer
1. $H_{0}: \mu \geq 45$
2. $H_{a}: \mu < 45$
Example $4$
In an issue of U. S. News and World Report, an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.
• $H_{0}: p \leq 0.066$
• $H_{a}: p > 0.066$
Exercise $4$
On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol ($=, \neq, \geq, <, \leq, >$) for the null and alternative hypotheses.
1. $H_{0}: p \_ 0.40$
2. $H_{a}: p \_ 0.40$
Answer
1. $H_{0}: p = 0.40$
2. $H_{a}: p > 0.40$
COLLABORATIVE EXERCISE
Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.
Review
In a hypothesis test, sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we:
1. Evaluate the null hypothesis, typically denoted with $H_{0}$. The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality $(=, \leq \text{or} \geq)$
2. Always write the alternative hypothesis, typically denoted with $H_{a}$ or $H_{1}$, using less than, greater than, or not equals symbols, i.e., $(\neq, >, \text{or} <)$.
3. If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis.
4. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.
Formula Review
$H_{0}$ and $H_{a}$ are contradictory.
If $H_{a}$ has: equal $(=)$ greater than or equal to $(\geq)$ less than or equal to $(\leq)$
then $H_{a}$ has: not equal $(\neq)$ or greater than $(>)$ or less than $(<)$ less than $(<)$ greater than $(>)$
• If $\alpha \leq p$-value, then do not reject $H_{0}$.
• If$\alpha > p$-value, then reject $H_{0}$.
$\alpha$ is preconceived. Its value is set before the hypothesis test starts. The $p$-value is calculated from the data.References
Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm.
Glossary
Hypothesis
a statement about the value of a population parameter, in case of two hypotheses, the statement assumed to be true is called the null hypothesis (notation $H_{0}$) and the contradictory statement is called the alternative hypothesis (notation $H_{a}$).
9.02: Null and Alternative Hypotheses
Exercise $5$
You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. What is the random variable? Describe in words.
Answer
The random variable is the mean Internet speed in Megabits per second.
Exercise $1$
You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. State the null and alternative hypotheses.
Exercise $1$
The American family has an average of two children. What is the random variable? Describe in words.
Answer
The random variable is the mean number of children an American family has.
Exercise $8$
The mean entry level salary of an employee at a company is \$58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses.
Exercise $9$
A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words.
Answer
The random variable is the proportion of people picked at random in Times Square visiting the city.
Exercise $10$
A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses.
Exercise $11$
In a population of fish, approximately 42% are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses.
Answer
1. $H_{0}: p = 0.42$
2. $H_{a}: p < 0.42$
Exercise $12$
Suppose that a recent article stated that the mean time spent in jail by a first–time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses be? The distribution of the population is normal.
1. $H_{0}$:_______
2. $H_{a}$:_______
Exercise $13$
A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time on death row could likely be 15 years, what would the null and alternative hypotheses be?
1. $H_{0}$:_________
2. $H_{a}$:_________
Answer
1. $H_{0}: \mu = 15$
2. $H_{a}: \mu \neq 15$
Exercise
The National Institute 9.2.14 of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population, what would the null and alternative hypotheses be?
1. $H_{0}$:_______
2. $H_{a}$:_______ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.02%3A_Null_and_Alternative_Hypotheses/9.2E%3A_Null_and_Alternative_Hypotheses_%28Exercises%29.txt |
When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis $H_{0}$ and the decision to reject or not. The outcomes are summarized in the following table:
ACTION $H_{0}$ is Actually True $H_{0}$ is Actually False
Do not reject $H_{0}$ Correct Outcome Type II error
Reject $H_{0}$ Type I Error Correct Outcome
The four possible outcomes in the table are:
1. The decision is not to reject $H_{0}$ when $H_{0}$ is true (correct decision).
2. The decision is to reject $H_{0}$ when $H_{0}$ is true (incorrect decision known as aType I error).
3. The decision is not to reject $H_{0}$ when, in fact, $H_{0}$ is false (incorrect decision known as a Type II error).
4. The decision is to reject $H_{0}$ when $H_{0}$ is false (correct decision whose probability is called the Power of the Test).
Each of the errors occurs with a particular probability. The Greek letters $\alpha$ and $\beta$ represent the probabilities.
• $\alpha =$ probability of a Type I error $= P(\text{Type I error}) =$ probability of rejecting the null hypothesis when the null hypothesis is true.
• $\beta =$ probability of a Type II error $= P(\text{Type II error}) =$ probability of not rejecting the null hypothesis when the null hypothesis is false.
$\alpha$ and $\beta$ should be as small as possible because they are probabilities of errors. They are rarely zero.
The Power of the Test is $1 - \beta$. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. The following are examples of Type I and Type II errors.
Example $1$: Type I vs. Type II errors
Suppose the null hypothesis, $H_{0}$, is: Frank's rock climbing equipment is safe.
• Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe.
• Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.
$\alpha =$ probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe.
$\beta =$ probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe.
Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)
Exercise $1$
Suppose the null hypothesis, $H_{0}$, is: the blood cultures contain no traces of pathogen $X$. State the Type I and Type II errors.
Answer
• Type I error: The researcher thinks the blood cultures do contain traces of pathogen $X$, when in fact, they do not.
• Type II error: The researcher thinks the blood cultures do not contain traces of pathogen $X$, when in fact, they do.
Example $2$
Suppose the null hypothesis, $H_{0}$, is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital.
• Type I error: The emergency crew thinks that the victim is dead when, in fact, the victim is alive.
• Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead.
$\alpha =$ probability that the emergency crew thinks the victim is dead when, in fact, he is really alive $= P(\text{Type I error})$.
$\beta =$ probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead $= P(\text{Type II error})$.
The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)
Exercise $2$
Suppose the null hypothesis, $H_{0}$, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II?
Answer
The error with the greater consequence is the Type II error: the patient will be thought well when, in fact, he is sick, so he will not get treatment.
Example $3$
It’s a Boy Genetic Labs claim to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, $H_{0}$, is: It’s a Boy Genetic Labs has no effect on gender outcome.
• Type I error: This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, $\alpha$.
• Type II error: This results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, $\beta$.
The error of greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy.
Exercise $3$
“Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence.
Answer
In this scenario, an appropriate null hypothesis would be $H_{0}$: the mean level of toxins is at most $800 \mu\text{g}$, $H_{0}: \mu_{0} \leq 800 \mu\text{g}$.
Type I error: The DMF believes that toxin levels are still too high when, in fact, toxin levels are at most $800 \mu\text{g}$. The DMF continues the harvesting ban.
Type II error: The DMF believes that toxin levels are within acceptable levels (are at least 800 μg) when, in fact, toxin levels are still too high (more than $800 \mu\text{g}$). The DMF lifts the harvesting ban. This error could be the most serious. If the ban is lifted and clams are still toxic, consumers could possibly eat tainted food.
In summary, the more dangerous error would be to commit a Type II error, because this error involves the availability of tainted clams for consumption.
Example $4$
A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious?
• Type I: A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%.
• Type II: A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%.
In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option.
Exercise $4$
Determine both Type I and Type II errors for the following scenario:
Assume a null hypothesis, $H_{0}$, that states the percentage of adults with jobs is at least 88%. Identify the Type I and Type II errors from these four statements.
1. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%
2. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%.
3. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%.
4. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%.
Answer
Type I error: c
Type I error: b
Summary
In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters $\alpha$ and $\beta$, for a Type I and a Type II error respectively. The power of the test, $1 - \beta$, quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable.
Formula Review
• $\alpha =$ probability of a Type I error $= P(\text{Type I error}) =$ probability of rejecting the null hypothesis when the null hypothesis is true.
• $\beta =$ probability of a Type II error $= P(\text{Type II error}) =$ probability of not rejecting the null hypothesis when the null hypothesis is false.
Glossary
Type 1 Error
The decision is to reject the null hypothesis when, in fact, the null hypothesis is true.
Type 2 Error
The decision is not to reject the null hypothesis when, in fact, the null hypothesis is false.
9.03: Outcomes and the Type I and Type II Errors
Exercise $5$
The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences. Answer Type I: The mean price of mid-sized cars is$32,000, but we conclude that it is not $32,000. Type II: The mean price of mid-sized cars is not$32,000, but we conclude that it is \$32,000.
Exercise $6$
A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences.
Exercise $7$
For Exercise 9.12, what are $\alpha$ and $\beta$ in words?
Answer
$\alpha =$ the probability that you think the bag cannot withstand -15 degrees F, when in fact it can
$\beta =$ the probability that you think the bag can withstand -15 degrees F, when in fact it cannot
Exercise $8$
In words, describe $1 - \beta$ For Exercise $\PageIndex{}$
Exercise $9$
A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, $H_{0}$, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences.
Answer
Type I: The procedure will go well, but the doctors think it will not.
Type II: The procedure will not go well, but the doctors think it will.
Exercise $10$
A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, $H_{0}$, is: the surgical procedure will go well. Which is the error with the greater consequence?
Exercise $11$
The power of a test is 0.981. What is the probability of a Type II error?
Answer
0.019
Exercise $12$
A group of divers is exploring an old sunken ship. Suppose the null hypothesis, $H_{0}$, is: the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences.
Exercise $13$
A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, $H_{0}$, is: the sample does not contain E-coli. The probability that the sample does not contain E-coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E-coli, but the microbiologist thinks it does not is 0.002. What is the power of this test?
Answer
0.998
Exercise $14$
A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, $H_{0}$, is: the sample contains E-coli. Which is the error with the greater consequence? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.03%3A_Outcomes_and_the_Type_I_and_Type_II_Errors/9.3E%3A_Outcomes_and_the_Type_I_and_Type_II_Errors_%28Exercises%29.txt |
Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's $t$-distribution. (Remember, use a Student's $t$-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually $n$ is large or the sample size is large).
If you are testing a single population mean, the distribution for the test is for means:
$\bar{X} - N\left(\mu_{x}, \frac{\sigma_{x}}{\sqrt{n}}\right)$
or
$t_{df}$
The population parameter is $\mu$. The estimated value (point estimate) for $\mu$ is $\bar{x}$, the sample mean.
If you are testing a single population proportion, the distribution for the test is for proportions or percentages:
$P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)$
The population parameter is $p$. The estimated value (point estimate) for $p$ is $p′$. $p' = \frac{x}{n}$ where $x$ is the number of successes and n is the sample size.
Assumptions
When you perform a hypothesis test of a single population mean $\mu$ using a Student's $t$-distribution (often called a $t$-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a $t$-test will work even if the population is not approximately normally distributed).
When you perform a hypothesis test of a single population mean $\mu$ using a normal distribution (often called a $z$-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.
When you perform a hypothesis test of a single population proportion $p$, you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are: there are a certain number $n$ of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success $p$. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities $np$ and $nq$ must both be greater than five $(np > 5$ and $nq > 5)$. Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with $\mu = p$ and $\sigma = \sqrt{\frac{pq}{n}}$. Remember that $q = 1 – p$.
Summary
In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.
When testing for a single population mean:
1. A Student's $t$-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation.
2. The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation.
When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of successes and the mean number of failures satisfy the conditions: $np > 5$ and $nq > 5$ where $n$ is the sample size, $p$ is the probability of a success, and $q$ is the probability of a failure.
Formula Review
If there is no given preconceived $\alpha$, then use $\alpha = 0.05$.
Types of Hypothesis Tests
• Single population mean, known population variance (or standard deviation): Normal test.
• Single population mean, unknown population variance (or standard deviation): Student's $t$-test.
• Single population proportion: Normal test.
• For a single population mean, we may use a normal distribution with the following mean and standard deviation. Means: $\mu = \mu_{\bar{x}}$ and $\sigma_{\bar{x}} = \frac{\sigma_{x}}{\sqrt{n}}$
• A single population proportion, we may use a normal distribution with the following mean and standard deviation. Proportions: $\mu = p$ and $\sigma = \sqrt{\frac{pq}{n}}$.
Glossary
Binomial Distribution
a discrete random variable (RV) that arises from Bernoulli trials. There are a fixed number, $n$, of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV Χ is defined as the number of successes in $n$ trials. The notation is: $X \sim B(n, p) \mu = np$ and the standard deviation is $\sigma = \sqrt{npq}$. The probability of exactly $x$ successes in $n$ trials is $P(X = x) = \binom{n}{x} p^{x}q^{n-x}$.
Normal Distribution
a continuous random variable (RV) with pdf $f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}$, where $\mu$ is the mean of the distribution, and $\sigma$ is the standard deviation, notation: $X \sim N(\mu, \sigma)$. If $\mu = 0$ and $\sigma = 1$, the RV is called the standard normal distribution.
Standard Deviation
a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: $s$ for sample standard deviation and $\sigma$ for population standard deviation.
Student's t-Distribution
investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student. The major characteristics of the random variable (RV) are:
• It is continuous and assumes any real values.
• The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution.
• It approaches the standard normal distribution as $n$ gets larger.
• There is a "family" of $t$-distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data items.
9.04: Distribution Needed for Hypothesis Testing
Exercise \(1\)
Which two distributions can you use for hypothesis testing for this chapter?
Answer
A normal distribution or a Student’s t-distribution
Exercise \(2\)
Which distribution do you use when you are testing a population mean and the standard deviation is known? Assume sample size is large.
Exercise \(3\)
Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large.
Answer
Use a Student’s \(t\)-distribution
Exercise \(4\)
A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal.
Exercise \(5\)
A population has a mean of 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test?
Answer
a normal distribution for a single population mean
Exercise \(6\)
It is thought that 42% of respondents in a taste test would prefer Brand A. In a particular test of 100 people, 39% preferred Brand A. What distribution should you use to perform a hypothesis test?
Exercise \(7\)
You are performing a hypothesis test of a single population mean using a Student’s \(t\)-distribution. What must you assume about the distribution of the data?
Answer
It must be approximately normally distributed.
Exercise \(8\)
You are performing a hypothesis test of a single population mean using a Student’s \(t\)-distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test?
Exercise \(9\)
You are performing a hypothesis test of a single population proportion. What must be true about the quantities of \(np\) and \(nq\)?
Answer
They must both be greater than five.
Exercise \(10\)
You are performing a hypothesis test of a single population proportion. You find out that \(np\) is less than five. What must you do to be able to perform a valid hypothesis test?
Exercise \(11\)
You are performing a hypothesis test of a single population proportion. The data come from which distribution?
Answer
binomial distribution | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.04%3A_Distribution_Needed_for_Hypothesis_Testing/9.4E%3A_Distribution_Needed_for_Hypothesis_Testing_%28Exercises%29.txt |
Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test.
Rare Events
Suppose you make an assumption about a property of the population (this assumption is the null hypothesis). Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur if the assumption is true, then you would conclude that your assumption about the population is probably incorrect. (Remember that your assumption is just an assumption—it is not a fact and it may or may not be true. But your sample data are real and the data are showing you a fact that seems to contradict your assumption.)
For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a$100 bill. The probability of this happening is $\frac{1}{200} = 0.005$. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A "rare event" has occurred (Didi getting the$100 bill) so Ali doubts the assumption about only one \$100 bill being in the basket.
Using the Sample to Test the Null Hypothesis
Use the sample data to calculate the actual probability of getting the test result, called the $p$-value. The $p$-value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample.
A large $p$-value calculated from the data indicates that we should not reject the null hypothesis. The smaller the $p$-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it.
Draw a graph that shows the $p$-value. The hypothesis test is easier to perform if you use a graph because you see the problem more clearly.
Example $1$
Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm. and the distribution of heights is normal.
• The null hypothesis could be $H_{0}: \mu \leq 15$
• The alternate hypothesis is $H_{a}: \mu > 15$
The words "is more than" translates as a "$>$" so "$\mu > 15$" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis.
Since $\sigma$ is known ($\sigma = 0.5 cm.$), the distribution for the population is known to be normal with mean $μ = 15$ and standard deviation
$\dfrac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{10}} = 0.16. \nonumber$
Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p-value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm.
The $p$-value, then, is the probability that a sample mean is the same or greater than 17 cm. when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means.
$p\text{-value} = P(\bar{x} > 17)$ which is approximately zero.
A $p$-value of approximately zero tells us that it is highly unlikely that a loaf of bread rises no more than 15 cm, on average. That is, almost 0% of all loaves of bread would be at least as high as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone), we conclude that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There is sufficient evidence that the true mean height for the population of the baker's loaves of bread is greater than 15 cm.
Exercise $1$
A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5.
• $H_{0}: \mu \leq 12$
• $H_{a}: \mu > 12$
The $p$-value is 0.0013
Draw a graph that shows the $p$-value.
Answer
$p\text{-value} = 0.0013$
Decision and Conclusion
A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the $p$-value and a preset or preconceived $\alpha$ (also called a "significance level"). A preset $\alpha$ is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem.
When you make a decision to reject or not reject $H_{0}$, do as follows:
• If $\alpha > p\text{-value}$, reject $H_{0}$. The results of the sample data are significant. There is sufficient evidence to conclude that $H_{0}$ is an incorrect belief and that the alternative hypothesis, $H_{a}$, may be correct.
• If $\alpha \leq p\text{-value}$, do not reject $H_{0}$. The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis,$H_{a}$, may be correct.
When you "do not reject $H_{0}$", it does not mean that you should believe that H0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of $H_{0}$.
Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem.
Example $2$
When using the $p$-value to evaluate a hypothesis test, it is sometimes useful to use the following memory device
• If the $p$-value is low, the null must go.
• If the $p$-value is high, the null must fly.
