MNET 315 Ch 11 Text Nonparametric Tests (Missing from book)

The Sign Test 11.1 584 CHAPTER 11 Nonparametric Tests What You Should Learn How to use the sign test to test a population median How to use the paired-sample sign test to test the difference between two population medians (dependent samples) The Sign Test for a Population Median The Paired-Sample Sign Test The Sign Test for a Population Median Many of the hypothesis tests studied so far have imposed one or more requirements for a population distribution. For instance, some tests require that a population must have a normal distribution, and other tests require that population variances be equal. What should you do when such requirements cannot be met? For these cases, statisticians have developed hypothesis tests that are “distribution free.” Such tests are called nonparametric tests. A nonparametric test is a hypothesis test that does not require any specific conditions concerning the shapes of population distributions or the values of population parameters. DEFINITION Nonparametric tests are usually easier to perform than corresponding parametric tests. They are, however, usually less efficient than parametric tests. Stronger evidence is required to reject a null hypothesis using the results of a nonparametric test. Consequently, whenever possible, you should use a parametric test. One of the easiest nonparametric tests to perform is the sign test. The only condition necessary to use a sign test is that the sample is randomly selected. The sign test is a nonparametric test that can be used to test a population median against a hypothesized value k. DEFINITION The sign test for a population median can be left-tailed, right-tailed, or two-tailed. The null and alternative hypotheses for each type of test are shown below. Left-tailed test: H 0 : median Ú k and H a : median 6 k Right-tailed test: H 0 : median … k and H a : median 7 k Two-tailed test: H 0 : median = k and H a : median ≠ k To use the sign test, first compare each entry in the sample with the hypothesized median k . When the entry is below the median, assign it a - sign; when the entry is above the median, assign it a + sign; and when the entry is equal to the median, assign it a 0. Then compare the number of + and - signs. (The 0’s are ignored.) When there is a large difference between the number of + signs and the number of - signs, it is likely that the median is different from the hypothesized value and you should reject the null hypothesis. Study Tip For many nonparametric tests, statisticians test the median instead of the mean.

SECTION 11.1 The Sign Test 587 Using the Sign Test An organization claims that the median annual attendance for museums in the United States is at least 39,000. A random sample of 125 museums reveals that the annual attendances for 79 museums were less than 39,000, the annual attendances for 42 museums were more than 39,000, and the annual attendances for 4 museums were 39,000. At a = 0.01, is there enough evidence to reject the organization’s claim? (Adapted from American Association of Museums) SOLUTION The claim is “the median annual attendance for museums in the United States is at least 39,000.” So, the null and alternative hypotheses are H 0 : median Ú 39,000 (Claim) and H a : median 6 39,000. Because n 7 25, use Table 4 in Appendix B, the Standard Normal Table, to find the critical value. Because the test is a left-tailed test with a = 0.01, the critical value is z 0 = - 2.33. Of the 125 museums, there are 79 - signs and 42 + signs. When the 0’s are ignored, the sample size is n = 79 + 42 = 121, and x = 42. With these values, the test statistic is z = 1 42 + 0.5 2 - 0.5 1 121 2 2 121 2 = - 18 5.5 ≈ - 3.27. The figure shows the location of the rejection region and the test statistic z . Because z is less than the critical value, it is in the rejection region. So, you reject the null hypothesis. z − 3 − 4 − 2 − 1 0 1 2 3 4 z 0 = − 2.33 α = 0.01 z ≈ − 3.27 Interpretation There is enough evidence at the 1% level of significance to reject the organization’s claim that the median annual attendance for museums in the United States is at least 39,000. TRY IT YOURSELF 2 An organization claims that the median age of museum workers in the United States is 46 years old. A random sample of 95 museum workers reveals that 57  museum workers were less than 46 years old, 34 museum workers were more than 46 years old, and 4 museum workers were 46 years old. At a = 0.10, can you reject the organization’s claim? (Adapted from American Association of Museums) Answer: Page T1 EXAMPLE 2 Picturing the World For recent college graduates in the United States, a financial analyst claims that the median auto loan is $21,883. A random sample of recent college graduates reveals that the loans for 42 graduates were less than $21,883 and the loans for 35 graduates were greater than $21,883. (Adapted from lendedu.com) Would you use a parametric test or a nonparametric test to test the claim that for recent college graduates in the United States, the median auto loan is $21,883? Explain your reasoning. Study Tip When performing a two-tailed sign test, remember to use only the left-tailed critical value.

11.1 EXERCISES 590 CHAPTER 11 Nonparametric Tests For Extra Help: MyLab Statistics Building Basic Skills and Vocabulary 1. What is a nonparametric test? How does a nonparametric test differ from a parametric test? What are the advantages and disadvantages of using a nonparametric test? 2. When the sign test is used, what population parameter is being tested? 3. Describe the test statistic for the sign test when the sample size n is less than or equal to 25 and when n is greater than 25. 4. In your own words, explain why the hypothesis test discussed in this section is called the sign test. 5. Explain how to use the sign test to test a population median. 6. List the two conditions that must be met in order to use the paired-sample sign test. Using and Interpreting Concepts Performing a Sign Test In Exercises 7–22, (a) identify the claim and state H 0 and H a , (b) find the critical value, (c) find the test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. 7. Credit Card Charges A financial service accountant claims that the median credit card balance of college students is more than $300. You randomly select the credit card accounts of 12 college students and record the balance for each account. The balances (in dollars) are listed below. At a = 0.01, can you support the accountant’s claim? (Adapted from Sallie Mae) 346.71 382.59 255.03 202.17 309.80 265.88 299.41 270.38 296.54 318.46 245.92 309.47 8. Temperature A meteorologist claims that the median daily high temperature  for the month of July in Pittsburgh is 83 ° Fahrenheit. The high temperatures (in degrees Fahrenheit) for 15 randomly selected July days in Pittsburgh are listed below. At a = 0.01, is there enough evidence to reject the meteorologist’s claim? (Adapted from U.S. National Oceanic and Atmospheric Administration) 74 79 81 86 90 79 81 83 81 74 78 76 84 82 85 9. Sales Prices of Homes A real estate agent claims that the median sales price of new privately owned one-family homes sold in a recent month is $253,000 or less. The sales prices (in dollars) of 10 randomly selected homes are listed below. At a = 0.05, is there enough evidence to reject the agent’s claim? (Adapted from National Association of Realtors) 262,600 300,100 269,200 249,400 183,400 253,500 325,600 223,500 241,300 271,300

SECTION 11.1 The Sign Test 593 22. SAT Scores A guidance counselor claims that students who take the SAT twice will improve their scores the second time they take the SAT. The table shows both math SAT scores for 12 students who took the SAT twice. At a = 0.01, can you support the guidance counselor’s claim? Student 1 2 3 4 5 6 Score on first SAT 440 510 420 450 620 450 Score on second SAT 440 570 510 470 610 450 Student 7 8 9 10 11 12 Score on first SAT 350 470 320 510 630 570 Score on second SAT 370 530 290 500 640 600 23. Feeling Your Age A research organization conducts a survey by randomly selecting adults and asking each, “How do you feel relative to your age?” The results are shown in the figure. (Adapted from Pew Research Center) Younger 11 Older 3 My age 9 (a) Use a sign test to test the null hypothesis that the proportion of adults who feel older is equal to the proportion of adults who feel younger. Assign a + sign to each adult who responded “older,” assign a - sign to each adult who responded “younger,” and assign a 0 to each adult who responded “my age.” Use a = 0.05. (b) What can you conclude? 24. Contacting Parents A research organization conducts a survey by randomly selecting adults and asking each, “How frequently do you contact your parents by phone?” The results are shown in the figure. (Adapted from Pew Research Center) Weekly 12 Daily 8 Other 6 (a) Use a sign test to test the null hypothesis that the proportion of adults who contact their parents by phone weekly is equal to the proportion of adults who contact their parents by phone daily. Assign a + sign to each adult who responded “weekly,” assign a - sign to each adult who responded “daily,” and assign a 0 to each adult who responded “other.” Use a = 0.05. (b) What can you conclude?

