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in Chapter 11, once a dependence (or interaction) is found, you must look within the table at the relative or conditional proportions for each level of classiﬁcation. For example, consider shift 1, which produced a total of 94 defects. These defects can be divided into types using the conditional proportions for this sample shown in the ﬁrst column of Table 14.5. If you follow the same procedure for the other two shifts, you can then compare the distributions of defect types for the three shifts, as shown in Table 14.5. Now compare the three sets of proportions (each sums to 1). It appears that shifts 1 and 2 produce defects in the same general order—types C, B, A, and D from most to least—though in differing proportions. Shift 3 shows a different pattern—the most 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 605 type C defects again but followed by types A, D, and B, in that order. Depending on which type of defect is the most important to the manufacturer, each shift should be cautioned separately about the reasons for producing too many defects. TABLE 14.5 ● Conditional Probabilities for Types of Defect Within Three Shifts Types of Defects A B C D Shift 3 2 3 3 2 6.28.27 14.32 41.35 9 1 1 6 9 2 0 5.17.05 9 1 1 9 6 1 1 5.16 4 9 2 1.22 4 9 4 5.48 4 9 1 3.14 4 9 Total 1.00 1.00 1.00 The Chi-Square Test of Independence applet can help you visualize the distribution of the observed frequencies. In Figure 14.2(a), the blue bars (blue in Figure 14.2(a)) represent categories that have an excess of defectives relative to expected and red cells (gray in Figure 14.2(a)) indicate a deﬁcit of defectives relative to expected. The intensity of the color reﬂects the magnitude of the discrepancy. button to view the expected distribution of In Figure 14.2(b), we used the defectives if the null hypothesis is true. The relative heights of the rectangles in each of the three columns correspond to the conditional distribution of defectives per shift given in Table 14.5. We will use this applet for the MyApplet Exercises at the end of the chapter.
(a) (b) FI GUR E 1 4. 2 Chi-Square Test of Independence applet ● 606 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA How Do I Determine the Appropriate Number of Degrees of Freedom? Remember the general procedure for determining degrees of freedom: 1. Start with k rc categories or cells in the contingency table. 2. Subtract one degree of freedom because all of the rc cell probabilities must sum to 1. 3. You had to estimate (r 1) row probabilities and (c 1) column probabilities to calculate the estimated expected cell counts. (The last one of the row and column probabilities is determined because the marginal row and column probabilities must also sum to 1.) Subtract (r 1) and (c 1) df. The total degrees of freedom for the r c contingency table are df rc 1 (r 1) (c 1) rc r c 1 (r 1)(c 1) EXAMPLE 14.5 A survey was conducted to evaluate the effectiveness of a new ﬂu vaccine that had been administered in a small community. The vaccine was provided free of charge in a two-shot sequence over a period of 2 weeks. Some people received the two-shot sequence, some appeared for only the ﬁrst shot, and others received neither. A survey of 1000 local residents the following spring provided the information shown in Table 14.6. Do the data present sufficient evidence to indicate that the vaccine was successful in reducing the number of ﬂu cases in the community? TABLE 14.6 ● 2 3 Contingency Table No Vaccine One Shot Flu No Flu Total 24 289 313 9 100 109 Two Shots Total 13 565 578 46 954 1000 Solution The success of the vaccine in reducing the number of ﬂu cases can be assessed in two parts: • If the vaccine is successful, the proportions of people who get the ﬂu should vary, depending on which of the three treatments they received. • Not only must this dependence exist, but the proportion of people who get the ﬂu should decrease as the amount of ﬂu prevention treatment increases— from zero to one to two shots. The ﬁrst part can be tested using the chi-square test with these hypotheses: H0 : No relationship between treatment and incidence of ﬂu Ha : Incidence of ﬂu depends on amount of ﬂu treatment As
usual, computer software packages can eliminate all of the tedious calculations and, if the data are entered correctly, provide the correct output containing the observed 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 607 Use the value of X2 and the p-value from the printout to test the hypothesis of independence. value of the test statistic and its p-value. Such a printout, generated by MINITAB, is shown in Figure 14.3. You can ﬁnd instructions for generating this printout in the section “My MINITAB ” at the end of this chapter. The observed value of the test statistic, X2 17.313, has a p-value of.000 and the results are declared highly signiﬁcant. That is, the null hypothesis is rejected. There is sufficient evidence to indicate a relationship between treatment and incidence of ﬂu. FI GUR E 1 4. 3 MINITAB output for Example 14.5 ● Chi-Square Test: No Vaccine, One Shot, Two Shots Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts No Vaccine One Shot Two Shots Total 1 24 9 13 46 14.40 5.01 26.59 6.404 3.169 6.944 2 289 100 565 954 298.60 103.99 551.41 0.309 0.153 0.335 Total 313 109 578 1000 Chi-Sq = 17.313, DF = 2, P-Value = 0.000 What is the nature of this relationship? To answer this question, look at Table 14.7 and Figure 14.4, which give the incidence of ﬂu in the sample for each of the three treatment groups. The answer is obvious. The group that received two shots was less susceptible to the ﬂu; only one ﬂu shot does not seem to decrease the susceptibility! TABLE 14.7 ● Incidence of Flu for Three Treatments No Vaccine One Shot Two Shots 2 4.08 3 1 3 1 3 9.02.08 8 7 5 09 1 FI GUR E 1 4. 4 Chi-Square Test of Independence applet ● 608 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA 14.4 EXERCISES BASIC TECHNIQUES 14.16 Calculate the value and give the number of degrees of freedom for
X2 for these contingency tables: a. Columns Rows 1 1 2 3 120 79 31 2 70 108 49 3 55 95 81 4 16 43 140 b. Columns Rows 1 1 2 35 120 2 16 92 3 84 206 14.17 Suppose that a consumer survey summarizes the responses of n 307 people in a contingency table that contains three rows and ﬁve columns. How many degrees of freedom are associated with the chi-square test statistic? 14.18 A survey of 400 respondents produced these cell counts in a 2 3 contingency table: Columns Rows 1 1 2 Total 37 66 103 2 34 57 91 3 93 113 206 Total 164 236 400 a. If you wish to test the null hypothesis of “independence”—that the probability that a response falls in any one row is independent of the column it falls in—and you plan to use a chi-square test, how many degrees of freedom will be associated with the x 2 statistic? b. Find the value of the test statistic. c. Find the rejection region for a.01. d. Conduct the test and state your conclusions. e. Find the approximate p-value for the test and inter- pret its value. 14.19 Gender Differences Male and female respondents to a questionnaire on gender differences were categorized into three groups according to their answers on the ﬁrst question: Group 1 Group 2 Group 3 Men Women 37 7 49 50 72 31 Use the MINITAB printout to determine whether there is a difference in the responses according to gender. Explain the nature of the differences, if any exist. MINITAB output for Exercise 14.19 Chi-Square Test: Group 1, Group 2, Group 3 Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Group 1 Group 2 Group 3 Total 1 37 49 72 158 28.26 63.59 66.15 2.703 3.346 0.517 2 7 50 31 88 15.74 35.41 36.85 4.853 6.007 0.927 Total 44 99 103 246 Chi-Sq = 18.352, DF = 2, P-Value = 0.000 APPLICATIONS 14.20 Mandatory Health Care In 2006, a new law passed in Massachusetts would require all residents to have health insurance. Low-income residents would get state subsidies to help pay insurance premiums, but everyone would pay something for health services. The plan would penalize people without any insurance and charge fees to employers who don�
�t provide coverage. An ABC News/Washington Post poll4 involving n 1027 adults nationwide asked the question, “Would you support or oppose this plan in your state?” The data that follows is based on the results of this study. Affiliation Support Oppose Unsure Democrats Independents Republicans 256 60 235 163 40 222 22 5 24 a. Are there signiﬁcant differences in the proportions of those surveyed who support, oppose, and are unsure about this plan among Democrats, Independents, and Republicans? Use a.05. b. If signiﬁcant differences exist, describe the nature of the differences by ﬁnding the proportions of those who support, oppose, and are unsure for each of the given affiliations. 14.21 Anxious Infants A study was conducted by Joseph Jacobson and Diane Wille to determine the effect of early child care on infant-mother attachment patterns.5 In the study, 93 infants were classiﬁed as either “secure” or “anxious” using the Ainsworth strange situation paradigm. In addition, the infants were classiﬁed according to the average number of hours per week that they spent in child care. The data are presented in the table. Low (0–3 hours) Moderate (4–19 hours) High (20–54 hours) Secure Anxious 24 11 35 10 5 8 a. Do the data provide sufficient evidence to indicate that there is a difference in attachment pattern for the infants depending on the amount of time spent in child care? Test using a.05. b. What is the approximate p-value for the test in part a? 14.22 Spending Patterns Is there a difference in the spending patterns of high school seniors depending on their gender? A study to investigate this question focused on 196 employed high school seniors. Students were asked to classify the amount of their earnings that they spent on their car during a given month: None or Only a Little Some About Half Most Almost All All or Male 73 Female 57 12 15 6 11 4 9 3 6 A portion of the MINITAB printout is given here. Use the printout to analyze the relationship between spending patterns and gender. Write a short paragraph explaining your statistical conclusions and their practical implications. Partial MINITAB output for Exercise 14.22 Chi-Square Test: None, Some, Half, Most, All Chi-Sq = 6.696, DF = 4, P
-Value = 0.153 2 cells with expected counts less than 5. 14.23 Waiting for a Prescription How long do you wait to have your prescriptions ﬁlled? According to USA Today, “about 3 in 10 Americans wait more than 20 minutes to have a prescription ﬁlled.”6 Suppose a comparison of waiting times for pharmacies in HMOs and pharmacies in drugstores produced the following results. Waiting Time 15 minutes 16–20 minutes 20 minutes Don’t know HMO Drugstores 75 44 21 10 119 21 37 23 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 609 a. Is there sufficient evidence to indicate that there is a difference in waiting times for pharmacies in HMOs and pharmacies in drugstores? Use a.01. b. If we consider only if the waiting time is more than 20 minutes, is there a signiﬁcant difference in waiting times between pharmacies in HMOs and pharmacies in drugstores at the 1% level of signiﬁcance? EX1424 14.24 The JFK Assassination More than 40 years after the assassination of John F. Kennedy, a FOX News poll shows most Americans disagree with the government’s conclusions about the killing. The Warren Commission found that Lee Harvey Oswald acted alone when he shot Kennedy, but many Americans are not so sure. Do you think that we know all the facts about the assassination of President John F. Kennedy or do you think there was a cover-up? Here are the results from a poll of 900 registered voters nationwide:7 We Know All the Facts There Was a Cover-Up (Not Sure) Democrats Republicans Independents 42 64 20 309 246 115 31 46 27 a. Do these data provide sufficient evidence to conclude that there is a difference in voters’ opinions about a possible cover-up depending on the political affiliation of the voter? Test using a.05. b. If there is a signiﬁcant difference in part a, describe the nature of these differences. EX1425 14.25 Telecommuting As an alternative to ﬂextime, many companies allow employees to do some of their work at home. Individuals in a random sample of 300 workers were classiﬁed according to salary and number of workdays per week spent at home. Workdays at Home per Week Salary Less Than One At Least One, but Not All All at Home Under $
25,000 $25,000 to $49,999 $50,000 to $74,999 Above $75,000 38 54 35 33 16 26 22 29 14 12 9 12 a. Do the data present sufficient evidence to indicate that salary is dependent on the number of workdays spent at home? Test using a.05. b. Use Table 5 in Appendix I to approximate the p-value for this test of hypothesis. Does the p-value conﬁrm your conclusions from part a? 610 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA 14.26 Telecommuting II An article in American Demographics addressed the same telecommuting issue (Exercise 14.25) in a EX1426A EX1426B slightly different way. They concluded that “people who work exclusively at home tend to be older and better educated than those who have to leave home to report to work.”8 Use the data below based on random samples of 300 workers each to either support or refute their conclusions. Use the appropriate test of hypothesis, and explain why you either agree or disagree with the American Demographics conclusions. Note that “Mixed” workers are those who reported working at home at least one full day in a typical week. Workers Age Non-Home Mixed Home 15–34 35–54 55 and over 73 85 22 23 40 12 12 23 10 Workers Education Non-Home Mixed Home Less than H.S. diploma H.S. graduate Some college/Assoc. degree B.A. or more 23 54 53 41 3 12 24 42 5 11 14 18 14.5 EXAMPLE 14.6 COMPARING SEVERAL MULTINOMIAL POPULATIONS: A TWO-WAY CLASSIFICATION WITH FIXED ROW OR COLUMN TOTALS An r c contingency table results when each of n experimental units is counted as falling into one of the rc cells of a multinomial experiment. Each cell represents a pair of category levels—row level i and column level j. Sometimes, however, it is not advisable to use this type of experimental design—that is, to let the n observations fall where they may. For example, suppose you want to study the opinions of American families about their income levels—say, low, medium, and high. If you randomly select n 1200 families for your survey, you may not ﬁnd any who classify themselves as low-income families! It might be better to decide
ahead of time to survey 400 families in each income level. The resulting data will still appear as a two-way classiﬁcation, but the column totals are ﬁxed in advance. In another ﬂu prevention experiment like Example 14.5, the experimenter decides to search the clinic records for 300 patients in each of the three treatment categories: no vaccine, one shot, and two shots. The n 900 patients will then be surveyed regarding their winter ﬂu history. The experiment results in a 2 3 table with the column totals ﬁxed at 300, shown in Table 14.8. By ﬁxing the column totals, the experimenter no longer has a multinomial experiment with 2 3 6 cells. Instead, there are three separate binomial experiments—call them 1, 2, and 3—each with a given probability pj of contracting the ﬂu and qj of not contracting the ﬂu. (Remember that for a binomial population, pj qj 1.) TABLE 14.8 ● Cases of Flu for Three Treatments No Vaccine One Shot Two Shots Total Flu No Flu Total 300 300 300 r1 r2 n 14.5 COMPARING SEVERAL MULTINOMIAL POPULATIONS: A TWO-WAY CLASSIFICATION WITH FIXED ROW OR COLUMN TOTALS ❍ 611 Suppose you used the chi-square test to test for the independence of row and column classiﬁcations. If a particular treatment (column level) does not affect the incidence of ﬂu, then each of the three binomial populations should have the same incidence of ﬂu so that p1 p2 p3 and q1 q2 q3. The 2 3 classiﬁcation in Example 14.6 describes a situation in which the chisquare test of independence is equivalent to a test of the equality of c 3 binomial proportions. Tests of this type are called tests of homogeneity and are used to compare several binomial populations. If there are more than two row categories with ﬁxed column totals, then the test of independence is equivalent to a test of the equality of c sets of multinomial proportions. You do not need to be concerned about the theoretical equivalence of the chi-square tests for these two experimental designs. Whether the columns (or rows) are ﬁxed or not,
the test statistic is calculated as Eˆ X2 S(Oij ij)2 where Eˆ Eˆij icj ij r n which has an approximate chi-square distribution in repeated sampling with df (r 1)(c 1). How Do I Determine the Appropriate Number of Degrees of Freedom? Remember the general procedure for determining degrees of freedom: 1. Start with the rc cells in the two-way table. 2. Subtract one degree of freedom for each of the c multinomial populations, whose column probabilities must add to one—a total of c df. 3. You had to estimate (r 1) row probabilities, but the column probabilities are ﬁxed in advance and did not need to be estimated. Subtract (r 1) df. The total degrees of freedom for the r c (ﬁxed-column) table are rc c (r 1) rc c r 1 (r 1)(c 1) EXAMPLE 14.7 A survey of voter sentiment was conducted in four midcity political wards to compare the fractions of voters who favor candidate A. Random samples of 200 voters were polled in each of the four wards with the results shown in Table 14.9. The values in parentheses in the table are the expected cell counts. Do the data present sufﬁcient evidence to indicate that the fractions of voters who favor candidate A differ in the four wards? TABLE 14.9 ● Voter Opinions in Four Wards Ward 1 2 3 4 Favor A 76 (59) Do Not Favor A 124 (141) Total 200 53 (59) 147 (141) 59 (59) 141 (141) 48 (59) 152 (141) 200 200 200 Total 236 564 800 612 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA Solution Since the column totals are ﬁxed at 200, the design involves four binomial experiments, each containing the responses of 200 voters from each of the four wards. To test the equality of the proportions who favor candidate A in all four wards, the null hypothesis H0 : p1 p2 p3 p4 is equivalent to the null hypothesis H0 : Proportion favoring candidate A is independent of ward and will be rejected if the test statistic X2 is too large. The observed value of the test statistic, X2 10.722, and its associated p-value,.013, are shown in Figure 14.