This memory aid relates a $p$-value less than the established alpha (the $p$ is low) as rejecting the null hypothesis and, likewise, relates a $p$-value higher than the established alpha (the $p$ is high) as not rejecting the null hypothesis.
Fill in the blanks.
Reject the null hypothesis when ______________________________________.
The results of the sample data _____________________________________.
Do not reject the null when hypothesis when __________________________________________.
The results of the sample data ____________________________________________.
Answer
Reject the null hypothesis when the $p$-value is less than the established alpha value. The results of the sample data support the alternative hypothesis.
Do not reject the null hypothesis when the $p$-value is greater than the established alpha value. The results of the sample data do not support the alternative hypothesis.
Exercise $2$
It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows:
• $H_{0}: p = 0.50, H_{a}: p > 0.50$
• $\alpha = 0.01$
• $p\text{-value} = 0.025$
Interpret the results and state a conclusion in simple, non-technical terms.
Answer
Since the $p$-value is greater than the established alpha value (the $p$-value is high), we do not reject the null hypothesis. There is not enough evidence to support It’s a Boy Genetics Labs' stated claim that their procedures improve the chances of a boy being born.
Review
When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind:
• $\alpha > p-value$, reject the null hypothesis
• $\alpha \leq p-value$, do not reject the null hypothesis
Glossary
Level of Significance of the Test
probability of a Type I error (reject the null hypothesis when it is true). Notation: $\alpha$. In hypothesis testing, the Level of Significance is called the preconceived $\alpha$ or the preset $\alpha$.
$p$-value
the probability that an event will happen purely by chance assuming the null hypothesis is true. The smaller the $p$-value, the stronger the evidence is against the null hypothesis.
9.05: Rare Events the Sample Decision and Conclusion
Exercise $3$
When do you reject the null hypothesis?
Exercise $4$
The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely?
Answer
The outcome of winning is very unlikely.
Exercise $5$
The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why?
Exercise $6$
It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 years. Do the data support the claim that the mean height is less than 73 inches? The $p$-value is almost zero. State the null and alternative hypotheses and interpret the $p$-value.
Answer
$H_{0}: \mu \geq 73$
$H_{a}: \mu \leq 73$
The $p$-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5% level. The data do support the claim.
Exercise $7$
The mean age of graduate students at a University is at most 31 years with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1% level? The $p$-value is 0.0264. State the null and alternative hypotheses and interpret the p-value.
Exercise $8$
Does the shaded region represent a low or a high $p$-value compared to a level of significance of 1%?
Answer
The shaded region shows a low $p$-value.
Exercise $9$
What should you do when $\alpha > p\text{-value}$?
Exercise $10$
What should you do if $\alpha = p\text{-value}$?
Answer
Do not reject $H_{0}$.
Exercise $11$
If you do not reject the null hypothesis, then it must be true. Is this statement correct? State why or why not in complete sentences.
Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.
Exercise $12$
Is this a test of means or proportions?
Answer
means
Exercise $13$
What symbol represents the random variable for this test?
Exercise $14$
In words, define the random variable for this test.
Answer
the mean time spent in jail for 26 first time convicted burglars
Exercise $15$
Is the population standard deviation known and, if so, what is it?
Exercise $16$
Calculate the following:
1. $\bar{x}$ _______
2. $\sigma$ _______
3. $s_{x}$ _______
4. $n$ _______
Answer
1. 3
2. 1.5
3. 1.8
4. 26
Exercise $17$
Since both $\sigma$ and $s_{x}$ are given, which should be used? In one to two complete sentences, explain why.
Exercise $18$
State the distribution to use for the hypothesis test.
Answer
$\bar{X} - N\left(2.5, \frac{1.5}{\sqrt{26}}\right)$
Exercise $19$
A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time on death row could likely be 15 years.
1. Is this a test of one mean or proportion?
2. State the null and alternative hypotheses.
$H{0}$: ____________________ $H_{a}$: ____________________
3. Is this a right-tailed, left-tailed, or two-tailed test?
4. What symbol represents the random variable for this test?
5. In words, define the random variable for this test.
6. Is the population standard deviation known and, if so, what is it?
7. Calculate the following:
1. $\bar{x}$ = _____________
2. $s$ = ____________
3. $n$ = ____________
8. Which test should be used?
9. State the distribution to use for the hypothesis test.
10. Find the $p$-value.
11. At a pre-conceived $\alpha = 0.05$, what is your:
1. Decision:
2. Reason for the decision:
3. Conclusion (write out in a complete sentence): | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.05%3A_Rare_Events_the_Sample_Decision_and_Conclusion/9.5E%3A_Rare_Events_the_Sample_Decision_and_Conclusion_%28Exercise.txt |
• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset $\alpha$.
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is $\alpha = 0.05$.
• When you calculate the $p$-value and draw the picture, the $p$-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, $H_{a}$, tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• $H_{a}$ never has a symbol that contains an equal sign.
• Thinking about the meaning of the $p$-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller $p$-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a $p$-value of 0.056 ($\alpha = 0.05$ is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.
The following examples illustrate a left-, right-, and two-tailed test.
Example $1$
$H_{0}: \mu = 5, H_{a}: \mu < 5$
Test of a single population mean. $H_{a}$ tells you the test is left-tailed. The picture of the $p$-value is as follows:
Exercise $1$
$H_{0}: \mu = 10, H_{a}: \mu < 10$
Assume the $p$-value is 0.0935. What type of test is this? Draw the picture of the $p$-value.
Answer
left-tailed test
Example $2$
$H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2$
This is a test of a single population proportion. $H_{a}$ tells you the test is right-tailed. The picture of the p-value is as follows:
Exercise $2$
$H_{0}: \mu \leq 1, H_{a}: \mu > 1$
Assume the $p$-value is 0.1243. What type of test is this? Draw the picture of the $p$-value.
Answer
right-tailed test
Example $3$
$H_{0}: \mu = 50, H_{a}: \mu \neq 50$
This is a test of a single population mean. $H_{a}$ tells you the test is two-tailed. The picture of the $p$-value is as follows.
Exercise $3$
$H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5$
Assume the p-value is 0.2564. What type of test is this? Draw the picture of the $p$-value.
Answer
two-tailed test
Full Hypothesis Test Examples
Example $4$
Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds.Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.
Answer
Set up the Hypothesis Test:
Since the problem is about a mean, this is a test of a single population mean.
$H_{0}: \mu = 16.43, H_{a}: \mu < 16.43$
For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "$<$" tells you this is left-tailed.
Determine the distribution needed:
Random variable: $\bar{X} =$ the mean time to swim the 25-yard freestyle.
Distribution for the test: $\bar{X}$ is normal (population standard deviation is known: $\sigma = 0.8$)
$\bar{X} - N \left(\mu, \frac{\sigma_{x}}{\sqrt{n}}\right)$ Therefore, $\bar{X} - N\left(16.43, \frac{0.8}{\sqrt{15}}\right)$
$\mu = 16.43$ comes from $H_{0}$ and not the data. $\sigma = 0.8$, and $n = 15$.
Calculate the $p-\text{value}$ using the normal distribution for a mean:
$p\text{-value} = P(\bar{x} < 16) = 0.0187$ where the sample mean in the problem is given as 16.
$p\text{-value} = 0.0187$ (This is called the actual level of significance.) The $p-\text{value}$ is the area to the left of the sample mean is given as 16.
Graph:
$\mu = 16.43$ comes from $H_{0}$. Our assumption is $\mu = 16.43$.
Interpretation of the $p-\text{value}$: If $H_{0}$ is true, there is a 0.0187 probability (1.87%) that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.
Compare $\alpha$ and the $p-\text{value}$:
$\alpha = 0.05 p\text{-value} = 0.0187 \alpha > p\text{-value}$
Make a decision: Since $\alpha > p\text{-value}$, reject $H_{0}$.
This means that you reject $\mu = 16.43$. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.
Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.
The p-value can easily be calculated.
Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Statsand press ENTER. Arrow down and enter 16.43 for $\mu_{0}$ (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow down to $\mu$ : (alternate hypothesis) and arrow over to $< \mu_{0}$. Press ENTER. Arrow down to Calculateand press ENTER. The calculator not only calculates the p-value ($p = 0.0187$) but it also calculates the test statistic (z-score) for the sample mean. $\mu < 16.43$ is the alternative hypothesis. Do this set of instructions again except arrow toDraw(instead of Calculate). Press ENTER. A shaded graph appears with $z = -2.08$ (test statistic) and $p = 0.0187$ ($p-\text{value}$). Make sure when you use Drawthat no other equations are highlighted in $Y =$ and the plots are turned off.
When the calculator does a $Z$-Test, the Z-Test function finds the p-value by doing a normal probability calculation using the central limit theorem:
$P(\bar{X} < 16)$ 2nd DISTR normcdf ($(−10^{99},16,16.43,\frac{0.8}{\sqrt{15}})$.
The Type I and Type II errors for this problem are as follows:
The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)
The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)
Exercise $4$
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset $\alpha = 0.05$. Assume the throw distances for footballs are normal.
First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.
Press STAT and arrow over to TESTS. Press 1: $Z$-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for $\mu_{0}$ (null hypothesis), 2 for $\sigma$, 45 for the sample mean, and 20 for $n$. Arrow down to $\mu$: (alternative hypothesis) and set it either as $<$, $\neq$, or $>$. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value but it also calculates the test statistic (z-score) for the sample mean. Select $<$, $\neq$, or $>$ for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and $p$-value. Make sure when you use Draw that no other equations are highlighted in $Y =$ and the plots are turned off.
Answer
Since the problem is about a mean, this is a test of a single population mean.
• $H_{0}: \mu = 40$
• $H_{a}: \mu > 40$
• $p = 0.0062$
Because $p < \alpha$, we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.
Historical Note
The traditional way to compare the two probabilities, $\alpha$ and the $p-\text{value}$, is to compare the critical value ($z$-score from $\alpha$) to the test statistic ($z$-score from data). The calculated test statistic for the $p$-value is –2.08. (From the Central Limit Theorem, the test statistic formula is $z = \frac{\bar{x}-\mu_{x}}{\left(\frac{\sigma_{x}}{\sqrt{n}}\right)}$. For this problem, $\bar{x} = 16$, $\mu_{x} = 16.43$ from the null hypotheses is, $\sigma_{x} = 0.8$, and $n = 15$.) You can find the critical value for $\alpha = 0.05$ in the normal table (see 15.Tables in the Table of Contents). The $z$-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The $z$-score is –1.645. Since –1.645 > –2.08 (which demonstrates that $\alpha > p-\text{value}$), reject $H_{0}$. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities $\alpha$ and the $p$-value is very common. For this problem, the $p-\text{value}$, 0.0187 is considerably smaller than $\alpha = 0.05$. You can be confident about your decision to reject. The graph shows $\alpha$, the $p-\text{value}$, and the test statistics and the critical value.
Example $5$
A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1).
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.
Answer
Set up the Hypothesis Test:
Since the problem is about a mean weight, this is a test of a single population mean.
• $H_{0}: \mu = 275$
• $H_{a}: \mu > 275$
This is a right-tailed test.
Calculating the distribution needed:
Random variable: $\bar{X} =$ the mean weight, in pounds, lifted by the football players.
Distribution for the test: It is normal because $\sigma$ is known.
• $\bar{X} - N\left(275, \frac{55}{\sqrt{30}}\right)$
• $\bar{x} = 286.2$ pounds (from the data).
• $\sigma = 55$ pounds (Always use $\sigma$ if you know it.) We assume $\mu = 275$ pounds unless our data shows us otherwise.
Calculate the p-value using the normal distribution for a mean and using the sample mean as input (see [link] for using the data as input):
$p\text{-value} = P(\bar{x} > 286.2) = 0.1323.\nonumber$
Interpretation of the p-value: If $H_{0}$ is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.
Compare $\alpha$ and the $p-\text{value}$:
$\alpha = 0.025 p-value = 0.1323$
Make a decision: Since $\alpha < p\text{-value}$, do not reject $H_{0}$.
Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.
The $p-\text{value}$ can easily be calculated.
Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for $\mu_{0}$, 55 for $\sigma$, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to $\mu$: and arrow over to $> \mu_{0}$. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the $p-\text{value}$ ($p = 0.1331$), a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. $\mu > 275$ is the alternative hypothesis. Do this set of instructions again except arrow toDraw (instead of Calculate). Press ENTER. A shaded graph appears with $z = 1.112$ (test statistic) and $p = 0.1331$ ($p-\text{value})$. Make sure when you use Drawthat no other equations are highlighted in $Y =$ and the plots are turned off.
Example $6$
Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.
Answer
Set up the hypothesis test:
A 5% level of significance means that $\alpha = 0.05$. This is a test of a single population mean.
$H_{0}: \mu = 65 H_{a}: \mu > 65$
Since the instructor thinks the average score is higher, use a "$>$". The "$>$" means the test is right-tailed.
Determine the distribution needed:
Random variable: $\bar{X} =$ average score on the first statistics test.
Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given $n = 10$ sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's $t$.
Use $t_{df}$. Therefore, the distribution for the test is $t_{9}$ where $n = 10$ and $df = 10 - 1 = 9$.
Calculate the $p$-value using the Student's $t$-distribution:
$p\text{-value} = P(\bar{x} > 67) = 0.0396$ where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.
Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.
Compare $\alpha$ and the $p-\text{value}$:
Since $α = 0.05$ and $p\text{-value} = 0.0396$. $\alpha > p\text{-value}$.
Make a decision: Since $\alpha > p\text{-value}$, reject $H_{0}$.
This means you reject $\mu = 65$. In other words, you believe the average test score is more than 65.
Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.
The $p\text{-value}$ can easily be calculated.
Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for $\mu_{0}$, the name of the list where you put the data, and 1 for Freq:. Arrow down to $\mu$: and arrow over to $> \mu_{0}$. Press ENTER. Arrow down to Calculate and pressENTER. The calculator not only calculates the $p\text{-value}$ (p = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. $\mu > 65$ is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with $t = 1.9781$ (test statistic) and $p = 0.0396$ ($p\text{-value}$). Make sure when you use Draw that no other equations are highlighted in $Y =$ and the plots are turned off.
Exercise $6$
It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of$1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4,$3, $2,$3, $1,$7, $2,$1, $1,$2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.
Answer
• $H_{0}: \mu = 5$
• $H_{a}: \mu < 5$
• $p = 0.0082$
Because $p < \alpha$, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week. • Type I Error: To conclude that the stock price is growing slower than$5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true). • Type II Error: To conclude that the stock price is growing at a rate of$5 a week when, in fact, the stock price is growing slower than \$5 a week (do not reject the null hypothesis when the null hypothesis is false).
Example $7$
Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.
Answer
Set up the hypothesis test:
The 1% level of significance means that α = 0.01. This is a test of a single population proportion.
$H_{0}: p = 0.50$ $H_{a}: p \neq 0.50$
The words "is the same or different from" tell you this is a two-tailed test.
Calculate the distribution needed:
Random variable: $P′ =$ the percent of of first-time brides who are younger than their grooms.
Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion.
$P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\nonumber$
Therefore,
$P' - N\left(0.5, \sqrt{\frac{0.5-0.5}{100}}\right)\nonumber$
where $p = 0.50, q = 1−p = 0.50$, and $n = 100$
Calculate the p-value using the normal distribution for proportions:
$p\text{-value} = P(p′ < 0.47 \space or \space p′ > 0.53) = 0.5485\nonumber$
where $x = 53, p' = \frac{x}{n} = \frac{53}{100} = 0.53\nonumber$.
Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion $p'$ is 0.53 or more OR 0.47 or less (see the graph in Figure).
$\mu = p = 0.50$ comes from $H_{0}$, the null hypothesis.
$p′ = 0.53$. Since the curve is symmetrical and the test is two-tailed, the $p′$ for the left tail is equal to $0.50 – 0.03 = 0.47$ where $\mu = p = 0.50$. (0.03 is the difference between 0.53 and 0.50.)
Compare $\alpha$ and the $p\text{-value}$:
Since $\alpha = 0.01$ and $p\text{-value} = 0.5485$. $\alpha < p\text{-value}$.
Make a decision: Since $\alpha < p\text{-value}$, you cannot reject $H_{0}$.
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.
The $p\text{-value}$ can easily be calculated.
Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter .5 for $p_{0}$, 53 for $x$ and 100 for $n$. Arrow down to Prop and arrow to not equals $p_{0}$. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the $p\text{-value}$ ($p = 0.5485$) and the test statistic ($z$-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with $z = 0.6$ (test statistic) and $p = 0.5485$ ($p\text{-value}$). Make sure when you useDraw that no other equations are highlighted in $Y =$ and the plots are turned off.
The Type I and Type II errors are as follows:
The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).
The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)
Exercise $7$
A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.
First, determine what type of test this is, set up the hypothesis test, find the $p\text{-value}$, sketch the graph, and state your conclusion.
Answer
Since the problem is about percentages, this is a test of single population proportions.
• $H_{0} : p = 0.85$
• $H_{a}: p \neq 0.85$
• $p = 0.7554$
Because $p > \alpha$, we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.
Example $8$
Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.
Answer
Set up the Hypothesis Test:
$H_{0}: p = 0.30, H_{a}: p \neq 0.30$
Determine the distribution needed:
The random variable is $P′ =$ proportion of households that have three cell phones.
The distribution for the hypothesis test is $P' - N\left(0.30, \sqrt{\frac{(0.30 \cdot 0.70)}{150}}\right)$
Exercise $8$.2
a. The value that helps determine the $p\text{-value}$ is $p′$. Calculate $p′$.
Answer
a. $p' = \frac{x}{n}$ where $x$ is the number of successes and $n$ is the total number in the sample.