596 CHAPTER 11 Nonparametric Tests Performing a Wilcoxon Signed-Rank Test A golf club manufacturer claims that golfers can lower their scores by using the manufacturer’s newly designed golf clubs. The table shows the scores of 10 golfers while using the old design and while using the new design on the same golf course. At a = 0.05, can you support the manufacturer’s claim? Golfer 1 2 3 4 5 6 7 8 9 10 Score (old design) 89 84 96 74 91 85 95 82 92 81 Score (new design) 83 83 92 76 91 80 87 85 90 77 SOLUTION The claim is “golfers can lower their scores.” To test this claim, use the null and alternative hypotheses below. H 0 : The new design does not lower scores. H a : The new design lowers scores. (Claim) This Wilcoxon signed-rank test is a one-tailed test with a = 0.05, and because one data pair has a difference of 0, n = 9 instead of 10. From Table 9 in Appendix B, the critical value is 8. To find the test statistic w s , complete a table as shown below. Score (old design) Score (new design) Difference Absolute value Rank Signed rank 89 83 6 6 8 8 84 83 1 1 1 1 96 92 4 4 5.5 5.5 74 76 - 2 2 2.5 - 2.5 91 91 0 0 — — 85 80 5 5 7 7 95 87 8 8 9 9 82 85 - 3 3 4 - 4 92 90 2 2 2.5 2.5 81 77 4 4 5.5 5.5 The sum of the negative ranks is - 2.5 + 1 - 4 2 = - 6.5. The sum of the positive ranks is 8 + 1 + 5.5 + 7 + 9 + 2.5 + 5.5 = 38.5. The test statistic is the smaller absolute value of these two sums. Because 0 - 6.5 0 6 0 38.5 0 , the test statistic is w s = 6.5. Because the test statistic is less than the critical value, that is, 6.5 6 8, you reject the null hypothesis. Interpretation There is enough evidence at the 5% level of significance to support the claim that golfers can lower their scores by using the newly designed clubs. EXAMPLE 1 Study Tip Do not assign a rank to any difference of 0. In the case of a tie between data entries, use the average of the corresponding ranks. For instance, when two data entries are tied for the fifth rank, use the average of 5 and 6, which is 5.5, as the rank for both entries. The next data entry will be assigned a rank of 7, not 6. When three entries are tied for the fifth rank, use the average of 5, 6, and 7, which is 6, as the rank for all three data entries. The next data entry will be assigned a rank of 8.

SECTION 11.2 The Wilcoxon Tests 599 The sample size for men is 10 and the sample size for women is 12. Because 10 6 12, n 1 = 10 and n 2 = 12. Before calculating the test statistic, you must find the values of R , m R , and s R . The table shows the combined data listed in ascending order and the corresponding ranks. Ordered data Sample Rank 77 F 1 78 M 2 84 F 3 85 F 4 86 M 5.5 86 F 5.5 87 F 7 90 F 8 91 F 9 93 M 10.5 93 F 10.5 94 M 12 95 M 13 97 F 14 98 M 15.5 98 F 15.5 99 M 17 100 F 18 101 M 19.5 101 F 19.5 114 M 21 117 M 22 Because the smaller sample is the sample of males, R is the sum of the male rankings. R = 2 + 5.5 + 10.5 + 12 + 13 + 15.5 + 17 + 19.5 + 21 + 22 = 138 Using n 1 = 10 and n 2 = 12, you can find m R and s R as follows. m R = n 1 1 n 1 + n 2 + 1 2 2 = 10 1 10 + 12 + 1 2 2 = 230 2 = 115 Study Tip Remember that in the case of a tie between data entries, use the average of the corresponding ranks.

602 CHAPTER 11 Nonparametric Tests 5. Earnings by Degree A college administrator claims that there is a difference in the earnings of people with bachelor’s degrees and those with advanced degrees. The table shows the earnings (in thousands of dollars) of a random sample of 11 people with bachelor’s degrees and 10 people with advanced degrees. At a = 0.05, is there enough evidence to support the administrator’s claim? (Adapted from U.S. Census Bureau) Bachelor’s degree 62 58 71 84 78 58 52 64 68 60 62 Advanced degree 88 91 99 85 90 91 98 98 95 87 6. Headaches A medical researcher wants to determine whether a new drug affects the number of headache hours experienced by headache sufferers. To do so, the researcher randomly selects seven patients and asks each to give the number of headache hours (per day) each experiences before and after taking the drug. The table shows the results. At a = 0.05, can the researcher conclude that the new drug affects the number of headache hours? Patient 1 2 3 4 5 6 7 Headache hours (before) 0.8 2.4 2.8 2.6 2.7 0.9 1.2 Headache hours (after) 1.6 1.3 1.6 1.4 1.5 1.6 1.7 7. Teacher Salaries A teacher’s union representative claims that there is a difference in the salaries earned by teachers in Wisconsin and Michigan. The table shows the salaries (in thousands of dollars) of a random sample of 11 teachers from Wisconsin and 12 teachers from Michigan. At a = 0.05, is there enough evidence to support the representative’s claim? (Adapted from National Education Association) Wisconsin 55 59 49 56 51 61 55 61 53 47 52 Michigan 61 65 55 62 57 67 61 67 59 53 58 76 8. Heart Rate A physician wants to determine whether an experimental medication affects an individual’s heart rate. The physician randomly selects 15 patients and measures the heart rate of each. The subjects then take the medication and have their heart rates measured after one hour. The table shows the results. At a = 0.05, can the physician conclude that the experimental medication affects an individual’s heart rate? Patient 1 2 3 4 5 6 7 8 Heart rate (before) 72 81 75 76 79 74 65 67 Heart rate (after) 73 80 75 79 74 76 73 67 Patient 9 10 11 12 13 14 15 Heart rate (before) 76 83 66 75 76 78 68 Heart rate (after) 74 77 70 77 76 75 74

SECTION 11.2 The Wilcoxon Tests 603 Extending Concepts Wilcoxon Signed-Rank Test for n + 30 When you are performing a Wilcoxon signed-rank test and the sample size n is greater than 30, you can use the Standard Normal Table and the formula below to find the test statistic. z = w s - n 1 n + 1 2 4 B n 1 n + 1 21 2 n + 1 2 24 In Exercises 9 and 10, perform the Wilcoxon signed-rank test using the test statistic for n 7 30. 9. Fuel Additive A petroleum engineer wants to know whether a certain fuel additive improves a car’s gas mileage. To decide, the engineer records the gas mileages (in miles per gallon) of 33 randomly selected cars with and without the fuel additive. The table shows the results. At a = 0.10, can the engineer conclude that the gas mileage is improved? Car 1 2 3 4 5 6 7 8 9 10 11 Without additive 36.4 36.4 36.6 36.6 36.8 36.9 37.0 37.1 37.2 37.2 36.7 With additive 36.7 36.9 37.0 37.5 38.0 38.1 38.4 38.7 38.8 38.9 36.3 Car 12 13 14 15 16 17 18 19 20 21 22 Without additive 37.5 37.6 37.8 37.9 37.9 38.1 38.4 40.2 40.5 40.9 35.0 With additive 38.9 39.0 39.1 39.4 39.4 39.5 39.8 40.0 40.0 40.1 36.3 Car 23 24 25 26 27 28 29 30 31 32 33 Without additive 32.7 33.6 34.2 35.1 35.2 35.3 35.5 35.9 36.0 36.1 37.2 With additive 32.8 34.2 34.7 34.9 34.9 35.3 35.9 36.4 36.6 36.6 38.3 10. Fuel Additive A petroleum engineer claims that a fuel additive improves gas mileage. The table shows the gas mileages (in miles per gallon) of 32 randomly selected cars measured with and without the fuel additive. Test the petroleum engineer’s claim at a = 0.05. Car 1 2 3 4 5 6 7 8 9 10 11 Without additive 34.0 34.2 34.4 34.4 34.6 34.8 35.6 35.7 30.2 31.6 32.3 With additive 36.6 36.7 37.2 37.2 37.3 37.4 37.6 37.7 34.2 34.9 34.9 Car 12 13 14 15 16 17 18 19 20 21 22 Without additive 33.0 33.1 33.7 33.7 33.8 35.7 36.1 36.1 36.6 36.6 36.8 With additive 34.9 35.7 36.0 36.2 36.5 37.8 38.1 38.2 38.3 38.3 38.7 Car 23 24 25 26 27 28 29 30 31 32 Without additive 37.1 37.1 37.2 37.9 37.9 38.0 38.0 38.4 38.8 42.1 With additive 38.8 38.9 39.1 39.1 39.2 39.4 39.8 40.3 40.8 43.2