5. The results are signiﬁcant (P.025); that is, H0 is rejected and you can conclude that there is a difference in the proportions of voters who favor candidate A among the four wards. F IG URE 14. 5 MINITAB output for Example 14.7 ● Chi-Square Test: Ward 1, Ward 2, Ward 3, Ward 4, Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Ward 1 Ward 2 Ward 3 Ward 4 Total 1 76 53 59 48 236 59.00 59.00 59.00 59.00 4.898 0.610 0.000 2.051 2 124 147 141 152 564 141.00 141.00 141.00 141.00 2.050 0.255 0.000 0.858 Total 200 200 200 200 800 Chi-Sq = 10.722 DF = 3, P-Value = 0.013 What is the nature of the differences discovered by the chi-square test? To answer this question, look at Table 14.10, which shows the sample proportions who favor candidate A in each of the four wards. It appears that candidate A is doing best in the ﬁrst ward and worst in the fourth ward. Is this of any practical signiﬁcance to the candidate? Possibly a more important observation is that the candidate does not have a plurality of voters in any of the four wards. If this is a two-candidate race, candidate A needs to increase his campaigning! TABLE 14.10 ● Proportions in Favor of Candidate A in Four Wards Ward 3 59/200.30 Ward 4 48/200.24 Ward 1 76/200.38 Ward 2 53/200.27 14.5 COMPARING SEVERAL MULTINOMIAL POPULATIONS: A TWO-WAY CLASSIFICATION WITH FIXED ROW OR COLUMN TOTALS ❍ 613 14.5 EXERCISES BASIC TECHNIQUES 14.27 Random samples of 200 observations were selected from each of three populations, and each observation was classiﬁed according to whether it fell into one of three mutually exclusive categories: Category Population 1 1 2 3 108 87 112 2 52 51 39 3 40 62 49 Total 200 200 200 You want to know whether the data provide sufficient evidence to indicate that the proportions of observations in the three categories depend on the population from which they were drawn
. a. Give the value of X2 for the test. b. Give the rejection region for the test for a.01. c. State your conclusions. d. Find the approximate p-value for the test and inter- pret its value. 14.28 Suppose you wish to test the null hypothesis that three binomial parameters pA, pB, and pC are equal versus the alternative hypothesis that at least two of the parameters differ. Independent random samples of 100 observations were selected from each of the populations. The data are shown in the table. Population A 24 76 B 19 81 C 33 67 Successes Failures Total 100 100 100 Total 76 224 300 a. Write the null and alternative hypotheses for testing the equality of the three binomial proportions. b. Calculate the test statistic and ﬁnd the approximate p-value for the test in part a. c. Use the approximate p-value to determine the statistical signiﬁcance of your results. If the results are statistically signiﬁcant, explore the nature of the differences in the three binomial proportions. APPLICATIONS 14.29 The Sandwich Generation How do Americans in the “sandwich generation” balance the demands of caring for older and younger relatives? In a telephone poll of Americans aged 45 to 55 years conducted by the New York Times,9 the number providing ﬁnancial support for their parents is listed in the next display. Provide Financial Support Yes White Americans African Americans Hispanic Americans Asian Americans 40 56 68 84 No 160 144 132 116 Is there a signiﬁcant difference in the proportion of individuals providing ﬁnancial support for their parents for these subpopulations of Americans? Use a.01. 14.30 Diseased Chickens A particular poultry disease is thought to be noncommunicable. To test this theory, 30,000 chickens were randomly partitioned into three groups of 10,000. One group had no contact with diseased chickens, one had moderate contact, and the third had heavy contact. After a 6-month period, data were collected on the number of diseased chickens in each group of 10,000. Do the data provide sufficient evidence to indicate a dependence between the amount of contact between diseased and nondiseased fowl and the incidence of the disease? Use a.05. No Contact Disease No Disease Total 87 9,913 10,000 Moderate Contact 89 9,911 10,000 Heavy Contact 124 9,876 10,
000 14.31 Long-Term Care A study conducted in northwest England made an assessment of EX1431 long-term care facilities that have residents with dementia.10 The homes included those that provided specialized services for elderly people with mental illness/health problems, known as “EMI homes,” as well as others classiﬁed as “non-EMI homes.” It was expected that the EMI homes would score higher on several measures of service quality for people with dementia. One measure included the structure of the home and the services provided, as given in the next table. Home Type Care Type EMI Non-EMI Total Nursing care Residential care Dual-registered Total 54 59 49 162 22 77 26 125 76 136 75 287 614 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA a. Describe the binomial experiments whose proportions are being compared in this experiment. recorded the number of family members in each of the households. The data are shown in the table. b. Do these data indicate that the type of care provided varies by the three types of home? Test at the a.01 level. c. Based upon the results of part b, explain the practical nature of the relationship between home type and care type. 14.32 Deep-Sea Research W.W. Menard has conducted research involving manganese nodules, a mineral-rich concoction found abundantly on the deepsea ﬂoor.11 In one portion of his report, Menard provides data relating the magnetic age of the earth’s crust to the “probability of ﬁnding manganese nodules.” The table gives the number of samples of the earth’s core and the percentage of those that contain manganese nodules for each of a set of magnetic-crust ages. Do the data provide sufficient evidence to indicate that the probability of ﬁnding manganese nodules in the deepsea earth’s crust is dependent on the magnetic-age classiﬁcation? Age Miocene—recent Oligocene Eocene Paleocene Late Cretaceous Early and Middle Cretaceous Jurassic Number of Samples Percentage with Nodules 389 140 214 84 247 1120 99 5.9 17.9 16.4 21.4 21.1 14.2 11.0 EX1433 14.33 How Big Is the Household? A local chamber of commerce surveyed 120 households in
their city—40 in each of three types of residence (apartment, duplex, or single residence)—and Type of Residence Family Members 1 2 3 4 or more Apartment Duplex Single Residence 8 16 10 6 20 8 10 2 1 9 14 16 Is there a signiﬁcant difference in the family size distributions for the three types of residence? Test using a.01. If there are signiﬁcant differences, describe the nature of these differences. EX1434 14.34 Churchgoing and Age A snapshot in USA Today indicates that there is a gap in church attendence between 20-year-olds and older Americans.12 Suppose that we randomly select 100 Americans in each of ﬁve age groups and record the numbers who say they attend church in a typical week. Attend Church Regularly? 20s 30s 40s 50s Yes No Source: Barna Research Group 31 69 42 58 47 53 48 52 60 53 47 a. Do the data indicate that the proportion of adults who attend church regularly differs depending on age? Test using a.05. b. If there are signﬁcant differences in part a, describe the nature of these differences by calculating the proportion of churchgoers in each age category. Where do the signiﬁcant differences appear to lie? 14.6 THE EQUIVALENCE OF STATISTICAL TESTS Remember that when there are only k 2 categories in a multinomial experiment, the experiment reduces to a binomial experiment where you record the number of successes x (or O1) in n (or O1 O2) trials. Similarly, the data that result from two binomial experiments can be displayed as a two-way classiﬁcation with r 2 and c 2, so that the chi-square test of homogeneity can be used to compare the two binomial proportions, p1 and p2. For these two situations, we have presented statistical tests for the binomial proportions based on the z-statistic of Chapter 9: 14.7 OTHER APPLICATIONS OF THE CHI-SQUARE TEST ❍ 615 • One sample: z pˆ p0 q0 p0 n • Two samples: z pˆ1 pˆ2 pˆqˆ 1 1 n n 2 1 k 2 Successes Failures r c 2 Sample 1 Sample 2 Successes Successes Failures Failures Why are there two different tests for the
same statistical hypothesis? Which one should you use? For these two situations, you can use either the z test or the chisquare test, and you will obtain identical results. For either the one- or two-sample test, we can prove algebraically that z 2 2 so that the test statistic z will be the square root (either positive or negative, depending on the data) of the chi-square statistic. Furthermore, we can show theoretically that the same relationship holds for the critical values in the z and x 2 tables in Appendix I, which produces identical p-values for the two equivalent tests. To test a one-tailed alternative hypothesis such as H0: p1 p2, ﬁrst determine whether pˆ1 pˆ2 0, that is, if the difference in sample proportions has the appropriate sign. If so, the appropriate critical value of x 2 from Table 5 will have one degree of freedom a right-tail area of 2a. For example, the critical x 2 value with 1 df and a.05 will be x 2.10 2.70554 1.6452. In summary, you are free to choose the test (z or X2 ) that is most convenient. Since most computer packages include the chi-square test, and most do not include the largesample z-tests, the chi-square test may be preferable to you! The one- and two-sample binomial tests from Chapter 9 are equivalent to chi-square tests— z 2 x 2. OTHER APPLICATIONS OF THE CHI-SQUARE TEST 14.7 The application of the chi-square test for analyzing count data is only one of many classiﬁcation problems that result in multinomial data. Some of these applications are quite complex, requiring complicated or calculationally difficult procedures for estimating the expected cell counts. However, several applications are used often enough to make them worth mentioning. • Goodness-of-ﬁt tests: You can design a goodness-of-ﬁt test to determine whether data are consistent with data drawn from a particular probability distribution—possibly the normal, binomial, Poisson, or other distributions. The cells of a sample frequency histogram correspond to the k cells of a multinomial experiment. Expected cell counts are calculated using the probabilities associated with the hypothesized probability distribution. • Time-dependent multinomials: You can use the chi-square statistic to investigate the rate of change of mult
inomial (or binomial) proportions over time. For example, suppose that the proportion of correct answers on a 100-question exam is recorded for a student, who then repeats the exam in 616 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA each of the next 4 weeks. Does the proportion of correct responses increase over time? Is learning taking place? In a process monitored by a quality control plan, is there a positive trend in the proportion of defective items as a function of time? • Multidimensional contingency tables: Instead of only two methods of classiﬁcation, you can investigate a dependence among three or more classiﬁcations. The two-way contingency table is extended to a table in more than two dimensions. The methodology is similar to that used for the r c contingency table, but the analysis is a bit more complex. • Log-linear models: Complex models can be created in which the logarithm of the cell probability (ln pij) is some linear function of the row and column probabilities. Most of these applications are rather complex and might require that you consult a professional statistician for advice before you conduct your experiment. In all statistical applications that use Pearson’s chi-square statistic, assumptions must be satisﬁed in order that the test statistic have an approximate chi-square probability distribution. ASSUMPTIONS • The cell counts O1, O2,..., Ok must satisfy the conditions of a multinomial experiment, or a set of multinomial experiments created by ﬁxing either the row or column totals. • The expected cell counts E1, E2,..., Ek should equal or exceed 5. You can usually be fairly certain that you have satisﬁed the ﬁrst assumption by carefully preparing and designing your experiment or sample survey. When you calculate the expected cell counts, if you ﬁnd that one or more is less than 5, these options are available to you: • Choose a larger sample size n. The larger the sample size, the closer the chisquare distribution will approximate the distribution of your test statistic X2. It may be possible to combine one or more of the cells with small expected cell counts, thereby satisfying the assumption. • Finally, make sure that you are calculating the degrees of freedom correctly and that you carefully evaluate the statistical and practical conclusions that can be drawn from your test. CH
APTER REVIEW Key Concepts and Formulas I. The Multinomial Experiment 1. There are n identical trials, and each outcome falls into one of k categories. 2. The probability of falling into category i is pi and remains constant from trial to trial. 3. The trials are independent, Spi 1, and we measure Oi, the number of observations that fall into each of the k categories. MY MINITAB ❍ 617 II. Pearson’s Chi-Square Statistic X2 S(Oi Ei)2 where Ei npi Ei which has an approximate chi-square distribution with degrees of freedom determined by the application. III. The Goodness-of-Fit Test 1. This is a one-way classiﬁcation with cell prob- abilities speciﬁed in H0. 2. Use the chi-square statistic with Ei npi calcu- lated with the hypothesized probabilities. 3. df k 1 (Number of parameters esti- mated in order to ﬁnd Ei) 4. If H0 is rejected, investigate the nature of the differences using the sample proportions. 3. If the null hypothesis of independence of classiﬁcations is rejected, investigate the nature of the dependence using conditional proportions within either the rows or columns of the contingency table. V. Fixing Row or Column Totals 1. When either the row or column totals are ﬁxed, the test of independence of classiﬁcations becomes a test of the homogeneity of cell probabilities for several multinomial experiments. 2. Use the same chi-square statistic as for contin- gency tables. 3. The large-sample z-tests for one and two binomial proportions are special cases of the chisquare statistic. IV. Contingency Tables VI. Assumptions 1. The cell counts satisfy the conditions of a multinomial experiment, or a set of multinomial experiments with ﬁxed sample sizes. 2. All expected cell counts must equal or exceed ﬁve in order that the chi-square approximation is valid. 1. A two-way classiﬁcation with n observations categorized into r c cells of a two-way table using two different methods of classiﬁcation is called a contingency table. 2. The test for independence of classiﬁcation methods uses the chi-square statistic X2 S
(Oij Êij)2 Êij with Êij ri cj and df (r 1)(c 1) n The Chi-Square Test Several procedures are available in the MINITAB package for analyzing categorical data. The appropriate procedure depends on whether the data represent a one-way classiﬁcation (a single multinomial experiment) or a two-way classiﬁcation or contingency table. If the raw categorical data have been stored in the MINITAB worksheet rather than the observed cell counts, you may need to tally or cross-classify the data to obtain the cell counts before continuing. For example, suppose you have recorded the gender (M or F) and the college status (Fr, So, Jr, Sr, G) for 100 statistics students. The MINITAB worksheet would contain two columns of 100 observations each. Each row would contain an individual’s gender in column 1 and college status in column 2. To obtain the observed cell counts (Oij) for the 2 5 contingency table, use Stat Tables Cross Tabulation and Chi-Square to generate the Dialog box shown in Figure 14.6. Under “Categorical Variables,” select “Gender” for the row variable and “Status” for the column variable. Leave the boxes marked “For Layers” and “Frequencies 618 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA are in:” blank. Make sure that the square labeled “Display Counts” is checked. Click the Chi-Square... button to display the dialog box in Figure 14.6. Check the boxes for “Chi-Square Analysis” and “Expected Cell Counts.” Click OK twice. This sequence of commands not only tabulates the contingency table, but also performs the chi-square test of independence and displays the results in the Session window shown in Figure 14.7. For the gender/college status data, the large p-value (P.153) indicates a nonsigniﬁcant result. There is insufficient evidence to indicate that a student’s gender is dependent on class status. If the observed cell counts in the contingency table have already been tabulated, simply enter the counts into c columns of the MINITAB worksheet, use Stat Tables Chi-Square Test (Two-Way Table
in Worksheet), and select the appropriate columns before clicking OK. For the gender/college status data, you can enter the counts into columns C3–C7 as shown in Figure 14.8. The resulting output will be labeled differently but will look exactly like the output in Figure 14.7. A simple test of a single multinomial experiment can be set up by considering whether the proportions of male and female statistics students are the same—that is, p1.5 and p2.5. In MINITAB 15, use Stat Tables Chi-Square Goodness-of-Fit Test (One Variable) to display the dialog box in Figure 14.9. If you have raw categorical data in a column, click the “Categorical data:” button and enter the “Gender” column in the cell. If you have summary values of observed counts for each category, choose “Observed counts.” Then enter the column containing the observed counts or type the observed counts for each category. For this test, we can choose “Equal proportions” to test H0: p1 p2.5. When you have different proportions for each category, use “Speciﬁc proportions.” You can F IG URE 14. 6 ● FI GUR E 1 4. 7 ● MY MINITAB ❍ 619 FI GUR E 1 4. 8 ● store the proportions for each category in a column, choose “Input column” and enter the column. If you want to type the proportion for each category, choose “Input constants” and type the proportions for the corresponding categories. Click OK. The resulting output will include several graphs along with the values for Oi and Ei for each category, the observed value of the test statistic, X2 1.44, and its p-value 0.230, which is not signiﬁcant. There is insufficient evidence to indicate a difference in the proportion of male and female statistics students. 620 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA If you are using a previous version of MINITAB, you will have to determine the observed and expected cell counts, and enter them into separate columns in the worksheet. Then use Calc Calculator and the expression SUM((‘O’- ‘E’)**2/‘E’) to calculate the observed value of the test
statistic. F IG URE 14. 9 ● Supplementary Exercises Starred (*) exercises are optional. 14.35 Floor Polish A manufacturer of ﬂoor polish conducted a consumer preference experiment to see whether a new ﬂoor polish A was superior to those produced by four competitors, B, C, D, and E. A sample of 100 housekeepers viewed ﬁve patches of ﬂooring that had received the ﬁve polishes, and each indicated the patch that he or she considered superior in appearance. The lighting, background, and so on were approximately the same for all ﬁve patches. The results of the survey are listed here: Polish Frequency A 27 B 17 C 15 D 22 E 19 Do these data present sufficient evidence to indicate a preference for one or more of the polished patches of ﬂoor over the others? If one were to reject the hypothesis of no preference for this experiment, would this imply that polish A is superior to the others? Can you suggest a better way of conducting the experiment? 14.36 Physical Fitness in the U.S. A survey was conducted to investigate the interest of middle-aged adults in physical ﬁtness programs in Rhode Island, Colorado, California, and Florida. The objective of the investigation was to determine whether adult participation in physical ﬁtness programs varies from one region of the United States to another. A random sample of people were interviewed in each state and these data were recorded: Rhode Island Colorado California Florida Participate 46 Do Not Participate 149 63 178 108 192 121 179 Do the data indicate a difference in adult participation in physical ﬁtness programs from one state to another? If so, describe the nature of the differences. 14.37 Fatal Accidents Accident data were analyzed to determine the numbers of fatal accidents for automobiles of three sizes. The data for 346 accidents are as follows: Small Medium Large Fatal Not Fatal 67 128 26 63 16 46 Do the data indicate that the frequency of fatal accidents is dependent on the size of automobiles? Write a short paragraph describing your statistical results and their practical implications. 14.38 Physicians and Medicare Patients An experiment was conducted to investigate the effect of general hospital experience on the attitudes of physicians toward Medicare patients. A random sample of 50 physicians who had just completed 4 weeks of service in a general hospital and 50 physicians who had not were categorized according to their concern for Medicare patients. The data are shown in the table. Do the data provide sufficient evidence to
indicate a change in “concern” after the general hospital experience? If so, describe the nature of the change. Hospital Service No Hospital Service Low High High Low Total 27 9 5 9 32 18 Partial MINITAB output for Exercise 14.38 Chi-Square Test: High, Low Chi-Sq = 6.752, DF = 1, P-Value = 0.009 14.39 Discovery-Based Teaching Two biology instructors set out to evaluate the EX1439 effects of discovery-based teaching compared to the standard lecture-based teaching approach in the laboratory.13 The standard lecture-based approach provided a list of instructions to follow at each step of the laboratory exercise, whereas the discovery-based approach SUPPLEMENTARY EXERCISES ❍ 621 asked questions rather than providing directions, and used small group reports to decide the best way to proceed in reaching the laboratory objective. One evaluation of the techniques involved written appraisals of both techniques by students at the end of the course. The comparison of the number of positive and negative responses for both techniques is given in the following table. Group Discovery Control Positive Evaluations Negative Evaluations 37 31 11 17 Total 48 48 a. Is there a signiﬁcant difference in the proportion of positive responses for each of the teaching methods? Use a.05. If so, how would you describe this difference? b. What is the approximate p-value for the test in part a? 14.40 Baby’s Sleeping Position Does a baby’s sleeping position affect the development of motor skills? In one study, 343 full-term infants were examined at their 4-month checkup for various developmental milestones, such as rolling over, grasping a rattle, and reaching for an object.14 The baby’s predominant sleep position—either prone (on the stomach) or supine (on the back) or side—was determined by a telephone interview with the parent. The sample results for 320 of the 343 infants for whom information was received are shown in the table. The researcher reported that infants who slept in the side or supine position were less likely to roll over at the 4-month checkup than infants who slept primarily in the prone position (P.001). Prone Supine or Side Number of Infants Number Who Roll Over 121 93 199 119 a. Use a large-sample z-test to conﬁrm or refute the researcher’s conclusion. b. Rewrite the sample data as a
2 2 contingency table. Use the chi-square test for homogeneity to conﬁrm or refute the researcher’s conclusion. c. Compare the results of parts a and b. Conﬁrm that the two test statistics are related as z2 X2 and that the critical values for rejecting H0 have the same relationship. 14.41 Refer to Exercise 14.40. Find the p-value for the large-sample z test in part a. Compare this p-value with the p-value for the chi-square test, shown in the partial MINITAB printout. 622 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA Partial MINITAB output for Exercise 14.41 Chi-Square Test: Prone, Side Chi-Sq = 9.795, DF = 1, P-Value = 0.002 14.42 Baby’s Sleeping Position II The researchers in Exercise 14.40 also measured several other developmental milestones and their relationship to the infant’s predominant sleep position.14 The results of their research are presented in the table for the 320 infants at their 4-month checkup. Milestone Score Prone Supine or Side P Pulls to sit with no head lag Grasps rattle Reaches for object Pass Fail Pass Fail Pass Fail 79 6 102 3 107 3 144 20 167 1 183 5.21.13.97 Use your knowledge of the analysis of categorical data to explain the experimental design(s) used by the researchers. What hypotheses were of interest to the researchers, and what statistical test would the researchers have used? Explain the conclusions that can be drawn from the three p-values in the last column of the table and the practical implications that can be drawn from the statistical results. Have any statistical assumptions been violated? 14.43 Flower Color and Shape A botanist performs a secondary cross of petunias involving independent factors that control leaf shape and ﬂower color, where the factor A represents red color, a represents white color, B represents round leaves, and b represents long leaves. According to the Mendelian model, the plants should exhibit the characteristics AB, Ab, aB, and ab in the ratio 9:3:3:1. Of 160 experimental plants, the following numbers were observed: AB 95 Ab 30 aB 28 ab 7 Is there sufficient evidence to refute the Mendelian model at the a.01 level? 14.44 Salmonella Is your holiday