$x = 43, n = 150$
$p′ = 43150$
Exercise $8$.3
b. What is a success for this problem?
Answer
b. A success is having three cell phones in a household.
Exercise $8$.4
c. What is the level of significance?
Answer
c. The level of significance is the preset $\alpha$. Since $\alpha$ is not given, assume that $\alpha = 0.05$.
Exercise $8$.5
d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
Calculate the $p\text{-value}$.
Answer
d. $p\text{-value} = 0.7216$
Exercise $8$.6
e. Make a decision. _____________(Reject/Do not reject) $H_{0}$ because____________.
Answer
e. Assuming that $\alpha = 0.05, \alpha < p\text{-value}$. The decision is do not reject $H_{0}$ because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.
Exercise $8$
Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p-value, state your conclusion, and identify the Type I and Type II errors.
Answer
• $H_{0}: p = 0.92$
• $H_{a}: p < 0.92$
• $p\text{-value} = 0.0046$
Because $p < 0.05$, we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.
• Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).
The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter $p$. The distribution for the test is normal. The estimated proportion $p′$ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived $\alpha = 0.01$, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!
Example $9$
My dog has so many fleas,
They do not come off with ease.
As for shampoo, I have tried many types
Even one called Bubble Hype,
Which only killed 25% of the fleas,
Unfortunately I was not pleased.
I've used all kinds of soap,
Until I had given up hope
Until one day I saw
An ad that put me in awe.
A shampoo used for dogs
Called GOOD ENOUGH to Clean a Hog
Guaranteed to kill more fleas.
I gave Fido a bath
And after doing the math
His number of fleas
Started dropping by 3's!
Before his shampoo
I counted 42.
At the end of his bath,
I redid the math
And the new shampoo had killed 17 fleas.
So now I was pleased.
Now it is time for you to have some fun
With the level of significance being .01,
You must help me figure out
Use the new shampoo or go without?
Answer
Set up the hypothesis test:
$H_{0}: p \leq 0.25$ $H_{a}: p > 0.25$
Determine the distribution needed:
In words, CLEARLY state what your random variable $\bar{X}$ or $P′$ represents.
$P′ =$ The proportion of fleas that are killed by the new shampoo
State the distribution to use for the test.
Normal:
$N\left(0.25, \sqrt{\frac{(0.25){1-0.25}}{42}}\right)\nonumber$
Test Statistic: $z = 2.3163$
Calculate the $p\text{-value}$ using the normal distribution for proportions:
$p\text{-value} = 0.0103\nonumber$
In one to two complete sentences, explain what the p-value means for this problem.
If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 $\left(\frac{17}{42}\right)$ or more.
Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the $p\text{-value}$.
Compare $\alpha$ and the $p\text{-value}$:
Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.
alpha decision reason for decision
0.01 Do not reject $H_{0}$ $\alpha < p\text{-value}$
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.
Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.
Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.
This test result is not very definitive since the $p\text{-value}$ is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.
Example $10$
The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.
1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95
Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.
Answer
Let’s follow a four-step process to answer this statistical question.
1. State the Question: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be
1. $H_{0}: \mu \leq 1$
2. $H_{a}: \mu > 1$
2. Plan: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
3. Do the calculations: We will input the sample data into the TI-83 as follows.
4. State the Conclusions: Since the $p\text{-value} (p = 0.036)$ is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.
Example $11$
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.
Answer
We will follow the four-step process.
1. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be
1. $H_{0}: p \leq 0.00034$
2. $H_{a}: p > 0.00034$
If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.
2. We will be testing a sample proportion with $x = 172$ and $n = 420,019$. The sample is sufficiently large because we have $np = 420,019(0.00034) = 142.8$, $nq = 420,019(0.99966) = 419,876.2$, two independent outcomes, and a fixed probability of success $p = 0.00034$. Thus we will be able to generalize our results to the population.
3. The associated TI results are
Figure $11$.
Figure $12$.
4. Since the $p\text{-value} = 0.0073$ is greater than our alpha value $= 0.005$, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.
Example $12$
According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.
Answer
We will follow the four-step plan.
1. We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
2. Since we are presented with proportions, we will use a one-proportion z-test. The hypotheses for the test will be
1. $H_{0}: p = 0.00078$
2. $H_{a}: p \neq 0.00078$
3. The following screen shots display the summary statistics from the hypothesis test.
Figure $13$.
Figure $14$.
4. Since the $p\text{-value}$, $p = 0.00063$, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.
Review
The hypothesis test itself has an established process. This can be summarized as follows:
1. Determine $H_{0}$ and $H_{a}$. Remember, they are contradictory.
2. Determine the random variable.
3. Determine the distribution for the test.
4. Draw a graph, calculate the test statistic, and use the test statistic to calculate the $p\text{-value}$. (A z-score and a t-score are examples of test statistics.)
5. Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion using English sentences.
Notice that in performing the hypothesis test, you use $\alpha$ and not $\beta$. $\beta$ is needed to help determine the sample size of the data that is used in calculating the $p\text{-value}$. Remember that the quantity $1 – \beta$ is called the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.
Exercise $8$
Assume $H_{0}: \mu = 9$ and $H_{a}: \mu < 9$. Is this a left-tailed, right-tailed, or two-tailed test?
Answer
This is a left-tailed test.
Exercise $9$
Assume $H_{0}: \mu \leq 6$ and $H_{a}: \mu > 6$. Is this a left-tailed, right-tailed, or two-tailed test?
Exercise $10$
Assume $H_{0}: p = 0.25$ and $H_{a}: p \neq 0.25$. Is this a left-tailed, right-tailed, or two-tailed test?
Answer
This is a two-tailed test.
Exercise $11$
Draw the general graph of a left-tailed test.
Exercise $12$
Draw the graph of a two-tailed test.
Answer
Exercise $13$
A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?
Exercise $14$
Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?
Answer
a right-tailed test
Exercise $15$
A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?
Exercise $16$
You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?
Answer
a left-tailed test
Exercise $17$
If the alternative hypothesis has a not equals ( $\neq$ ) symbol, you know to use which type of test?
Exercise $18$
Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?
Answer
This is a left-tailed test.
Exercise $19$
Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?
Exercise $20$
Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?
Answer
This is a two-tailed test.
Glossary
Central Limit Theorem
Given a random variable (RV) with known mean $\mu$ and known standard deviation $\sigma$. We are sampling with size $n$ and we are interested in two new RVs - the sample mean, $\bar{X}$, and the sample sum, $\sum X$. If the size $n$ of the sample is sufficiently large, then $\bar{X} - N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$ and $\sum X - N \left(n\mu, \sqrt{n}\sigma\right)$. If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal $n$ times the population mean. The standard deviation of the distribution of the sample means, $\frac{\sigma}{\sqrt{n}}$, is called the standard error of the mean.
9.07: Hypothesis Testing of a Single Mean and Single Proportion (Worksheet)
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will select the appropriate distributions to use in each case.
• The student will conduct hypothesis tests and interpret the results.
Television Survey
In a recent survey, it was stated that Americans watch television on average four hours per day. Assume that $\sigma = 2$. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower.
1. $H_{0}$: _____________
2. $H_{a}$: _____________
3. In words, define the random variable. __________ = ______________________
4. The distribution to use for the test is _______________________.
5. Determine the test statistic using your data.
6. Draw a graph and label it appropriately.Shade the actual level of significance.
1. Graph:
2. Determine the $p\text{-value}$.
7. Do you or do you not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Language Survey
About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3%.
1. $H_{0}$: ___________
2. $H_{a}$: ___________
3. In words, define the random variable. __________ = _______________
4. The distribution to use for the test is ________________
5. Determine the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
2. Determine the $p\text{-value}$.
7. Do you or do you not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Jeans Survey
Suppose that young adults own an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal.
1. $H_{0}$: _____________
2. $H_{a}$: _____________
3. In words, define the random variable. __________ = ______________________
4. The distribution to use for the test is _______________________.
5. Determine the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
2. Determine the $p\text{-value}$.
7. Do you or do you not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.06%3A_Additional_Information_and_Full_Hypothesis_Test_Examples.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
9.2: Null and Alternative Hypotheses
Q 9.2.1
Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.
State the null hypothesis, $H_{0}$, and the alternative hypothesis. $H_{a}$, in terms of the appropriate parameter $(\mu \text{or} p)$.
1. The mean number of years Americans work before retiring is 34.
2. At most 60% of Americans vote in presidential elections.
3. The mean starting salary for San Jose State University graduates is at least $100,000 per year. 4. Twenty-nine percent of high school seniors get drunk each month. 5. Fewer than 5% of adults ride the bus to work in Los Angeles. 6. The mean number of cars a person owns in her lifetime is not more than ten. 7. About half of Americans prefer to live away from cities, given the choice. 8. Europeans have a mean paid vacation each year of six weeks. 9. The chance of developing breast cancer is under 11% for women. 10. Private universities' mean tuition cost is more than$20,000 per year.
S 9.2.1
1. $H_{0}: \mu = 34; H_{a}: \mu \neq 34$
2. $H_{0}: p \leq 0.60; H_{a}: p > 0.60$
3. $H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000$
4. $H_{0}: p = 0.29; H_{a}: p \neq 0.29$
5. $H_{0}: p = 0.05; H_{a}: p < 0.05$
6. $H_{0}: \mu \leq 10; H_{a}: \mu > 10$
7. $H_{0}: p = 0.50; H_{a}: p \neq 0.50$
8. $H_{0}: \mu = 6; H_{a}: \mu \neq 6$
9. $H_{0}: p ≥ 0.11; H_{a}: p < 0.11$
10. $H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000$
Q 9.2.2
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:
1. $p < 0.30$
2. $p \leq 0.30$
3. $p \geq 0.30$
4. $p > 0.30$
Q 9.2.3
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:
1. $p = 0.20$
2. $p > 0.20$
3. $p < 0.20$
4. $p \leq 0.20$
c
Q 9.2.4
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:
1. $H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5$
2. $H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5$
3. $H_{0}: \mu = 4.75, H_{a}: \mu > 4.75$
4. $H_{0}: \mu = 4.5, H_{a}: \mu > 4.5$
9.3: Outcomes and the Type I and Type II Errors
Q 9.3.1
State the Type I and Type II errors in complete sentences given the following statements.
1. The mean number of years Americans work before retiring is 34.
2. At most 60% of Americans vote in presidential elections.
3. The mean starting salary for San Jose State University graduates is at least $100,000 per year. 4. Twenty-nine percent of high school seniors get drunk each month. 5. Fewer than 5% of adults ride the bus to work in Los Angeles. 6. The mean number of cars a person owns in his or her lifetime is not more than ten. 7. About half of Americans prefer to live away from cities, given the choice. 8. Europeans have a mean paid vacation each year of six weeks. 9. The chance of developing breast cancer is under 11% for women. 10. Private universities mean tuition cost is more than$20,000 per year.
S 9.3.1
1. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
2. Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
3. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least$100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than$100,000.
4. Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
5. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
6. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
7. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
8. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
9. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
10. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most$20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than$20,000.
Q 9.3.2
For statements a-j in Exercise 9.109, answer the following in complete sentences.
1. State a consequence of committing a Type I error.
2. State a consequence of committing a Type II error.
Q 9.3.3
When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?
1. To conclude the drug is safe when in, fact, it is unsafe.
2. Not to conclude the drug is safe when, in fact, it is safe.
3. To conclude the drug is safe when, in fact, it is safe.
4. Not to conclude the drug is unsafe when, in fact, it is unsafe.
b
Q 9.3.4
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.
1. at least 20%, when in fact, it is less than 20%.
2. 20%, when in fact, it is 20%.
3. less than 20%, when in fact, it is at least 20%.
4. less than 20%, when in fact, it is less than 20%.
Q 9.3.4
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?
The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours
1. is more than seven hours.
2. is at most seven hours.
3. is at least seven hours.
4. is less than seven hours.
d
Q 9.3.5
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:
1. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
2. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
3. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
4. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher
9.4: Distribution Needed for Hypothesis Testing
Q 9.4.1
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is $\bar{X} \sim$ ________________
1. $N\left(7.24, \frac{1.93}{\sqrt{22}}\right)$
2. $N\left(7.24, 1.93\right)$
3. $t_{22}$
4. $t_{21}$
d
9.5: Rare Events, the Sample, Decision and Conclusion
Q 9.5.1
The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.
1. Is this a test of one mean or proportion?
2. State the null and alternative hypotheses.
$H_{0}$ : ____________________ $H_{a}$ : ____________________
3. Is this a right-tailed, left-tailed, or two-tailed test?
4. What symbol represents the random variable for this test?
5. In words, define the random variable for this test.
6. Calculate the following:
1. $x =$ ________________
2. $n =$ ________________
3. $p′ =$ _____________
7. Calculate $\sigma_{x} =$ __________. Show the formula set-up.
8. State the distribution to use for the hypothesis test.
9. Find the $p\text{-value}$.
10. At a pre-conceived $\alpha = 0.05$, what is your:
1. Decision:
2. Reason for the decision:
3. Conclusion (write out in a complete sentence):
9.6: Additional Information and Full Hypothesis Test Examples
For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link]. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
Note
If you are using a Student's $t$-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)
Q 9.6.1.
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using $\alpha = 0.05$, is the data highly inconsistent with the claim?
S 9.6.1
1. $H_{0}: \mu \geq 50,000$
2. $H_{a}: \mu < 50,000$
3. Let $\bar{X} =$ the average lifespan of a brand of tires.
4. normal distribution
5. $z = -2.315$
6. $p\text{-value} = 0.0103$
7. Check student’s solution.
1. alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The $p\text{-value}$ is less than 0.05.
4. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
8. $(43,537, 49,463)$
Q 9.6.2
From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?
The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level? S 9.6.3 1. $H_{0}: \mu = 1.00$ 2. $H_{a}: \mu \neq 1.00$ 3. Let $\bar{X} =$ the average cost of a daily newspaper. 4. normal distribution 5. $z = –0.866$ 6. $p\text{-value} = 0.3865$ 7. Check student’s solution. 1. $\alpha: 0.01$ 2. Decision: Do not reject the null hypothesis. 3. Reason for decision: The $p\text{-value}$ is greater than 0.01. 4. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is$1. The mean cost could be $1. 8. $(0.84, 1.06)$ Q 9.6.4 An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? Q 9.6.5 The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let $x =$ the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten? S 9.6.5 1. $H_{0}: \mu = 10$ 2. $H_{a}: \mu \neq 10$ 3. Let $\bar{X}$ the mean number of sick days an employee takes per year. 4. Student’s t-distribution 5. $t = –1.12$ 6. $p\text{-value} = 0.300$ 7. Check student’s solution. 1. $\alpha: 0.05$ 2. Decision: Do not reject the null hypothesis. 3. Reason for decision: The $p\text{-value}$ is greater than 0.05. 4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten. 8. $(4.9443, 11.806)$ Q 9.6.6 In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level? Q 9.6.7 Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think? S 9.6.7 1. $H_{0}: p \geq 0.6$ 2. $H_{a}: p < 0.6$ 3. Let $P′ =$ the proportion of students who feel more enriched as a result of taking Elementary Statistics. 4. normal for a single proportion 5. 1.12 6. $p\text{-value} = 0.1308$ 7. Check student’s solution. 1. $\alpha: 0.05$ 2. Decision: Do not reject the null hypothesis. 3. Reason for decision: The $p\text{-value}$ is greater than 0.05. 4. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched. 8. Confidence Interval: $(0.409, 0.654)$ The “plus-4s” confidence interval is $(0.411, 0.648)$ Q 9.6.8 A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief. Q 9.6.9 Refer to Exercise 9.119. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four. S 9.6.9 1. $H_{0}: \mu = 4$ 2. $H_{a}: \mu \neq 4$ 3. Let $\bar{X}$ the average I.Q. of a set of brown trout. 4. two-tailed Student's t-test 5. $t = 1.95$ 6. $p\text{-value} = 0.076$ 7. Check student’s solution. 1. $\alpha: 0.05$ 2. Decision: Reject the null hypothesis. 3. Reason for decision: The $p\text{-value}$ is greater than 0.05 4. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. 8. $(3.8865,5.9468)$ Q 9.6.10 According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7? Q 9.6.11 A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results. S 9.6.11 1. $H_{0}: p \geq 0.13$ 2. $H_{a}: p < 0.13$ 3. Let $P′ =$ the proportion of Americans who have seen or sensed angels 4. normal for a single proportion 5. –2.688 6. $p\text{-value} = 0.0036$ 7. Check student’s solution. 1. alpha: 0.05 2. Decision: Reject the null hypothesis. 3. Reason for decision: The $p\text{-value}$e is less than 0.05. 4. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%. 8. $(0, 0.0623)$. The“plus-4s” confidence interval is (0.0022, 0.0978) Q 9.6.12 The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55. Q 9.6.13 Use the “Lap time” data for Lap 4 (see [link]) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given. S 9.6.13 1. $H_{0}: \mu \geq 129$ 2. $H_{a}: \mu < 129$ 3. Let $\bar{X} =$ the average time in seconds that Terri finishes Lap 4. 4. Student's t-distribution 5. $t = 1.209$ 6. 0.8792 7. Check student’s solution. 1. $\alpha: 0.05$ 2. Decision: Do not reject the null hypothesis. 3. Reason for decision: The $p\text{-value}$ is greater than 0.05. 4. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds. 8. $(128.63, 130.37)$ Q 9.6.14 Use the “Initial Public Offering” data (see [link]) to test the claim that the mean offer price was$18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.
Note
The following questions were written by past students. They are excellent problems!
Q 9.6.15
"Asian Family Reunion," by Chau Nguyen
Every two years it comes around.