604 CASE STUDY College Ranks 604 CHAPTER 11 Nonparametric Tests Each year, Forbes and the Center for College Affordability and Productivity (CCAP) release a list of the best colleges in the United States. Over 600 colleges and universities are ranked according to factors that fall into one of five categories. 1. Postgraduate success, which is based on salary of alumni by school and the alumni who appear on CCAP’s America’s Leaders list 2. Student debt, which is based on three components: average federal student loan debt load, student loan default rates, and predicted versus actual percent of students taking federal loans 3. Student satisfaction, which is based on student retention rates and student evaluations of professors 4. Graduation rate, which is based on how many students actually finish their degrees in four years and the actual versus predicted rate 5. Academic success, which is based on students who have won competitive scholarships and fellowships, and students who have gone on to earn Ph.D.s The table shows the student populations for randomly selected colleges by region on the 2016 list. EXERCISES 1. Construct a side-by-side box-and-whisker plot for the four regions. Do any of the median student populations appear to be the same? Do any appear to be different? In Exercises 2 –5, use the sign test to test the claim. What can you conclude? Use a = 0.05. 2. The median student population at a college in the Northeast is less than or equal to 7000. 3. The median student population at a college in the Midwest is greater than or equal to 8000. 4. The median student population at a college in the South is 10,000. 5. The median student population at a college in the West is different from 8000. In Exercises 6 and 7, use the Wilcoxon rank sum test to test the claim. Use a = 0.01. 6. There is no difference between student populations for colleges in the Midwest and colleges in the West. 7. There is a difference between student populations for colleges in the Northeast and colleges in the South. Student populations Northeast Midwest South West 1,805 24,766 6,621 1,498 9,181 2,948 14,769 1,394 14,317 1,459 29,175 1,144 2,113 3,688 15,984 8,132 20,445 3,418 2,850 12,820 1,632 14,747 27,511 50,320 5,123 14,906 24,932 31,354 755 5,931 49,610 2,127 15,117 2,791 10,033 19,934 18,090 11,458 1,575 31,332

The Kruskal-Wallis Test 11.3 SECTION 11.3 The Kruskal-Wallis Test 605 What You Should Learn How to use the Kruskal-Wallis test to determine whether three or more samples were selected from populations having the same distribution The Kruskal-Wallis Test The Kruskal-Wallis Test In Section 10.4, you learned how to use one-way ANOVA techniques to compare the means of three or more populations. When using one-way ANOVA, you should verify that each independent sample is selected from a population that is normally, or approximately normally, distributed. When you cannot verify that the populations are normal, you can still compare the distributions of three or more populations. To do so, you can use the Kruskal-Wallis test. The Kruskal-Wallis test is a nonparametric test that can be used to determine whether three or more independent samples were selected from populations having the same distribution. DEFINITION For a Kruskal-Wallis test, the null and alternative hypotheses are always similar to these statements. H 0 : All of the populations have the same distribution. H a : At least one population has a distribution that is different from the others. The conditions for using the Kruskal-Wallis test are that the samples must be random and independent, and the size of each sample must be at least 5. If these conditions are met, then the sampling distribution for the Kruskal-Wallis test is approximated by a chi-square distribution with k - 1 degrees of freedom, where k is the number of samples. You can calculate the Kruskal-Wallis test statistic using the formula below. For three or more independent samples, the test statistic for the Kruskal-Wallis test is H = 12 N 1 N + 1 2 a R 2 1 n 1 + R 2 2 n 2 + c + R 2 k n k b - 3 1 N + 1 2 where k is the number of samples, n i is the size of the i th sample, N is the sum of the sample sizes, and R i is the sum of the ranks of the i th sample. Test Statistic for the Kruskal-Wallis Test Performing a Kruskal-Wallis test consists of combining and ranking the sample data. The data are then separated according to sample and the sum of the ranks of each sample is calculated.

606 CHAPTER 11 Nonparametric Tests These sums are then used to calculate the test statistic H , which is an approximation of the variance of the rank sums. When the samples are selected from populations having the same distribution, the sums of the ranks will be approximately equal, H will be small, and you should fail to reject the null hypothesis. When the samples are selected from populations not having the same distribution, the sums of the ranks will be quite different, H will be large, and you should reject the null hypothesis. Because you only reject the null hypothesis when H is significantly large, the Kruskal-Wallis test is always a right-tailed test. Performing a Kruskal-Wallis Test In Words In Symbols 1. Verify that the samples are random and independent, and each sample size is at least 5. 2. Identify the claim. State the null State H 0 and H a . and alternative hypotheses. 3. Specify the level of significance. Identify a . 4. Identify the degrees d.f. = k - 1 of freedom. 5. Determine the critical value Use Table 6 in Appendix B. and the rejection region. 6. Find the sum of the ranks for each sample. a. List the combined data in ascending order. b. Rank the combined data. 7. Find the test statistic and sketch H = 12 N 1 N + 1 2 # a R 2 1 n 1 + R 2 2 n 2 + c + R 2 k n k b - 3 1 N + 1 2 the sampling distribution. 8. Make a decision to reject or fail If H is in the rejection region, to reject the null hypothesis. then reject H 0 . Otherwise, fail to reject H 0 . 9. Interpret the decision in the context of the original claim. GUIDELINES

SECTION 11.3 The Kruskal-Wallis Test 607 Performing a Kruskal-Wallis Test You want to compare the numbers of crimes reported in three police precincts in a city. To do so, you randomly select 10 weeks for each precinct and record the numbers of crimes reported. The table shows the results. At a = 0.01, can you conclude that the distribution of the numbers of crimes reported in at least one precinct is different from the others? Number of crimes reported for the week 101st Precinct (Sample 1) 106th Precinct (Sample 2) 113th Precinct (Sample 3) 60 65 69 52 55 51 49 64 70 52 66 61 50 53 67 48 58 65 57 50 62 45 54 59 44 70 60 56 62 63 SOLUTION You want to test the claim that the distribution of the numbers of crimes reported in at least one precinct is different from the others. The null and alternative hypotheses are as follows. H 0 : The distribution of the numbers of crimes reported is the same in all three precincts. H a : The distribution of the numbers of crimes reported in at least one precinct is different from the others. (Claim) The test is a right-tailed test with a = 0.01 and d.f. = k - 1 = 3 - 1 = 2. From Table 6 in Appendix B, the critical value is x 2 0 = 9.210. The rejection region is x 2 7 9.210. Before calculating the test statistic, you must find the sum of the ranks for each sample. The table shows the combined data listed in ascending order and the corresponding ranks. Ordered data Sample Rank 44 101st 1 45 101st 2 48 101st 3 49 101st 4 50 101st 5.5 50 106th 5.5 51 113th 7 52 101st 8.5 52 101st 8.5 53 106th 10 Ordered data Sample Rank 54 106th 11 55 106th 12 56 101st 13 57 101st 14 58 106th 15 59 113th 16 60 101st 17.5 60 113th 17.5 61 113th 19 62 106th 20.5 Ordered data Sample Rank 62 113th 20.5 63 113th 22 64 106th 23 65 106th 24.5 65 113th 24.5 66 106th 26 67 113th 27 69 113th 28 70 106th 29.5 70 113th 29.5 EXAMPLE 1