turkey safe? A “new federal survey found that 13% of turkeys are contaminated with the salmonella bacteria responsible for 1.3 million illnesses and about 500 deaths in a year in the US.”15 Use the table that follows to determine if there is a signiﬁcant difference in the contamination rate at three processing plants. One hundred turkeys were randomly selected from each of the processing lines at these three plants. Salmonella Present Plant Sample Size 1 2 3 42 23 22 100 100 100 Is there a signiﬁcant difference in the rate of salmonella contamination among these three processing plants? If there is a signiﬁcant difference, describe the nature of these differences. Use a.01. 14.45 An Arthritis Drug A study to determine the effectiveness of a drug (serum) for arthritis resulted in the comparison of two groups, each consisting of 200 arthritic patients. One group was inoculated with the serum; the other received a placebo (an inoculation that appears to contain serum but actually is nonactive). After a period of time, each person in the study was asked to state whether his or her arthritic condition had improved. These are the results: Treated Untreated Improved Not Improved 117 83 74 126 You want to know whether these data present sufficient evidence to indicate that the serum was effective in improving the condition of arthritic patients. a. Use the chi-square test of homogeneity to compare the proportions improved in the populations of treated and untreated subjects. Test at the 5% level of signiﬁcance. b. Test the equality of the two binomial proportions using the two-sample z-test of Section 9.6. Verify that the squared value of the test statistic z2 X2 from part a. Are your conclusions the same as in part a? 14.46 Parking at the University A survey was conducted to determine student, faculty, and administration attitudes about a new university parking policy. The distribution of those favoring or opposing the policy is shown in the table. Do the data provide sufficient evidence to indicate that attitudes about the parking policy are independent of student, faculty, or administration status? Student Faculty Administration Favor Oppose 252 139 107 81 43 40 14.47* The chi-square test used in Exercise 14.45 is equivalent to the two-tailed z-test of Section 9.6 provided a is the same for the two tests. Show algebraically that the
chi-square test statistic X2 is the square of the test statistic z for the equivalent test. 14.48 Fitting a Binomial Distribution You can use a goodness-of-ﬁt test to determine whether all of the criteria for a binomial experiment have actually been met in a given application. Suppose that an experiment consisting of four trials was repeated 100 times. The number of repetitions on which a given number of successes was obtained is recorded in the table: Possible Results (number of successes) Number of Times Obtained 0 1 2 3 4 11 17 42 21 9 Estimate p (assuming that the experiment was binomial), obtain estimates of the expected cell frequencies, and test for goodness of ﬁt. To determine the appropriate number of degrees of freedom for X2, note that p was estimated by a linear combination of the observed frequencies. 14.49 Antibiotics and Infection Infections sometimes occur when blood transfusions are given during surgical operations. An experiment was conducted to determine whether the injection of antibodies reduced the probability of infection. An examination of the records of 138 patients produced the data shown in the table. Do the data provide sufficient evidence to indicate that injections of antibodies affect the likelihood of transfusional infections? Test by using a.05. Infection No Infection Antibody No Antibody 4 11 78 45 14.50 German Manufacturing U.S. labor unions have traditionally been content to leave the management of the company to managers and corporate executives. But in Europe, worker participation in management decision making is an accepted idea that is continually spreading. To study the effect of worker participation in managerial decision making, 100 workers were interviewed in each of two separate German manufacturing plants. One plant had active worker participation in managerial decision making; the other did not. Each selected worker was asked whether he or she generally approved of the managerial decisions made within the ﬁrm. The results of the interviews are shown in the table: Participation No Participation Generally Approve Do Not Approve 73 27 51 49 SUPPLEMENTARY EXERCISES ❍ 623 a. Do the data provide sufficient evidence to indicate that approval or disapproval of management’s decisions depends on whether workers participate in decision making? Test by using the X2 test statistic. Use a.05. b. Do these data support the hypothesis that workers in a ﬁrm with participative decision making more generally approve of the ﬁrm’s managerial decisions than those employed by ﬁrms without particip
ative decision making? Test by using the z-test presented in Section 9.6. This problem requires a one-tailed test. Why? 14.51 Three Entrances An occupant-traffic study was conducted to aid in the remodeling of an office building that contains three entrances. The choice of entrance was recorded for a sample of 200 persons who entered the building. Do the data in the table indicate that there is a difference in preference for the three entrances? Find a 95% conﬁdence interval for the proportion of persons favoring entrance 1. Entrance Number Entering 1 83 2 61 3 56 14.52 Homeschool Teachers Parents who are concerned about public school environments and curricula are turning to homeschooling in order to control the content and atmosphere of the learning environments of their children. Although employment as a public school teacher requires a bachelor’s degree in education or a subject area, the educational background of homeschool teachers is quite varied. The educational background of a sample of n 500 parents involved in homeschooling their children in 2003 are provided in the ﬁrst table that follows, along with the corresponding percentages for parents who homeschooled in 1999. The education levels for U.S. citizens in general are given in the second table.16 Parent’s Education 2003 1999 Percentages High school or less Some college/technical Bachelor’s degree Graduate/professional degree 121 153 127 99 18.9 33.7 25.1 22.3 Education Level % U.S. Population, 2003 High school or less Some college Bachelor’s degree or higher 47.5 25.3 27.2 624 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA a. Is there a signiﬁcant change in the educational backgrounds of parents who homeschooled their children in 2003 compared with 1999? Use a.01. b. If there is a signiﬁcant change in the educational backgrounds of these parents, how would you describe that change? c. Using the second table, can we determine if homeschool teachers have the same educational backgrounds as the U.S. population in general? If not, which groups are underrepresented and which are overrepresented? 14.53 Are You a Good Driver? How would you rate yourself as a driver? According to a survey conducted by the Field Institute, most Californians think they are good drivers but have little respect for others’ driving ability. The data
show the distribution of opinions according to gender for two different questions, the ﬁrst rating themselves as drivers and the second rating others as drivers.17 Although not stated in the source, we assume that there were 100 men and 100 women in the surveyed group. Rating Self as a Driver Gender Excellent Good Fair Male Female 43 44 48 53 9 3 Rating Others As Drivers Gender Excellent Good Fair Poor/Very Poor Male Female 4 3 42 48 41 35 13 14 a. Is there sufficient evidence to indicate that there is a difference in the self-ratings between male and female drivers? Find the approximate p-value for the test. b. Is there sufficient evidence to indicate that there is a difference in the ratings of other drivers between male and female drivers? Find the approximate p-value for the test. c. Have any of the assumptions necessary for the analysis used in parts a and b been violated? What affect might this have on the validity of your conclusions? 14.54 Vehicle Colors Each model year seems to introduce new colors and different hues for a wide array of vehicles, from luxury cars, to full-size or intermediate models, to compacts and sports cars, to light trucks. However, white and silver/gray continue to make the top ﬁve or six colors across all of these categories of vehicles. The top ﬁve colors and their percentage of the market share for compact/sports cars are shown in the following table.18 Color Percent Silver 20 Gray 17 Blue 16 Black 14 White/ Pearl 10 To verify the ﬁgures, a random sample consisting of 250 compact/sports cars was taken and the color of the vehicles recorded. The sample provided the following counts for the categories given above: 60, 51, 43, 35, and 30, respectively. a. Is any category missing in the classiﬁcation? How many vehicles belong to that category? b. Is there sufficient evidence to indicate that our percentages of compact/sports cars differ from those given? Find the approximate p-value for the test. 14.55 Funny Cards When you choose a greeting card, do you always look for a humorous card, or does it depend on the occasion? A comparison sponsored by two of the nation’s leading manufacturers of greeting cards indicated a slight difference in the proportions of humorous designs made for three different occasions: Father’s Day, Mother’s Day, and Valentine’s Day.19 To test the accuracy of their comparison, random samples of 500
greeting cards purchased at a local card store in the week prior to each holiday were entered into a computer database, and the results in the table were obtained. Do the data indicate that the proportions of humorous greeting cards vary for these three holidays? (HINT: Remember to include a tabulation for all 1500 greeting cards.) Holiday Father’s Day Mother’s Day Valentine’s Day Percent Humorous 20 25 24 EX1456 14.56 Good Tasting Medicine Pﬁzer Canada Inc. is a pharmaceutical company that makes azithromycin, an antibiotic in a cherry-ﬂavored suspension used to treat bacterial infections in children. To compare the taste of their product with three competing medications, Pﬁzer tested 50 healthy children and 20 healthy adults. Among other taste-testing measures, they recorded the number of tasters who rated each of the four antibiotic suspensions as the best tasting.20 The results are shown in the table. Is there a difference in the perception of the best taste between adults and children? If so, what is the nature of the difference, and why is it of practical importance to the pharmaceutical company? Flavor of Antibiotic Banana Cherry* Wild Fruit Strawberry-Banana Children Adults 14 4 20 14 7 0 *Azithromycin produced by Pﬁzer Canada Inc. 9 2 EX1457 14.57 Rugby Injuries Knee injuries are a major problem for athletes in many contact sports. However, athletes who play certain positions are more prone to get knee injuries than other players, and their injuries tend to be more severe. The prevalence and patterns of knee injuries among women collegiate rugby players were investigated using a sample questionnaire, to which 42 rugby clubs responded.21 A total of 76 knee injuries were classiﬁed by type as well as the position (forward or back) of the player. Type of Knee Injury Position Forward Back Meniscal MCL Tear Tear 13 12 14 9 ACL Tear 7 14 Patella Dislocation PCL Tear 3 2 1 1 MINITAB output for Exercise 14.57 Chi-Square Test: Men Tear, MCL Tear, ACL Tear, Patella, PCL Tear Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Men Tear MCL Tear ACL Tear Patella PCL Tear Total 1 13 14 7 3 1 38 12.50 11.50 10.50 2.50 1.
00 0.020 0.543 1.167 0.100 0.000 2 12 9 14 2 1 38 12.50 11.50 10.50 2.50 1.00 0.020 0.543 1.167 0.100 0.000 Total 25 23 21 5 2 76 Chi-Sq = 3.660, DF = 4, P-Value = 0.454 4 cells with expected counts less than 5.0 a. Use the MINITAB printout to determine whether there is a difference in the distribution of injury types for rugby backs and forwards. Have any of the assumptions necessary for the chi-square test been violated? What effect will this have on the magnitude of the test statistic? b. The investigators report a signiﬁcant difference in the proportion of MCL tears for the two positions (P.05) and a signiﬁcant difference in the SUPPLEMENTARY EXERCISES ❍ 625 proportion of ACL tears (P.05), but indicate that all other injuries occur with equal frequency for the two positions. Do you agree with those conclusions? Explain. EX1458 14.58 Favorite Fast Foods The number of Americans who visit fast-food restaurants regularly has grown steadily over the past decade. For this reason, marketing experts are interested in the demographics of fast-food customers. Is a customer’s preference for a fast-food chain affected by the age of the customer? If so, advertising might need to target a particular age group. Suppose a random sample of 500 fast-food customers aged 16 and older was selected, and their favorite fast-food restaurants along with their age groups were recorded, as shown in the table: Age Group McDonald’s Burger King Wendy’s Other 16–21 21–30 30–49 50 75 89 54 21 34 42 52 25 10 19 28 7 6 10 18 10 Use an appropriate method to determine whether or not a customer’s fast-food preference is dependent on age. Write a short paragraph presenting your statistical conclusions and their practical implications for marketing experts. 14.59 Catching a Cold Is your chance of getting a cold inﬂuenced by the number of social contacts you have? A recent study by Sheldon Cohen, a psychology professor at Carnegie Mellon University, seems to show that the more social relationships you have, the less susceptible you are to colds.22 A group of 276 healthy men and women were grouped according to their number of relationships (
such as parent, friend, church member, neighbor). They were then exposed to a virus that causes colds. An adaptation of the results is shown in the table. Number of Relationships Three or Fewer Four or Five Six or More Cold No Cold Total 49 31 80 43 57 100 34 62 96 a. Do the data provide sufficient evidence to indicate that susceptibility to colds is affected by the number of relationships you have? Test at the 5% signiﬁcance level. 626 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA b. Based on the results of part a, describe the nature of the relationship between the two categorical variables: cold incidence and number of social relationships. Do your observations agree with the author’s conclusions? EX1460 14.60 Crime and Educational Achievement A criminologist studying criminal offenders who have a record of one or more arrests is interested in knowing whether the educational achievement level of the offender inﬂuences the frequency of arrests. He has classiﬁed his data using four educational level classiﬁcations: A: completed 6th grade or less B: completed 7th, 8th, or 9th grade C: completed 10th, 11th, or 12th grade D: education beyond 12th grade The contingency table shows the number of offenders in each educational category, along with the number of times they have been arrested. Educational Achievement Number of Arrests A 1 2 3 or more 55 15 7 B 40 25 8 C 43 18 12 D 30 22 10 Do the data present sufficient evidence to indicate that the number of arrests is dependent on the educational achievement of a criminal offender? Test using a.05. 14.61 More Business on the Weekends A department store manager claims that her store has twice as many customers on Fridays and Saturdays than on any other day of the week (the store is closed on Sundays). That is, the probability that a customer visits the store Friday is 2/8, the probability that a customer visits the store Saturday is 2/8, while the probability that a customer visits the store on each of the remaining weekdays is 1/8. During an average week, the following numbers of customers visited the store: Day Number of Customers Monday Tuesday Wednesday Thursday Friday Saturday 95 110 125 75 181 214 Can the manager’s claim be refuted at the a.05 level of signiﬁcance? Exercises 14.62 Use the Chi
-Square Probabilities applet to ﬁnd the value of x 2 with the following area a to its right: a. a.05, df 15 b. a.01, df 11 14.63 Use the Chi-Square Probabilities applet to ﬁnd the rejection region for a chi-square test of speciﬁed probabilities for a goodness-of-ﬁt test involving k categories for the following cases: a. k 14, a.005 b. k 3, a.05 14.64 Use the Chi-Square Probabilities applet to calculate the p-value for the following chi-square tests: a. X2.81, df 3 b. X2 25.40, df 13 14.65 Three hundred people were surveyed, and were asked to select their preferred brand of laptop computer, given that the prices were equivalent. The results are shown in the table. Brand I Brand II Brand III 115 120 65 Use the ﬁrst Goodness-of-Fit applet to determine if consumers have a preference for one of the three brands. If a signiﬁcant difference exists, describe the difference in practical terms. Use a.01. 14.66 In Exercise 14.13, the color distribution of M&M’S milk chocolate candies was given. Use the third Goodness-of-Fit applet to verify the results of Exercise 14.13. Do the data substantiate the percentages reported by Mars, Incorporated? Describe the nature of the differences, if there are any. 14.67 Refer to the color distribution given in Exercise 14.13. Using an individual-sized bag of milk chocolate M&M’S, count the number of M&M’S in MYAPPLET EXERCISES ❍ 627 each of the six colors. Use the third Goodness-of-Fit applet to determine if the percentages reported by Mars, Incorporated can be substantiated. Describe the nature of the differences, if there are any. 14.68 Repeat the instructions in Exercise 14.67 with another individual bag of M&M’S. Are your conclusions the same? 14.69 Opinion and Political Affiliation A group of 306 people were interviewed to determine their opinion concerning a particular current U.S. foreign policy issue. At the same time, their political affiliation was recorded. Do the data in the table present sufficient evidence to indicate a dependence
between party affiliation and the opinion expressed for the sampled population? Use the third Chi-Square Test of Independence applet. Approve Do Not Approve No Opinion Republicans Democrats 114 87 53 27 17 8 14.70 A study of the purchase decisions of three stock portfolio managers, A, B, and C, was conducted to compare the numbers of stock purchases that resulted in proﬁts over a time period less than or equal to 1 year. One hundred randomly selected purchases were examined for each of the managers. Do the data provide evidence of differences among the rates of successful purchases for the three managers? Use the third Chi-Square Test of Independence applet. A 63 37 B 71 29 C 55 45 Proﬁt No proﬁt 628 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA CASE STUDY Libraries Can a Marketing Approach Improve Library Services? Carole Day and Del Lowenthal studied the responses of young adults in their evaluation of library services.23 Of the n 200 young adults involved in the study, n1 152 were students and n2 48 were nonstudents. The table presents the percents and numbers of favorable responses for each group to seven questions in which the atmosphere, staff, and design of the library were examined. Question Student Favorable n1 152 Nonstudent Favorable n2 48 3 4 5 6 7 11 13 77 91.4 Libraries are friendly 79.6% Libraries are dull Library staff are helpful Library staff are less 60.5 helpful to teenagers Libraries are so quiet they feel uncomfortable Libraries should be more brightly decorated Libraries are badly signposted 45.4 75.6 29 121 117 139 92 115 44 69 56.2% 58.3 87.5 45.8 52.05 18.8 43.8 27 28 42 22 25 9 21 P (x 2).01.05 NS.01.01 NS NS Source: Data from C. Day and D. Lowenthal, “The Use of Open Group Discussions in Marketing Library Services to Young Adults,” by C. Day and D. Lowenthal, British Journal of Educational Psychology, 62(1992): 324–340. The entry in the last column labeled P(x 2) is the p-value for testing the hypothesis of no difference in the proportion of students and nonstudents who answer each question favorably. Hence, each question gives rise to a 2 2 contingency table. 1. Perform a test of homogeneity for each question and verify
the reported p-value of the test. 2. Questions 3, 4, and 7 are concerned with the atmosphere of the library; questions 5 and 6 are concerned with the library staff; and questions 11 and 13 are concerned with the library design. How would you summarize the results of your analyses regarding these seven questions concerning the image of the library? 3. With the information given, is it possible to do any further testing concerning the proportion of favorable versus unfavorable responses for two or more questions simultaneously? 15 Nonparametric Statistics GENERAL OBJECTIVE In Chapters 8–10, we presented statistical techniques for comparing two populations by comparing their respective population parameters (usually their population means). The techniques in Chapters 8 and 9 are applicable to data that are at least quantitative, and the techniques in Chapter 10 are applicable to data that have normal distributions. The purpose of this chapter is to present several statistical tests for comparing populations for the many types of data that do not satisfy the assumptions speciﬁed in Chapters 8–10. CHAPTER INDEX ● The Friedman Fr -test (15.7) ● The Kruskal–Wallis H-test (15.6) ● Parametric versus nonparametric tests (15.1) ● The rank correlation coefficient (15.8) ● The sign test for a paired experiment (15.3) ● The Wilcoxon rank sum test: Independent random sam- ples (15.2) ● The Wilcoxon signed-rank test for a paired experiment (15.5) © Don Carstens/Brand X/CORBIS How’s Your Cholesterol Level? What is your cholesterol level? Many of us have become more health conscious in the last few years as we read the nutritional labels on the food products we buy and choose foods that are low in fat and cholesterol and high in ﬁber. The case study at the end of this chapter involves a taste-testing experiment to compare three types of egg substitutes, using nonparametric techniques. 629 630 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS INTRODUCTION 15.1 When sample sizes are small and the original populations are not normal, use nonparametric techniques. Some experiments generate responses that can be ordered or ranked, but the actual value of the response cannot be measured numerically except with an arbitrary scale that you might create. It may be that you are able to tell only whether one observation is larger than another. Perhaps you can rank a whole set of observations