We all get together from different towns.
In my honest opinion,
It's not a typical family reunion.
Not forty, or fifty, or sixty,
But how about seventy companions!
The kids would play, scream, and shout
One minute they're happy, another they'll pout.
The teenagers would look, stare, and compare
From how they look to what they wear.
The men would chat about their business
That they make more, but never less.
Money is always their subject
And there's always talk of more new projects.
The women get tired from all of the chats
They head to the kitchen to set out the mats.
Some would sit and some would stand
Eating and talking with plates in their hands.
Then come the games and the songs
And suddenly, everyone gets along!
With all that laughter, it's sad to say
That it always ends in the same old way.
They hug and kiss and say "good-bye"
And then they all begin to cry!
I say that 60 percent shed their tears
But my mom counted 35 people this year.
She said that boys and men will always have their pride,
So we won't ever see them cry.
I myself don't think she's correct,
So could you please try this problem to see if you object?
S 9.6.15
1. $H_{0}: p = 0.60$
2. $H_{a}: p < 0.60$
3. Let $P′ =$ the proportion of family members who shed tears at a reunion.
4. normal for a single proportion
5. –1.71
6. 0.0438
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for decision: $p\text{-value} < \alpha$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the $p\text{-value}$ and alpha are quite close, so other tests should be done.
8. We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. $(0.3829, 0.6171)$. The“plus-4s” confidence interval (see chapter 8) is $(0.3861, 0.6139)$
Note that here the “large-sample” $1 - \text{PropZTest}$ provides the approximate $p\text{-value}$ of 0.0438. Whenever a $p\text{-value}$ based on a normal approximation is close to the level of significance, the exact $p\text{-value}$ based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.
Q 9.6.16
"The Problem with Angels," by Cyndy Dowling
Although this problem is wholly mine,
The catalyst came from the magazine, Time.
On the magazine cover I did find
The realm of angels tickling my mind.
Inside, 69% I found to be
In angels, Americans do believe.
Then, it was time to rise to the task,
Ninety-five high school and college students I did ask.
Viewing all as one group,
Random sampling to get the scoop.
So, I asked each to be true,
"Do you believe in angels?" Tell me, do!
Hypothesizing at the start,
Totally believing in my heart
That the proportion who said yes
Would be equal on this test.
Lo and behold, seventy-three did arrive,
Out of the sample of ninety-five.
Now your job has just begun,
Solve this problem and have some fun.
Q 9.6.17
"Blowing Bubbles," by Sondra Prull
Studying stats just made me tense,
I had to find some sane defense.
Some light and lifting simple play
To float my math anxiety away.
Blowing bubbles lifts me high
Takes my troubles to the sky.
POIK! They're gone, with all my stress
Bubble therapy is the best.
The label said each time I blew
The average number of bubbles would be at least 22.
I blew and blew and this I found
From 64 blows, they all are round!
But the number of bubbles in 64 blows
Varied widely, this I know.
20 per blow became the mean
They deviated by 6, and not 16.
From counting bubbles, I sure did relax
But now I give to you your task.
Was 22 a reasonable guess?
Find the answer and pass this test!
S 9.6.17
1. $H_{0}: \mu \geq 22$
2. $H_{a}: \mu < 22$
3. Let $\bar{X} =$ the mean number of bubbles per blow.
4. Student's t-distribution
5. –2.667
6. $p\text{-value} = 0.00486$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for decision: The $p\text{-value}$ is less than 0.05.
4. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
8. $(18.501, 21.499)$
Q 9.6.18
"Dalmatian Darnation," by Kathy Sparling
A greedy dog breeder named Spreckles
Bred puppies with numerous freckles
The Dalmatians he sought
Possessed spot upon spot
The more spots, he thought, the more shekels.
His competitors did not agree
That freckles would increase the fee.
They said, “Spots are quite nice
But they don't affect price;
One should breed for improved pedigree.”
The breeders decided to prove
This strategy was a wrong move.
Breeding only for spots
Would wreak havoc, they thought.
His theory they want to disprove.
They proposed a contest to Spreckles
Comparing dog prices to freckles.
In records they looked up
One hundred one pups:
Dalmatians that fetched the most shekels.
They asked Mr. Spreckles to name
An average spot count he'd claim
To bring in big bucks.
Said Spreckles, “Well, shucks,
It's for one hundred one that I aim.”
Said an amateur statistician
Who wanted to help with this mission.
“Twenty-one for the sample
Standard deviation's ample:
They examined one hundred and one
Dalmatians that fetched a good sum.
They counted each spot,
Mark, freckle and dot
And tallied up every one.
Instead of one hundred one spots
They averaged ninety six dots
Can they muzzle Spreckles’
Obsession with freckles
Based on all the dog data they've got?
Q 9.6.19
"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall
As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.
One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most$1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:
Price per box of Mac and Cheese:
• 5 stores @ $2.02 • 15 stores @$0.25
• 3 stores @ $1.29 • 6 stores @$0.35
• 4 stores @ $2.27 • 7 stores @$1.50
• 5 stores @ $1.89 • 8 stores @ 0.75. I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most$1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)
S 9.6.19
1. $H_{0}: \mu \leq 1$
2. $H_{a}: \mu > 1$
3. Let $\bar{X} =$ the mean cost in dollars of macaroni and cheese in a certain town.
4. Student's $t$-distribution
5. $t = 0.340$
6. $p\text{-value} = 0.36756$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The $p\text{-value}$ is greater than 0.05
4. Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than$1.
8. $(0.8291, 1.241)$
Q 9.6.20
"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi
THE CHARACTERS (in order of appearance):
• HAMLET, Prince of Denmark and student of Statistics
• POLONIUS, Hamlet’s tutor
• HOROTIO, friend to Hamlet and fellow student
Scene: The great library of the castle, in which Hamlet does his lessons
Act I
(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)
POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!
HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.
POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.
HORATIO (to Hamlet): What should we do, my Lord?
HAMLET: Go to thy purpose, Horatio.
HORATIO: To what end, my Lord?
HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.
(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)
Act II
(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)
POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?
HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.
POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)
HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)
(Curtain falls)
Q 9.6.21
"Untitled," by Stephen Chen
I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.
So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?
S 9.6.21
1. $H_{0}: p = 0.01$
2. $H_{a}: p > 0.01$
3. Let $P′ =$ the proportion of errors generated
4. Normal for a single proportion
5. 2.13
6. 0.0165
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis
3. Reason for decision: The $p\text{-value}$ is less than 0.05.
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.
8. Confidence interval: $(0, 0.094)$.
The“plus-4s” confidence interval is $(0.004, 0.144)$.
Q 9.6.22
"Japanese Girls’ Names"
by Kumi Furuichi
It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.
However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.
I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.
Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.
Q 9.6.23
"Phillip’s Wish," by Suzanne Osorio
My nephew likes to play
Chasing the girls makes his day.
He asked his mother
If it is okay
To get his ear pierced.
She said, “No way!”
To poke a hole through your ear,
Is not what I want for you, dear.
He argued his point quite well,
Says even my macho pal, Mel,
Has gotten this done.
It’s all just for fun.
C’mon please, mom, please, what the hell.
Again Phillip complained to his mother,
Saying half his friends (including their brothers)
Are piercing their ears
And they have no fears
He wants to be like the others.
She said, “I think it’s much less.
We must do a hypothesis test.
And if you are right,
I won’t put up a fight.
But, if not, then my case will rest.”
We proceeded to call fifty guys
To see whose prediction would fly.
Nineteen of the fifty
Said piercing was nifty
And earrings they’d occasionally buy.
Then there’s the other thirty-one,
Who said they’d never have this done.
So now this poem’s finished.
Will his hopes be diminished,
Or will my nephew have his fun?
S 9.6.23
1. $H_{0}: p = 0.50$
2. $H_{a}: p < 0.50$
3. Let $P′ =$ the proportion of friends that has a pierced ear.
4. normal for a single proportion
5. –1.70
6. $p\text{-value} = 0.0448$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis
3. Reason for decision: The $p\text{-value}$ is less than 0.05. (However, they are very close.)
4. Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
8. Confidence Interval: $(0.245, 0.515)$: The “plus-4s” confidence interval is $(0.259, 0.519)$.
Q 9.6.24
"The Craven," by Mark Salangsang
Once upon a morning dreary
In stats class I was weak and weary.
Pondering over last night’s homework
Whose answers were now on the board
This I did and nothing more.
While I nodded nearly napping
Suddenly, there came a tapping.
As someone gently rapping,
Rapping my head as I snore.
Quoth the teacher, “Sleep no more.”
“In every class you fall asleep,”
The teacher said, his voice was deep.
“So a tally I’ve begun to keep
Of every class you nap and snore.
The percentage being forty-four.”
“My dear teacher I must confess,
While sleeping is what I do best.
The percentage, I think, must be less,
A percentage less than forty-four.”
This I said and nothing more.
“We’ll see,” he said and walked away,
And fifty classes from that day
He counted till the month of May
The classes in which I napped and snored.
The number he found was twenty-four.
At a significance level of 0.05,
Please tell me am I still alive?
Or did my grade just take a dive
Plunging down beneath the floor?
Upon thee I hereby implore.
Q 9.6.25
Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.
S 9.6.25
1. $H_{0}: p = 0.40$
2. $H_{a}: p < 0.40$
3. Let $P′ =$ the proportion of schoolmates who fear public speaking.
4. normal for a single proportion
5. –1.01
6. $p\text{-value} = 0.1563$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The $p\text{-value}$ is greater than 0.05.
4. Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
8. Confidence Interval: $(0.3241, 0.4240)$: The “plus-4s” confidence interval is $(0.3257, 0.4250)$.
Q 9.6.26
Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.
Q 9.6.27
According to an article in Bloomberg Businessweek, New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.
S 9.6.27
1. $H_{0}: p = 0.14$
2. $H_{a}: p < 0.14$
3. Let $P′ =$ the proportion of NYC residents that smoke.
4. normal for a single proportion
5. –0.2756
6. $p\text{-value} = 0.3914$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The $p\text{-value}$ is greater than 0.05.
4. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
8. Confidence Interval: $(0.0502, 0.2070)$: The “plus-4s” confidence interval (see chapter 8) is $(0.0676, 0.2297)$.
Q 9.6.28
The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.
Q 9.6.29
Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than$69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of$7,489. Conduct a hypothesis test.
S 9.6.29
1. $H_{0}: \mu = 69,110$
2. $H_{0}: \mu > 69,110$
3. Let $\bar{X} =$ the mean salary in dollars for California registered nurses.
4. Student's t-distribution
5. $t = 1.719$
6. $p\text{-value}: 0.0466$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for decision: The $p\text{-value}$ is less than 0.05.
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds \$69,110.
8. $(68,757, 73,485)$
Q 9.6.30
La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.
Q 9.6.31
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?
After conducting the test, your decision and conclusion are
1. Reject $H_{0}$: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
2. Do not reject $H_{0}$: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
3. Do not reject $H_{0}$: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
4. Reject $H_{0}$: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
c
Q 9.6.32
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.
At a 1% level of significance, an appropriate conclusion is:
1. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
2. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
3. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
4. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.
Q 9.6.33
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
At a significance level of $a = 0.05$, what is the correct conclusion?
1. There is enough evidence to conclude that the mean number of hours is more than 4.75
2. There is enough evidence to conclude that the mean number of hours is more than 4.5
3. There is not enough evidence to conclude that the mean number of hours is more than 4.5
4. There is not enough evidence to conclude that the mean number of hours is more than 4.75
S 9.6.33
c
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten exercises, answer each question.
State the null and alternate hypothesis.
State the $p\text{-value}$.
State $\alpha$.
What is your decision?
Write a conclusion.
Answer any other questions asked in the problem.
Q 9.6.34
According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.
Q 9.6.35
A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?
S 9.6.35
1. $H_{0}: p = 0.488$ $H_{a}: p \neq 0.488$
2. $p\text{-value} = 0.0114$
3. $\alpha = 0.05$
4. Reject the null hypothesis.
5. At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
6. The survey does not appear to be accurate.
Q 9.6.36
Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using $\alpha = 0.05$, is the AAA proportion accurate?
Q 9.6.37
The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the $\alpha = 0.05$ level in Kentucky? Are the results applicable across the country? Why?
S 9.6.37
1. $H_{0}: p = 0.517$ $H_{0}: p \neq 0.517$
2. $p\text{-value} = 0.9203$.
3. $\alpha = 0.05$.
4. Do not reject the null hypothesis.
5. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
6. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.
Q 9.6.38
For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use $\alpha = 0.01$ level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?
Q 9.6.39
The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the $\alpha = 0.05 level$, can it be concluded that the mean rainfall was below the reported average? What if $\alpha = 0.01$? Assume the amount of summer rainfall follows a normal distribution.
S 9.6.39
1. $H_{0}: \mu \geq 11.52$ $H_{a}: \mu < 11.52$
2. $p\text{-value} = 0.000002$ which is almost 0.
3. $\alpha = 0.05$.
4. Reject the null hypothesis.
5. At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
6. We would make the same conclusion if alpha was 1% because the $p\text{-value}$ is almost 0.
Q 9.6.40
A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the $\alpha = 0.10$ level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?
Q 9.6.41
A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals
3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1
At the $\alpha = 0.05$ level can it be concluded that the sample mean is higher than 5.8 visits per year?
S 9.6.42
1. $H_{0}: \mu \leq 5.8$ $H_{a}: \mu > 5.8$
2. $p\text{-value} = 0.9987$
3. $\alpha = 0.05$
4. Do not reject the null hypothesis.
5. At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.
Q 9.6.42
According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:
5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2
At $\alpha = 0.05$ level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?
Q 9.6.43
The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?
S 9.6.43
1. $H_{0}: \mu \geq 150$ $H_{0}: \mu < 150$
2. $p\text{-value} = 0.0622$
3. $\alpha = 0.01$
4. Do not reject the null hypothesis.
5. At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
6. The student academic group’s claim appears to be correct. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.E%3A_Hypothesis_Testing_with_One_Sample_%28Exercises%29.txt |
You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is still the same, just expanded. To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions.
• 10.1: Prelude to Hypothesis Testing with Two Samples
This chapter deals with the following hypothesis tests: Independent groups (samples are independent) Test of two population means. Test of two population proportions. Matched or paired samples (samples are dependent) Test of the two population proportions by testing one population mean of differences.
• 10.2: Two Population Means with Unknown Standard Deviations
The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples.
• 10.3: Two Population Means with Known Standard Deviations
Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations.
• 10.4: Comparing Two Independent Population Proportions
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.
• 10.5: Matched or Paired Samples
When using a hypothesis test for matched or paired samples, the following characteristics should be present: Simple random sampling is used. Sample sizes are often small. Two measurements (samples) are drawn from the same pair of individuals or objects. Differences are calculated from the matched or paired samples. The differences form the sample that is used for the hypothesis test. Either the matched pairs have differences that come from a population that is normal or the number of difference
• 10.6: Hypothesis Testing for Two Means and Two Proportions (Worksheet)
A statistics Worksheet: The student will select the appropriate distributions to use in each case. The student will conduct hypothesis tests and interpret the results.
• 10.E: Hypothesis Testing with Two Samples (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
10: Hypothesis Testing with Two Samples
Learning Objectives
By the end of this chapter, the student should be able to:
• Classify hypothesis tests by type.
• Conduct and interpret hypothesis tests for two population means, population standard deviations known.
• Conduct and interpret hypothesis tests for two population means, population standard deviations unknown.
• Conduct and interpret hypothesis tests for two population proportions.
• Conduct and interpret hypothesis tests for matched or paired samples.
Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years.
There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores.
You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is still the same, just expanded.
To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions.
This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p-values. TI-83+ and TI-84 instructions are included as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests.
This chapter deals with the following hypothesis tests:
Independent groups (samples are independent)
• Test of two population means.
• Test of two population proportions.
Matched or paired samples (samples are dependent)
• Test of the two population proportions by testing one population mean of differences. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.01%3A_Prelude_to_Hypothesis_Testing_with_Two_Samples.txt |
1. The two independent samples are simple random samples from two distinct populations.
2. For the two distinct populations:
• if the sample sizes are small, the distributions are important (should be normal)
• if the sample sizes are large, the distributions are not important (need not be normal)
The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch $t$-test. The degrees of freedom formula was developed by Aspin-Welch.
The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, $\bar{X}_{1} - \bar{X}_{2}$, and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.
Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, $\bar{X}_{1} - \bar{X}_{2}$.
The standard error is:
$\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}$
The test statistic (t-score) is calculated as follows:
$\dfrac{(\bar{x}-\bar{x}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}$
where:
• $s_{1}$ and $s_{2}$, the sample standard deviations, are estimates of $\sigma_{1}$ and $\sigma_{1}$, respectively.
• $\sigma_{1}$ and $\sigma_{2}$ are the unknown population standard deviations.
• $\bar{x}_{1}$ and $\bar{x}_{2}$ are the sample means. $\mu_{1}$ and $\mu_{2}$ are the population means.
The number of degrees of freedom ($df$) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The $df$ are not always a whole number. The test statistic calculated previously is approximated by the Student's t-distribution with $df$ as follows:
Degrees of freedom
$df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1}-1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2} + \left(\dfrac{1}{n_{2}-1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}$
When both sample sizes $n_{1}$ and $n_{2}$ are five or larger, the Student's t approximation is very good. Notice that the sample variances $(s_{1})^{2}$ and $(s_{2})^{2}$ are not pooled. (If the question comes up, do not pool the variances.)
It is not necessary to compute the degrees of freedom by hand. A calculator or computer easily computes it.