608 CHAPTER 11 Nonparametric Tests The sum of the ranks for each sample is as follows. R 1 = 1 + 2 + 3 + 4 + 5.5 + 8.5 + 8.5 + 13 + 14 + 17.5 = 77 R 2 = 5.5 + 10 + 11 + 12 + 15 + 20.5 + 23 + 24.5 + 26 + 29.5 = 177 R 3 = 7 + 16 + 17.5 + 19 + 20.5 + 22 + 24.5 + 27 + 28 + 29.5 = 211 Using these sums and the values n 1 = 10, n 2 = 10, n 3 = 10, and N = 30, the test statistic is H = 12 30 1 30 + 1 2 a 77 2 10 + 177 2 10 + 211 2 10 b - 3 1 30 + 1 2 ≈ 12.521. The figure shows the location of the rejection region and the test statistic H . Because H is in the rejection region, you reject the null hypothesis. 2 4 6 8 10 12 14 χ 2 H ≈ 12.521 α = 0.01 0 2 χ = 9.210 Interpretation There is enough evidence at the 1% level of significance to support the claim that the distribution of the numbers of crimes reported in at least one precinct is different from the others. TRY IT YOURSELF 1 You want to compare the salaries of veterinarians who work in Texas, Florida, and California. To compare the salaries, you randomly select several veterinarians in each state and record their salaries. The table shows the salaries (in thousands of dollars). At a = 0.05, can you conclude that the distribution of the veterinarians’ salaries in at least one state is different from the others? (Adapted from U.S. Bureau of Labor Statistics) Sample salaries (in thousands of dollars) TX (Sample 1) FL (Sample 2) CA (Sample 3) 85.3 143.3 111.3 149.9 135.9 83.4 97.9 121.6 126.8 91.0 80.4 146.1 89.6 116.6 154.0 147.7 106.7 160.2 63.3 84.7 57.6 74.8 95.0 113.2 118.7 105.3 131.0 101.1 Answer: Page T1 Picturing the World The randomly collected data below were used to compare the water temperatures (in degrees Fahrenheit) of cities bordering the Gulf of Mexico. (Adapted from National Oceanographic Data Center) Cedar Key, FL (Sample 1) Eugene Island, LA (Sample 2) Dauphin Island, AL (Sample 3) 62 51 63 69 55 51 77 57 54 59 63 60 60 74 75 75 82 80 83 85 70 65 60 78 79 64 82 86 76 84 82 83 86 At A = 0.05, can you conclude that at least one temperature distribution is different from the others?

11.3 EXERCISES SECTION 11.3 The Kruskal-Wallis Test 609 For Extra Help: MyLab Statistics Building Basic Skills and Vocabulary 1. What are the conditions for using a Kruskal-Wallis test? 2. Explain why the Kruskal-Wallis test is always a right-tailed test. Using and Interpreting Concepts Performing a Kruskal-Wallis Test In Exercises 3 – 6, (a) identify the claim and state H 0 and H a , (b) find the critical value and identify the rejection region, (c) find the test statistic H, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. 3. Home Insurance The table shows the annual premiums for a random sample of home insurance policies in Connecticut, Massachusetts, and Virginia. At a = 0.05, can you conclude that the distribution of the annual premiums in at least one state is different from the others? (Adapted from National Association of Insurance Commissioners) State Annual premium (in dollars) Connecticut 1303 1098 1263 1413 1538 1179 1320 Massachusetts 1382 1302 1257 1572 1387 1166 1034 Virginia 1035 950 766 845 1132 838 755 4. Hourly Rates A researcher wants to determine whether there is a difference in the hourly pay rates for registered nurses in Indiana, Kentucky, and Ohio. The researcher randomly selects several registered nurses in each state and records the hourly pay rate for each. The table shows the results. At a = 0.05, can the researcher conclude that the distribution of the hourly pay rates of registered nurses in at least one state is different from the others? (Adapted from U.S. Bureau of Labor Statistics) State Hourly pay rate (in dollars) Indiana 28.83 29.28 27.68 28.43 31.27 26.13 30.47 Kentucky 27.77 26.40 28.92 31.02 29.37 32.42 25.42 Ohio 27.84 32.24 33.64 33.91 27.34 29.89 5. Annual Salaries The table shows the annual salaries for a random sample of private industry workers in Kentucky, North Carolina, South Carolina, and West Virginia. At a = 0.10, can you conclude that the distribution of the annual salaries of private industry workers in at least one state is different from the others? (Adapted from U.S. Bureau of Labor Statistics) State Annual salary (in thousands of dollars) Kentucky 39.9 41.6 50.5 62.1 38.3 32.9 39.9 North Carolina 48.8 47.2 41.9 59.6 40.8 44.9 48.8 South Carolina 35.4 43.0 49.1 48.5 40.3 41.7 35.4 West Virginia 34.8 45.9 36.6 45.1 50.3 38.1 34.8

610 CHAPTER 11 Nonparametric Tests 6. Caffeine Content The table shows the amounts of caffeine (in milligrams) in 16-ounce servings for a random sample of beverages. At a = 0.01, can you conclude that the distribution of the amounts of caffeine in at least one beverage is different from the others? (Adapted from Center for Science in the Public Interest) Beverage Amount of caffeine in 16-ounce serving (in milligrams) Coffees 320 300 206 150 266 Soft drinks 95 96 56 51 71 72 47 Energy drinks 200 141 160 152 154 166 Teas 100 106 42 15 32 10 Extending Concepts Comparing Two Tests In Exercises 7 and 8, (a) perform a Kruskal-Wallis test. (b) perform a one-way ANOVA test, assuming that each population is normally distributed and the population variances are equal. (c) Compare the results. 7. Hospital Patient Stays An insurance underwriter claims that the number of days patients spend in the hospital is different in at least one region of the United States. The table shows the numbers of days randomly selected patients spent in the hospital in four U.S. regions. At a = 0.01, can you support the underwriter’s claim? (Adapted from U.S. National Center for Health Statistics) Region Number of days Northeast 8 6 6 3 5 11 3 8 1 6 Midwest 5 4 3 9 1 4 6 3 4 7 South 5 8 1 5 8 7 5 1 West 2 3 6 6 5 4 3 6 5 8. Energy Consumption The table shows the energy consumed (in millions of Btu) in one year for a random sample of households from four U.S. regions. At a = 0.01, can you conclude that the energy consumed is different in at least one region? (Adapted from U.S. Energy Information Administration) Region Energy consumed (in millions of Btu) Northeast 61 95 140 127 93 97 84 123 89 163 Midwest 59 158 169 140 95 187 123 104 88 37 72 South 86 35 67 86 142 69 65 62 West 81 39 85 35 113 46 125 70 77 63

Rank Correlation 11.4 SECTION 11.4 Rank Correlation 611 What You Should Learn How to use the Spearman rank correlation coefficient to determine whether the correlation between two variables is significant The Spearman Rank Correlation Coefficient The Spearman Rank Correlation Coefficient In Section 9.1, you learned how to measure the strength of the relationship between two variables using the Pearson correlation coefficient r . Two requirements for the Pearson correlation coefficient are that the variables are linearly related and that the variables have a bivariate normal distribution. When these requirements cannot be met, you can examine the relationship between two variables using the nonparametric equivalent to the Pearson correlation coefficient—the Spearman rank correlation coefficient. The Spearman rank correlation coefficient has several advantages over the Pearson correlation coefficient. For instance, the Spearman rank correlation coefficient can be used to describe the relationship between linear or nonlinear data. The Spearman rank correlation coefficient can be used for data at the ordinal level. And, the Spearman rank correlation coefficient is easier to calculate by hand than the Pearson correlation coefficient. The Spearman rank correlation coefficient r s is a measure of the strength of the relationship between two variables. The Spearman rank correlation coefficient is calculated using the ranks of paired sample data entries. If there are no ties in the ranks of either variable, then the formula for the Spearman rank correlation coefficient is r s = 1 - 6 Σ d 2 n ( n 2 - 1 ) where n is the number of paired data entries and d is the difference between the ranks of a paired data entry. If there are ties in the ranks and the number of ties is small relative to the number of data pairs, then the formula can still be used to approximate r s . DEFINITION The values of r s range from - 1 to 1, inclusive. When the ranks of corresponding data pairs are exactly identical, r s is equal to 1. When the ranks are in “reverse” order, r s is equal to - 1. When the ranks of corresponding data pairs have no relationship, r s is equal to 0. After calculating the Spearman rank correlation coefficient, you can determine whether the correlation between the variables is significant. You can make this determination by performing a hypothesis test for the population correlation coefficient r s . The null and alternative hypotheses for this test are listed below. H 0 : r s = 0 (There is no correlation between the variables.) H a : r s ≠ 0 (There is a significant correlation between the variables.) Table 10 in Appendix B lists the critical values for the Spearman rank correlation coefficient for selected levels of significance and sample sizes. The test statistic for the hypothesis test is the Spearman rank correlation coefficient r s .