without actually knowing the exact numerical values of the measurements. Here are a few examples: • The sales abilities of four sales representatives are ranked from best to worst. • The edibility and taste characteristics of ﬁve brands of raisin bran are rated on an arbitrary scale of 1 to 5. • Five automobile designs are ranked from most appealing to least appealing. How can you analyze these types of data? The small-sample statistical methods presented in Chapters 10–13 are valid only when the sampled population(s) are normal or approximately so. Data that consist of ranks or arbitrary scales from 1 to 5 do not satisfy the normality assumption, even to a reasonable degree. In some applications, the techniques are valid if the samples are randomly drawn from populations whose variances are equal. When data do not appear to satisfy these and similar assumptions, an alternative method of analysis can be used—nonparametric statistical methods. Nonparametric methods generally specify hypotheses in terms of population distributions rather than parameters such as means and standard deviations. Parametric assumptions are often replaced by more general assumptions about the population distributions, and the ranks of the observations are often used in place of the actual measurements. Research has shown that nonparametric statistical tests are almost as capable of detecting differences among populations as the parametric methods of preceding chapters when normality and other assumptions are satisﬁed. They may be, and often are, more powerful in detecting population differences when these assumptions are not satisﬁed. For this reason, some statisticians advocate the use of nonparametric procedures in preference to their parametric counterparts. We will present nonparametric methods appropriate for comparing two or more populations using either independent or paired samples. We will also present a measure of association that is useful in determining whether one variable increases as the other increases or whether one variable decreases as the other increases. THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES 15.2 In comparing the means of two populations based on independent samples, the pivotal statistic was the difference in the sample means. If you are not certain that the assumptions required for a two-sample t-test are satisﬁed, one alternative is to replace the values of the observations by their ranks and proceed as though the ranks were the actual observations. Two different nonparametric tests use a test statistic based on these sample ranks: • Wilcoxon rank sum test • Mann-Whitney U-test 15.2 THE W
ILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 631 They are equivalent in that they use the same sample information. The procedure that we will present is the Wilcoxon rank sum test, proposed by Frank Wilcoxon, which is based on the sum of the ranks of the sample that has the smaller sample size. Assume that you have n1 observations from population 1 and n2 observations from population 2. The null hypothesis to be tested is that the two population distributions are identical versus the alternative hypothesis that the population distributions are different. These are the possibilities for the two populations: • • • If H0 is true and the observations have come from the same or identical populations, then the observations from both samples should be randomly mixed when jointly ranked from small to large. The sum of the ranks of the observations from sample 1 should be similar to the sum of the ranks from sample 2. If, on the other hand, the observations from population 1 tend to be smaller than those from population 2, then these observations would have the smaller ranks because most of these observations would be smaller than those from population 2. The sum of the ranks of these observations would be “small.” If the observations from population 1 tend to be larger than those in population 2, these observations would be assigned larger ranks. The sum of the ranks of these observations would tend to be “large.” For example, suppose you have n1 3 observations from population 1—2, 4, and 6—and n2 4 observations from population 2—3, 5, 8, and 9. Table 15.1 shows seven observations ordered from small to large. TABLE 15.1 ● Seven Observations in Order y3 Observation x1 y1 y2 x2 x3 Data Rank y4 9 7 The smallest observation, x1 2, is assigned rank 1; the next smallest observation, y1 3, is assigned rank 2; and so on. The sum of the ranks of the observations from sample 1 is 1 3 5 9, and the rank sum from sample 2 is 2 4 6 7 19. How do you determine whether the rank sum of the observations from sample 1 is signiﬁcantly small or signiﬁcantly large? This depends on the probability distribution of the sum of the ranks of one of the samples. Since the ranks for n1 n2 N observations are the ﬁrst N integers, the sum of these
ranks can be shown to be N(N 1)/2. In this simple example, the sum of the N 7 ranks is 1 2 3 4 5 6 7 7(8)/2 or 28. Hence, if you know the rank sum for one of the samples, you can ﬁnd the other by subtraction. In our example, notice that the rank sum for sample 1 is 9, whereas the second rank sum is (28 9) 19. This means that only one of the two rank sums is needed for the test. To simplify the tabulation of critical values for this test, you should use the rank sum from the smaller sample as the test statistic. What happens if two or more observations are equal? Tied observations are assigned the average of the ranks that the observations would have had if they had been slightly different in value. To implement the Wilcoxon rank sum test, suppose that independent random samples of size n1 and n2 are selected from populations 1 and 2, respectively. Let n1 represent the smaller of the two sample sizes, and let T1 represent the sum of the ranks 632 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS of the observations in sample 1. If population 1 lies to the left of population 2, T1 will be “small.” T1 will be “large” if population 1 lies to the right of population 2. FORMULAS FOR THE WILCOXON RANK SUM STATISTIC (FOR INDEPENDENT SAMPLES) Let T1 Sum of the ranks for the ﬁrst sample 1 n1(n1 n2 1) T1 T * T * 1 is the value of the rank sum for n1 if the observations had been ranked from large to small. (It is not the rank sum for the second sample.) Depending on the nature of the alternative hypothesis, one of these two values will be chosen as the test statistic, T. Table 7 in Appendix I can be used to locate critical values for the test statistic for four different values of one-tailed tests with a.05,.025,.01, and.005. To use Table 7 for a two-tailed test, the values of a are doubled—that is, a.10,.05,.02, and.01. The tabled entry gives the value of a such that P(T a) a. To see how to locate a critical value for the Wilcoxon rank
sum test, suppose that n1 8 and n2 10 for a one-tailed test with a.05. You can use Table 7(a), a portion of which is reproduced in Table 15.2. Notice that the table is constructed assuming that n1 n2. It is for this reason that we designate the population with the smaller sample size as population 1. Values of n1 are shown across the top of the table, and values of n2 are shown down the left side. The entry—a 56, shaded—is the critical value for rejection of H0. The null hypothesis of equality of the two distributions should be rejected if the observed value of the test statistic T is less than or equal to 56. TABLE 15.2 ● Table 7 in Appendix I A Portion of the 5% Left-Tailed Critical Values, n2 10 4 9 10 4 10 11 12 13 14 15 16 17 n1 5 19 20 21 23 24 26 6 7 8 28 29 31 33 35 39 41 43 45 51 54 56 THE WILCOXON RANK SUM TEST Let n1 denote the smaller of the two sample sizes. This sample comes from population 1. The hypotheses to be tested are H0 : The distributions for populations 1 and 2 are identical versus one of three alternative hypotheses: 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 633 Ha : The distributions for populations 1 and 2 are different (a two-tailed test) Ha : The distribution for population 1 lies to the left of that for population 2 (a left-tailed test) Ha : The distribution for population 1 lies to the right of that for population 2 (a right-tailed test) 1. Rank all n1 n2 observations from small to large. 2. Find T1, the rank sum for the observations in sample 1. This is the test statistic for a left-tailed test. 3. Find T * 1 n1(n1 n2 1) T1, the sum of the ranks of the observations from population 1 if the assigned ranks had been reversed from large to small. (The value of T * 1 is not the sum of the ranks of the observations in sample 2.) This is the test statistic for a right-tailed test. 4. The test statistic for a two-tailed test is T, the minimum of T1 and T * 1. 5. H0 is rejected if the observed test statistic is less than or
equal to the critical value found using Table 7 in Appendix I. We illustrate the use of Table 7 with the next example. EXAMPLE 15.1 The wing stroke frequencies of two species of Euglossine bees were recorded for a sample of n1 4 Euglossa mandibularis Friese (species 1) and n2 6 Euglossa imperialis Cockerell (species 2).1 The frequencies are listed in Table 15.3. Can you conclude that the distributions of wing strokes differ for these two species? Test using a.05. TABLE 15.3 ● Wing Stroke Frequencies for Two Species of Bees Species 1 Species 2 235 225 190 188 180 169 180 185 178 182 Solution You ﬁrst need to rank the observations from small to large, as shown in Table 15.4. TABLE 15.4 ● Wing Stroke Frequencies Ranked from Small to Large Data Species Rank 169 178 180 180 182 185 188 190 225 235 10 634 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS The hypotheses to be tested are H0 : The distributions of the wing stroke frequencies are the same for the two species versus Ha : The distributions of the wing stroke frequencies differ for the two species Since the sample size for individuals from species 1, n1 4, is the smaller of the two sample sizes, you have T1 7 8 9 10 34 and 1 n1(n1 n2 1) T1 4(4 6 1) 34 10 T * For a two-tailed test, the test statistic is T 10, the smaller of T1 34 and T * 1 10. For this two-tailed test with a.05, you can use Table 7(b) in Appendix I with n1 4 and n2 6. The critical value of T such that P(T a) a/2.025 is 12, and you should reject the null hypothesis if the observed value of T is 12 or less. Since the observed value of the test statistic—T 10—is less than 12, you can reject the hypothesis of equal distributions of wing stroke frequencies at the 5% level of signiﬁcance. A MINITAB printout of the Wilcoxon rank sum test (called Mann–Whitney by MINITAB) for these data is given in Figure 15.1. You will ﬁnd instructions for generating this output in the section “My MINITAB ” at the end of this chapter. Notice that the
rank sum of the ﬁrst sample is given as W 34.0, which agrees with our calculations. With a reported p-value of.0142 calculated by MINITAB, you can reject the null hypothesis at the 5% level. F IG URE 15. 1 Printout for Example 15.1 ● Mann-Whitney Test and CI: Species 1, Species 2 N Median Species 1 4 207.50 Species 2 6 180.00 Point estimate for ETA1-ETA2 is 30.50 95.7 Percent CI for ETA1-ETA2 is (5.99,56.01) W = 34.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0142 The test is significant at 0.0139 (adjusted for ties) Normal Approximation for the Wilcoxon Rank Sum Test Table 7 in Appendix I contains critical values for sample sizes of n1 n2 3, 4,..., 15. Provided n1 is not too small,† approximations to the probabilities for the Wilcoxon rank sum statistic T can be found using a normal approximation to the distribution of T. It can be shown that the mean and variance of T are n2 1) and s 2 mT n1(n1 2 n2 1) T n1n2(n1 2 1 †Some researchers indicate that the normal approximation is adequate for samples as small as n1 n2 4. 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 635 The distribution of mT z T sT is approximately normal with mean 0 and standard deviation 1 for values of n1 and n2 as small as 10. If you try this approximation for Example 15.1, you get 6 1) 22 n2 1) 4(4 mT n1(n1 2 2 and 6 1) 22 n2 1) 4(6)(4 T n1n2(n1 s 2 12 2 1 The p-value for this test is 2P(T 34). If you use a.5 correction for continuity in calculating the value of z because n1 and n2 are both small,† you have (34 mT z T ).5 2.45 sT 2 2 22 The p-value for this test is 2P(T 34) 2P(z 2.45)
2(.0071).0142 the value reported on the MINITAB printout in Figure 15.1. THE WILCOXON RANK SUM TEST FOR LARGE SAMPLES: n1 W 10 and n2 W 10 1. Null hypothesis: H0 : The population distributions are identical 2. Alternative hypothesis: Ha : The two population distributions are not identical (a two-tailed test). Or Ha : The distribution of population 1 is shifted to the right (or left) of the distribution of population 2 (a one-tailed test). 3. Test statistic: z 4. Rejection region: T n1(n1 n2 1)/2 1)/12 n1n2(n 1 n2 a. For a two-tailed test, reject H0 if z za/2 or z za/2. b. For a one-tailed test in the right tail, reject H0 if z za. c. For a one-tailed test in the left tail, reject H0 if z za. Or reject H0 if p-value a. Tabulated values of z are found in Table 3 of Appendix I. EXAMPLE 15.2 An experiment was conducted to compare the strengths of two types of kraft papers: one a standard kraft paper of a speciﬁed weight and the other the same standard kraft paper treated with a chemical substance. Ten pieces of each type of paper, randomly selected from production, produced the strength measurements shown in Table 15.5. Test the null hypothesis of no difference in the distributions of strengths for the two †Since the value of T 34 lies to the right of the mean 22, the subtraction of.5 in using the normal approximation takes into account the lower limit of the bar above the value 34 in the probability distribution of T. 636 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS types of paper versus the alternative hypothesis that the treated paper tends to be stronger (i.e., its distribution of strength measurements is shifted to the right of the corresponding distribution for the untreated paper). Strength Measurements (and Their Ranks) TABLE 15.5 ● for Two Types of Paper Standard 1 Treated 2 1.21 (2) 1.43 (12) 1.35 (6) 1.51 (17) 1.39 (9) 1.17 (1) 1.48 (14) 1.42 (11) 1.29 (3
.5) 1.40 (10) Rank sum 1.49 (15) 1.37 (7.5) 1.67 (20) 1.50 (16) 1.31 (5) 1.29 (3.5) 1.52 (18) 1.37 (7.5) 1.44 (13) 1.53 (19) T1 85.5 1 n1(n1 n2 1) T1 210 85.5 124.5 T * Solution Since the sample sizes are equal, you are at liberty to decide which of the two samples should be sample 1. Choosing the standard treatment as the ﬁrst sample, you can rank the 20 strength measurements, and the values of T1 and T * 1 are shown at the bottom of the table. Since you want to detect a shift in the standard (1) measurements to the left of the treated (2) measurements, you conduct a left-tailed test: H0 : No difference in the strength distributions Ha : Standard distribution lies to the left of the treated distribution and use T T1 as the test statistic, looking for an unusually small value of T. To ﬁnd the critical value for a one-tailed test with a.05, index Table 7(a) in Appendix I with n1 n2 10. Using the tabled entry, you can reject H0 when T 82. Since the observed value of the test statistic is T 85.5, you are not able to reject H0. There is insufficient evidence to conclude that the treated kraft paper is stronger than the standard paper. To use the normal approximation to the distribution of T, you can calculate n2 1) 10( mT n1(n1 21) 105 2 2 and n2 1) 10(1 T n1n2(n1 )(21) 175 0 s 2 2 1 1 2 with sT 175 13.23. Then mT 85. z T 105 1.47 5 sT.23 3 1 The one-tailed p-value corresponding to z 1.47 is p-value P(z 1.47).5.4292.0708 which is larger than a.05. The conclusion is the same. You cannot conclude that the treated kraft paper is stronger than the standard paper. 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 637 When should the
Wilcoxon rank sum test be used in preference to the two-sample unpaired t-test? The two-sample t-test performs well if the data are normally distributed with equal variances. If there is doubt concerning these assumptions, a normal probability plot could be used to assess the degree of nonnormality, and a two-sample F-test of sample variances could be used to check the equality of variances. If these procedures indicate either nonnormality or inequality of variance, then the Wilcoxon rank sum test is appropriate. 15.2 EXERCISES BASIC TECHNIQUES 1 as the test statistic? 15.1 Suppose you want to use the Wilcoxon rank sum test to detect a shift in distribution 1 to the right of distribution 2 based on samples of size n1 6 and n2 8. a. Should you use T1 or T * b. What is the rejection region for the test if a.05? c. What is the rejection region for the test if a.01? 15.2 Refer to Exercise 15.1. Suppose the alternative hypothesis is that distribution 1 is shifted either to the left or to the right of distribution 2. a. Should you use T1 or T * b. What is the rejection region for the test if a.05? c. What is the rejection region for the test if a.01? 15.3 Observations from two random and independent samples, drawn from populations 1 and 2, are given here. Use the Wilcoxon rank sum test to determine whether population 1 is shifted to the left of population 2. 1 as the test statistic? Sample 1 Sample. State the null and alternative hypotheses to be tested. b. Rank the combined sample from smallest to largest. Calculate T1 and T * 1. c. What is the rejection region for a.05? d. Do the data provide sufficient evidence to indicate that population 1 is shifted to the left of population 2? 15.4 Independent random samples of size n1 20 and n2 25 are drawn from nonnormal populations 1 and 2. The combined sample is ranked and T1 252. Use the large-sample approximation to the Wilcoxon rank sum test to determine whether there is a difference in the two population distributions. Calculate the p-value for the test. 15.5 Suppose you wish to detect a shift in distribution 1 to the right of distribution 2 based on sample sizes n1 12 and n2 14. If T1 193
, what do you conclude? Use a.05. APPLICATIONS 15.6 Alzheimer’s Disease In some tests of healthy, elderly men, a new drug has restored their memory almost to that of young people. It will soon be tested on patients with Alzheimer’s disease, the fatal brain disorder that destroys the mind. According to Dr. Gary Lynch of the University of California, Irvine, the drug, called ampakine CX-516, accelerates signals between brain cells and appears to signiﬁcantly sharpen memory.2 In a preliminary test on students in their early 20s and on men aged 65–70, the results were particularly striking. After being given mild doses of this drug, the 65–70-year-old men scored nearly as high as the young people. The accompanying data are the numbers of nonsense syllables recalled after 5 minutes for 10 men in their 20s and 10 men aged 65–70. Use the Wilcoxon rank sum test to determine whether the distributions for the number of nonsense syllables recalled are the same for these two groups. 20s 65–70s 15.7 Alzheimer’s, continued Refer to Exercise 15.6. Suppose that two more groups of 10 men each are tested on the number of nonsense syllables they can remember after 5 minutes. However, this time the 65–70-year-olds are given a mild dose of ampakine CX-516. Do the data provide sufficient evidence to conclude that this drug improves memory in men aged 65–70 compared with that of 20-year-olds? Use an appropriate level of a. 20s 65–70s 11 10 7 3 10 6 3 638 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS 15.8 Dissolved Oxygen Content The observations in the table are dissolved oxygen contents in water. The higher the dissolved oxygen content, the greater the ability of a river, lake, or stream to support aquatic life. In this experiment, a pollution-control inspector suspected that a river community was releasing semitreated sewage into a river. To check this theory, ﬁve randomly selected specimens of river water were selected at a location above the town and another ﬁve below. These are the dissolved oxygen readings (in parts per million): Above Town Below Town 4.8 5.0 5.2 4.7 5.0 4.9 4.9 4.8 5.1 4.9 a. Use a one
-tailed Wilcoxon rank sum test with a.05 to conﬁrm or refute the theory. b. Use a Student’s t-test (with a.05) to analyze the data. Compare the conclusion reached in part a. EX1509 15.9 Eye Movement In an investigation of the visual scanning behavior of deaf children, measurements of eye movement were taken on nine deaf and nine hearing children. The table gives the eye-movement rates and their ranks (in parentheses). Does it appear that the distributions of eyemovement rates for deaf children and hearing children differ? Deaf Children Hearing Children 2.75 (15) 2.14 (11) 3.23 (18) 2.07 (10) 2.49 (14) 2.18 (12) 3.16 (17) 2.93 (16) 2.20 (13) Rank Sum 126.89 (1) 1.43 (7) 1.06 (4) 1.01 (3).94 (2) 1.79 (8) 1.12 (5.5) 2.01 (9) 1.12 (5.5) 45 15.10 Color TVs The table lists the life (in months) of service before failure of a color television circuit board for 8 television sets manufactured by ﬁrm A and 10 sets manufactured by ﬁrm B. Use the Wilcoxon rank sum test to analyze the data, and test to see whether the life of service before failure of the circuit boards differs for the circuit boards produced by the two manufacturers. Firm A B Life of Circuit Board (months) 32 41 25 39 40 36 31 47 35 45 29 34 37 48 39 44 43 33 15.11 Weights of Turtles The weights of turtles caught in two different lakes were mea- EX1511 sured to compare the effects of the two lake environments on turtle growth. All the turtles were the same age and were tagged before being released into the lakes. The weights for n1 10 tagged turtles caught in lake 1 and n2 8 caught in lake 2 are listed here: Lake Weight (oz) 1 2 14.1 15.2 12.2 13.0 13.9 14.5 14.1 13.6 14.7 12.4 13.8 14.0 16.1 12.7 15.3 11.9 12.5 13.8 Do the data provide sufficient evidence to indicate a difference in the distributions of weights for
the tagged turtles exposed to the two lake environments? Use the Wilcoxon rank sum test with a.05 to answer the question. EX1512 15.12 Chemotherapy Cancer treatment by means of chemicals—chemotherapy—kills both cancerous and normal cells. In some instances, the toxicity of the cancer drug—that is, its effect on normal cells—can be reduced by the simultaneous injection of a second drug. A study was conducted to determine whether a particular drug injection reduced the harmful effects of a chemotherapy treatment on the survival time for rats. Two randomly selected groups of 12 rats were used in an experiment in which both groups, call them A and B, received the toxic drug in a dose large enough to cause death, but in addition, group B received the antitoxin, which was to reduce the toxic effect of the chemotherapy on normal cells. The test was terminated at the end of 20 days, or 480 hours. The survival times for the two groups of rats, to the nearest 4 hours, are shown in the table. Do the data provide sufficient evidence to indicate that rats receiving the antitoxin tend to survive longer after chemotherapy than those not receiving the antitoxin? Use the Wilcoxon rank sum test with a.05. Chemotherapy Only A Chemotherapy plus Drug B 84 128 168 92 184 92 76 104 72 180 144 120 140 184 368 96 480 188 480 244 440 380 480 196 15.3 THE SIGN TEST FOR A PAIRED EXPERIMENT ❍ 639 THE SIGN TEST FOR A PAIRED EXPERIMENT 15.3 The sign test is a fairly simple procedure that can be used to compare two populations when the samples consist of paired observations. This type of experimental design is called the paired-difference or matched pairs design, which you used to compare the average wear for two types of tires in Section 10.5. In general, for each pair, you measure whether the ﬁrst response—say, A—exceeds the second response— say, B. The test statistic is x, the number of times that A exceeds B in the n pairs of observations. When the two population distributions are identical, the probability that A exceeds B equals p.5, and x, the number of times that A exceeds B, has a binomial distribution. Only pairs without ties are included in the test. Hence, you can test the hypothesis of identical population distributions by testing H0 : p.5 versus either a one- or two-tailed alternative. Critical values for
the rejection region or exact p-values can be found using the cumulative binomial tables in Appendix I. THE SIGN TEST FOR COMPARING TWO POPULATIONS 1. Null hypothesis: H0 : The two population distributions are identical and P(A exceeds B) p.5 2. Alternative hypothesis: a. Ha : The population distributions are not identical and p.5 b. Ha : The population of A measurements is shifted to the right of the popu- lation of B measurements and p.5 c. Ha : The population of A measurements is shifted to the left of the popula- tion of B measurements and p.5 3. Test statistic: For n, the number of pairs with no ties, use x, the number of times that (A B) is positive. 4. Rejection region: a. For the two-tailed test Ha : p.5, reject H0 if x xL or x xU, where P(x xL) a/2 and P(x xU) a/2 for x having a binomial distribution with p.5. b. For Ha : p.5, reject H0 if x xU with P(x xU) a. c. For Ha : p.5, reject H0 if x xL with P(x xL) a. Or calculate the p-value and reject H0 if the p-value a. One problem that may occur when you are conducting a sign test is that the measurements associated with one or more pairs may be equal and therefore result in tied observations. When this happens, delete the tied pairs and reduce n, the total number of pairs. The following example will help you understand how the sign test is constructed and used. EXAMPLE 15.3 The numbers of defective electrical fuses produced by two production lines, A and B, were recorded daily for a period of 10 days, with the results shown in Table 15.6. The response variable, the number of defective fuses, has an exact binomial distribution with a large number of fuses produced per day. Although this variable will have an 640 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS approximately normal distribution, the plant supervisor would prefer a quick-and-easy statistical test to determine whether one production line tends to produce more defectives than the other. Use the sign test to test the appropriate hypothesis. TABLE 15.6 ● Defective Fuses From Two Production Lines Day Line A Line B Sign