Example $1$: Independent groups
The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table $1$. Each populations has a normal distribution.
Table $1$
Sample Size Average Number of Hours Playing Sports Per Day Sample Standard Deviation
Girls 9 2 0.8660.866
Boys 16 3.2 1.00
Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.
Answer
The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, $\mu_{g}$ is the population mean for girls and $\mu_{b}$ is the population mean for boys. This is a test of two independent groups, two population means.
Random variable: $\bar{X}_{g} - \bar{X}_{b} =$ difference in the sample mean amount of time girls and boys play sports each day.
• $H_{0}: \mu_{g} = \mu_{b}$
• $H_{0}: \mu_{g} - \mu_{b} = 0$
• $H_{a}: \mu_{g} \neq \mu_{b}$
• $H_{a}: \mu_{g} - \mu_{b} \neq 0$
The words "the same" tell you $H_{0}$ has an "=". Since there are no other words to indicate $H_{a}$, assume it says "is different." This is a two-tailed test.
Distribution for the test: Use $t_{df}$ where $df$ is calculated using the $df$ formula for independent groups, two population means. Using a calculator, $df$ is approximately 18.8462. Do not pool the variances.
Calculate the p-value using a Student's t-distribution: $p\text{-value} = 0.0054$
Graph:
$s_{g} = 0.866$
$s_{b} = 1$
So,
$\bar{x}_{g} - \bar{x}_{b} = 2 - 3.2 = -1.2$
Half the $p\text{-value}$ is below –1.2 and half is above 1.2.
Make a decision: Since $\alpha > p\text{-value}$, reject $H_{0}$. This means you reject $\mu_{g} = \mu_{b}$. The means are different.
Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, $\sqrt{0.866}$ for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The $p\text{-value}$ is $p = 0.0054$, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.
Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).
Exercise $1$
Two samples are shown in Table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.
Table $2$
Sample Size Sample Mean Sample Standard Deviation
Population A 25 5 1
Population B 16 4.7 1.2
Answer
The $p\text{-value}$ is $0.4125$, which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.
When the sum of the sample sizes is larger than $30 (n_{1} + n_{2} > 30)$ you can use the normal distribution to approximate the Student's $t$.
Example $2$
A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.
1. Is this a test of two means or two proportions?
2. Are the populations standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
6. Is this test right-, left-, or two-tailed?
7. What is the $p\text{-value}$?
8. Do you reject or not reject the null hypothesis?
Solutions
1. two means
2. unknown
3. Student's t
4. $\bar{X}_{A} - \bar{X}_{B}$
5. $H_{0}: \mu_{A} \leq \mu_{B}$ and $H_{a}: \mu_{A} > \mu_{B}$
6. right
7. g. 0.1928
8. h. Do not reject.
9. i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.
Exercise $2$
A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.
1. Are the population standard deviations known?
2. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?
Answer
1. They are unknown.
2. The $p\text{-value} = 0.0878$. At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.
Example $3$
A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table $3$ and Table $4$.
Table $3$: Online Class
67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7
70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5
94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4
Table $4$: Face-to-face Class
77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2
69.4 57.6 69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8
98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4
Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:
1. Is this a test of two means or two proportions?
2. Are the population standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
6. Is this test right, left, or two tailed?
7. What is the $p\text{-value}$?
8. Do you reject or not reject the null hypothesis?
9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
(See the conclusion in Example, and write yours in a similar fashion)
Be careful not to mix up the information for Group 1 and Group 2!
Answer
1. two means
2. unknown
3. Student's $t$
4. $\bar{X}_{1} - \bar{X}_{2}$
1. $H_{0}: \mu_{1} = \mu_{2}$ Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
2. $H_{a}: \mu_{1} < \mu_{2}$ Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
5. left-tailed
6. $p\text{-value} = 0.0011$
Figure $3$.
7. Reject the null hypothesis
8. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class.
At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.
First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to $\mu_{1}$: and arrow to $\neq \mu_{1}$ (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.
Cohen's Standards for Small, Medium, and Large Effect Sizes
Cohen's $d$ is a measure of effect size based on the differences between two means. Cohen’s $d$, named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.
Table $5$: Cohen's Standard Effect Sizes
Size of effect $d$
Small 0.2
medium 0.5
Large 0.8
Cohen's $d$ is the measure of the difference between two means divided by the pooled standard deviation: $d = \dfrac{\bar{x}_{2}-\bar{x}_{2}}{s_{\text{pooled}}}$ where $s_{pooled} = \sqrt{\dfrac{(n_{1}-1)s^{2}_{1} + (n_{2}-1)s^{2}_{2}}{n_{1}+n_{2}-2}}$
Example $4$
Calculate Cohen’s d for Example. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.
Answer
$\mu_{1} = 4 s_{1} = 1.5 n_{1} = 11$
$\mu_{2} = 3.5 s_{2} = 1 n_{2} = 9$
$d = 0.384$
The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.
Example $5$
Calculate Cohen’s $d$ for Example. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.
Answer
$d = 0.834$; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.
Example 10.2.6
Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in Table and Table, respectively.
Northeast
94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6
77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5
76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0
West
126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6
116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6
78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23.4 21.5
Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:
1. Is this a test of two means or two proportions?
2. Are the population standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
6. Is this test right, left, or two tailed?
7. What is the $p\text{-value}$?
8. Do you reject or not reject the null hypothesis?
9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
10. Calculate Cohen’s d and interpret it.
Answer
1. two means
2. unknown
3. Student’s-t
4. $\bar{X}_{1} - \bar{X}_{2}$
1. $H_{0}: \mu_{1} = \mu_{2}$ Null hypothesis: the means of the weighted alphas are equal.
2. $H_{a}: \mu_{1} \neq \mu_{2}$ Alternative hypothesis : the means of the weighted alphas are not equal.
5. two-tailed
6. $p\text{-value} = 0.8787$
7. Do not reject the null hypothesis
8. This indicates that the trends in stocks are about the same in the top 30 banks in each region.
Figure $4$.
5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean weighted alphas for the banks in the northeast and the west are different
9. $d = 0.040$, Very small, because 0.040 is less than Cohen’s value of 0.2 for small effect size. The size of the difference of the means of the weighted alphas for the two regions of banks is small indicating that there is not a significant difference between their trends in stocks.
Review
Two population means from independent samples where the population standard deviations are not known
• Random Variable: $\bar{X}_{1} - \bar{X}_{2} =$ the difference of the sampling means
• Distribution: Student's t-distribution with degrees of freedom (variances not pooled)
Formula Review
Standard error: $SE = \sqrt{\dfrac{(s_{1}^{2})}{n_{1}} + \dfrac{(s_{2}^{2})}{n_{2}}}$
Test statistic (t-score): $t = \dfrac{(\bar{x}_{1}-\bar{x}_{2}) - (\mu_{1}-\mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}$
Degrees of freedom:
$df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1} - 1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2}} + \left(\dfrac{1}{n_{2} - 1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}$
where:
• $s_{1}$ and $s_{2}$ are the sample standard deviations, and n1 and n2 are the sample sizes.
• $x_{1}$ and $x_{2}$ are the sample means.
Cohen’s $d$ is the measure of effect size:
$d = \dfrac{\bar{x}_{1} - \bar{x}_{2}}{s_{\text{pooled}}}$
where
$s_{\text{pooled}} = \sqrt{\dfrac{(n_{1} - 1)s^{2}_{1} + (n_{2} - 1)s^{2}_{2}}{n_{1} + n_{2} - 2}}$
Glossary
Degrees of Freedom ($df$)
the number of objects in a sample that are free to vary.
Standard Deviation
A number that is equal to the square root of the variance and measures how far data values are from their mean; notation: $s$ for sample standard deviation and $\sigma$ for population standard deviation.
Variable (Random Variable)
a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters $X, Y, Z,$... Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters $x, y, z,$.... For example, if $X$ is the number of children in a family, then $x$ represents a specific integer 0, 1, 2, 3, .... Variables in statistics differ from variables in intermediate algebra in the two following ways.
• The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if $X =$ hair color, then the domain is {black, blond, gray, green, orange}.
• We can tell what specific value x of the random variable $X$ takes only after performing the experiment. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.02%3A_Two_Population_Means_with_Unknown_Standard_Deviations.txt |
Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is $\bar{X}_{1} - \bar{X}_{2}$. The normal distribution has the following format:
Normal distribution is:
$\bar{X}_{1} - \bar{X}_{2} \sim{N}\left[\mu_{1} - \mu_{2}, \sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\right] \label{eq1}$
The standard deviation is:
$\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\label{eq2}$
The test statistic (z-score) is:
$z = \dfrac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}} \label{eq3}$
Example $1$
Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table.
Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation
1 3 0.33
2 2.9 0.36
Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.
Answer
This is a test of two independent groups, two population means, population standard deviations known.
Random Variable: $\bar{X}_{1} - \bar{X}_{2} =$ difference in the mean number of months the competing floor waxes last.
• $H_{0}: \mu_{1} \leq \mu_{2}$
• $H_{a}: \mu_{1} > \mu_{2}$
The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into $H_{a}$. Therefore, this is a right-tailed test.
Distribution for the test: The population standard deviations are known so the distribution is normal. Using Equation \ref{eq1}, the distribution is:
$\bar{X}_{1} - \bar{X}_{2} \sim{N} \left(0, \sqrt{\dfrac{0.33^{2}}{20} + \dfrac{0.36^{2}}{20}}\right)$
Since $\mu_{1} \leq \mu_{2}$ then $\mu_{1} - \mu_{2} \leq 0$ and the mean for the normal distribution is zero.
Calculate the $p\text{-value}$ using the normal distribution: $p\text{-value} = 0.1799$
Graph:
$\bar{X}_{1} - \bar{X}_{2} = 3 – 2.9 = 0.1$
Compare $\alpha$ and the $p\text{-value}$: $\alpha = 0.05$ and $p\text{-value} = 0.1799$. Therefore, $\alpha < p\text{-value}$.
Make a decision: Since $\alpha < p\text{-value}$, do not reject $H_{0}$.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.
Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to $\mu1$: and arrow to $> \mu_{2}$. Press ENTER. Arrow down to Calculate and press ENTER. The $p\text{-value}$ is $p = 0.1799$ and the test statistic is 0.9157. Do the procedure again, but instead of Calculatedo Dra.
Exercise $1$
The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance.
Engine Sample Mean Number of RPM Population Standard Deviation
1 1,500 50
2 1,600 60
Answer
The $p\text{-value}$ is almost 0, so we reject the null hypothesis. There is sufficient evidence to conclude that Engine 2 runs at a higher RPM than Engine 1.
Example $2$: Age of Senators
An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years.
Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance.
Answer
This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators
Random variable: $\bar{X}_{1} - \bar{X}_{2} =$ difference in the mean age of Democratic and Republican U.S. senators.
• $H_{0}: \mu_{1} \leq \mu_{2} H_{0}: \mu_{1} - \mu_{2} \leq 0$
• $H_{a}: \mu_{1} > \mu_{2} H_{a}: \mu_{1} - \mu_{2} > 0$
The words "older than" translates as a “>” symbol and goes into $H_{a}$. Therefore, this is a right-tailed test.
Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is: $\bar{X}_{1} - \bar{X}_{2} \sim N\left[0, \sqrt{\dfrac{(9.55)^{2}}{30} + \dfrac{(10.17)^{2}}{30}}\right]$
Since $\mu_{1} \leq \mu_{2}, \mu_{1} - \mu_{2} \leq 0$ and the mean for the normal distribution is zero.
(Calculating the p-value using the normal distribution gives $p\text{-value} = 0.4040$)
Graph:
Compare $\alpha$ and the $p\text{-value}$: $\alpha = 0.05$ and $p\text{-value} = 0.4040$. Therefore, $\alpha < p\text{-value}$.
Make a decision: Since $\alpha < p\text{-value}$, do not reject $H_{0}$.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators.
Review
A hypothesis test of two population means from independent samples where the population standard deviations are known will have these characteristics:
• Random variable: $\overline{X}_1−\overline{X}_2 =$ the difference of the means
• Distribution: normal distribution
Formula Review
Normal Distribution:
$\bar{X}_{1} - \bar{X}_{2} = \sim N\left[\mu_{1} - \mu_{2}, \sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\right]$
Generally $\mu_{1} - \mu_{2} = 0$.
Test Statistic (z-score):
$z = \dfrac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}}$
Generally $\mu_{1} - \mu_{2} = 0$.
where:
$\sigma_{1}$ and $\sigma_{2}$ are the known population standard deviations. $n_{1}$ and $n_{1}$ are the sample sizes. $\bar{x}_{1}$ and $\bar{x}_{2}$ are the sample means. $\mu_{1}$ and $\mu_{2}$ are the population means | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.03%3A_Two_Population_Means_with_Known_Standard_Deviations.txt |
When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present:
1. The two independent samples are simple random samples that are independent.
2. The number of successes is at least five, and the number of failures is at least five, for each of the samples.
3. Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results.
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.
The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, $H_{0}: p_{A} = p_{B}$. To conduct the test, we use a pooled proportion, $p_{c}$.
The pooled proportion is calculated as follows:
$p_{c} = \dfrac{x_{A} + x_{B}}{n_{A} + n_{B}}$
The distribution for the differences is:
$P\_{A} - P'_{b} N\left[0, \sqrt{p_{c}(1 - p_{c})\left(\dfrac{1}{n_{A}} + \dfrac{1}{n_{B}}\right)}\right]$
The test statistic (z-score) is:
$z = \dfrac{(p'_{A} - p'_{B}) - (p_{A} - p_{B})}{\sqrt{p_{c}(1 - p_{c})\left(\dfrac{1}{n_{A}} + \dfrac{1}{n_{B}}\right)}}$
Example $1$
Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.
Answer
The problem asks for a difference in proportions, making it a test of two proportions.
Let $A$ and $B$ be the subscripts for medication A and medication B, respectively. Then $p_{A}$ and $p_{B}$ are the desired population proportions.
Random Variable:$P′_{A} – P′_{B} =$ difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B.
$H_{0}: p_{A} = p_{B}$
$p_{A} - p_{B} = 0$
$H_{a}: p_{A} \neq p_{B}$
$p_{A} - p_{B} \neq 0$
The words "is a difference" tell you the test is two-tailed.
Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:
$p_{c} = \dfrac{x_{A} + x_{B}}{n_{A} + n_{B}} = \dfrac{20 + 12}{200 + 200} = 800 1 - p_{c} = 0.92$
$P'_{A} - P'_{B} - N\left[0, \sqrt{(0.08)(0.92)\left(\dfrac{1}{200} + \dfrac{1}{200}\right)}\right]$
$P'_{A} - P'_{B}$ follows an approximate normal distribution.
Calculate the p-value using the normal distribution: $p\text{-value} = 0.1404$.
Estimated proportion for group A: $p'_{A} = \dfrac{x_{A}}{n_{A}} = \dfrac{20}{200} = 0.1$
Estimated proportion for group B: $p'_{B} = \dfrac{x_{B}}{n_{B}} = \dfrac{12}{200} = 0.06$
Graph:
$P′_{A} - P′_{B} = 0.1 – 0.06 = 0.04$.
Half the $p\text{-value}$ is below -0.04, and half is above 0.04.
Compare $\alpha$ and the $p\text{-value}: \alpha = 0.01$ and the $p\text{-value} = 0.1404$. $\alpha < p\text{-value}$.
Make a decision: Since $\alpha < p\text{-value}$, do not reject $H_{0}$.
Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B.
Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw.
Exercise $1$
Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance.
Answer
The $p\text{-value}$ is 0.0379, so we can reject the null hypothesis. At the 5% significance level, the data support that there is a difference in the pressure tolerances between the two valves.
Example $2$: Sexting
A research study was conducted about gender differences in “sexting.” The researcher believed that the proportion of girls involved in “sexting” is less than the proportion of boys involved. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table. Is the proportion of girls sending sexts less than the proportion of boys “sexting?” Test at a 1% level of significance.
Males Females
Sent “sexts” 183 156
Total number surveyed 2231 2169
Answer
This is a test of two population proportions. Let M and F be the subscripts for males and females. Then $p_{M}$ and $p_{F}$ are the desired population proportions.
Random variable:$p'_{F} - p'_{M} =$ difference in the proportions of males and females who sent “sexts.”
$H_{a}: p_{F} = p_{m} H_{0}: p_{F} - p_{M} = 0$
$H_{a}: p_{F} < p_{m} H_{a}: p_{F} - p_{M} < 0$
The words "less than" tell you the test is left-tailed.
Distribution for the test: Since this is a test of two population proportions, the distribution is normal:
$p_{C} = \dfrac{x_{F} + x_{M}}{n_{F} + n_{M}} = \dfrac{156 + 183}{2169 + 2231} = 0.077$
$1 - p_{C} = 0.923$
Therefore,
$p'_{F} - p'_{M} \sim N\left(0, \sqrt{(0.077)(0.923)\left(\dfrac{1}{2169} + \dfrac{1}{2231}\right)}\right)$
$p′_{F} – p′_{M}$ follows an approximate normal distribution.
Calculate the $p\text{-value}$ using the normal distribution:
$p\text{-value} = 0.1045$
Estimated proportion for females: 0.0719
Estimated proportion for males: 0.082
Graph:
Decision: Since $\alpha < p\text{-value}$, Do not reject $H_{0}$
Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending “sexts” is less than the proportion of boys sending “sexts.”
Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The $p\text{-value}$ is $P = 0.1045$ and the test statistic is $z = -1.256$.
Example $3$
Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5% have an iPhone. Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the proportion of white iPhone owners greater than the proportion of African American iPhone owners?