612 CHAPTER 11 Nonparametric Tests Testing the Significance of the Spearman Rank Correlation Coefficient In Words In Symbols 1. Identify the claim. State the null State H 0 and H a . and alternative hypotheses. 2. Specify the level of significance. Identify a . 3. Determine the critical value. Use Table 10 in Appendix B. 4. Find the test statistic. r s = 1 - 6 Σ d 2 n ( n 2 - 1 ) 5. Make a decision to reject or fail If 0 r s 0 is greater than the to reject the null hypothesis. critical value, then reject H 0 . Otherwise, fail to reject H 0 . 6. Interpret the decision in the context of the original claim. GUIDELINES The Spearman Rank Correlation Coefficient The table shows the school enrollments of males and females for a random sample of 10 colleges. At a = 0.05, can you conclude that there is a significant correlation between the number of males and the number of females enrolled at a college? Male Female 1786 2182 4246 4415 1419 1537 1188 1236 2394 2182 1079 919 4049 4209 3595 3741 1102 1086 1345 1282 SOLUTION The claim is “there is a significant correlation between the number of males and the number of females enrolled at a college.” The null and alternative hypotheses are listed below. H 0 : r s = 0 (There is no correlation between the number of males and the number of females enrolled at a college.) H a : r s ≠ 0 (There is a significant correlation between the number of males and the number of females enrolled at a college.) (Claim) EXAMPLE 1

SECTION 11.4 Rank Correlation 613 Each data set has 10 entries. Because a = 0.05 and n = 10, the critical value is 0.648. Before calculating the test statistic, you must find Σ d 2 , the sum of the squares of the differences of the ranks of the data sets. You can use a table to calculate Σ d 2 , as shown below. Male Rank Female Rank d d 2 1786 6 2182 6.5 - 0.5 0.25 4246 10 4415 10 0 0 1419 5 1537 5 0 0 1188 3 1236 3 0 0 2394 7 2182 6.5 0.5 0.25 1079 1 919 1 0 0 4049 9 4209 9 0 0 3595 8 3741 8 0 0 1102 2 1086 2 0 0 1345 4 1282 4 0 0 Σ d 2 = 0.5 When n = 10 and Σ d 2 = 0.5, the test statistic is r s = 1 - 6 Σ d 2 n ( n 2 - 1 ) = 1 - 6 1 0.5 2 10 ( 10 2 - 1 ) ≈ 0.997. Because 0 r s 0 ≈ 0.997 7 0.648, you reject the null hypothesis. Interpretation There is enough evidence at the 5% level of significance to conclude that there is a significant correlation between the number of males and the number of females enrolled at a college. TRY IT YOURSELF 1 The table shows the prices (in dollars per bushel) received for oat and wheat for a random sample of seven U.S. farmers. At a = 0.10, can you conclude that there is a significant correlation between the oat and wheat prices? (Adapted from U.S. Department of Agriculture) Oat Wheat 1.84 3.67 1.97 3.49 2.03 3.68 2.25 3.88 2.35 3.91 2.31 4.02 2.40 4.15 Answer: Page T1 Picturing the World The table shows the retail prices (in dollars per pound) for ground beef and fresh whole chicken for a random sample of nine U.S. grocery stores. (Adapted from U.S. Bureau of Labor Statistics) Beef Chicken 3.69 1.44 3.66 1.42 3.65 1.48 3.68 1.50 3.60 1.47 3.55 1.46 3.55 1.41 3.56 1.47 3.59 1.46 Does a significant correlation exist between ground beef and chicken prices in U.S. grocery stores? Use A = 0.10. Study Tip Remember that in the case of a tie between data entries, use the average of the corresponding ranks.

11.4 EXERCISES 614 CHAPTER 11 Nonparametric Tests For Extra Help: MyLab Statistics Building Basic Skills and Vocabulary 1. What are some advantages of the Spearman rank correlation coefficient over the Pearson correlation coefficient? 2. Describe the ranges of the Spearman rank correlation coefficient and the Pearson correlation coefficient. 3. What does it mean when r s is equal to 1? What does it mean when r s is equal to - 1? What does it mean when r s is equal to 0? 4. Explain, in your own words, what r s and r s represent in Example 1. Using and Interpreting Concepts Testing a Claim In Exercises 5 – 8, (a) identify the claim and state H 0 and H a , (b) find the critical value, (c) find the test statistic r s , (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. 5. Farming Expenses In an agricultural report, a commodities analyst claims that there is a significant correlation between purchased seed expenses and fertilizer and lime expenses in the farming business. The table shows the total purchased seed expenses and fertilizer and lime expenses for farms in eight randomly selected states for a recent year. At a = 0.05, is there enough evidence to support the analyst’s claim? (Source: U.S. Department of Agriculture) State Purchased seed expenses (in millions of dollars) Fertilizer and lime expenses (in millions of dollars) Arkansas 490 480 California 1530 2060 Florida 490 480 Kentucky 266 402 Michigan 741 642 North Carolina 380 470 Ohio 879 858 Washington 360 560 6. Exercise Machines The table shows the overall scores and the prices for a random sample of nine different models of elliptical exercise machines. The overall score represents the ergonomics, exercise range, ease of use, construction, heart-rate monitoring, and safety. At a = 0.05, can you conclude that there is a significant correlation between the overall score and the price? (Source: Consumer Reports) Overall score 77 75 73 71 Price (in dollars) 3700 1700 1300 900 Overall score 66 66 64 62 58 Price (in dollars) 1000 1400 1800 1000 700

SECTION 11.4 Rank Correlation 615 7. Crop Prices The table shows the prices (in dollars per bushel) received for barley and corn for a random sample of nine U.S. farmers. At a = 0.05, can you conclude that there is a significant correlation between the barley and corn prices? (Adapted from U.S. Department of Agriculture) Barley 4.89 4.52 4.85 4.97 5.12 4.91 5.08 4.98 4.87 Corn 3.21 3.22 3.29 3.23 3.33 3.40 3.44 3.49 3.43 8. Vacuum Cleaners The table shows the overall scores and the prices for a random sample of 12 different models of vacuum cleaners. The overall score represents cleaning, airflow, handling, noise, and emissions. At a = 0.10, can you conclude that there is a significant correlation between the overall score and the price? (Source: Consumer Reports) Overall score 65 71 69 47 55 38 Price (in dollars) 150 200 550 350 470 90 Overall score 47 47 47 57 34 65 Price (in dollars) 80 130 210 190 300 260 Test Scores and GNI In Exercises 9 –12, use the table below. The table shows the average achievement scores of 15-year-olds in science and mathematics along with the gross national incomes (GNI) of nine randomly selected countries for a recent year. (The GNI is a measure of the total value of goods and services produced by the economy of a country.) (Source: Organization for Economic Cooperation and Development; The World Bank) Country Science average Mathematics average GNI (in billions of dollars) Canada 528 516 1,529 France 495 493 2,458 Germany 509 506 3,437 Italy 481 490 1,815 Japan 538 532 4,549 Mexico 416 408 1,143 Spain 493 486 1,192 Sweden 493 494 503 United States 496 470 18,496 9. Science and GNI At a = 0.10, can you conclude that there is a significant correlation between science achievement scores and GNI? 10. Math and GNI At a = 0.10, can you conclude that there is a significant correlation between mathematics achievement scores and GNI? 11. Science and Math At a = 0.10, can you conclude that there is a significant correlation between science and mathematics achievement scores? 12. Writing a Summary Use the results from Exercises 9 –11 to write a summary about the correlation (or lack of correlation) between test scores and GNI.