of Difference 1 2 3 4 5 6 7 8 9 10 170 164 140 184 174 142 191 169 161 200 201 179 159 195 177 170 183 179 170 212 Solution For this paired-difference experiment, x is the number of times that the observation for line A exceeds that for line B in a given day. If there is no difference in the distributions of defectives for the two production lines, then p, the proportion of days on which A exceeds B, is.5, which is the hypothesized value in a test of the binomial parameter p. Very small or very large values of x, the number of times that A exceeds B, are contrary to the null hypothesis. Since n 10 and the hypothesized value of p is.5, Table 1 of Appendix I can be used to ﬁnd the exact p-value for the test of H0 : p.5 versus Ha : p.5 The observed value of the test statistic—which is the number of “plus” signs in the table—is x 1, and the p-value is calculated as p-value 2P(x 1) 2(.011).022 The fairly small p-value.022 allows you to reject H0 at the 5% level. There is signiﬁcant evidence to indicate that the number of defective fuses is not the same for the two production lines; in fact, line B produces more defectives than line A. In this example, the sign test is an easy-to-calculate rough tool for detecting faulty production lines and works perfectly well to detect a signiﬁcant difference using only a minimum amount of information. Normal Approximation for the Sign Test When the number of pairs n is large, the critical values for rejection of H0 and the approximate p-values can be found using a normal approximation to the distribution of x, which was discussed in Section 6.4. Because the binomial distribution is perfectly symmetric when p.5, this approximation works very well, even for n as small as 10. For n 25, you can conduct the sign test by using the z statistic, n p n x.5 z n q 5 np x. as the test statistic. In using z, you are testing the null hypothesis p.5 versus the alternative p.5 for a two-tailed test or versus the alternative p.5 (or p.5) for a one-tailed test. The tests use the familiar rejection regions
of Chapter 9. 15.3 THE SIGN TEST FOR A PAIRED EXPERIMENT ❍ 641 SIGN TEST FOR LARGE SAMPLES: n W 25 1. Null hypothesis: H0 : p.5 (one treatment is not preferred to a second treat- ment) 2. Alternative hypothesis: Ha : p.5, for a two-tailed test (NOTE: We use the two-tailed test as an example. Many analyses might require a one-tailed test.) 5n. 3. Test statistic: z n 5 x. EXAMPLE 15.4 4. Rejection region: Reject H0 if z za/2 or z za/2, where za/2 is the z-value from Table 3 in Appendix I corresponding to an area of a/2 in the upper tail of the normal distribution. A production superintendent claims that there is no difference between the employee accident rates for the day versus the evening shifts in a large manufacturing plant. The number of accidents per day is recorded for both the day and evening shifts for n 100 days. It is found that the number of accidents per day for the evening shift xE exceeded the corresponding number of accidents on the day shift xD on 63 of the 100 days. Do these results provide sufficient evidence to indicate that more accidents tend to occur on one shift than on the other or, equivalently, that P(xE xD) 1/2? Solution This study is a paired-difference experiment, with n 100 pairs of observations corresponding to the 100 days. To test the null hypothesis that the two distributions of accidents are identical, you can use the test statistic n x.5 z n 5. where x is the number of days in which the number of accidents on the evening shift exceeded the number of accidents on the day shift. Then for a.05, you can reject the null hypothesis if z 1.96 or z 1.96. Substituting into the formula for z, you get n x (.5 100) ( ).5 3 2.60 1 z 0 n.5 5 5 1 0 63. Since the calculated value of z exceeds za/2 1.96, you can reject the null hypothesis. The data provide sufficient evidence to indicate a difference in the accident rate distributions for the day versus evening shifts. When should the sign test be used in preference to the paired t-test? When only the direction of the difference in the measurement is given, only the sign test can be used
. On the other hand, when the data are quantitative and satisfy the normality and constant variance assumptions, the paired t-test should be used. A normal probability plot can be used to assess normality, while a plot of the residuals (di d) can reveal large deviations that might indicate a variance that varies from pair to pair. When there are doubts about the validity of the assumptions, statisticians often recommend that both tests be performed. If both tests reach the same conclusions, then the parametric test results can be considered to be valid. 642 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS 15.3 EXERCISES BASIC TECHNIQUES 15.13 Suppose you wish to use the sign test to test Ha : p.5 for a paired-difference experiment with n 25 pairs. a. State the practical situation that dictates the alterna- tive hypothesis given. b. Use Table 1 in Appendix I to ﬁnd values of a (a.15) available for the test. 15.14 Repeat the instructions of Exercise 15.13 for Ha : p.5. 15.15 Repeat the instructions of Exercises 15.13 and 15.14 for n 10, 15, and 20. EX1516 15.16 A paired-difference experiment was conducted to compare two populations. The data are shown in the table. Use a sign test to determine whether the population distributions are different. Population 1 1 2 8.9 8.8 2 8.1 7.4 3 9.3 9.0 Pairs 4 7.7 7.8 5 10.4 9.9 6 8.3 8.1 7 7.4 6.9 a. State the null and alternative hypotheses for the test. b. Determine an appropriate rejection region with a.01. c. Calculate the observed value of the test statistic. d. Do the data present sufficient evidence to indicate that populations 1 and 2 are different? APPLICATIONS 15.17 Property Values In Exercise 10.45, you compared the property evaluations of two EX1517 tax assessors, A and B. Their assessments for eight properties are shown in the table: Property Assessor A Assessor B 1 2 3 4 5 6 7 8 76.3 88.4 80.2 94.7 68.7 82.8 76.1 79.0 75.1 86.8 77.3 90.6 69.1 81.0 75.3 79.1 a
. Use the sign test to determine whether the data present sufficient evidence to indicate that one of the assessors tends to be consistently more conservative than the other; that is, P(xA xB) 1/2. Test by using a value of a near.05. Find the p-value for the test and interpret its value. b. Exercise 10.45 uses the t statistic to test the null hypothesis that there is no difference in the mean property assessments between assessors A and B. Check the answer (in the answer section) for Exercise 10.45 and compare it with your answer to part a. Do the test results agree? Explain why the answers are (or are not) consistent. EX1518 15.18 Gourmet Cooking Two gourmets, A and B, rated 22 meals on a scale of 1 to 10. The data are shown in the table. Do the data provide sufficient evidence to indicate that one of the gourmets tends to give higher ratings than the other? Test by using the sign test with a value of a near.05. Meal A B Meal 10 11 12 13 14 15 16 17 18 19 20 21 22 10 8 6 3 4 2 3 a. Use the binomial tables in Appendix I to ﬁnd the exact rejection region for the test. b. Use the large-sample z statistic. (NOTE: Although the large-sample approximation is suggested for n 25, it works fairly well for values of n as small as 15.) c. Compare the results of parts a and b. 15.19 Lead Levels in Blood A study reported in the American Journal of Public Health (Science News)—the ﬁrst to follow blood lead levels in lawabiding handgun hobbyists using indoor ﬁring ranges—documents a signiﬁcant risk of lead poisoning.3 Lead exposure measurements were made on 17 members of a law enforcement trainee class before, during, and after a 3-month period of ﬁrearm instruction at a state-owned indoor ﬁring range. No trainee had elevated blood lead levels before the training, but 15 of the 17 ended their training with blood lead levels deemed “elevated” by the Occupational Safety and Health Administration (OSHA). If the use of an indoor 15.4 A COMPARISON OF STATISTICAL TESTS ❍ 643 ﬁring range causes no increase in blood lead levels, then p, the probability that
a person’s blood lead level increases, is less than or equal to.5. If, however, use of the indoor ﬁring range causes an increase in a person’s blood lead levels, then p.5. Use the sign test to determine whether using an indoor ﬁring range has the effect of increasing a person’s blood lead level with a.05. (HINT: The normal approximation to binomial probabilities is fairly accurate for n 17.) 15.20 Recovery Rates Clinical data concerning the effectiveness of two drugs in treating EX1520 a particular disease were collected from ten hospitals. The numbers of patients treated with the drugs varied from one hospital to another. You want to know whether the data present sufficient evidence to indicate a higher recovery rate for one of the two drugs. a. Test using the sign test. Choose your rejection region so that a is near.05. b. Why might it be inappropriate to use the Student’s t-test in analyzing the data? Drug A Drug B Hospital Number in Group Number Recovered Percentage Recovered Hospital Number in Group Number Recovered Percentage Recovered 1 2 3 4 5 6 7 8 9 10 84 63 56 77 29 48 61 45 79 62 63 44 48 57 20 40 42 35 57 48 75.0 69.8 85.7 74.0 69.0 83.3 68.9 77.8 72.2 77. 10 96 83 91 47 60 27 69 72 89 46 82 69 73 35 42 22 52 57 76 37 85.4 83.1 80.2 74.5 70.0 81.5 75.4 79.2 85.4 80.4 A COMPARISON OF STATISTICAL TESTS 15.4 The experiment in Example 15.3 is designed as a paired-difference experiment. If the assumptions of normality and constant variance, s 2 d, for the differences were met, would the sign test detect a shift in location for the two populations as efficiently as the paired t-test? Probably not, since the t-test uses much more information than the sign test. It uses not only the sign of the difference, but also the actual values of the differences. In this case, we would say that the sign test is not as efficient as the paired t-test. However, the sign test might be more efficient if the usual assumptions were not met. When two different statistical tests can both be used to test a hypothesis based on the same data, it is
natural to ask, Which is better? One way to answer this question would be to hold the sample size n and a constant for both procedures and compare b, the probability of a Type II error. Statisticians, however, prefer to examine the power of a test. Definition Power 1 b P(reject H0 when Ha is true) Since b is the probability of failing to reject the null hypothesis when it is false, the power of the test is the probability of rejecting the null hypothesis when it is false and some speciﬁed alternative is true. It is the probability that the test will do what it was designed to do—that is, detect a departure from the null hypothesis when a departure exists. 644 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS Probably the most common method of comparing two test procedures is in terms of the relative efficiency of a pair of tests. Relative efficiency is the ratio of the sample sizes for the two test procedures required to achieve the same a and b for a given alternative to the null hypothesis. In some situations, you may not be too concerned whether you are using the most powerful test. For example, you might choose to use the sign test over a more powerful competitor because of its ease of application. Thus, you might view tests as microscopes that are used to detect departures from an hypothesized theory. One need not know the exact power of a microscope to use it in a biological investigation, and the same applies to statistical tests. If the test procedure detects a departure from the null hypothesis, you are delighted. If not, you can reanalyze the data by using a more powerful microscope (test), or you can increase the power of the microscope (test) by increasing the sample size. THE WILCOXON SIGNED-RANK TEST FOR A PAIRED EXPERIMENT 15.5 A signed-rank test proposed by Frank Wilcoxon can be used to analyze the paireddifference experiment of Section 10.5 by considering the paired differences of two treatments, 1 and 2. Under the null hypothesis of no differences in the distributions for 1 and 2, you would expect (on the average) half of the differences in pairs to be negative and half to be positive; that is, the expected number of negative differences between pairs would be n/2 (where n is the number of pairs). Furthermore, it follows that positive and negative differences of equal absolute magnitude should occur with equal probability. If one were to order the differences according to
their absolute values and rank them from smallest to largest, the expected rank sums for the negative and positive differences would be equal. Sizable differences in the sums of the ranks assigned to the positive and negative differences would provide evidence to indicate a shift in location between the distributions of responses for the two treatments, 1 and 2. If distribution 1 is shifted to the right of distribution 2, then more of the differences are expected to be positive, and this results in a small number of negative differences. Therefore, to detect this one-sided alternative, use the rank sum T —the sum of the ranks of the negative differences—and reject the null hypothesis for signiﬁcantly small values of T. Along these same lines, if distribution 1 is shifted to the left of distribution 2, then more of the differences are expected to be negative, and the number of positive differences is small. Hence, to detect this one-sided alternative, use T —the sum of the ranks of the positive differences—and reject the null hypothesis if T is signiﬁcantly small. CALCULATING THE TEST STATISTIC FOR THE WILCOXON SIGNED-RANK TEST 1. Calculate the differences (x1 x2) for each of the n pairs. Differences equal to 0 are eliminated, and the number of pairs, n, is reduced accordingly. 2. Rank the absolute values of the differences by assigning 1 to the smallest, 2 to the second smallest, and so on. Tied observations are assigned the average of the ranks that would have been assigned with no ties. 3. Calculate the rank sum for the negative differences and label this value T. Similarly, calculate T, the rank sum for the positive differences. 15.5 THE WILCOXON SIGNED-RANK TEST FOR A PAIRED EXPERIMENT ❍ 645 For a two-tailed test, use the smaller of these two quantities T as a test statistic to test the null hypothesis that the two population relative frequency histograms are identical. The smaller the value of T, the greater is the weight of evidence favoring rejection of the null hypothesis. Therefore, you will reject the null hypothesis if T is less than or equal to some value—say, T0. To detect the one-sided alternative, that distribution 1 is shifted to the right of distribution 2, use the rank sum T of the negative differences and reject the null hypothesis for small values of T —say, T T0. If you wish to
detect a shift of distribution 2 to the right of distribution 1, use the rank sum T of the positive differences as a test statistic and reject the null hypothesis for small values of T —say, T T0. The probability that T is less than or equal to some value T0 has been calculated for a combination of sample sizes and values of T0. These probabilities, given in Table 8 in Appendix I, can be used to ﬁnd the rejection region for the T test. An abbreviated version of Table 8 is shown in Table 15.7. Across the top of the table you see the number of differences (the number of pairs) n. Values of a for a one-tailed test appear in the ﬁrst column of the table. The second column gives values of a for a two-tailed test. Table entries are the critical values of T. You will recall that the critical value of a test statistic is the value that locates the boundary of the rejection region. For example, suppose you have n 7 pairs and you are conducting a two-tailed test of the null hypothesis that the two population relative frequency distributions are identical. Checking the n 7 column of Table 15.7 and using the second row (corresponding to a.05 for a two-tailed test), you see the entry 2 (shaded). This value is T0, the critical value of T. As noted earlier, the smaller the value of T, the greater is the evidence to reject the null hypothesis. Therefore, you will reject the null hypothesis for all values of T less than or equal to 2. The rejection region for the Wilcoxon signed-rank test for a paired experiment is always of the form: Reject H0 if T T0, where T0 is the critical value of T. The rejection region is shown symbolically in Figure 15.2. TABLE 15.7 ● Critical Values of T An Abbreviated Version of Table 8 in Appendix I; One-Sided a.050 a.025 a.010 a.005 Two-Sided a.10 a.05 a.02 a.01 One-Sided a.050 a.025 a.010 a.005 Two-Sided a.10 a.05 a.02 a.01 ● FI GUR E 1 5. 2 Rejection region for the Wilcoxon signed-rank test for a paired experiment (reject H0 if T T0) 0 1 2 T0 Rejection region
10 n 11 11 8 5 3 14 11 7 5 n 12 n 13 n 14 n 15 n 16 n 17 17 14 10 7 21 17 13 10 26 21 16 13 36 30 24 19 41 35 28 23 30 25 20 16 T 646 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS WILCOXON SIGNED-RANK TEST FOR A PAIRED EXPERIMENT 1. Null hypothesis: H0 : The two population relative frequency distributions are identical 2. Alternative hypothesis: Ha : The two population relative frequency distributions differ in location (a two-tailed test). Or Ha : The population 1 relative frequency distribution is shifted to the right of the relative frequency distribution for population 2 (a one-tailed test). 3. Test statistic a. For a two-tailed test, use T, the smaller of the rank sum for positive and the rank sum for negative differences. b. For a one-tailed test (to detect the alternative hypothesis described above), use the rank sum T of the negative differences. 4. Rejection region a. For a two-tailed test, reject H0 if T T0, where T0 is the critical value given in Table 8 in Appendix I. b. For a one-tailed test (to detect the alternative hypothesis described above), use the rank sum T of the negative differences. Reject H0 if T T0.† 1). NOTE: It can be shown that T T n(n 2 An experiment was conducted to compare the densities of cakes prepared from two different cake mixes, A and B. Six cake pans received batter A, and six received batter B. Expecting a variation in oven temperature, the experimenter placed an A and a B cake side by side at six different locations in the oven. Test the hypothesis of no difference in the population distributions of cake densities for two different cake batters. Solution The data (density in ounces per cubic inch) and differences in density for six pairs of cakes are given in Table 15.8. The boxplot of the differences in Figure 15.3 shows fairly strong skewing and a very large difference in the right tail, which indicates that the data may not satisfy the normality assumption. The sample of differences is too small to make valid decisions about normality and constant variance. In this situation, Wilcoxon’s signed-rank test may be the prudent test to use. As with other nonparametric tests, the null hypothesis to be tested is that the two population frequency distributions of
cake densities are identical. The alternative hypothesis, which implies a two-tailed test, is that the distributions are different. Because the amount of data is small, you can conduct the test using a.10. From Table 8 in Appendix I, the critical value of T for a two-tailed test, a.10, is T0 2. Hence, you can reject H0 if T 2. †To detect a shift of distribution 2 to the right of distribution 1, use the rank sum T of the positive differences as the test statistic and reject H0 if T T0. EXAMPLE 15.5 15.5 THE WILCOXON SIGNED-RANK TEST FOR A PAIRED EXPERIMENT ❍ 647 TABLE 15.8 ● Densities of Six Pairs of Cakes xA.135.102.098.141.131.144 xB.129.120.112.152.135.163 Difference (xA xB) Rank.006.018.014.011.004.019 2 5 4 3 1 6 FI GUR E 1 5. 3 Box plot of differences for Example 15.5 ● 0.020 0.015 0.010 0.005 Differences 0.000 0.005 The differences (x1 x2) are calculated and ranked according to their absolute values in Table 15.8. The sum of the positive ranks is T 2, and the sum of the negative ranks is T 19. The test statistic is the smaller of these two rank sums, or T 2. Since T 2 falls in the rejection region, you can reject H0 and conclude that the two population frequency distributions of cake densities differ. A MINITAB printout of the Wilcoxon signed-rank test for these data is given in Figure 15.4. You will ﬁnd instructions for generating this output in the section “My MINITAB ” at the end of this chapter. You can see that the value of the test statistic agrees with the other calculations, and the p-value indicates that you can reject H0 at the 10% level of signiﬁcance. FI GUR E 1 5. 4 MINITAB printout for Example 15.5 ● Wilcoxon Signed Rank Test: Difference Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median Difference 6 6 2.0 0.093 -0
.01100 Normal Approximation for the Wilcoxon Signed-Rank Test Although Table 8 in Appendix I has critical values for n as large as 50, T, like the Wilcoxon signed-rank test, will be approximately normally distributed when the null 648 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS hypothesis is true and n is large—say, 25 or more. This enables you to construct a large-sample z-test, where 1) E(T) n(n 4 2n 1) T n(n 1 )( s 2 2 4 Then the z statistic (T ) z T E sT 1) T n(n 4 n(n 1 ) 4 2 (2n 1) can be used as a test statistic. Thus, for a two-tailed test and a.05, you can reject the hypothesis of identical population distributions when z 1.96. A LARGE-SAMPLE WILCOXON SIGNED-RANK TEST FOR A PAIRED EXPERIMENT: n W 25 1. Null hypothesis: H0 : The population relative frequency distributions 1 and 2 are identical. 2. Alternative hypothesis: Ha : The two population relative frequency distributions differ in location (a two-tailed test). Or Ha : The population 1 relative frequency distribution is shifted to the right (or left) of the relative frequency distribution for population 2 (a one-tailed test). 3. Test statistic: z T [n(n 1)/4] [n(n 1)(2n 1)]/24 4. Rejection region: Reject H0 if z za/2 or z za/2 for a two-tailed test. For a one-tailed test, place all of a in one tail of the z distribution. To detect a shift in distribution 1 to the right of distribution 2, reject H0 when z za. To detect a shift in the opposite direction, reject H0 if z za. Tabulated values of z are given in Table 3 in Appendix I. 15.5 EXERCISES BASIC TECHNIQUES 15.21 Suppose you wish to detect a difference in the locations of two population distributions based on a paired-difference experiment consisting of n 30 pairs. a. Give the null and alternative hypotheses for the 15.22 Refer to Exercise 15.21. Suppose you wish to detect only a shift in distribution 1 to the right of distribution 2. a. Give the null and alternative hypotheses for the
Wilcoxon signed-rank test. Wilcoxon signed-rank test. b. Give the test statistic. c. Give the rejection region for the test for a.05. d. If T 249, what are your conclusions? [NOTE: T T n(n 1)/2.] b. Give the test statistic. c. Give the rejection region for the test for a.05. d. If T 249, what are your conclusions? [NOTE: T T n(n 1)/2.] 15.5 THE WILCOXON SIGNED-RANK TEST FOR A PAIRED EXPERIMENT ❍ 649 15.23 Refer to Exercise 15.21. Conduct the test using the large-sample z-test. Compare your results with the nonparametric test results in Exercise 15.22, part d. a. Do the data provide sufficient evidence to indicate a difference in the monthly breakdown rates for the two machines? Test by using a value of a near.05. 15.24 Refer to Exercise 15.22. Conduct the test using the large-sample z-test. Compare your results with the nonparametric test results in Exercise 15.21, part d. 15.25 Refer to Exercise 15.16 and data set EX1516. The data in this table are from a paired-difference experiment with n 7 pairs of observations. Population 1 2 1 8.9 8.8 2 8.1 7.4 3 9.3 9.0 Pairs 4 7.7 7.8 5 10.4 9.9 6 8.3 8.1 7 7.4 6.9 a. Use Wilcoxon’s signed-rank test to determine whether there is a signiﬁcant difference between the two populations. b. Compare the results of part a with the result you got in Exercise 15.16. Are they the same? Explain. APPLICATIONS 15.26 Property Values II In Exercise 15.17, you used the sign test to determine whether the data provided sufficient evidence to indicate a difference in the distributions of property assessments for assessors A and B. a. Use the Wilcoxon signed-rank test for a paired experiment to test the null hypothesis that there is no difference in the distributions of property assessments between assessors A and B. Test by using a value of a near.05. b. Compare the conclusion of the test in part a with the conclusions derived from the t-test in Exercise 10.