Answer
This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then pW and pA are the desired population proportions.
Random variable: $p′_{W} – p′_{A} =$ difference in the proportions of Android and iPhone users.
$H_{0}: p_{W} = p_{A} H_{0}: p_{W} – p_{A} = 0$
$H_{a}: p_{W} > p_{A} H_{a}: p_{W} – p_{A} < 0$
The words "more popular" indicate that the test is right-tailed.
Distribution for the test: The distribution is approximately normal:
$p_{C} = \dfrac{x_{W} + x_{A}}{n_{W} + n_{A}} = \dfrac{134 + 12}{1343 + 232} = 0.0927$
$1 - p_{C} = 0.9073$
Therefore,
$p'_{W} - p'_{A} \sim N\left(0, \sqrt{(0.0927)(0.9073)\left(\dfrac{1}{1343} + \dfrac{1}{232}\right)}\right)$
$p'_{W} - p'_{A}$ follows an approximate normal distribution.
Calculate the $p\text{-value}$ using the normal distribution:
$p\text{-value} = 0.0077$
Estimated proportion for group A: 0.10
Estimated proportion for group B: 0.05
Graph:
Decision: Since $\alpha > p\text{-value}$, reject the $H_{0}$.
Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans.
TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The P-value is P = 0.0092 and the test statistic is Z = 2.33.
Example $3$
A concerned group of citizens wanted to know if the proportion of forcible rapes in Texas was different in 2011 than in 2010. Their research showed that of the 113,231 violent crimes in Texas in 2010, 7,622 of them were forcible rapes. In 2011, 7,439 of the 104,873 violent crimes were in the forcible rape category. Test at a 5% significance level. Answer the following questions:
1. Is this a test of two means or two proportions?
2. Which distribution do you use to perform the test?
3. What is the random variable?
4. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols.
5. Is this test right-, left-, or two-tailed?
6. What is the $p\text{-value}$?
7. Do you reject or not reject the null hypothesis?
8. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________.
Solutions
a. two proportions
b. normal for two proportions
c. Subscripts: 1 = 2010, 2 = 2011 $P′_{1} - P′_{2}$
d. Subscripts: 1 = 2010, 2 = 2011 $H_{0}: p_{1} = p_{2} H_{0}: p_{1} − p_{2} = 0$ $H_{0}: p_{1} \neq p_{2} H_{0}: p_{1} − p_{2} \neq 0$
e. two-tailed
f. $p\text{-value} = 0.00086$
g. Reject the $H_{0}$.
h. At the 5% significance level, from the sample data, there is sufficient evidence to conclude that there is a difference between the proportion of forcible rapes in 2011 and 2010.
Review
• Test of two population proportions from independent samples.
• Random variable: $\hat{p}_{A} - \hat{p}_{B} =$ difference between the two estimated proportions
• Distribution: normal distribution
Formula Review
Pooled Proportion:
$p_{c} = \dfrac{x_{F} + x_{M}}{n_{F} + n_{M}}$
Distribution for the differences:
$p'_{A} - p'_{B} \sim N\left[0, \sqrt{p_{c}(1-p_{c})\left(\dfrac{1}{n_{A}} + \dfrac{1}{n_{B}}\right)}\right]$
where the null hypothesis is $H_{0}: p_{A} = p_{B}$ or $H_{0}: p_{A} - p_{B} = 0$.
Test Statistic (z-score):
$z = \dfrac{(p'_{A} - p'_{B})}{\sqrt{p_{c}(1-p_{c})\left(\dfrac{1}{n_{A}} + \dfrac{1}{n_{B}}\right)}}$
where the null hypothesis is $H_{0}: p_{A} = p_{B}$ or $H_{0}: p_{A} - p_{B} = 0$.
and
• $p'_{A}$ and $p'_{B}$ are the sample proportions, $p_{A}$ and $p_{B}$ are the population proportions,
• $P_{c}$ is the pooled proportion, and $n_{A}$ and $n_{B}$ are the sample sizes.
Glossary
Pooled Proportion
estimate of the common value of $p_{1}$ and $p_{2}$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.04%3A_Comparing_Two_Independent_Population_Proportions.txt |
When using a hypothesis test for matched or paired samples, the following characteristics should be present:
1. Simple random sampling is used.
2. Sample sizes are often small.
3. Two measurements (samples) are drawn from the same pair of individuals or objects.
4. Differences are calculated from the matched or paired samples.
5. The differences form the sample that is used for the hypothesis test.
6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.
In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, $\mu_{d}$, is then tested using a Student's $t$-test for a single population mean with $n - 1$ degrees of freedom, where $n$ is the number of differences.
The test statistic ($t$-score) is:
$t = \dfrac{\bar{x}_{d} - \mu_{d}}{\left(\dfrac{s_{d}}{\sqrt{n}}\right)}$
Example $1$
A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table. A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.
Subject: A B C D E F G H
Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6
After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0
Answer
Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.")
After Data Before Data Difference
6.8 6.6 0.2
2.4 6.5 -4.1
7.4 9 -1.6
8.5 10.3 -1.8
8.1 11.3 -3.2
6.1 8.1 -2
3.4 6.3 -2.9
2 11.6 -9.6
The data for the test are the differences: $\{0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6\}$
The sample mean and sample standard deviation of the differences are: $\bar{x}_{d} = -3.13$ and $s_{d} = 2.91$ Verify these values.
Let $\mu_{d}$ be the population mean for the differences. We use the subscript dd to denote "differences."
Random variable:
$\bar{X}_{d} =$ the mean difference of the sensory measurements
$H_{0}: \mu_{d} \geq 0$
The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. $\mu_{d}$ is the population mean of the differences.
$H_{a}: \mu_{d} < 0$
The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.
Distribution for the test:
The distribution is a Student's t with $df = n - 1 = 8 - 1 = 7$. Use $t_{7}$. (Notice that the test is for a single population mean.)
Calculate the p-value using the Student's-t distribution:
$p\text{-value} = 0.0095$
Graph:
$\bar{X}_{d}$ is the random variable for the differences.
The sample mean and sample standard deviation of the differences are:
$\bar{x}_{d} = -3.13$
$s_{d} = 2.91$
Compare $\alpha$ and the $p\text{-value}$
$\alpha = 0.05$ and $p\text{-value} = 0.0095$. $\alpha > p\text{-value}$
Make a decision
Since $\alpha > p\text{-value}$, reject $H_{0}$. This means that $\mu_{d} < 0$ and there is improvement.
Conclusion
At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.
For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after - before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list.
Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for $\mu_{0}$, the name of the list where you put the data, and 1 for Freq:. Arrow down to $\mu$: and arrow over to < $\mu_{0}$. Press ENTER. Arrow down to Calculate and press ENTER. The $p\text{-value}$ is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate). Press ENTER.
Exercise $1$
A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level.
Subject A B C D E F G H I
Before 209 210 205 198 216 217 238 240 222
After 199 207 189 209 217 202 211 223 201
Answer
The $p\text{-value}$ is 0.0130, so we can reject the null hypothesis. There is enough evidence to suggest that the diet lowers cholesterol.
Example $2$
A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:
Weight (in pounds) Player 1 Player 2 Player 3 Player 4
Amount of weight lifted prior to the class 205 241 338 368
Amount of weight lifted after the class 295 252 330 360
The coach wants to know if the strength development class makes his players stronger, on average.
Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: $\{90, 11, -8, -8\}$. Assume the differences have a normal distribution.
Using the differences data, calculate the sample mean and the sample standard deviation.
$\bar{x}_{d} = 21.3$
and
$s_{d} = 46.7$
The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative.
Using the difference data, this becomes a test of a single __________ (fill in the blank).
Define the random variable: $\bar{X}$ mean difference in the maximum lift per player.
The distribution for the hypothesis test is $t_{3}$.
• $H_{0}: \mu_{d} \leq 0$,
• $H_{a}: \mu_{d} > 0$
Graph:
Calculate the $p\text{-value}$: The $p\text{-value}$ is 0.2150
Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because $\alpha < p\text{-value}$.
What is the conclusion?
At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.
Exercise $2$
A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in Table. Are the scores, on average, higher after the class? Test at a 5% level.
SAT Scores Student 1 Student 2 Student 3 Student 4
Score before class 1840 1960 1920 2150
Score after class 1920 2160 2200 2100
Answer
The $p\text{-value}$ is 0.0874, so we decline to reject the null hypothesis. The data do not support that the class improves SAT scores significantly.
Example $3$
Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in Table.
Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7
Dominant Hand 30 26 34 17 19 26 20
Weaker Hand 28 14 27 18 17 26 16
Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant.
Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: $\{2, 12, 7, –1, 2, 0, 4\}$. The differences have a normal distribution.
Using the differences data, calculate the sample mean and the sample standard deviation. $\bar{x} = 3.71$, $s_{d} = 4.5$.
Random variable: $\bar{X} =$ mean difference in the distances between the hands.
Distribution for the hypothesis test: $t_{6}$
$H_{0}: \mu_{d} = 0 H_{a}: \mu_{d} \neq 0$
Graph:
Calculate the p-value: The $p\text{-value}$ is 0.0716 (using the data directly).
(test statistic = 2.18. $p\text{-value} = 0.0719$ using $(\bar{x}_{d} = 3.71, s_{d} = 4.5$.
Decision: Assume $\alpha = 0.05$. Since $\alpha < p\text{-value}$, Do not reject $H_{0}$.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put.
Exercise $3$
Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table. Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level.
Player 1 Player 2 Player 3 Player 4 Player 5
Dominant Hand 120 111 135 140 125
Off-hand 105 109 98 111 99
Answer
The $p\text{-level}$ is 0.0230, so we can reject the null hypothesis. The data show that the players do not throw the same distance with their off-hands as they do with their dominant hands.
Review
A hypothesis test for matched or paired samples (t-test) has these characteristics:
• Test the differences by subtracting one measurement from the other measurement
• Random Variable: $x_{d} =$ mean of the differences
• Distribution: Student’s t-distribution with $n - 1$ degrees of freedom
• If the number of differences is small (less than 30), the differences must follow a normal distribution.
• Two samples are drawn from the same set of objects.
• Samples are dependent.
Formula Review
Test Statistic (t-score): $t = \dfrac{\bar{x}_{d}}{\left(\dfrac{s_{d}}{\sqrt{n}}\right)}$
where:
$x_{d}$ is the mean of the sample differences. $\mu_{d}$ is the mean of the population differences. $s_{d}$ is the sample standard deviation of the differences. $n$ is the sample size. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.05%3A_Matched_or_Paired_Samples.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
• The student will select the appropriate distributions to use in each case.
• The student will conduct hypothesis tests and interpret the results.
Supplies:
• the business section from two consecutive days’ newspapers
• three small packages of M&Ms®
• five small packages of Reese's Pieces®
Increasing Stocks Survey
Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks, and 32 NASDAQ stocks and complete the following statements.
1. $H_{0}$: _________
2. $H_{a}$: _________
3. In words, define the random variable.
4. The distribution to use for the test is _____________.
5. Calculate the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
2. Calculate the $p\text{-value}$.
7. Do you reject or not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Decreasing Stocks Survey
Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day.
1. $H_{0}$: ________
2. $H_{a}$: ________
3. In words, define the random variable.
4. The distribution to use for the test is _____________.
5. Calculate the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
2. Calculate the $p\text{-value}$:
7. Do you reject or not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Candy Survey
Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands.
1. $H_{0}$: ________
2. $H_{a}$: ________
3. In words, define the random variable.
4. What distribution should be used for this test?
5. Calculate the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
2. Calculate the $p\text{-value}$.
7. Do you reject or not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Shoe Survey
Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample.
1. $H_{0}$: ________
2. $H_{a}$: ________
3. In words, define the random variable.
4. The distribution to use for the test is ________________.
5. Calculate the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
2. Calculate the $p\text{-value}$.
7. Do you reject or not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.06%3A_Hypothesis_Testing_for_Two_Means_and_Two_Proportions_%28Worksheet%29.txt |
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
10.2: Two Population Means with Unknown Standard Deviations
Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for
1. independent group means, population standard deviations, and/or variances known
2. independent group means, population standard deviations, and/or variances unknown
3. matched or paired samples
4. single mean
5. two proportions
6. single proportion
Exercise 10.2.3
It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal.
Answer
two proportions
Exercise 10.2.4
A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this.
Exercise 10.2.5
A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted.
Answer
matched or paired samples
Exercise 10.2.6
S 10.3.3
Subscripts: 1 = boys, 2 = girls
1. $H_{0}: \mu_{1} \leq \mu_{2}$
2. $H_{a}: \mu_{1} > \mu_{2}$
3. The random variable is the difference in the mean auto insurance costs for boys and girls.
4. normal
5. test statistic: $z = 2.50$
6. $p\text{-value}: 0.0062$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for Decision: $p\text{-value} < \alpha$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.
Q 10.3.4
A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and$1,011, respectively. The population standard deviations are known to be $254 and$87, respectively. Conduct a hypothesis test to determine if the means are statistically the same.
Q 10.3.5
Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim.
S 10.3.5
Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans
1. $H_{0}: \mu_{1} \geq \mu_{2}$
2. $H_{a}: \mu_{1} < \mu_{2}$
3. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans.
4. normal
5. test statistic: 6.36
6. $p\text{-value}: 0$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for decision: $p\text{-value} < \alpha$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans.
Q 10.3.6
A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test.
Q 10.3.7
One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). Table contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife).
Wife’s Score 2 2 3 3 4 2 1 1 2 4
Husband’s Score 2 2 1 3 2 1 1 1 2 4
S 10.3.7
1. $H_{0}: \mu_{d} = 0$
2. $H_{a}: \mu_{d} < 0$
3. The random variable $X_{d}$ is the average difference between husband’s and wife’s satisfaction level.
4. $t_{9}$
5. test statistic: $t = –1.86$
6. $p\text{-value}: 0.0479$
7. Check student’s solution
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis, but run another test.
3. Reason for Decision: $p\text{-value} < \alpha$
4. Conclusion: This is a weak test because alpha and the p-value are close. However, there is insufficient evidence to conclude that the mean difference is negative.
10.4: Comparing Two Independent Population Proportions
Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1.
Exercise 10.4.2
Is this a test of means or proportions?
Exercise 10.4.3
What is the random variable?
Answer
$P'_{OS_{1}} - P'_{OS_{2}} =$ difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2.
Exercise 10.4.4
State the null and alternative hypotheses.
Exercise 10.4.5
What is the $p\text{-value}$?
Answer
0.1018
Exercise 10.4.6
What can you conclude about the two operating systems?
Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota.
Exercise 10.4.7
Is this a test of means or proportions?
Answer
proportions
Exercise 10.4.8
State the null and alternative hypotheses.
1. $H_{0}$: _________
2. $H_{a}$: _________
Exercise 10.4.9
Is this a right-tailed, left-tailed, or two-tailed test? How do you know?
Answer
right-tailed
Exercise 10.4.10
What is the random variable of interest for this test?
Exercise 10.4.11
In words, define the random variable for this test.
Answer
The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.
Exercise 10.4.12
Which distribution (normal or Student's t) would you use for this hypothesis test?
Exercise 10.4.13
Explain why you chose the distribution you did for the Exercise 10.56.
Answer
Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.
Exercise 10.4.14
Calculate the test statistic.
Exercise 10.4.15
Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the $p\text{-value}$.
Answer
Check student’s solution.
Exercise 10.4.16
Find the $p\text{-value}$.
Exercise 10.4.17
At a pre-conceived $\alpha = 0.05$, what is your:
1. Decision:
2. Reason for the decision:
3. Conclusion (write out in a complete sentence):
Answer
1. Reject the null hypothesis.
2. $p\text{-value} < \alpha$
3. At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.
Exercise 10.4.18
Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not?
DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link]. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
NOTE
If you are using a Student's t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)
Q 10.4.1
A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them.
Q 10.4.2
We are interested in whether the proportions of female suicide victims for ages 15 to 24 are the same for the whites and the blacks races in the United States. We randomly pick one year, 1992, to compare the races. The number of suicides estimated in the United States in 1992 for white females is 4,930. Five hundred eighty were aged 15 to 24. The estimate for black females is 330. Forty were aged 15 to 24. We will let female suicide victims be our population.
S 10.4.2
1. $H_{0}: P_{W} = P_{B}$
2. $H_{a}: P_{W} \neq P_{B}$
3. The random variable is the difference in the proportions of white and black suicide victims, aged 15 to 24.
4. normal for two proportions
5. test statistic: –0.1944
6. $p\text{-value}: 0.8458$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for decision: $p\text{-value} > \alpha$
4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of white and black female suicide victims, aged 15 to 24, are different.
Q 10.4.3
Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, $\left(\frac{(larger + smaller dimension}{larger dimension}\right)$ was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation?
Q 10.4.4
A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different?
S 10.4.4
Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College
1. $H_{0}: p_{1} = p_{2}$
2. $H_{a}: p_{1} \neq p_{2}$
3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.
4. normal for two proportions
5. test statistic: 4.29
6. $p\text{-value}: 0.00002$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for decision: p-value < alpha
4. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.
Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system . It is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test.
• “2011” subscript: 2011 group.
• “2010” subscript: 2010 group
Q 10.4.5
This is:
1. a test of two proportions
2. a test of two independent means
3. a test of a single mean
4. a test of matched pairs.
Q 10.4.6
An appropriate null hypothesis is:
1. $p_{2011} \leq p_{2010}$
2. $p_{2011} \geq p_{2010}$
3. $\mu_{2011} \leq \mu_{2010}$
4. $p_{2011} > p_{2010}$
a
Q 10.4.7
The $p\text{-value}$ is 0.0022. At a 1% level of significance, the appropriate conclusion is
1. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
2. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
3. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
4. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
Q 10.4.8
Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders.