616 CHAPTER 11 Nonparametric Tests Extending Concepts Testing the Spearman Rank Correlation Coefficient for n + 30 When you are testing the significance of the Spearman rank correlation coefficient and the sample size n is greater than 30, you can use the expression below to find the critical value. { z 2 n - 1 , z corresponds to the level of significance In Exercises 13 and 14, test the Spearman rank correlation coefficient. 13. Work Injuries The table shows the average hours worked per week and the numbers of on-the-job injuries for a random sample of U.S. companies in a recent year. At a = 0.10, can you conclude that there is a significant correlation between average hours worked and the number of on-the-job injuries? Hours worked 46 43 41 40 41 42 45 45 42 45 44 44 Injuries 22 25 18 17 20 22 28 29 24 26 26 25 Hours worked 45 46 47 47 46 46 49 50 50 42 41 42 Injuries 27 29 29 30 29 29 30 30 30 23 22 23 Hours worked 41 41 41 41 40 39 38 39 39 Injuries 21 19 18 18 17 16 16 16 16 14. Work Injuries in Construction The table shows the average hours worked per week and the numbers of on-the-job injuries for a random sample of U.S. construction companies in a recent year. At a = 0.05, can you conclude that there is a significant correlation between average hours worked and the number of on-the-job injuries? Hours worked 38 38 37 38 38 40 39 39 39 40 39 41 Injuries 11 11 9 10 10 17 15 14 14 16 15 17 Hours worked 41 42 41 41 41 42 42 42 42 41 41 39 Injuries 17 21 18 18 18 22 21 19 21 18 17 12 Hours worked 38 38 39 39 36 37 36 37 37 37 37 Injuries 12 11 13 12 6 6 6 6 7 8 7

The Runs Test 11.5 SECTION 11.5 The Runs Test 617 What You Should Learn How to use the runs test to determine whether a data set is random The Runs Test for Randomness The Runs Test for Randomness In obtaining a sample of data, it is important for the data to be selected randomly. But how do you know whether the sample data are truly random? One way to test for randomness in a data set is to use a runs test for randomness. Before using a runs test for randomness, you must first know how to determine the number of runs in a data set. A run is a sequence of data having the same characteristic. Each run is preceded by and followed by data with a different characteristic or by no data at all. The number of data in a run is called the length of the run. DEFINITION Finding the Number of Runs A liquid-dispensing machine has been designed to fill one-liter bottles. A quality control inspector decides whether each bottle is filled to an acceptable level and passes inspection 1 P 2 or fails inspection 1 F 2 . Determine the number of runs for each sequence and find the length of each run. 1. P P P P P P P P F F F F F F F F 2. P F P F P F P F P F P F P F P F 3. P P F F F F P F F F P P P P P P SOLUTION 1. There are two runs. The first 8 P ’s form a run of length 8 and the first 8 F ’s form another run of length 8, as shown below. P P P P P P P P F F F F F F F F 1st run 2nd run 2. There are 16 runs each of length 1, as shown below. P F P F P F P F P F P F P F P F 1st run 2nd run… …16th run 3. There are 5 runs, the first of length 2, the second of length 4, the third of length 1, the fourth of length 3, and the fifth of length 6, as shown below. P P F F F F P F F F P P P P P P 1st run 2nd run 3rd run 4th run 5th run EXAMPLE 1

618 CHAPTER 11 Nonparametric Tests TRY IT YOURSELF 1 A machine produces engine parts. An inspector measures the diameter of each engine part and determines whether the part passes inspection 1 P 2 or fails inspection 1 F 2 . The results are shown below. Determine the number of runs in the sequence and find the length of each run. P P P F P F P P P P F F P F P P F F F P P P F P P P Answer: Page T1 When each value in a set of data can be categorized into one of two separate categories, you can use the runs test for randomness to determine whether the data are random. The runs test for randomness is a nonparametric test that can be used to determine whether a sequence of sample data is random. DEFINITION The runs test for randomness considers the number of runs in a sequence of sample data in order to test whether a sequence is random. When a sequence has too few or too many runs, it is usually not random. For instance, the sequence P P P P P P P P F F F F F F F F from Example 1, part 1, has too few runs (only 2 runs). The sequence P F P F P F P F P F P F P F P F from Example 1, part 2, has too many runs (16 runs). So, these sample data are probably not random. You can use a hypothesis test to determine whether the number of runs in a sequence of sample data is too high or too low. The runs test is a two-tailed test, and the null and alternative hypotheses are listed below. H 0 : The sequence of data is random. H a : The sequence of data is not random. When using the runs test, let n 1 represent the number of data that have one characteristic and let n 2 represent the number of data that have the second characteristic. It does not matter which characteristic you choose to be represented by n 1 . Let G represent the number of runs. n 1 = number of data with one characteristic n 2 = number of data with the other characteristic G = number of runs Table 12 in Appendix B lists the critical values for the runs test for selected values of n 1 and n 2 at the a = 0.05 level of significance. (In this text, you will use only the a = 0.05 level of significance when performing runs tests.) When n 1 or n 2 is greater than 20, you can use the standard normal distribution to find the critical values.

SECTION 11.5 The Runs Test 619 You can calculate the test statistic for the runs test as follows. When n 1 … 20 and n 2 … 20, the test statistic for the runs test is G , the number of runs. When n 1 7 20 or n 2 7 20, the test statistic for the runs test is z = G - m G s G where m G = 2 n 1 n 2 n 1 + n 2 + 1 and s G = B 2 n 1 n 2 1 2 n 1 n 2 - n 1 - n 2 2 1 n 1 + n 2 2 2 1 n 1 + n 2 - 1 2 . Test Statistic for the Runs Test Performing a Runs Test for Randomness In Words In Symbols 1. Identify the claim. State the null State H 0 and H a . and alternative hypotheses. 2. Specify the level of significance. Identify a . (Use a = 0.05 for the runs test.) 3. Determine the number of data that Determine n 1 , n 2 , and G . have each characteristic and the number of runs. 4. Determine the critical values. When n 1 … 20 and n 2 … 20, use Table 12 in Appendix B. When n 1 7 20 or n 2 7 20, use Table 4 in Appendix B. 5. Find the test statistic. When n 1 … 20 and n 2 … 20, use G . When n 1 7 20 or n 2 7 20, use z = G - m G s G 6. Make a decision to reject or fail If G is less than or equal to to reject the null hypothesis. the lower critical value or greater than or equal to the upper critical value, then reject H 0 . Otherwise, fail to reject H 0 . Or, if z is in the rejection region, then reject H 0 . Otherwise, fail to reject H 0 . 7. Interpret the decision in the context of the original claim. GUIDELINES

620 CHAPTER 11 Nonparametric Tests Using the Runs Test As people enter a concert, an usher records where they are sitting. The results for 13 people are shown, where L represents a lawn seat and P represents a pavilion seat. At a = 0.05, can you conclude that the sequence of seat locations is not random? L L L P P L P P P L L P L SOLUTION The claim is “the sequence of seat locations is not random.” To test this claim, use the null and alternative hypotheses below. H 0 : The sequence of seat locations is random. H a : The sequence of seat locations is not random. (Claim) To find the critical values, first determine n 1 , the number of L ’s; n 2 the number of P ’s; and G , the number of runs. L L L P P L P P P L L P L 1st 2nd 3rd 4th 5th 6th 7th run run run run run run run n 1 = number of L ’s = 7 n 2 = number of P ’s = 6 G = number of runs = 7 Because n 1 … 20, n 2 … 20, and a = 0.05, use Table 12 to find the lower critical value 3 and the upper critical value 12. The test statistic is the number of runs G = 7. Because the test statistic G is between the critical values 3 and 12, you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 5% level of significance to support the claim that the sequence of seat locations is not random. So, it appears that the sequence of seat locations is random. TRY IT YOURSELF 2 The genders of 15 students as they enter a classroom are shown below, where F represents a female and M represents a male. At a = 0.05, can you conclude that the sequence of genders is not random? M F F F M M F F M F M M F F F Answer: Page T1 EXAMPLE 2