43 and the sign test in Exercise 15.17. Explain why these test conclusions are (or are not) consistent. 15.27 Machine Breakdowns The number of machine breakdowns per month was EX1527 recorded for 9 months on two identical machines, A and B, used to make wire rope: Month 14 7 10 9 6 13 6 7 B 7 12 9 15 12 6 12 5 13 b. Can you think of a reason the breakdown rates for the two machines might vary from month to month? 15.28 Gourmet Cooking II Refer to the comparison of gourmet meal ratings in Exercise 15.18, and use the Wilcoxon signed-rank test to determine whether the data provide sufficient evidence to indicate a difference in the ratings of the two gourmets. Test by using a value of a near.05. Compare the results of this test with the results of the sign test in Exercise 15.18. Are the test conclusions consistent? 15.29 Traffic Control Two methods for controlling traffic, A and B, were used at each EX1529 of n 12 intersections for a period of 1 week, and the numbers of accidents that occurred during this time period were recorded. The order of use (which method would be employed for the ﬁrst week) was selected in a random manner. You want to know whether the data provide sufficient evidence to indicate a difference in the distributions of accident rates for traffic control methods A and B. Intersection Method B A Intersection Method 10 11 12. Analyze using a sign test. b. Analyze using the Wilcoxon signed-rank test for a paired experiment. 15.30 Jigsaw Puzzles Eight people were asked to perform a simple puzzle-assembly EX1530 task under normal conditions and under stressful conditions. During the stressful time, a mild shock was delivered to subjects 3 minutes after the start of the experiment and every 30 seconds thereafter until the task was completed. Blood pressure readings were taken under both conditions. The data in the table are the highest readings during the experiment. Do the data present sufficient evidence to indicate higher blood pressure readings under stressful conditions? Analyze the data using the Wilcoxon signed-rank test for a paired experiment. a. What three testing procedures can be used to test for differences in the distribution of recall scores with and without imagery? What assumptions are required for the parametric procedure? Do these data satisfy these assumptions? b. Use both the sign test and the Wilcoxon signed- rank test to test for differences in the distributions
of recall scores under these two conditions. c. Compare the results of the tests in part b. Are the conclusions the same? If not, why not? 650 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS Subject Normal Stressful 1 2 3 4 5 6 7 8 126 117 115 118 118 128 125 120 130 118 125 120 121 125 130 120 15.31 Images and Word Recall A psychology class performed an experiment to EX1531 determine whether a recall score in which instructions to form images of 25 words were given differs from an initial recall score for which no imagery instructions were given. Twenty students participated in the experiment with the results listed in the table. Student With Imagery Without Imagery Student With Imagery Without Imagery 1 2 3 4 5 6 7 8 9 10 20 24 20 18 22 19 20 19 17 21 5 9 5 9 6 11 8 11 7 9 11 12 13 14 15 16 17 18 19 20 17 20 20 16 24 22 25 21 19 23 8 16 10 12 7 9 21 14 12 13 THE KRUSKAL–WALLIS H-TEST FOR COMPLETELY RANDOMIZED DESIGNS 15.6 Just as the Wilcoxon rank sum test is the nonparametric alternative to Student’s t-test for a comparison of population means, the Kruskal–Wallis H-test is the nonparametric alternative to the analysis of variance F-test for a completely randomized design. It is used to detect differences in locations among more than two population distributions based on independent random sampling. The procedure for conducting the Kruskal–Wallis H-test is similar to that used for the Wilcoxon rank sum test. Suppose you are comparing k populations based on independent random samples n1 from population 1, n2 from population 2,..., nk from population k, where n1 n2 nk n The ﬁrst step is to rank all n observations from the smallest (rank 1) to the largest (rank n). Tied observations are assigned a rank equal to the average of the ranks they would have received if they had been nearly equal but not tied. You then calculate the rank sums T1, T2,..., Tk for the k samples and calculate the test statistic 2 S T 2 1 3(n 1) H i ni n(n 1) 15.6 THE KRUSKAL–WALLIS H-TEST FOR COMPLETELY RANDOMIZED DESIG
NS ❍ 651 which is proportional to S ni (Ti T)2, the sum of squared deviations of the rank means about the grand mean T n(n 1)/2n (n 1)/2. The greater the differences in locations among the k population distributions, the larger is the value of the H statistic. Thus, you can reject the null hypothesis that the k population distributions are identical for large values of H. How large is large? It can be shown (proof omitted) that when the sample sizes are moderate to large—say, each sample size is equal to ﬁve or larger—and when H0 is true, the H statistic will have approximately a chi-square distribution with (k 1) degrees of freedom. Therefore, for a given value of a, you can reject H0 when the H statistic exceeds x 2 a (see Figure 15.5). FI GUR E 1 5. 5 Approximate distribution of the H statistic when H0 is true ● f(H) Chi-square distribution 0 α χ 2 α H Rejection region EXAMPLE 15.6 The data in Table 15.9 were collected using a completely randomized design. They are the achievement test scores for four different groups of students, each group taught by a different teaching technique. The objective of the experiment is to test the hypothesis of no difference in the population distributions of achievement test scores versus the alternative that they differ in location; that is, at least one of the distributions is shifted above the others. Conduct the test using the Kruskal–Wallis H-test with a.05. TABLE 15.9 ● Test Scores (and Ranks) from Four Teaching Techniques 1 65 (3) 87 (19) 73 (8) 79 (12.5) 81 (15.5) 69 (5.5) Rank Sum T1 63.5 2 75 (9) 69 (5.5) 83 (17.5) 81 (15.5) 72 (7) 79 (12.5) 90(22) T2 89 4 94 (23) 89 (21) 80 (14) 88 (20) 3 59 (1) 78 (11) 67 (4) 62 (2) 83 (17.5) 76 (10) T3 45.5 T4 78 Solution Before you perform a nonparametric analysis on these data, you can use a one-way analysis of variance to provide the two plots in Figure 15.6. It appears
that technique 4 has a smaller variance than the other three and that there is a marked deviation in the right tail of the normal probability plot. These deviations could be considered minor and either a parametric or nonparametric analysis could be used. 652 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS Normal Probability Plot of the Residuals (response is Scores) ● FI GU RE 15.6 A normal probability plot and a residual plot following a one-way analysis of variance for Example 15.6 t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 20 10 0 Residual 10 20 Residuals versus the Fitted Values (response is Scores) 15 10 5 0 5 10 l a u d i s e R 70 75 80 Fitted Value 85 90 In the Kruskal–Wallis H-test procedure, the ﬁrst step is to rank the n 23 observations from the smallest (rank 1) to the largest (rank 23). These ranks are shown in parentheses in Table 15.9. Notice how the ties are handled. For example, two observations at 69 are tied for rank 5. Therefore, they are assigned the average 5.5 of the two ranks (5 and 6) that they would have occupied if they had been slightly different. The rank sums T1, T2, T3, and T4 for the four samples are shown in the bottom row of the table. Substituting rank sums and sample sizes into the formula for the H statistic, you get 2 S T 2 1 3(n 1) H i n n(n 1) i 1 ( 23 (63 2 24) 3(24) 8)2.5)2 9)2.5)2 (7 (45 (8 4 6 7 6 79.775102 72 7.775102 The rejection region for the H statistic for a.05 includes values of H x 2.05, where.05 is based on (k 1) (4 1) 3 df. The value of x2 given in Table 5 in x 2.05 7.81473. The observed value of the H statistic, H 7.775102, Appendix I is x 2 does not fall into the rejection region for the test. Therefore, there is insufficient evidence to indicate differences in the distributions of achievement test scores for the four teaching techniques. 15.6 THE KRUSKAL–WALLIS H-TEST FOR
COMPLETELY RANDOMIZED DESIGNS ❍ 653 A MINITAB printout of the Kruskal–Wallis H-test for these data is given in Figure 15.7. Notice that the p-value,.051, is only slightly greater than the 5% level necessary to declare statistical signiﬁcance. ● FI GUR E 1 5. 7 MINITAB printout for the Kruskal–Wallis test for Example 15.6 Kruskal-Wallis Test: Scores versus Techniques Kruskal-Wallis Test on Scores Techniques N Median Ave Rank Z 1 6 76.00 10.6 -0.60 2 7 79.00 12.7 0.33 3 6 71.50 7.6 -1.86 4 4 88.50 19.5 2.43 Overall 23 12.0 H = 7.78 DF = 3 P = 0.051 H = 7.79 DF = 3 P = 0.051 (adjusted for ties) * NOTE * One or more small samples EXAMPLE 15.7 Compare the results of the analysis of variance F-test and the Kruskal–Wallis H-test for testing for differences in the distributions of achievement test scores for the four teaching techniques in Example 15.6. Solution The MINITAB printout for a one-way analysis of variance for the data in Table 15.9 is given in Figure 15.8. The analysis of variance shows that the F-test for testing for differences among the means for the four techniques is signiﬁcant at the.028 level. The Kruskal–Wallis H-test did not detect a shift in population distributions at the.05 level of signiﬁcance. Although these conclusions seem to be far apart, the test results do not differ strongly. The p-value.028 corresponding to F 3.77, with df1 3 and df2 19, is slightly less than.05, in contrast to the p-value.051 for H 7.78, df 3, which is slightly greater than.05. Someone viewing the p-values for the two tests would see little difference in the results of the F- and H-tests. However, if you adhere to the choice of a.05, you cannot reject H0 using the H-test. FI GUR E 1 5. 8 MINITAB printout for Example 15.7 ● One
-way ANOVA: Scores versus Techniques Source DF SS MS F P Techniques 3 712.6 237.5 3.77 0.028 Error 19 1196.6 63.0 Total 22 1909.2 THE KRUSKAL–WALLIS H-TEST FOR COMPARING MORE THAN TWO POPULATIONS: COMPLETELY RANDOMIZED DESIGN (INDEPENDENT RANDOM SAMPLES) 1. Null hypothesis: H0 : The k population distributions are identical. 2. Alternative hypothesis: Ha : At least two of the k population distributions differ in location. 2 S T 2 1 3(n 1) 3. Test statistic: H i n n(n 1) i (continued) 654 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS THE KRUSKAL–WALLIS H-TEST FOR COMPARING MORE THAN TWO POPULATIONS: COMPLETELY RANDOMIZED DESIGN (INDEPENDENT RANDOM SAMPLES) (continued) where ni Sample size for population i Ti Rank sum for population i n Total number of observations n1 n2 nk 4. Rejection region for a given a: H x 2 a with (k 1) df Assumptions • All sample sizes are greater than or equal to ﬁve. • Ties take on the average of the ranks that they would have occupied if they had not been tied. The Kruskal–Wallis H-test is a valuable alternative to a one-way analysis of variance when the normality and equality of variance assumptions are violated. Again, normal probability plots of residuals and plots of residuals per treatment group are helpful in determining whether these assumptions have been violated. Remember that a normal probability plot should appear as a straight line with a positive slope; residual plots per treatment groups should exhibit the same spread above and below the 0 line. 15.6 EXERCISES BASIC TECHNIQUES 15.32 Three treatments were compared using a completely randomized design. The data are EX1533 15.33 Four treatments were compared using a completely randomized design. The data are EX1532 shown in the table. Treatment 1 26 29 23 24 28 26 2 27 31 30 28 29 32 30 33 3 25 24 27 22 24 20 21 shown here: Treatment 1 124 167 135 160 159 144 133 2 147 121 136 114 129 117 109 3 141 144 139 162 155 150 4 117 128 102 119 128 123 Do the data
provide sufficient evidence to indicate a difference in location for at least two of the population distributions? Test using the Kruskal–Wallis H statistic with a.05. Do the data provide sufficient evidence to indicate a difference in location for at least two of the population distributions? Test using the Kruskal–Wallis H statistic with a.05. 15.6 THE KRUSKAL–WALLIS H-TEST FOR COMPLETELY RANDOMIZED DESIGNS ❍ 655 APPLICATIONS 15.34 Swampy Sites II Exercise 11.13 presents data (see data set EX1113) on the rates of growth of vegetation at four swampy underdeveloped sites. Six plants were randomly selected at each of the four sites to be used in the comparison. The data are the mean leaf length per plant (in centimeters) for a random sample of ten leaves per plant. Location Mean Leaf Length (cm) 1 2 3 4 5.7 6.2 5.4 3.7 6.3 5.3 5.0 3.2 6.1 5.7 6.0 3.9 6.0 6.0 5.6 4.0 5.8 5.2 4.9 3.5 6.2 5.5 5.2 3.6 a. Do the data present sufficient evidence to indicate differences in location for at least two of the distributions of mean leaf length corresponding to the four locations? Test using the Kruskal–Wallis H-test with a.05. b. Find the approximate p-value for the test. c. You analyzed this same set of data in Exer- cise 11.13 using an analysis of variance. Find the p-value for the F-test used to compare the four location means in Exercise 11.13. d. Compare the p-values in parts b and c and explain the implications of the comparison. 15.35 Heart Rate and Exercise Exercise 11.60 presented data (data set EX1160) on the heart rates for samples of 10 men randomly selected from each of four age groups. Each man walked a treadmill at a ﬁxed grade for a period of 12 minutes, and the increase in heart rate (the difference before and after exercise) was recorded (in beats per minute). The data are shown in the table. 10–19 20–39 40–59 60–69 29 33 26 27 39 35 33 29 36 22 24 27 33 31 21 28 24 34 21 32 37 25
22 33 28 26 30 34 27 33 28 29 34 36 21 20 25 24 33 32 Total 309 275 295 282 a. Do the data present sufficient evidence to indicate differences in location for at least two of the four age groups? Test using the Kruskal–Wallis H-test with a.01. b. Find the approximate p-value for the test in part a. c. Since the F-test in Exercise 11.60 and the H-test in part a are both tests to detect differences in location of the four heart-rate populations, how do the test results compare? Compare the p-values for the two tests and explain the implications of the comparison. 15.36 pH Levels in Water A sampling of the acidity of rain for ten randomly selected EX1536 rainfalls was recorded at three different locations in the United States: the Northeast, the Middle Atlantic region, and the Southeast. The pH readings for these 30 rainfalls are shown in the table. (NOTE: pH readings range from 0 to 14; 0 is acid, 14 is alkaline. Pure water falling through clean air has a pH reading of 5.7.) Northeast Middle Atlantic Southeast 4.45 4.02 4.13 3.51 4.42 3.89 4.18 3.95 4.07 4.29 4.60 4.27 4.31 3.88 4.49 4.22 4.54 4.76 4.36 4.21 4.55 4.31 4.84 4.67 4.28 4.95 4.72 4.63 4.36 4.47 a. Do the data present sufficient evidence to indicate differences in the levels of acidity in rainfalls in the three different locations? Test using the Kruskal– Wallis H-test. b. Find the approximate p-value for the test in part a and interpret it. EX1537 15.37 Advertising Campaigns The results of an experiment to investigate product recognition for three advertising campaigns were reported in Example 11.14. The responses were the percentage of 400 adults who were familiar with the newly advertised product. The normal probability plot indicated that the data were not approximately normal and another method of analysis should be used. Is there a signiﬁcant difference among the three population distributions from which these samples came? Use an appropriate nonparametric method to answer this question. Campaign 2.28.41.34.39.27 3.21.30.26.33.