S 10.4.9
Test: two independent sample proportions.
Random variable: $p′_{1} - p′_{2}$
Distribution:
$H_{0}: p_{1} = p_{2}$
$H_{a}: p_{1} \neq p_{2}$
The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users.
Graph: two-tailed
$p\text{-value}: 0.0033$
Decision: Reject the null hypothesis.
Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older.
Q 10.4.10
are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table. Test at the 1% level of significance.
Number who are obese Sample size
Men 42,769 155,525
Women 67,169 248,775
Q 10.4.11
Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. Table details the number of tablet owners for each age group. Test at the 1% level of significance.
16–29 year olds 30 years old and older
Own a Tablet 69 231
Sample Size 628 2,309
S 10.4.11
Test: two independent sample proportions
Random variable: $p′_{1} - p′_{2}$
Distribution:
$H_{0}: p_{1} = p_{2}$
$H_{a}: p_{1} > p_{2}$
A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.
Graph: right-tailed
$p\text{-value}: 0.2354$
Decision: Do not reject the $H_{0}$.
Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.
Q 10.4.12
A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5% level of significance.
Q 10.4.13
While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey.
S 10.4.13
Subscripts: 1: men; 2: women
1. $H_{0}: p_{1} \leq p_{2}$
2. $H_{a}: p_{1} > p_{2}$
3. $P'_{1} - P\_{2}$ is the difference between the proportions of men and women who enjoy shopping for electronic equipment.
4. normal for two proportions
5. test statistic: 0.22
6. $p\text{-value}: 0.4133$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Do not reject the null hypothesis.
3. Reason for Decision: $p\text{-value} > \alpha$
4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.
Q 10.4.14
We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of$4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of$10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software.
Q 10.4.15
Joan Nguyen recently claimed that the proportion of college-age males with at least one pierced ear is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe that the proportion of males has reached the proportion of females?
S 10.4.15
1. $H_{0}: p_{1} = p_{2}$
2. $H_{a}: p_{1} \neq p_{2}$
3. $P'_{1} - P\_{2}$ is the difference between the proportions of men and women that have at least one pierced ear.
4. normal for two proportions
5. test statistic: –4.82
6. $p\text{-value}: 0$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for Decision: $p\text{-value} < \alpha$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.
Q 10.4.16
Use the data sets found in [link] to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds?
Q 10.4.17
"To Breakfast or Not to Breakfast?" by Richard Ayore
In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come.
If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine.
Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!”
And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table, solve our problem.
Work hours with breakfast Work hours without breakfast
8 6
7 5
9 5
5 4
9 7
8 7
10 7
7 5
6 6
9 5
S 10.4.17
1. $H_{0}: \mu_{d} = 0$
2. $H_{a}: \mu_{d} > 0$
3. The random variable $X_{d}$ is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.
4. $t_{9}$
5. test statistic: 4.8963
6. $p\text{-value}: 0.0004$
7. Check student’s solution.
1. $\alpha: 0.05$
2. Decision: Reject the null hypothesis.
3. Reason for Decision:$p\text{-value} < \alpha$
4. Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.
10.5: Matched or Paired Samples
Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level.
Installation A B C D E F G H
Before 3 6 4 2 5 8 2 6
After 1 5 2 0 1 0 2 2
Exercise 10.5.4
What is the random variable?
Answer
the mean difference of the system failures
Exercise 10.5.5
State the null and alternative hypotheses.
Exercise 10.5.6
What is the $p\text{-value}$?
Answer
0.0067
Exercise 10.5.7
Draw the graph of the $p\text{-value}$.
Exercise 10.5.8
What conclusion can you draw about the software patch?
Answer
With a $p\text{-value} 0.0067$, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.
Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level.
Subject A B C D E F
Before 3 4 3 2 4 5
After 4 5 6 4 5 7
Exercise 10.5.9
State the null and alternative hypotheses.
Exercise 10.5.10
What is the $p\text{-value}$?
Answer
0.0021
Exercise 10.5.11
What is the sample mean difference?
Exercise 10.5.12
Draw the graph of the $p\text{-value}$.
Answer
Exercise 10.5.13
What conclusion can you draw about the juggling class?
Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level.
Patient A B C D E F
Before 161 162 165 162 166 171
After 158 159 166 160 167 169
Exercise 10.5.14
State the null and alternative hypotheses.
Answer
$H_{0}: \mu_{d} \geq 0$
$H_{a}: \mu_{d} < 0$
Exercise 10.5.15
What is the test statistic?
Exercise 10.5.16
What is the $p\text{-value}$?
Answer
0.0699
Exercise 10.5.17
What is the sample mean difference?
Exercise 10.5.18
What is the conclusion?
Answer
We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.
Bringing It Together
Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test.
1. independent group means, population standard deviations and/or variances known
2. independent group means, population standard deviations and/or variances unknown
3. matched or paired samples
4. single mean
5. two proportions
6. single proportion
Exercise 10.5.19
A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet.
Exercise 10.5.20
A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.
Answer
e
Exercise 10.5.21
The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females.
Exercise 10.5.22
A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.
Answer
d
Exercise 10.5.23
A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively.
Exercise 10.5.24
According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this.
Answer
f
Exercise 10.5.25
According to a recent study, U.S. companies have a mean maternity-leave of six weeks.
Exercise 10.5.26
A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally.
Answer
e
Exercise 10.5.27
A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected:
Pre-course score Post-course score
1 300
960 920
1010 1100
840 880
1100 1070
1250 1320
860 860
1330 1370
790 770
990 1040
1110 1200
740 850
Exercise 10.5.28
University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked.
Answer
f
Exercise 10.5.29
Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown Table. Determine the appropriate test and best distribution to use for that test.
Left-handed Right-handed
Sample size 41 41
Sample mean 97.5 98.1
Sample standard deviation 17.5 19.2
1. Two independent means, normal distribution
2. Two independent means, Student’s-t distribution
3. Matched or paired samples, Student’s-t distribution
4. Two population proportions, normal distribution
Exercise 10.5.30
A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as Table.
Player 1 Player 2 Player 3 Player 4
Mean score before class 83 78 93 87
Mean score after class 80 80 86 86
This is:
1. a test of two independent means.
2. a test of two proportions.
3. a test of a single mean.
4. a test of a single proportion.
Answer
a
DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
NOTE
If you are using a Student's t-distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)
Q 10.5.1
Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table. Do you think that their cholesterol levels were significantly lowered?
Starting cholesterol level Ending cholesterol level
140 140
220 230
110 120
240 220
200 190
180 150
190 200
360 300
280 300
260 240
S 10.5.1
$p\text{-value} = 0.1494$
At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.
Use the following information to answer the next two exercises. A new AIDS prevention drug was tried on a group of 224 HIV positive patients. Forty-five patients developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four years. We want to test whether the method of treatment reduces the proportion of patients that develop AIDS after four years or if the proportions of the treated group and the untreated group stay the same.
Let the subscript $t =$ treated patient and $ut =$ untreated patient.
Q 10.5.2
The appropriate hypotheses are:
1. $H_{0}: p_{t} < p_{ut}$ and $H_{a}: p_{t} \geq p_{ut}$
2. $H_{0}: p_{t} \leq p_{ut}$ and $H_{a}: p_{t} > p_{ut}$
3. $H_{0}: p_{t} = p_{ut}$ and $H_{a}: p_{t} \neq p_{ut}$
4. $H_{0}: p_{t} = p_{ut}$ and $H_{a}: p_{t} < p_{ut}$
Q 10.5.3
If the $p\text{-value}$ is 0.0062 what is the conclusion (use $\alpha = 0.05$)?
1. The method has no effect.
2. There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.
3. There is sufficient evidence to conclude that the method increases the proportion of HIV positive patients who develop AIDS after four years.
4. There is insufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.
S 10.5.3
b
Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: $\bar{x}_{d} = -10.2$ $s_{d} = 8.4$. Using the data, test the hypothesis that the blood pressure has decreased after the training.
Q 10.5.4
The distribution for the test is:
1. $t_{5}$
2. $t_{6}$
3. $N(-10.2, 8.4)$
4. $N\left(-10.2, \frac{8.4}{\sqrt{6}}\right)$
Q 10.5.5
If $\alpha = 0.05$, the $p\text{-value}$ and the conclusion are
1. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training.
2. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training.
3. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.
4. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training.
c
Q 10.5.6
A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows.
Player 1 Player 2 Player 3 Player 4
Mean score before class 83 78 93 87
Mean score after class 80 80 86 86
The correct decision is:
1. Reject $H_{0}$.
2. Do not reject the $H_{0}$.
Q 10.5.7
A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. The group compared the estimates of new female breast cancer cases by southern state in 2012 and in 2013. The results are in Table.
Southern States 2012 2013
Alabama 3,450 3,720
Arkansas 2,150 2,280
Florida 15,540 15,710
Georgia 6,970 7,310
Kentucky 3,160 3,300
Louisiana 3,320 3,630
Mississippi 1,990 2,080
North Carolina 7,090 7,430
Oklahoma 2,630 2,690
South Carolina 3,570 3,580
Tennessee 4,680 5,070
Texas 15,050 14,980
Virginia 6,190 6,280
S 10.5.7
Test: two matched pairs or paired samples (t-test)
Random variable: $\bar{X}_{d}$
Distribution: $t_{12}$
$H_{0}: \mu_{d} = 0 H_{a}: \mu_{d} > 0$
The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.
Graph: right-tailed
$p\text{-value}: 0.0004$
Decision: Reject $H_{0}$
Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.
Q 10.5.8
A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in Table. Test at the 1% level of significance.
Cities Hyatt Regency prices in dollars Hilton prices in dollars
Atlanta 107 169
Boston 358 289
Chicago 209 299
Dallas 209 198
Denver 167 169
Indianapolis 179 214
Los Angeles 179 169
New York City 625 459
Philadelphia 179 159
Washington, DC 245 239
Q 10.5.9
A politician asked his staff to determine whether the underemployment rate in the northeast decreased from 2011 to 2012. The results are in Table.
Northeastern States 2011 2012
Connecticut 17.3 16.4
Delaware 17.4 13.7
Maine 19.3 16.1
Maryland 16.0 15.5
Massachusetts 17.6 18.2
New Hampshire 15.4 13.5
New Jersey 19.2 18.7
New York 18.5 18.7
Ohio 18.2 18.8
Pennsylvania 16.5 16.9
Rhode Island 20.7 22.4
Vermont 14.7 12.3
West Virginia 15.5 17.3
S 10.5.9
Test: matched or paired samples (t-test)
Difference data: $\{–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8\}$
Random Variable: $\bar{X}_{d}$
Distribution: $H_{0}: \mu_{d} = 0 H_{a}: \mu_{d} < 0$
The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.
Graph: left-tailed.
$p\text{-value}: 0.1207$
Decision: Do not reject $H_{0}$.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/10%3A_Hypothesis_Testing_with_Two_Samples/10.E%3A_Hypothesis_Testing_with_Two_Samples_%28Exercises%29.txt |
A chi-squared test is any statistical hypothesis test in which the sampling distribution of the test statistic is a chi-square distribution when the null hypothesis is true.
• 11.1: Prelude to The Chi-Square Distribution
You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution.
• 11.2: Facts About the Chi-Square Distribution
The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population.
• 11.3: Goodness-of-Fit Test
In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.
• 11.4: Test of Independence
Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test.
• 11.5: Test for Homogeneity
The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.
• 11.6: Comparison of the Chi-Square Tests
You have seen the Chi-square test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which Chi-square test is the appropriate one to use.
• 11.7: Test of a Single Variance
A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). A test of a single variance may be right-tailed, left-tailed, or two-tailed
• 11.8: Lab 1- Chi-Square Goodness-of-Fit (Worksheet)
A statistics Worksheet: The student will evaluate data collected to determine if they fit either the uniform or exponential distributions.
• 11.9: Lab 2- Chi-Square Test of Independence (Worksheet)
A statistics Worksheet: The student will evaluate if there is a significant relationship between favorite type of snack and gender.
• 11.E: The Chi-Square Distribution (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected].
11: The Chi-Square Distribution
CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Interpret the chi-square probability distribution as the sample size changes.
• Conduct and interpret chi-square goodness-of-fit hypothesis tests.
• Conduct and interpret chi-square test of independence hypothesis tests.
• Conduct and interpret chi-square homogeneity hypothesis tests.
• Conduct and interpret chi-square single variance hypothesis tests.
Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test.
You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution.
In this chapter, you will learn the three major applications of the chi-square distribution:
1. the goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example
2. the test of independence, which determines if events are independent, such as in the movie example
3. the test of a single variance, which tests variability, such as in the coffee example
Though the chi-square distribution depends on calculators or computers for most of the calculations, there is a table available (see [link]). TI-83+ and TI-84 calculator instructions are included in the text.
COLLABORATIVE CLASSROOM EXERCISE
Look in the sports section of a newspaper or on the Internet for some sports data (baseball averages, basketball scores, golf tournament scores, football odds, swimming times, and the like). Plot a histogram and a boxplot using your data. See if you can determine a probability distribution that your data fits. Have a discussion with the class about your choice.
11.02: Facts About the Chi-Square Distribution
The notation for the chi-square distribution is:
$\chi \sim \chi^{2}_{df}$
where $df =$ degrees of freedom which depends on how chi-square is being used. (If you want to practice calculating chi-square probabilities then use $df = n - 1$. The degrees of freedom for the three major uses are each calculated differently.)
For the $\chi^{2}$ distribution, the population mean is $\mu = df$ and the population standard deviation is
$\sigma = \sqrt{2(df)}.$
The random variable is shown as $\chi^{2}$, but may be any upper case letter. The random variable for a chi-square distribution with $k$ degrees of freedom is the sum of $k$ independent, squared standard normal variables.
$\chi^{2} = (Z_{1})^{2} + ... + (Z_{k})^{2}$
1. The curve is nonsymmetrical and skewed to the right.
2. There is a different chi-square curve for each $df$.
1. The test statistic for any test is always greater than or equal to zero.
2. When $df > 90$, the chi-square curve approximates the normal distribution. For $\chi \sim \chi^{2}_{1,000}$ the mean, $\mu = df = 1,000$ and the standard deviation, $\mu = \sqrt{2(1,000)}$. Therefore, $X \sim N(1,000, 44.7)$, approximately.
3. The mean, $\mu$, is located just to the right of the peak.
Review
The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population.
An important parameter in a chi-square distribution is the degrees of freedom $df$ in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom.
The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom $df$. For $df > 90$, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests.
Formula Review
$\chi^{2} = (Z_{1})^{2} + (Z_{2})^{2} + ... + (Z_{df})^{2}$ chi-square distribution random variable
$\mu_{\chi^{2}} = df$ chi-square distribution population mean
$\sigma_{\chi^{2}} = \sqrt{2(df)}$ Chi-Square distribution population standard deviation
Exercise $1$
If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation?
Answer
mean $= 25$ and standard deviation $= 7.0711$
Exercise $2$
If $df > 90$, the distribution is _____________. If $df = 15$, the distribution is ________________.
Exercise $3$
When does the chi-square curve approximate a normal distribution?
Answer
when the number of degrees of freedom is greater than 90
Exercise $4$
Where is $\mu$ located on a chi-square curve?
Exercise $5$
Is it more likely the df is 90, 20, or two in the graph?
Answer
$df = 2$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/11%3A_The_Chi-Square_Distribution/11.01%3A_Prelude_to_The_Chi-Square_Distribution.txt |
In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.
The test statistic for a goodness-of-fit test is:
$\sum_k \frac{(O - E)^{2}}{E}$
where:
• $O =$ observed values (data)
• $E =$ expected values (from theory)
• $k =$ the number of different data cells or categories
The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are $n$ terms of the form $\frac{(O - E)^{2}}{E}$.
The number of degrees of freedom is $df = (\text{number of categories} - 1)$.
The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.
The expected value for each cell needs to be at least five in order for you to use this test.
Example 11.3.1
Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to the table below.
Number of absences per term Expected number of students
0–2 50
3–5 30
6–8 12
9–11 6
12+ 2
A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in the table below displays the results of that survey.
Number of absences per term Actual number of students
0–2 35
3–5 40
6–8 20
9–11 1
12+ 4
Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.
• $H_{0}$: Student absenteeism fits faculty perception.
The alternative hypothesis is the opposite of the null hypothesis.
• $H_{a}$: Student absenteeism does not fit faculty perception.
Exercise $1$.1
a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?
Answer
a. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that group with the "9–11" group to create new tables where the number of students for each entry are at least five. The new results are in the table below.
Number of absences per term Expected number of students
0–2 50
3–5 30
6–8 12
9+ 8
Number of absences per term Actual number of students
0–2 35
3–5 40
6–8 20
9+ 5
Exercise $1$.2
b. What is the number of degrees of freedom ($df$)?
Answer
b. There are four "cells" or categories in each of the new tables.
$df = \text{number of cells} - 1 = 4 - 1 = 3$
Exercise $1$
A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in the table below.
Number produced Number defective
0–100 5
101–200 6
201–300 7
301–400 8
401–500 10
A random sample was taken to determine the actual number of defects. The table below shows the results of the survey.
Number produced Number defective
0–100 5
101–200 7
201–300 8
301–400 9
401–500 11
State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.
Answer
$H_{0}$:The number of defects fits expectations.
$H_{a}$:The number of defects does not fit expectations.
$df = 4$
Example 11.3.2
Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in the table below. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.