SECTION 11.5 The Runs Test 621 Using the Runs Test You want to determine whether the selection of recently hired employees in a large company is random with respect to gender. The genders of 36 recently hired employees are shown below, where F represents a female and M represents a male. At a = 0.05, can you conclude that the sequence of employees is not random? M M F F F F M M M M M M F F F F F M M M M M M M F F F M M M M F M M F M SOLUTION The claim is “the sequence of employees is not random.” To test this claim, use the null and alternative hypotheses below. H 0 : The sequence of employees is random. H a : The sequence of employees is not random. (Claim) To find the critical values, first determine n 1 , the number of F ’s; n 2 , the number of M ’s; and G , the number of runs. M M F F F F M M M M M M 1st run 2nd run 3rd run F F F F F M M M M M M M 4th run 5th run F F F M M M M F M M F M 6th 7th 8th 9th 10th 11th run run run run run run n 1 = number of F ’s = 14 n 2 = number of M ’s = 22 G = number of runs = 11 Because n 2 7 20, use Table 4 in Appendix B to find the critical values. Because the test is a two-tailed test with a = 0.05, the critical values are - z 0 = - 1.96. and z 0 = 1.96. Before calculating the test statistic, find the values of m G and s G , as follows. m G = 2 n 1 n 2 n 1 + n 2 + 1 = 2 1 14 21 22 2 14 + 22 + 1 = 616 36 + 1 ≈ 18.11 EXAMPLE 3

622 CHAPTER 11 Nonparametric Tests s G = B 2 n 1 n 2 1 2 n 1 n 2 - n 1 - n 2 2 1 n 1 + n 2 2 2 1 n 1 + n 2 - 1 2 = B 2 1 14 21 22 2 [2 1 14 21 22 2 - 14 - 22] 1 14 + 22 2 2 1 14 + 22 - 1 2 ≈ 2.81 You can find the test statistic as follows. z = G - m G s G ≈ 11 - 18.11 2.81 ≈ - 2.53 The figure shows the location of the rejection regions and the test statistic z . Because z is in the rejection region, you reject the null hypothesis. z − 3 − 2 − 1 0 1 2 3 − z 0 = − 1.96 z ≈ − 2.53 z 0 = 1.96 α = 0.025 1 2 α α = 0.025 1 2 1 − = 0.95 Interpretation There is enough evidence at the 5% level of significance to support the claim that the sequence of employees with respect to gender is not random. TRY IT YOURSELF 3 Let S represent a day in a small town in which it snowed and let N represent a day in the same town in which it did not snow. The snowfall results for the entire month of January are shown below. At a = 0.05, can you conclude that the sequence is not random? N N N S S N N S N S N N N N N S N S N S N N S N S S N N N N N Answer: Page T1 When n 1 or n 2 is greater than 20, you can also use a P @ value to perform a hypothesis test for the randomness of the data. In Example 3, you can calculate the P @ value to be 0.0114. Because P 6 a , you reject the null hypothesis. Picturing the World The sequence shows the National Football League conference of each winning team for the first 51 Super Bowls, where A represents the American Football Conference and N represents the National Football Conference. (Source: National Football League) N N A A A N A A A A A N A A A N N A N N N N N N N N N N N N N A A N A A N A A A A N A N N N A N A A A At A = 0.05, can you conclude that the sequence of conferences of Super Bowl winning teams is random?

11.5 EXERCISES SECTION 11.5 The Runs Test 623 For Extra Help: MyLab Statistics Building Basic Skills and Vocabulary 1. In your own words, explain why the hypothesis test discussed in this section is called the runs test. 2. Describe the test statistic for the runs test when the sample sizes n 1 and n 2 are less than or equal to 20 and when either n 1 or n 2 is greater than 20. Using and Interpreting Concepts Finding the Number of Runs In Exercises 3 – 6, determine the number of runs in the sequence. Then find the length of each run. 3. T F T F T T T F F F T F 4. U U D D U D U U D D U D U U 5. M F M F M F F F F F F M M M F F M M M M 6. A A A B B B A B B A A A A A A B A A B A B B 7. Find the values of n 1 and n 2 in Exercise 3. 8. Find the values of n 1 and n 2 in Exercise 4. 9. Find the values of n 1 and n 2 in Exercise 5. 10. Find the values of n 1 and n 2 in Exercise 6. Finding Critical Values In Exercises 11–14, use the sequence and Table 12 in Appendix B to determine the number of runs that are considered too high and the number of runs that are considered too low for the data to be in random order. 11. T F T F T F T F T F T F 12. M F M M M M M M F F M M 13. N S S S N N N N N S N S N S S N N N 14. X X X X X X X Y Y Y Y Y Y Y Y Y Y Y Y Y Y Performing a Runs Test In Exercises 15 – 20, (a) identify the claim and state H 0 and H a , (b) find the critical values, (c) find the test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Use a = 0.05. 15. Coin Toss A coach records the results of the coin toss at the beginning of each football game for a season. The results are shown, where H represents heads and T represents tails. The coach claimed the tosses were not random. Test the coach’s claim. H T T T H T H H T T T T H T H H 16. Senate The sequence shows the majority party of the U.S. Senate after each election for a recent group of years, where R represents the Republican party and D represents the Democratic party. Can you conclude that the sequence is not random? (Source: U.S. Senate) R D D D R R R R R R R D D D D D D D R D D R D D D D D D D D D D D D D R R R D D D D R R R D R R D D D D R R

624 CHAPTER 11 Nonparametric Tests 17. Baseball The sequence shows the Major League Baseball league of each World Series winning team from 1969 to 2016, where N represents the National League and A represents the American League. Can you conclude that the sequence of leagues of World Series winning teams is not random? (Source: Major League Baseball) N A N A A A N N A A N N N N A A A N A N A N A A A N A N A A A N A N A A N A N A N N N A N A N 18. Number Generator A number generator outputs the sequence of digits shown, where O represents an odd digit and E represents an even digit. Test the claim that the digits were not randomly generated. O O O E E E E O O O O O E E E E O O E E E E O O O O E E E E O O 19. Dog Identifications A team of veterinarians record, in order, the genders of every dog that is microchipped at their pet hospital in one month. The genders of recently microchipped dogs are shown, where F represents a female and M represents a male. A veterinarian claims that the microchips are random by gender. Do you have enough evidence to reject the doctor’s claim? M M F M F F F F F M M M F F F M F F F F F M F F F M F F F 20. Golf Tournament A golf tournament official records whether each past winner is American-born ( A ) or foreign-born ( F ). The results are shown for every year the tournament has existed. Can you conclude that the sequence is not random? F F A F F A F F A F F A F F A F F A F F F F F F A F F A F F A F F A F F A F A F F A F F F F F A F F F F F A F F F A Extending Concepts Runs Test with Quantitative Data In Exercises 21–23, use the following information to perform a runs test. You can also use the runs test for randomness with quantitative data. First, calculate the median. Then assign a + sign to those values above the median and a - sign to those values below the median. Ignore any values that are equal to the median. Use a = 0.05. 21. Daily High Temperatures The sequence shows the daily high temperatures (in degrees Fahrenheit) for a city during the month of July. Test the claim that the daily high temperatures do not occur randomly. 84 87 92 93 95 84 82 83 81 87 92 98 99 93 84 85 86 92 91 95 84 92 83 81 87 92 98 89 93 84 85 22. Exam Scores The sequence shows the exam scores of a class based on the order in which the students finished the test. Test the claim that the scores occur randomly. 83 94 80 76 92 89 65 75 82 87 90 91 81 99 97 72 72 89 90 92 87 76 74 66 88 81 90 92 89 76 80 23. Use technology to generate a sequence of 30 numbers from 1 to 99, inclusive. Test the claim that the sequence of numbers is not random.

USES AND ABUSES Statistics in the Real World Uses and Abuses 625 EXERCISES 1. Insufficient Evidence Give an example of a nonparametric test in which there is not enough evidence to reject the null hypothesis. 2. Using an Inappropriate Test Discuss the nonparametric tests described in this chapter and match each test with its parametric counterpart, which you studied in earlier chapters. Uses Nonparametric Tests Before you could perform many of the hypothesis tests you learned about in previous chapters, you had to ensure that certain conditions about the population were satisfied. For instance, before you could perform a t @ test, you had to verify that the population was normally distributed or the sample size was at least 30. One advantage of the nonparametric tests shown in this chapter is that they are distribution free. That is, they do not require any particular information about the population or populations being tested. Another advantage of nonparametric tests is that they are easier to perform than their parametric counterparts. This means that they are easier to understand and quicker to use. Nonparametric tests can often be used when data are at the nominal or ordinal level. Abuses Insufficient Evidence Stronger evidence is needed to reject a null hypothesis in a nonparametric test than in a corresponding parametric test. That is, when you are trying to support a claim represented by the alternative hypothesis, you might need a larger sample when performing a nonparametric test. When the outcome of a nonparametric test results in failure to reject the null hypothesis, you should investigate the sample size used. It may be that a larger sample will produce different results. Using an Inappropriate Test In general, when information about the population (such as the condition of normality) is known, it is more efficient to use a parametric test. When information about the population is not known, however, nonparametric tests can be helpful.