31 1.33.29.21.32.25 656 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS THE FRIEDMAN Fr-TEST FOR RANDOMIZED BLOCK DESIGNS 15.7 The Friedman Fr-test, proposed by Nobel Prize–winning economist Milton Friedman, is a nonparametric test for comparing the distributions of measurements for k treatments laid out in b blocks using a randomized block design. The procedure for conducting the test is very similar to that used for the Kruskal–Wallis H-test. The ﬁrst step in the procedure is to rank the k treatment observations within each block. Ties are treated in the usual way; that is, they receive an average of the ranks occupied by the tied observations. The rank sums T1, T2,..., Tk are then obtained and the test statistic 12 S T i Fr 1) bk(k 2 3b(k 1) is calculated. The value of the Fr statistic is at a minimum when the rank sums are equal—that is, T1 T2 Tk—and increases in value as the differences among the rank sums increase. When either the number k of treatments or the number b of blocks is larger than ﬁve, the sampling distribution of Fr can be approximated by a chi-square distribution with (k 1) df. Therefore, as for the Kruskal–Wallis H-test, the rejection region for the Fr-test consists of values of Fr for which Fr x 2 a Suppose you wish to compare the reaction times of people exposed to six different stimuli. A reaction time measurement is obtained by subjecting a person to a stimulus and then measuring the time until the person presents some speciﬁed reaction. The objective of the experiment is to determine whether differences exist in the reaction times for the stimuli used in the experiment. To eliminate the person-to-person variation in reaction time, four persons participated in the experiment and each person’s reaction time (in seconds) was measured for each of the six stimuli. The data are given in Table 15.10 (ranks of the observations are shown in parentheses). Use the Friedman Fr-test to determine whether the data present sufficient evidence to indicate differences in the distributions of reaction times for the six stimuli. Test using a.05. EXAMPLE 15.8 TABLE 15.10 ● Reaction Times to Six Stimuli Stimulus Subject A 1 2 3 4 Rank
Sum.6 (2.5).7 (3.5).9 (3).5 (2) T1 11 B.9 (6) 1.1 (6) 1.3 (6).7 (5) T2 23 C.8 (5).7 (3.5) 1.0 (4.5).8 (6) T3 19 D E.7 (4).8 (5) 1.0 (4.5).6 (3.5) T4 17.5 (1).5 (1.5).7 (1).4 (1) T5 4.5 F.6 (2.5).5 (1.5).8 (2).6 (3.5) T6 9.5 Solution In Figure 15.9, the plot of the residuals for each of the six stimuli reveals that stimuli 1, 4, and 5 have variances somewhat smaller than the other stimuli. Furthermore, the normal probability plot of the residuals reveals a change in the slope of the line following the ﬁrst three residuals, as well as curvature in the upper portion of the plot. It appears that a nonparametric analysis is appropriate for these data. 15.7 THE FRIEDMAN Fr-TEST FOR RANDOMIZED BLOCK DESIGNS ❍ 657 Residuals versus Stimulus (response is Time) 1 2 3 4 5 6 Stimulus Normal Probability Plot of the Residuals (response is Time) ● FI GUR E 1 5. 9 A plot of treatments versus residuals and a normal probability plot of residuals for Example 15..10 0.05 0.00 0.05 0.10 0.15 0.20 t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 0.2 0.1 0.0 Residual 0.1 0.2 You wish to test H0 : The distributions of reaction times for the six stimuli are identical versus the alternative hypothesis Ha : At least two of the distributions of reaction times for the six stimuli differ in location Table 15.10 shows the ranks (in parentheses) of the observations within each block and the rank sums for each of the six stimuli (the treatments). The value of the Fr statistic for these data is 2 3b(k 1) 12 S T i Fr 1) bk(k
2 1 [(11)2 (23)2 (19)2 (9.5)2] 3(4)(7) 6 )(7) (4)( 100.75 84 16.75 Since the number k 6 of treatments exceeds ﬁve, the sampling distribution of Fr can be approximated by a chi-square distribution with (k 1) (6 1) 5 df. Therefore, for a.05, you can reject H0 if.05 11.0705 x 2 Fr x 2 where.05 658 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS This rejection region is shown in Figure 15.10. Since the observed value Fr 16.75.05 11.0705, it falls in the rejection region. You can therefore reject H0 exceeds x2 and conclude that the distributions of reaction times differ in location for at least two stimuli. The MINITAB printout of the Friedman Fr-test for the data is given in Figure 15.11. F IG URE 15. 10 Rejection region for Example 15.8 ● ) f(Fr α =.05 0 χ2.05 = 11.0705 Fr Rejection region Observed value of Fr F IG URE 15. 11 MINITAB printout for Example 15.8 ● Friedman Test: Time versus Stimulus blocked by Subject S = 16.75 DF = 5 P = 0.005 S = 17.37 DF = 5 P = 0.004 (adjusted for ties) Est Sum of Stimulus N Median Ranks 1 4 0.6500 11.0 2 4 1.0000 23.0 3 4 0.8000 19.0 4 4 0.7500 17.0 5 4 0.5000 4.5 6 4 0.6000 9.5 Grand median = 0.7167 EXAMPLE 15.9 Find the approximate p-value for the test in Example 15.8. Solution Consulting Table 5 in Appendix I with 5 df, you ﬁnd that the observed value of Fr 16.75 exceeds the table value x 2.005 16.7496. Hence, the p-value is very close to, but slightly less than,.005. THE FRIEDMAN Fr-TEST FOR A RANDOMIZED BLOCK DESIGN 1. Null hypothesis: H0 : The k population distributions are identical 2. Alternative hypothesis: Ha : At least two of the k population distributions dif- fer in location 12 S T i 3
. Test statistic: Fr 1) bk(k 2 3b(k 1) 15.7 THE FRIEDMAN Fr-TEST FOR RANDOMIZED BLOCK DESIGNS ❍ 659 where b Number of blocks k Number of treatments Ti Rank sum for treatment i, i 1, 2,..., k 4. Rejection region: Fr x 2 a, where x 2 a is based on (k 1) df Assumption: Either the number k of treatments or the number b of blocks is greater than ﬁve. 15.7 EXERCISES BASIC TECHNIQUES 15.38 A randomized block design is used to compare three treatments in six blocks. Treatment 1 3.2 2.8 4.5 2.5 3.7 2.4 2 3.1 3.0 5.0 2.7 4.1 2.4 3 2.4 1.7 3.9 2.6 3.5 2.0 EX1538 Block 1 2 3 4 5 6 a. Use the Friedman Fr-test to detect differences in location among the three treatment distributions. Test using a.05. b. Find the approximate p-value for the test in part a. c. Perform an analysis of variance and give the ANOVA table for the analysis. a. Use the Friedman Fr-test to detect differences in location among the four treatment distributions. Test using a.05. b. Find the approximate p-value for the test in part a. c. Perform an analysis of variance and give the ANOVA table for the analysis. d. Give the value of the F statistic for testing the equality of the four treatment means. e. Give the approximate p-value for the F statistic in part d. f. Compare the p-values for the tests in parts a and d, and explain the practical implications of the comparison. d. Give the value of the F statistic for testing the APPLICATIONS equality of the three treatment means. e. Give the approximate p-value for the F statistic in part d. f. Compare the p-values for the tests in parts a and d, and explain the practical implications of the comparison. 15.39 A randomized block design is used to compare four treatments in eight blocks. EX1539 Treatment Block 1 2 3 4 5 6 7 8 1 89 93 91 85 90 86 87 93 2 81 86 85 79 84 78 80 86 3 84 86 87 80 85 83 83 88 4 85 88 86 82
85 84 82 90 EX1540 15.40 Supermarket Prices In a comparison of the prices of items at ﬁve supermarkets, six items were randomly selected and the price of each was recorded for each of the ﬁve supermarkets. The objective of the study was to see whether the data indicated differences in the levels of prices among the ﬁve supermarkets. The prices are listed in the table. Kash n’ Karry.33 Item Celery Colgate toothpaste 1.28 Campbell’s beef soup 1.05 Crushed pineapple Mueller’s spaghetti Heinz ketchup.83.68 1.41 Publix WinnDixie Albertsons Food 4 Less.34 1.49 1.19.95.79 1.69.69.59 1.44 1.37 1.23 1.19.95.87.83 1.79.69 1.65.58 1.28 1.10.84.69 1.49 660 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS a. Does the distribution of the prices differ from one supermarket to another? Test using the Friedman Fr-test with a.05. b. Find the approximate p-value for the test and inter- pret it. 15.41 Toxic Chemicals An experiment was conducted to compare the effects of three toxic EX1541 chemicals, A, B, and C, on the skin of rats. One-inch squares of skin were treated with the chemicals and then scored from 0 to 10 depending on the degree of irritation. Three adjacent 1-inch squares were marked on the backs of eight rats, and each of the three chemicals was applied to each rat. Thus, the experiment was blocked on rats to eliminate the variation in skin sensitivity from rat to rat Rats. Do the data provide sufficient evidence to indicate a difference in the toxic effects of the three chemicals? Test using the Friedman Fr-test with a.05. b. Find the approximate p-value for the test and inter- pret it. EX1542 15.42 Good Tasting Medicine In a study of the palatability of antibiotics in children, Dr. Doreen Matsui and colleagues used a voluntary sample of healthy children to assess their reactions to the taste of four antibiotics.4 The children’s response was measured on a 10-centimeter (cm) visual analog scale incorporating the use of faces, from sad (low score) to happy (high score). The minimum score was 0
and the maximum was 10. For the accompanying data (simulated from the results of Matsui’s report), each of ﬁve children was asked to taste each of four antibiotics and rate them using the visual (faces) analog scale from 0 to 10 cm. Antibiotic Child 1 2 3 4 5 1 4.8 8.1 5.0 7.9 3.9 2 2.2 9.2 2.6 9.4 7.4 3 6.8 6.6 3.6 5.3 2.1 4 6.2 9.6 6.5 8.5 2.0 a. What design is used in collecting these data? b. Using an appropriate statistical package for a twoway classiﬁcation, produce a normal probability plot of the residuals as well as a plot of residuals versus antibiotics. Do the usual analysis of variance assumptions appear to be satisﬁed? c. Use the appropriate nonparametric test to test for differences in the distributions of responses to the tastes of the four antibiotics. d. Comment on the results of the analysis of variance in part b compared with the nonparametric test in part c. RANK CORRELATION COEFFICIENT 15.8 In the preceding sections, we used ranks to indicate the relative magnitude of observations in nonparametric tests for the comparison of treatments. We will now use the same technique in testing for a relationship between two ranked variables. Two common rank correlation coefficients are the Spearman rs and the Kendall t. We will present the Spearman rs because its computation is identical to that for the sample correlation coefficient r of Chapters 3 and 12. Suppose eight elementary school science teachers have been ranked by a judge according to their teaching ability and all have taken a “national teachers’ examination.” The data are listed in Table 15.11. Do the data suggest an agreement between the judge’s ranking and the examination score? That is, is there a correlation between ranks and test scores? TABLE 15.11 ● Ranks and Test Scores for Eight Teachers Teacher Judge’s Rank Examination Score 15.8 RANK CORRELATION COEFFICIENT ❍ 661 44 72 69 70 93 82 67 80 The two variables of interest are rank and test score. The former is already in rank form, and the test scores can be ranked similarly, as shown in Table 15.12. The ranks for tied observations are obtained by averaging the ranks that the tied observations
would have had if no ties had been observed. The Spearman rank correlation coefficient rs is calculated by using the ranks of the paired measurements on the two variables x and y in the formula for r (see Chapter 12). TABLE 15.12 ● Ranks of Data in Table 15.11 Teacher Judge’s Rank, xi Test Rank, yi SPEARMAN’S RANK CORRELATION COEFFICIENT S x y rs Syy S x x where xi and yi represent the ranks of the ith pair of observations and (S yi) Sxy S (xi x)( yi y) S xi yi (S xi) n Sxx S (xi x)2 S x i Syy S (yi y)2 S y i 2 (S xi)2 n 2 (S yi)2 n When there are no ties in either the x observations or the y observations, the expression for rs algebraically reduces to the simpler expression S 2 d i 6 where di (xi yi) rs 1 n2 n( 1) If the number of ties is small in comparison with the number of data pairs, little error results in using this shortcut formula. 662 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS EXAMPLE 15.10 Calculate rs for the data in Table 15.12. Solution The differences and squares of differences between the two rankings are provided in Table 15.13. Substituting values into the formula for rs, you have S 6 rs 1 n2 n( 2 d i 1) 4) 14 ( 6.714 1 4 6 8( 1) TABLE 15.13 ● Differences and Squares of Differences for the Teacher Ranks Teacher 1 2 3 4 5 6 7 8 Total xi 7 4 2 6 1 3 8 5 yi 1 5 3 4 8 7 2 6 di 36 1 1 4 49 16 36 1 144 The Spearman rank correlation coefficient can be used as a test statistic to test the hypothesis of no association between two populations. You can assume that the n pairs of observations (xi, yi) have been randomly selected and, therefore, no association between the populations implies a random assignment of the n ranks within each sample. Each random assignment (for the two samples) represents a simple event associated with the experiment, and a value of rs can be calculated for each. Thus, it is possible to calculate the probability that rs assumes a large absolute value due solely to chance and thereby
suggests an association between populations when none exists. The rejection region for a two-tailed test is shown in Figure 15.12. If the alternative hypothesis is that the correlation between x and y is negative, you would reject H0 for negative values of rs that are close to 1 (in the lower tail of Figure 15.12). Similarly, if the alternative hypothesis is that the correlation between x and y is positive, you would reject H0 for large positive values of rs (in the upper tail of Figure 15.12). ● –1 F IG URE 15. 12 Rejection region for a twotailed test of the null hypothesis of no association, using Spearman’s rank correlation test Rejection region –r0 0 r0 1 Rejection region rs = Spearmans rank correlation coefficient The critical values of rs are given in Table 9 in Appendix I. An abbreviated version is shown in Table 15.14. Across the top of Table 15.14 (and Table 9 in Appendix I) are the recorded values of a that you might wish to use for a one-tailed test of the null hypothesis of no association between x and y. The number of rank pairs n appears at the left side of the table. The table entries give the critical value r0 for a one-tailed test. Thus, P(rs r0) a. 15.8 RANK CORRELATION COEFFICIENT ❍ 663 For example, suppose you have n 8 rank pairs and the alternative hypothesis is that the correlation between the ranks is positive. You would want to reject the null hypothesis of no association for only large positive values of rs, and you would use a one-tailed test. Referring to Table 15.14 and using the row corresponding to n 8 and the column for a.05, you read r0.643. Therefore, you can reject H0 for all values of rs greater than or equal to.643. The test is conducted in exactly the same manner if you wish to test only the alternative hypothesis that the ranks are negatively correlated. The only difference is that you would reject the null hypothesis if rs.643. That is, you use the negative of the tabulated value of r0 to get the lower-tail critical value. An Abbreviated Version of Table 9 in Appendix I; TABLE 15.14 ● for Spearman’s Rank Correlation Test a.01 a.025 a.05 n —.886.786.738.683.648.
623.591.566.545.525 —.943.893.833.783.745.736.703.673 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20.900.829.714.643.600.564.523.497.475.457.441.425.412.399.388.377 a.005 — — —.881.833.794.818.780.745 To conduct a two-tailed test, you reject the null hypothesis if rs r0 or rs r0. The value of a for the test is double the value shown at the top of the table. For example, if n 8 and you choose the.025 column, you will reject H0 if rs.738 or rs.738. The a-value for the test is 2(.025).05. SPEARMAN’S RANK CORRELATION TEST 1. Null hypothesis: H0 : There is no association between the rank pairs. 2. Alternative hypothesis: Ha : There is an association between the rank pairs (a two-tailed test). Or Ha : The correlation between the rank pairs is positive or negative (a one-tailed test). S x y 3. Test statistic: rs Syy S x x where xi and yi represent the ranks of the ith pair of observations. 4. Rejection region: For a two-tailed test, reject H0 if rs r0 or rs r0, where r0 is given in Table 9 in Appendix I. Double the tabulated probability to obtain the value of a for the two-tailed test. For a one-tailed test, reject H0 if rs r0 (for an upper-tailed test) or rs r0 (for a lower-tailed test). The a-value for a one-tailed test is the value shown in Table 9 in Appendix I. 664 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS EXAMPLE 15.11 TABLE 15.15 Test the hypothesis of no association between the populations for Example 15.10. Solution The critical value of rs for a one-tailed test with a.05 and n 8 is.643. You may assume that a correlation between the judge’s rank and the teachers’ test scores could not possibly be positive. (A low rank means good teaching and should be associated with a high test score if the judge and the test measure teaching ability
.) The alternative hypothesis is that the population rank correlation coefficient rs is less than 0, and you are concerned with a one-tailed statistical test. Thus, a for the test is the tabulated value for.05, and you can reject the null hypothesis if rs.643. The calculated value of the test statistic, rs.714, is less than the critical value for a.05. Hence, the null hypothesis is rejected at the a.05 level of signiﬁcance. It appears that some agreement does exist between the judge’s rankings and the test scores. However, it should be noted that this agreement could exist when neither provides an adequate yardstick for measuring teaching ability. For example, the association could exist if both the judge and those who constructed the teachers’ examination had a completely erroneous, but similar, concept of the characteristics of good teaching. What exactly does rs measure? Spearman’s correlation coefficient detects not only a linear relationship between two variables but also any other monotonic relationship (either y increases as x increases or y decreases as x increases). For example, if you calculated rs for the two data sets in Table 15.15, both would produce a value of rs 1 because the assigned ranks for x and y in both cases agree for all pairs (x, y). It is important to remember that a signiﬁcant value of rs indicates a relationship between x and y that is either increasing or decreasing, but is not necessarily linear. ● Twin Data Sets With rs 1 y x 2 x x y log10(x 16 25 36 10 100 1000 10,000 100,000 1,000,000 1 2 3 4 5 6 15.8 EXERCISES BASIC TECHNIQUES 15.43 Give the rejection region for a test to detect positive rank correlation if the number of pairs of ranks is 16 and you have these a-values: a. a.05 15.44 Give the rejection region for a test to detect negative rank correlation if the number of pairs of ranks is 12 and you have these a-values: a. a.05 b. a.01 b. a.01 15.45 Give the rejection region for a test to detect rank correlation if the number of pairs of ranks is 25 and you have these a-values: b. a.01 a. a.05 15.46 The following paired observations were obtained on two variables x and y: x y 1.2 1.0.8 1
.3 2.1.1 3.5.8 2.7.2 1.5.6 a. Calculate Spearman’s rank correlation coefficient rs. b. Do the data present sufficient evidence to indicate a correlation between x and y? Test using a.05. 15.49 Tennis Racquets The data shown in the accompanying table give measures of EX1549 bending stiffness and twisting stiffness as determined by engineering tests on 12 tennis racquets. 15.8 RANK CORRELATION COEFFICIENT ❍ 665 APPLICATIONS 15.47 Rating Political Candidates A political scientist wished to examine the rela- EX1547 tionship between the voter image of a conservative political candidate and the distance (in miles) between the residences of the voter and the candidate. Each of 12 voters rated the candidate on a scale of 1 to 20. Voter Rating Distance 1 2 3 4 5 6 7 8 9 10 11 12 12 7 5 19 17 12 9 18 3 8 15 4 75 165 300 15 180 240 120 60 230 200 130 130 a. Calculate Spearman’s rank correlation coefficient rs. b. Do these data provide sufficient evidence to indicate a negative correlation between rating and distance? EX1548 15.48 Competitive Running Is the number of years of competitive running experience related to a runner’s distance running performance? The data on nine runners, obtained from the study by Scott Powers and colleagues, are shown in the table:5 Runner Years of Competitive Running 10-Kilometer Finish Time (min 13 5 7 12 6 4 5 3 33.15 33.33 33.50 33.55 33.73 33.86 33.90 34.15 34.90 a. Calculate the rank correlation coefficient between years of competitive running x and a runner’s ﬁnish time y in the 10-kilometer race. b. Do the data provide sufficient evidence to indicate a rank correlation between y and x? Test using a.05. Racquet Bending Stiffness Twisting Stiffness 1 2 3 4 5 6 7 8 9 10 11 12 419 407 363 360 257 622 424 359 346 556 474 441 227 231 200 211 182 304 384 194 158 225 305 235 a. Calculate the rank correlation coefficient rs between bending stiffness and twisting stiffness. b. If a racquet has bending stiffness, is it also likely to have twisting stiffness? Use the rank correlation coefficient to determine whether there is a signi�
�cant positive relationship between bending stiffness and twisting stiffness. Use a.05. 15.50 Student Ratings A school principal suspected that a teacher’s attitude toward a ﬁrst-grader depended on his original judgment of the child’s ability. The principal also suspected that much of that judgment was based on the ﬁrst-grader’s IQ score, which was usually known to the teacher. After three weeks of teaching, a teacher was asked to rank the nine children in his class from 1 (highest) to 9 (lowest) as to his opinion of their ability. Calculate rs for these teacher–IQ ranks: Teacher IQ 15.51 Student Ratings, continued Refer to Exercise 15.50. Do the data provide sufficient evidence to indicate a positive correlation between the teacher’s ranks and the ranks of the IQs? Use a.05. EX1552 15.52 Art Critics Two art critics each ranked 10 paintings by contemporary (but anonymous) artists in accordance with their appeal to the respective critics. The ratings are shown in the table. Do the critics seem to agree on their ratings of contemporary art? That is, do the data provide sufficient evidence to indicate a positive correlation between critics A and B? Test by using a value of a near.05. Calculate rs. Do the data provide sufficient evidence to indicate an association between the grader’s ratings and the moisture contents of the leaves? 15.54 Social Skills Training A social skills training program was implemented with seven EX1554 mildly challenged students in a study to determine whether the program caused improvements in pre/post measures and behavior ratings. For one such test, the pre- and posttest scores for the seven students are given in the table: Student Pretest Posttest Earl Ned Jasper Charlie Tom Susie Lori 101 89 112 105 90 91 89 113 89 121 99 104 94 99 a. Use a nonparametric test to determine whether there is a signiﬁcant positive relationship between the pre- and posttest scores. b. Do these results agree with the results of the para- metric test in Exercise 12.51? 666 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS Painting Critic A Critic 10 6 4 9 1 2 7 3 8 5 10 5 6 10 2 3 8 1 7 4 9 15.53 Rating Tobacco Leaves An experiment was conducted to study the relationship EX1553 between the ratings of a tobacco leaf grader and the moisture