Day of the Week Employees were Most Absent
Monday Tuesday Wednesday Thursday Friday
Number of Absences 15 12 9 9 15
Answer
The null and alternative hypotheses are:
• $H_{0}$: The absent days occur with equal frequencies, that is, they fit a uniform distribution.
• $H_{a}$: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.
If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: $15 + 12 + 9 + 9 + 15 = 60$), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected ($E$) values. The values in the table are the observed ($O$) values or data.
This time, calculate the $\chi^{2}$ test statistic by hand. Make a chart with the following headings and fill in the columns:
• Expected ($E$) values $(12, 12, 12, 12, 12)$
• Observed ($O$) values $(15, 12, 9, 9, 15)$
• $(O – E)$
• $(O – E)^{2}$
• $\frac{(O - E)^{2}}{E}$
Now add (sum) the last column. The sum is three. This is the $\chi^{2}$ test statistic.
To find the p-value, calculate $P(\chi^{2} > 3)$. This test is right-tailed. (Use a computer or calculator to find the p-value. You should get $p\text{-value} = 0.5578$.)
The $dfs$ are the $\text{number of cells} - 1 = 5 - 1 = 4$
Press 2nd DISTR. Arrow down to $\chi^{2}$cdf. Press ENTER. Enter(3,10^99,4). Rounded to four decimal places, you should see 0.5578, which is the $p\text{-value}$.
Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)
The decision is not to reject the null hypothesis.
Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.
TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example Example has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList). Enter the list name and press ENTER.
Exercise $2$
Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in the table below.
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Number of Students 11 8 10 7 10 5 5
From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?
Answer
$df = 6$
$p\text{-value} = 0.6093$
We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.
Example 11.3.3
One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in the table below.
Number of Televisions Percent
0 10
1 16
2 55
3 11
4+ 8
The table contains expected ($E$) percents.
A random sample of 600 families in the far western United States resulted in the data in the table below.
Number of Televisions Frequency
Total = 600
0 66
1 119
2 340
3 60
4+ 15
The table contains observed ($O$) frequency values.
Exercise $3$.1
At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole?
Answer
This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.
The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in the table below.
Number of Televisions Percent Expected Frequency
0 10 (0.10)(600) = 60
1 16 (0.16)(600) = 96
2 55 (0.55)(600) = 330
3 11 (0.11)(600) = 66
over 3 8 (0.08)(600) = 48
Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter $0.10*600$.
$H_{0}$: The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population.
$H_{a}$: The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population.
Distribution for the test: $\chi^{2}_{4}$ where $df = (\text{the number of cells}) - 1 = 5 - 1 = 4$.
Note 11.3.3.1
$df \neq 600 - 1$
Calculate the test statistic: $\chi^{2} = 29.65$
Graph:
Probability statement: $p\text{-value} = P(\chi^{2} > 29.65) = 0.000006$
Compare α and the p-value:
$\alpha = 0.01$
$p\text{-value} = 0.000006$
So, $\alpha > p\text{-value}$.
Make a decision: Since $\alpha > p\text{-value}$, reject $H_{0}$.
This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.
Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole.
Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of Example). Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10*600, .16*600, .55*600, .11*600, .08*600. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006(rounded to six decimal places), which is the p-value.
The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.
Exercise $3$
The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in the table below.
Number of Pets Percent
0 18
1 25
2 30
3 18
4+ 9
A random sample of 1,000 students from the Eastern United States resulted in the data in the table below.
Number of Pets Frequency
0 210
1 240
2 320
3 140
4+ 90
At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the $p\text{-value}$?
Answer
$p\text{-value} = 0.0036$
We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole.
Example 11.3.4
Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level.
Answer
This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is ${HH, HT, TH, TT}$. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins ($20 HH, 27 HT, 30 TH, 23 TT$) fit the expected distribution?"
Random Variable: Let $X =$ the number of heads in one flip of the two coins. $X$ takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three. Since $X =$ the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.
$H_{0}$: The coins are fair.
$H_{a}$: The coins are not fair.
Distribution for the test: $\chi^{2}_{2}$ where $df = 3 - 1 = 2$.
Calculate the test statistic: $\chi^{2} = 2.14$
Graph:
Probability statement: $p\text{-value} = P(\chi^{2} > 2.14) = 0.3430$
Compare α and the p-value:
$\alpha = 0.05$
$p\text{-value} = 0.3430$
$\alpha < p\text{-value}$.
Make a decision: Since $\alpha < p\text{-value}$, do not reject $H_{0}$.
Conclusion: There is insufficient evidence to conclude that the coins are not fair.
Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum".Enter L3. Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 2.14,1E99,2). Rounded to four places, you should see .3430, which is the p-value.
The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.
Exercise $4$
Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. The table below shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?
MDG Region Adult Literacy Rate (%)
Developed Regions 99.0
Commonwealth of Independent States 99.5
Northern Africa 67.3
Sub-Saharan Africa 62.5
Latin America and the Caribbean 91.0
Eastern Asia 93.8
Southern Asia 61.9
South-Eastern Asia 91.9
Western Asia 84.5
Oceania 66.4
Answer
$df = 9$
$\chi^{2} \text{ test statistic} = 26.38$
Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4. Into L2, put the expected frequencies 82, 82, 82, 82, 82, 82, 82, 82, 82, 82. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum". Enter L3. Rounded to two decimal places, you should see 26.38. Press 2nd DISTR. Arrow down to 7:χ2cdf(or press 7). Press ENTER. Enter 26.38,1E99,9). Rounded to four places, you should see .0018, which is the p-value.
The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.
Review
To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five.
Formula Review
$\sum_k \frac{(O - E)^{2}}{E}$ goodness-of-fit test statistic where:
$O$: observed values
$E$: expected value
$k$: number of different data cells or categories
$df = k - 1$ degrees of freedom
Determine the appropriate test to be used in the next three exercises.
Exercise $5$
An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.
Exercise $6$
An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.
Answer
a goodness-of-fit test
Exercise $7$
A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.
Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in the table below.
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10
The actual distribution for a class of 20 is in the table below.
Grade Frequency
A 7
B 7
C 5
D 1
Exercise $8$
$df =$ ______
Answer
3
Exercise $9$
State the null and alternative hypotheses.
Exercise $10$
$\chi^{2} \text{test statistic} =$ ______
Answer
2.04
Exercise $11$
$p\text{-value} =$ ______
Exercise $12$
At the 5% significance level, what can you conclude?
Answer
We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.
Use the following information to answer the next nine exercises: The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in the table below.
Ethnicity Number of Cases
White 2,229
Hispanic 1,157
Black/African-American 457
Asian, Pacific Islander 232
Total = 4,075
The percentage of each ethnic group in Santa Clara County is as in the table below.
Ethnicity Percentage of total county population Number expected (round to two decimal places)
White 42.9% 1748.18
Hispanic 26.7%
Black/African-American 2.6%
Asian, Pacific Islander 27.8%
Total = 100%
Exercise $13$
If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.
Perform a goodness-of-fit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County.
Exercise $14$
$H_{0}$: _______
Answer
$H_{0}$: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.
Exercise $15$
$H_{a}$: _______
Exercise $16$
Is this a right-tailed, left-tailed, or two-tailed test?
Answer
right-tailed
Exercise $17$
degrees of freedom = _______
Exercise $18$
$\chi^{2} \text{test statistic}$ = _______
Answer
88,621
Exercise $19$
$p\text{-value} =$ _______
Exercise $20$
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the $p\text{-value}$.
Let $\alpha = 0.05$
Decision: ________________
Reason for the Decision: ________________
Conclusion (write out in complete sentences): ________________
Answer
Graph: Check student’s solution.
Decision: Reject the null hypothesis.
Reason for the Decision: $p\text{-value} < \alpha$
Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.
Exercise $21$
Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/11%3A_The_Chi-Square_Distribution/11.03%3A_Goodness-of-Fit_Test.txt |
Tests of independence involve using a contingency table of observed (data) values.
The test statistic for a test of independence is similar to that of a goodness-of-fit test:
$\sum_{(i \cdot j)} \frac{(O-E)^{2}}{E}$
where:
• $O =$ observed values
• $E =$ expected values
• $i =$ the number of rows in the table
• $j =$ the number of columns in the table
There are $i \cdot j$ terms of the form $\frac{(O-E)^{2}}{E}$.
The expected value for each cell needs to be at least five in order for you to use this test.
A test of independence determines whether two factors are independent or not. You first encountered the term independence in Probability Topics. As a review, consider the following example.
Example $1$
Suppose $A =$ a speeding violation in the last year and $B =$ a cell phone user while driving. If $A$ and $B$ are independent then $P(A \text{ AND } B) = P(A)P(B)$. $A \text{ AND } B$ is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not.
Let $y =$ expected number of drivers who used a cell phone while driving and received speeding violations.
If $A$ and $B$ are independent, then $P(A \text{ AND } B) = P(A)P(B)$. By substitution,
$\frac{y}{755} = \left(\frac{70}{755}\right)\left(\frac{305}{755}\right) \nonumber$
Solve for $y$:
$y = \frac{(70)(305)}{755} = 28.3 \nonumber$
About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations.
In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent). If we do a test of independence using the example, then the null hypothesis is:
$H_{0}$: Being a cell phone user while driving and receiving a speeding violation are independent events.
If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation.
The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.
The number of degrees of freedom for the test of independence is:
$df = (\text{number of columns} - 1)(\text{number of rows} - 1) \nonumber$
The following formula calculates the expected number ($E$):
$E = \frac{\text{(row total)(column total)}}{\text{total number surveyed}} \nonumber$
Exercise $1$
A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll?
Answer
About 16 students are expected to be music students and on the honor roll.
Example $2$
In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In Table $1$ is a sample of the adult volunteers and the number of hours they volunteer per week.
Table $1$: Number of Hours Worked Per Week by Volunteer Type (Observed). The table contains observed (O) values (data).
Type of Volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row Total
Community College Students 111 96 48 255
Four-Year College Students 96 133 61 290
Nonstudents 91 150 53 294
Column Total 298 379 162 839
Is the number of hours volunteered independent of the type of volunteer?
Answer
The observed table and the question at the end of the problem, "Is the number of hours volunteered independent of the type of volunteer?" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is always right-tailed.
• $H_{0}$: The number of hours volunteered is independent of the type of volunteer.
• $H_{a}$: The number of hours volunteered is dependent on the type of volunteer.
The expected results are in Table $2$.
Table $2$: Number of Hours Worked Per Week by Volunteer Type (Expected). The table contains expected($E$) values (data).
Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours
Community College Students 90.57 115.19 49.24
Four-Year College Students 103.00 131.00 56.00
Nonstudents 104.42 132.81 56.77
For example, the calculation for the expected frequency for the top left cell is
$E = \frac{(\text{row total})(\text{column total})}{\text{total number surveyed}} = \frac{(255)(298)}{839} = 90.57 \nonumber$
Calculate the test statistic: $\chi^{2} = 12.99$ (calculator or computer)
Distribution for the test: $\chi^{2}_{4}$
$df = (3 \text{ columns} – 1)(3 \text{ rows} – 1) = (2)(2) = 4 \nonumber$
Graph:
Probability statement: $p\text{-value} = P(\chi^{2} > 12.99) = 0.0113$
Compare $\alpha$ and the $p\text{-value}$: Since no $\alpha$ is given, assume $\alpha = 0.05$. $p\text{-value} = 0.0113$. $\alpha > p\text{-value}$.
Make a decision: Since $\alpha > p\text{-value}$, reject $H_{0}$. This means that the factors are not independent.
Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another.
For the example in Table, if there had been another type of volunteer, teenagers, what would the degrees of freedom be?
USING THE TI-83, 83+, 84, 84+ CALCULATOR
Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from Table. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. If necessary, use the arrow keys to move the cursor after Observed: and press 2nd MATRX. Press 1:[A] to select matrix A. It is not necessary to enter expected values. The matrix listed after Expected: can be blank. Arrow down to Calculate. Press ENTER. The test statistic is 12.9909 and the p-value = 0.0113. Do the procedure a second time, but arrow down to Draw instead of calculate.
Exercise $2$
The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. Table $3$ shows the results:
Table $3$
Industry Sector 2000 2010 2020 Total
Nonagriculture wage and salary 13,243 13,044 15,018 41,305
Goods-producing, excluding agriculture 2,457 1,771 1,950 6,178
Services-providing 10,786 11,273 13,068 35,127
Agriculture, forestry, fishing, and hunting 240 214 201 655
Nonagriculture self-employed and unpaid family worker 931 894 972 2,797
Secondary wage and salary jobs in agriculture and private household industries 14 11 11 36
Secondary jobs as a self-employed or unpaid family worker 196 144 152 492
Total 27,867 27,351 31,372 86,590
We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom.
Answer
• $H_{0}$: The number of jobs is independent of the year.
• $H_{a}$: The number of jobs is dependent on the year.
$df = 12$
Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to c:$\chi^{2}$-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 227.73 and the $p\text{-value} = 5.90E - 42 = 0$. Do the procedure a second time but arrow down to Draw instead of calculate.
Example $3$
De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. Table shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.
Need to Succeed in School vs. Anxiety Level
Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
Row Total
High Need 35 42 53 15 10 155
Medium Need 18 48 63 33 31 193
Low Need 4 5 11 15 17 52
Column Total 57 95 127 63 58 400
1. How many high anxiety level students are expected to have a high need to succeed in school?
2. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?
3. $E = \frac{(\text{row total})(\text{column total})}{\text{total surveyed}} =$ ________
4. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________.
Solution
a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.
$E = \frac{(\text{row total})(\text{column total})}{\text{total surveyed}} = \frac{155 \cdot 57}{400} = 22.09$
The expected number of students who have a high anxiety level and a high need to succeed in school is about 22.
b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.
c. $E = \frac{(\text{row total})(\text{column total})}{\text{total surveyed}} = 8.19$
d. 8
Exercise $3$
Refer back to the information in Note. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020?
Answer
12,727, 14,965
Review
To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5.
Formula Review
Test of Independence
• The number of degrees of freedom is equal to $(\text{number of columns - 1})(\text{number of rows - 1})$.
• The test statistic is $\sum_{(i \cdot j)} \frac{(O-E)^{2}}{E}$ where $O =$ observed values, $E =$ expected values, $i =$ the number of rows in the table, and $j =$ the number of columns in the table.
• If the null hypothesis is true, the expected number $E = \frac{(\text{row total})(\text{column total})}{\text{total surveyed}}$.
Determine the appropriate test to be used in the next three exercises.
Exercise $4$
A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups.
Answer
a test of independence
Exercise $5$
The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. He takes a random sample of 100 players from different organizations.
Exercise $6$
A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. She takes a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing.
Answer
a test of independence
Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table $4$ shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel.
Table $4$
Traveling Distance Third class Second class First class Total
1–100 miles 21 14 6 41
101–200 miles 18 16 8 42
201–300 miles 16 17 15 48
301–400 miles 12 14 21 47
401–500 miles 6 6 10 22
Total 73 67 60 200
Exercise $7$
State the hypotheses.
• $H_{0}$: _______
• $H_{a}$: _______
Exercise $8$
$df =$ _______
Answer
8
Exercise $9$
How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets?
Exercise $10$
How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets?
Answer
6.6
Exercise $11$
What is the test statistic?
Exercise $12$
What is the $p\text{-value}$?
Answer
0.0435
Exercise $13$
What can you conclude at the 5% level of significance?
Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans and 7,650 whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites.
Exercise $14$
Complete the table.
Table $5$: Smoking Levels by Ethnicity (Observed)
Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White TOTALS
1-10
11-20
21-30
31+
TOTALS
Answer
Table $\PageIndex{5B}$
Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White Totals
1-10 9,886 2,745 12,831 8,378 7,650 41,490
11-20 6,514 3,062 4,932 10,680 9,877 35,065
21-30 1,671 1,419 1,406 4,715 6,062 15,273
31+ 759 788 800 2,305 3,970 8,622
Totals 18,830 8,014 19,969 26,078 27,559 10,0450
Exercise $15$
State the hypotheses.
• $H_{0}$: _______
• $H_{a}$: _______
Exercise $16$
Enter expected values in Table. Round to two decimal places.
Calculate the following values:
Answer
Table $6$
Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White
1-10 7777.57 3310.11 8248.02 10771.29 11383.01
11-20 6573.16 2797.52 6970.76 9103.29 9620.27
21-30 2863.02 1218.49 3036.20 3965.05 4190.23
31+ 1616.25 687.87 1714.01 2238.37 2365.49
Exercise $17$
$df =$ _______
Exercise $18$
$\chi^{2} \text{test statistic} =$ ______
Answer
10,301.8
Exercise $19$
$p\text{-value} =$ ______
Exercise $20$
Is this a right-tailed, left-tailed, or two-tailed test? Explain why.
Answer
right
Exercise $21$
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the $p\text{-value}$.
State the decision and conclusion (in a complete sentence) for the following preconceived levels of $\alpha$.
Exercise $22$
$\alpha = 0.05$
1. Decision: ___________________
2. Reason for the decision: ___________________
3. Conclusion (write out in a complete sentence): ___________________
Answer
1. Reject the null hypothesis.
2. $p\text{-value} < \alpha$
3. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.
Exercise $23$
$\alpha = 0.05$
1. Decision: ___________________
2. Reason for the decision: ___________________
3. Conclusion (write out in a complete sentence): ___________________
Glossary
Contingency Table
a table that displays sample values for two different factors that may be dependent or contingent on one another; it facilitates determining conditional probabilities. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(OpenStax)/11%3A_The_Chi-Square_Distribution/11.04%3A_Test_of_Independence.txt |
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