626 CHAPTER 11 Nonparametric Tests Chapter Summary 11 Example(s) Review Exercises What Did You Learn? Section 11.1 How to use the sign test to test a population median z = 1 x + 0.5 2 - 0.5 n 1 n 2 1, 2 1– 3, 6 How to use the paired-sample sign test to test the difference between two population medians (dependent samples) 3 4, 5 Section 11.2 How to use the Wilcoxon signed-rank test and the Wilcoxon rank sum test to determine whether two samples are selected from populations having the same distribution z = R - m R s R , m R = n 1 1 n 1 + n 2 + 1 2 2 , s R = B n 1 n 2 1 n 1 + n 2 + 1 2 12 1, 2 7, 8 Section 11.3 How to use the Kruskal-Wallis test to determine whether three or more samples were selected from populations having the same distribution H = 12 N 1 N + 1 2 a R 2 1 n 1 + R 2 2 n 2 + c + R 2 k n k b - 3 1 N + 1 2 1 9, 10 Section 11.4 How to use the Spearman rank correlation coefficient to determine whether the correlation between two variables is significant r s = 1 - 6 Σ d 2 n ( n 2 - 1 ) 1 11, 12 Section 11.5 How to use the runs test to determine whether a data set is random G = number of runs, z = G - m G s G , m G = 2 n 1 n 2 n 1 + n 2 + 1, s G = B 2 n 1 n 2 1 2 n 1 n 2 - n 1 - n 2 2 1 n 1 + n 2 2 2 1 n 1 + n 2 - 1 2 1– 3 13, 14 The table summarizes parametric and nonparametric tests. Always use the parametric test when the conditions for that test are satisfied. Test application Parametric test Nonparametric test One-sample tests z @ test for a population mean t @ test for a population mean Sign test for a population median Two-sample tests Dependent samples Independent samples t @ test for the difference between means z @ test for the difference between means t @ test for the difference between means Paired-sample sign test Wilcoxon signed-rank test Wilcoxon rank sum test Tests involving three or more samples One-way ANOVA Kruskal-Wallis test Correlation Pearson correlation coefficient Spearman rank correlation coefficient Randomness (No parametric test) Runs test

Review Exercises 627 Review Exercises 11 Section 11.1 In Exercises 1– 6, use a sign test to test the claim by doing the following. (a) Identify the claim and state H 0 and H a . (b) Find the critical value. (c) Find the test statistic. (d) Decide whether to reject or fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim. 1. A store manager claims that the median number of customers per day is no more than 650. The numbers of customers per day for 17 randomly selected days are listed below. At a = 0.01, can you reject the manager’s claim? 675 665 601 642 554 653 639 650 645 550 677 569 650 660 682 689 590 2. A company claims that the median credit score for U.S. adults is at least 710. The credit scores for 13 randomly selected U.S. adults are listed below. At a = 0.05, can you reject the company’s claim? (Adapted from Fair Isaac Corporation) 750 782 805 695 700 706 625 589 690 772 745 704 710 3. A government agency claims that the median sentence length for all federal prisoners is 2 years. In a random sample of 180 federal prisoners, 65 have sentence lengths that are less than 2 years, 109 have sentence lengths that are more than 2 years, and 6 have sentence lengths that are 2 years. At a = 0.10, can you reject the agency’s claim? (Adapted from U.S. Sentencing Commission) 4. In a study testing the effects of calcium supplements on blood pressure in men, 10 randomly selected men were given a calcium supplement for 12 weeks. The table shows the measurements for each subject’s diastolic blood pressure taken before and after the 12-week treatment period. At a = 0.05, can you reject the claim that there was no reduction in diastolic blood pressure? (Adapted from the American Medical Association) Patient 1 2 3 4 5 Before treatment 107 110 123 129 112 After treatment 100 114 105 112 115 Patient 6 7 8 9 10 Before treatment 111 107 112 136 102 After treatment 116 106 102 125 104

628 CHAPTER 11 Nonparametric Tests 5. In a study testing the effects of an herbal supplement on blood pressure in men, 11 randomly selected men were given an herbal supplement for 12 weeks. The table shows the measurements for each subject’s diastolic blood pressure taken before and after the 12-week treatment period. At a = 0.05, can you reject the claim that there was no reduction in diastolic blood pressure? (Adapted from The Journal of the American Medical Association) Patient 1 2 3 4 5 6 Before treatment 123 109 112 102 98 114 After treatment 124 97 113 105 95 119 Patient 7 8 9 10 11 Before treatment 119 112 110 117 130 After treatment 114 114 121 118 133 6. An association claims that the median annual salary of lawyers is $118,160. In a random sample of 125 lawyers, 76 were paid less than $118,160, and 49 were paid more than $118,160. At a = 0.05, can you reject the association’s claim? (Adapted from U.S. Bureau of Labor Statistics) Section 11.2 In Exercises 7 and 8, use a Wilcoxon test to test the claim by doing the following. (a) Identify the claim and state H 0 and H a . (b) Decide whether to use a Wilcoxon signed-rank test or a Wilcoxon rank sum test. (c) Find the critical value(s). (d) Find the test statistic. (e) Decide whether to reject or fail to reject the null hypothesis. (f ) Interpret the decision in the context of the original claim. 7. A career placement advisor claims that there is a difference in the total times required to earn a doctorate degree by female and male graduate students. The table shows the total times (in years) to earn a doctorate for a random sample of 12 female and 12 male graduate students. At a = 0.01, can you support the advisor’s claim? (Adapted from Survey of Earned Doctorates) Female 9 11 9 12 11 8 10 13 6 6 8 9 Male 8 7 8 10 9 7 7 9 10 8 9 7

Review Exercises 629 8. A medical researcher claims that a new drug affects the number of headache hours experienced by headache sufferers. The numbers of headache hours (per day) experienced by eight randomly selected patients before and after taking the drug are shown in the table. At a = 0.05, can you support the researcher’s claim? Patient 1 2 3 4 5 6 7 8 Headache hours (before) 0.9 2.3 2.7 2.4 2.9 1.9 1.2 3.1 Headache hours (after) 1.4 1.5 1.4 1.8 1.3 0.6 0.7 1.9 Section 11.3 In Exercises 9 and 10, use the Kruskal-Wallis test to test the claim by doing the following. (a) Identify the claim and state H 0 and H a . (b) Find the critical value and identify the rejection region. (c) Find the test statistic H. (d) Decide whether to reject or fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim. 9. The table shows the ages for a random sample of doctorate recipients in three fields of study. At a = 0.01, can you conclude that the distribution of the ages of the doctorate recipients in at least one field of study is different from the others? (Adapted from Survey of Earned Doctorates) Field of study Age Life sciences 31 32 34 31 30 32 35 31 32 34 29 Physical sciences 30 31 32 31 30 29 31 30 32 33 30 Social sciences 32 35 31 33 34 31 35 36 32 30 33 10. The table shows the starting salaries for a random sample of college graduates in four fields of engineering. At a = 0.05, can you conclude that the distribution of the starting salaries in at least one field of engineering is different from the others? (Adapted from National Association of Colleges and Employers) Field of engineering Starting salary (in thousands of dollars) Chemical 68.4 65.9 71.7 70.5 64.3 69.9 67.5 65.7 69.4 71.1 Computer 68.2 67.6 65.8 66.4 69.5 72.6 67.0 70.2 68.5 66.4 Electrical 66.9 65.5 66.1 64.4 67.6 67.3 68.9 68.1 67.1 67.4 Mechanical 65.5 64.8 65.6 63.7 65.6 65.3 68.1 68.6 64.9 62.7