content of the tobacco leaves. Twelve leaves were rated by the grader on a scale of 1 to 10, and corresponding readings of moisture content were made. Leaf Grader’s Rating Moisture Content 1 2 3 4 5 6 7 8 9 10 11 12 9 6 7 7 5 8 2 6 1 10 9 3.22.16.17.14.12.19.10.12.05.20.16.09 SUMMARY 15.9 The nonparametric tests presented in this chapter are only a few of the many nonparametric tests available to experimenters. The tests presented here are those for which tables of critical values are readily available. Nonparametric statistical methods are especially useful when the observations can be rank ordered but cannot be located exactly on a measurement scale. Also, nonparametric methods are the only methods that can be used when the sampling designs have been correctly adhered to, but the data are not or cannot be assumed to follow the prescribed one or more distributional assumptions. We have presented a wide array of nonparametric techniques that can be used when either the data are not normally distributed or the other required assumptions are not met. One-sample procedures are available in the literature; however, we have concentrated on analyzing two or more samples that have been properly selected using random and independent sampling as required by the design involved. The nonparametric analogues of the parametric procedures presented in Chapters 10–14 are straightforward and fairly simple to implement: • The Wilcoxon rank sum test is the nonparametric analogue of the two-sample t-test. • The sign test and the Wilcoxon signed-rank tests are the nonparametric analogues of the paired-sample t-test. CHAPTER REVIEW ❍ 667 • The Kruskal–Wallis H-test is the rank equivalent of the one-way analysis of variance F-test. • The Friedman Fr-test is the rank equivalent of the randomized block design two-way analysis of variance F-test. • Spearman’s rank correlation rs is the rank equivalent of Pearson’s correlation coefficient. These and many more nonparametric procedures are available as alternatives to the parametric tests presented earlier. It is important to keep in mind that when the assumptions required of the sampled populations are relaxed, our ability to detect signiﬁcant differences in one or more population characteristics is decreased. CHAPTER REVIEW Key Concepts and Formulas I. Nonparametric Methods III
. Sign Test for a Paired Experiment 1. These methods can be used when the data cannot be measured on a quantitative scale, or when 2. The numerical scale of measurement is arbi- trarily set by the researcher, or when 3. The parametric assumptions such as normality or constant variance are seriously violated. II. Wilcoxon Rank Sum Test: Indepen- dent Random Samples 1. Jointly rank the two samples. Designate the smaller sample as sample 1. Then T1 Rank sum of sample 1 1 n1(n1 n2 1) T1 T * 2. Use T1 to test for population 1 to the left of population 2. Use T * 1 to test for population 1 to the right of population 2. Use the smaller of T1 and T * 1 to test for a difference in the locations of the two populations. 3. Table 7 of Appendix I has critical values for the rejection of H0. 4. When the sample sizes are large, use the nor- mal approximation: n2 1) mT n1(n1 2 n2 1) T n1n2(n1 s 2 12 mT z T sT 1. Find x, the number of times that observation A exceeds observation B for a given pair. 2. To test for a difference in two populations, test H0 : p.5 versus a one- or two-tailed alternative. 3. Use Table 1 of Appendix I to calculate the p-value for the test. 4. When the sample sizes are large, use the nor- mal approximation: x 5n. z n 5. IV. Wilcoxon Signed-Rank Test: Paired Experiment 1. Calculate the differences in the paired observations. Rank the absolute values of the differences. Calculate the rank sums T and T for the positive and negative differences, respectively. The test statistic T is the smaller of the two rank sums. 2. Table 8 in Appendix I has critical values for the rejection of H0 for both one- and two-tailed tests. 3. When the sample sizes are large, use the nor- mal approximation: z T [n(n 1)/4] [n(n 1)(2n 1)]/24 3. For block sizes of ﬁve or greater, the rejection region for Fr is based on the chi-square distribution with (k 1) degrees of freedom. VII. Spearman’s Rank Correlation Coefficient 1. Rank the
responses for the two variables from smallest to largest. 2. Calculate the correlation coefficient for the ranked observations: S x y rs or Syy S x x S 6 rs 1 n2 n( 2 d i 1) if there are no ties 3. Table 9 in Appendix I gives critical values for rank correlations signiﬁcantly different from 0. 4. The rank correlation coefficient detects not only signiﬁcant linear correlation but also any other monotonic relationship between the two variables. 668 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS V. Kruskal–Wallis H-Test: Completely Randomized Design 1. Jointly rank the n observations in the k sam- ples. Calculate the rank sums, Ti rank sum of sample i, and the test statistic 2 S T 1 2 3(n 1) H i n(n n 1) i 2. If the null hypothesis of equality of distribu- tions is false, H will be unusually large, resulting in a one-tailed test. 3. For sample sizes of ﬁve or greater, the rejection region for H is based on the chi-square distribution with (k 1) degrees of freedom. VI. The Friedman Fr-Test: Randomized Block Design 1. Rank the responses within each block from 1 to k. Calculate the rank sums, T1, T2,..., Tk, and the test statistic 12 S T i Fr 1) bk(k 2 3b(k 1) 2. If the null hypothesis of equality of treatment distributions is false, Fr will be unusually large, resulting in a one-tailed test. Nonparametric Procedures Many nonparametric procedures are available in the MINITAB package, including most of the tests discussed in this chapter. The Dialog boxes are all familiar to you by now, and we will discuss the tests in the order presented in the chapter. To implement the Wilcoxon rank sum test for two independent random samples, enter the two sets of sample data into two columns (say, C1 and C2) of the MINITAB worksheet. The Dialog box in Figure 15.13 is generated using Stat Nonparametrics Mann-Whitney. Select C1 and C2 for the First and Second Samples, and indicate the appropriate conﬁdence coefficient (for a conﬁdence interval) and alternative hypothesis. Clicking
OK will generate the output in Figure 15.1. The sign test and the Wilcoxon signed-rank test for paired samples are performed in exactly the same way, with a change only in the last command of the sequence. Even the Dialog boxes are identical! Enter the data into two columns of the MINITAB worksheet (we used the cake mix data in Section 15.5). Before you can implement either test, you must generate a column of differences using Calc Calculator, as shown in Figure 15.14. Use Stat Nonparametrics 1-Sample Sign or Stat Nonparametrics 1-Sample Wilcoxon to generate the appropriate Dialog box MY MINITAB ❍ 669 FI GUR E 1 5. 13 ● FI GUR E 1 5. 14 ● shown in Figure 15.15. Remember that the median is the value of a variable such that 50% of the values are smaller and 50% are larger. Hence, if the two population distributions are the same, the median of the differences will be 0. This is equivalent to the null hypothesis H0 : P(positive difference) P(negative difference).5 670 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS F IG URE 15. 15 ● used for the sign test. Select the column of differences for the Variables box, and select the test of the median equals 0 with the appropriate alternative. Click OK to obtain the printout for either of the two tests. The Session window printout for the sign test, shown in Figure 15.16, indicates a nonsigniﬁcant difference in the distributions of densities for the two cake mixes. Notice that the p-value (.2188) is not the same as the p-value for the Wilcoxon signed-rank test (.093 from Figure 15.4). However, if you are testing at the 5% level, both tests produce nonsigniﬁcant differences. F IG URE 15. 16 ● The procedures for implementing the Kruskal–Wallis H-test for k independent samples and Friedman’s Fr-test for a randomized block design are identical to the procedures used for their parametric equivalents. Review the methods described in the section “MyMINITAB ” in Chapter 11. Once you have entered the data as explained in that section, the commands Stat Nonparametrics Kruskal–Wallis or Stat Nonparametrics Friedman will generate a Dial
og box in which you specify the Response column and the Factor column, or the response column, the treatment column and the block column, respectively. Click OK to obtain the outputs for these nonparametric tests. SUPPLEMENTARY EXERCISES ❍ 671 Finally, you can generate the nonparametric rank correlation coefficient rs if you enter the data into two columns and rank the data using Data Rank. For example, the data on judge’s rank and test scores were entered into columns C6 and C7 of our MINITAB worksheet. Since the judge’s ranks are already in rank order, we need only to rank C7 by selecting “Exam Score” and storing the ranks in C8 [named “Rank ( y)” in Figure 15.17]. The commands Stat Basic Statistics Correlation will now produce the rank correlation coefficient when C6 and C8 are selected. However, the p-value that you see in the output does not produce exactly the same test as the critical values in Table 15.14. You should compare your value of rs with the tabled value to check for a signiﬁcant association between the two variables. FI GUR E 1 5. 17 ● Supplementary Exercises 15.55 Response Times An experiment was conducted to compare the response times for EX1555 two different stimuli. To remove natural person-toperson variability in the responses, both stimuli were presented to each of nine subjects, thus permitting an analysis of the differences between stimuli within each person. The table lists the response times (in seconds). Subject Stimulus 1 Stimulus.4 7.8 5.6 12.1 6.9 4.2 8.8 7.7 6.4 10.3 8.9 4.1 14.7 8.7 7.1 11.3 5.2 7.8 a. Use the sign test to determine whether sufficient evidence exists to indicate a difference in the mean response times for the two stimuli. Use a rejection region for which a.05. b. Test the hypothesis of no difference in mean response times using Student’s t-test. 15.56 Response Times, continued Refer to Exercise 15.55. Test the hypothesis that no difference exists in the distributions of response times for the two stimuli, using the Wilcoxon signed-rank test. Use a rejection region for which a is as near as possible to the a achieved in Exercise 15.55, part a. 15
.57 Identical Twins To compare two junior high schools, A and B, in academic EX1557 effectiveness, an experiment was designed requiring 672 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS the use of 10 sets of identical twins, each twin having just completed the sixth grade. In each case, the twins in the same set had obtained their schooling in the same classrooms at each grade level. One child was selected at random from each pair of twins and assigned to school A. The remaining children were sent to school B. Near the end of the ninth grade, a certain achievement test was given to each child in the experiment. The test scores are shown in the table. Twin Pair School A School 10 67 80 65 70 86 50 63 81 86 60 39 75 69 55 74 52 56 72 89 47 a. Test (using the sign test) the hypothesis that the two schools are the same in academic effectiveness, as measured by scores on the achievement test, versus the alternative that the schools are not equally effective. b. Suppose it was known that junior high school A had a superior faculty and better learning facilities. Test the hypothesis of equal academic effectiveness versus the alternative that school A is superior. 15.58 Identical Twins II Refer to Exercise 15.57. What answers are obtained if Wilcoxon’s signed-rank test is used in analyzing the data? Compare with your earlier answers. 15.59 Paper Brightness The coded values for a measure of brightness in paper (light EX1559 reﬂectivity), prepared by two different processes, are given in the table for samples of nine observations drawn randomly from each of the two processes. Do the data present sufficient evidence to indicate a difference in the brightness measurements for the two processes? Use both a parametric and a nonparametric test and compare your results. Process A B Brightness 6.1 9.1 9.2 8.2 8.7 8.6 8.9 6.9 7.6 7.5 7.1 7.9 9.5 8.3 8.3 7.8 9.0 8.9 15.60 Precision Instruments Assume (as in the case of measurements produced by two well-calibrated measuring instruments) the means of two populations are equal. Use the Wilcoxon rank sum statistic for testing hypotheses concerning the population variances as follows: a. Rank the combined sample. b. Number the ranked observations “from the outside in”; that is, number
the smallest observation 1, the largest 2, the next-to-smallest 3, the next-to-largest 4, and so on. This sequence of numbers induces an ordering on the symbols A (population A items) and B (population B items). If s 2 would expect to ﬁnd a preponderance of A’s near the ﬁrst of the sequences, and thus a relatively small “sum of ranks” for the A observations. c. Given the measurements in the table produced by well-calibrated precision instruments A and B, test at near the a.05 level to determine whether the more expensive instrument B is more precise than A. (Note that this implies a one-tailed test.) Use the Wilcoxon rank sum test statistic. A s 2 B, one Instrument A Instrument B 1060.21 1060.34 1060.27 1060.36 1060.40 1060.24 1060.28 1060.32 1060.30 d. Test using the equality of variance F-test. 15.61 Meat Tenderizers An experiment was conducted to compare the tenderness of EX1561 meat cuts treated with two different meat tenderizers, A and B. To reduce the effect of extraneous variables, the data were paired by the speciﬁc meat cut, by applying the tenderizers to two cuts taken from the same steer, by cooking paired cuts together, and by using a single judge for each pair. After cooking, each cut was rated by a judge on a scale of 1 to 10, with 10 corresponding to the most tender meat. The data are shown for a single judge. Do the data provide sufficient evidence to indicate that one of the two tenderizers tends to receive higher ratings than the other? Would a Student’s t-test be appropriate for analyzing these data? Explain. Tenderizer Cut Shoulder roast Chuck roast Rib steak Brisket Club steak Round steak Rump roast Sirloin steak Sirloin tip steak T-bone steak 10 15.62 Interviewing Job Prospects A large corporation selects college graduates for EX1562 employment using both interviews and a psychological achievement test. Interviews conducted at the home office of the company are far more expensive than the tests that can be conducted on campus. Consequently, the personnel office was interested in determining whether the test scores were correlated with interview ratings and whether tests could be substituted for interviews. The idea was not to eliminate interviews but to
reduce their number. To determine whether the measures were correlated, 10 prospects were ranked during interviews and tested. The paired scores are as listed here: Subject Interview Rank Test Score 1 2 3 4 5 6 7 8 9 10 8 5 10 3 6 1 4 7 9 2 74 81 66 83 66 94 96 70 61 86 Calculate the Spearman rank correlation coefficient rs. Rank 1 is assigned to the candidate judged to be the best. 15.63 Interviews, continued Refer to Exercise 15.62. Do the data present sufficient evidence to indicate that the correlation between interview rankings and test scores is less than zero? If this evidence does exist, can you say that tests can be used to reduce the number of interviews? 15.64 Word Association Experiments A comparison of reaction times for two different stimuli in a psychological word-association experiment produced the accompanying results when applied to a random sample of 16 people: Stimulus Reaction Time (sec Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Use an appropriate nonparametric test and explain your conclusions. EX1565 15.65 Math and Art The table gives the scores of a group of 15 students in mathematics and art. Use Wilcoxon’s signed-rank test to determine whether the median scores for these students differ signiﬁcantly for the two subjects. SUPPLEMENTARY EXERCISES ❍ 673 Student Math Art Student Math Art 1 2 3 4 5 6 7 8 22 37 36 38 42 58 58 60 53 68 42 49 51 65 51 71 9 10 11 12 13 14 15 62 65 66 56 66 67 62 55 74 68 64 67 73 65 15.66 Math and Art, continued Refer to Exercise 15.65. Compute Spearman’s rank correlation coefficient for these data and test H0 : no association between the rank pairs at the 10% level of signiﬁcance. 15.67 Yield of Wheat Exercise 11.68 presented an analysis of variance of the yields of ﬁve different varieties of wheat, observed on one plot each at each of six different locations (see data set EX1168). The data from this randomized block design are listed here: Varieties 1 A B C D E 35.3 30.7 38.2 34.9 32.4 Location 2 31.0 32.2 33.4 36.1 28.9 3 32.7 31.4 33.6 35.2 29.2 4 36
.8 31.7 37.1 38.3 30.7 5 37.2 35.0 37.3 40.2 33.9 6 33.1 32.7 38.2 36.0 32.1 a. Use the appropriate nonparametric test to deter- mine whether the data provide sufficient evidence to indicate a difference in the yields for the ﬁve different varieties of wheat. Test using a.05. b. Exercise 11.68 presented a computer printout of the analysis of variance for comparing the mean yields for the ﬁve varieties of wheat. How do the results of the analysis of variance F test compare with the test in part a? Explain. 15.68 Learning to Sell In Exercise 11.61, you compared the numbers of sales per trainee after completion of one of four different sales training programs (see data set EX1161). Six trainees completed training program 1, eight completed program 2, and so on. The numbers of sales per trainee are shown in the table. Training Program 1 2 3 4 74 87 80 83 78 78 84 86 92 69 73 99 86 90 93 94 85 97 91 81 63 71 65 86 79 73 70 Total 482 735 402 588 674 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS a. Do the data present sufficient evidence to indicate that the distribution of number of sales per trainee differs from one training program to another? Test using the appropriate nonparametric test. b. How do the test results in part a compare with the results of the analysis of variance F-test in Exercise 11.61? 15.69 Pollution from Chemical Plants In Exercise 11.66, you performed an analysis of variance to compare the mean levels of effluents in water at four different industrial plants (see data set EX1166). Five samples of liquid waste were taken at the output of each of four industrial plants. The data are shown in the table. Plant Polluting Effluents (lb/gal of waste) A B C D 1.65 1.70 1.40 2.10 1.72 1.85 1.75 1.95 1.50 1.46 1.38 1.65 1.37 2.05 1.65 1.88 1.60 1.80 1.55 2.00 a. Do the data present sufficient evidence to indicate a difference in the levels of pollutants for the four different industrial plants? Test using the appropriate nonparametric test. b
. Find the approximate p-value for the test and inter- pret its value. c. Compare the test results in part a with the analysis of variance test in Exercise 11.66. Do the results agree? Explain. 15.70 AIDS Research Scientists have shown that a newly developed vaccine can shield rhesus monkeys from infection by a virus closely related to the AIDScausing human immunodeﬁciency virus (HIV). In their work, Ronald C. Resrosiers and his colleagues at the New England Regional Primate Research Center gave each of n 6 rhesus monkeys ﬁve inoculations with the simian immunodeﬁciency virus (SIV) vaccine. One week after the last vaccination, each monkey received an injection of live SIV. Two of the six vaccinated monkeys showed no evidence of SIV infection for as long as a year and a half after the SIV injection.6 Scientists were able to isolate the SIV virus from the other four vaccinated monkeys, although these animals showed no sign of the disease. Does this information contain sufficient evidence to indicate that the vaccine is effective in protecting monkeys from SIV? Use a.10. 15.71 Heavy Metal An experiment was performed to determine whether there is an EX1571 accumulation of heavy metals in plants that were grown in soils amended with sludge and whether there is an accumulation of heavy metals in insects feeding on those plants.7 The data in the table are cadmium concentrations (in mg/kg) in plants grown under six different rates of application of sludge for three different harvests. The rates of application are the treatments. The three harvests represent time blocks in the two-way design. Rate Control 1 2 3 4 5 Harvest 2 153.7 199.6 210.7 179.0 203.7 236.1 1 162.1 199.8 220.0 194.4 204.3 218.9 3 200.4 278.2 294.8 341.1 330.2 344.2 a. Based on the MINITAB normal probability plot and the plot of residuals versus rates, are you willing to assume that the normality and constant variance assumptions are satisﬁed? MINITAB residual plots for Exercise 15.71 Residuals versus Rate (response is Cadmium) l a u d i s e R 40 30 20 10 0 10 20 30 40 t n e c r e P 99 95 90 80 70 60 50
40 30 20 10 5 1 0 1 2 3 4 5 Rate Normal Probability Plot of the Residuals (response is Cadmium) 40 30 20 10 0 10 20 30 40 50 Residual b. Using an appropriate method of analysis, analyze the data to determine whether there are signiﬁcant differences among the responses due to rates of application. 15.72 Refer to Exercise 15.71. The data in this table are the cadmium concentrations EX1572 found in aphids that fed on the plants grown in soil amended with sludge. Rate Control 1 2 3 4 5 1 16.2 16.9 12.7 31.3 38.5 20.6 Harvest 2 55.8 119.4 171.9 128.4 182.0 191.3 3 65.8 181.1 184.6 196.4 163.7 242.8 a. Use the MINITAB normal probability plot of the residuals and the plot of residuals versus rates of application to assess whether the assumptions of normality and constant variance are reasonable in this case. b. Based on your conclusions in part a, use an appropriate statistical method to test for signiﬁcant differences in cadmium concentrations for the six rates of application. MINITAB residual plots for Exercise 15.72 Residuals versus Rate (response is Cadmium) 0 1 2 3 4 5 Rate Normal Probability Plot of the Residuals (response is Cadmium) l a u d i s e R 50 25 0 25 50 t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 75 50 25 0 Residual 25 50 SUPPLEMENTARY EXERCISES ❍ 675 15.73 Rating Teaching Applicants Before ﬁlling several new teaching positions at the high school, the principal formed a review board consisting of ﬁve teachers who were asked to interview the 12 applicants and rank them in order of merit. Seven of the 12 applicants held college degrees but had limited teaching experience. Of the remaining ﬁve applicants, all had college degrees and substantial experience. The review board’s rankings are given in the table. Limited Experience Substantial Experience 4 6 7 9 10 11 12 1 2 3 5 8 Do these rankings indicate that the review board considers experience a prime factor in the selection of the best candidates? Test using a.05. 15.74 Contaminants in Chemicals A
manufacturer uses a large amount of a certain EX1574 chemical. Since there are just two suppliers of this chemical, the manufacturer wishes to test whether the percentage of contaminants is the same for the two sources against the alternative that there is a difference in the percentages of contaminants for the two suppliers. Data from independent random samples are shown below: Supplier 1 2.86.69.72 1.18.45 1.41.65 1.13.65.50 1.04.41.55.40.22.09.16.26.58.16.07.36.20.15 a. Use the Wilcoxon rank sum test to determine whether there is a difference in the contaminant percentages for the two suppliers. Use a.05. b. Use the large-sample approximation to the Wilcoxon rank sum test to determine whether there is a difference in the contaminant percentages for the two suppliers. Use a.05. Compare your conclusions to the conclusions from part a. 15.75 Lighting in the Classroom The productivity of 35 students was observed and measured both before and after the installation of new lighting in their classroom. The productivity of 21 of the 35 students was observed to have improved, whereas the productivity of the others appeared to show no perceptible gain as a 676 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS result of the new lighting. Use the normal approximation to the sign test to determine whether or not the new lighting was effective in increasing student productivity at the 5% level of signiﬁcance. EX1576 15.76 Reducing Cholesterol A drug was developed for reducing cholesterol levels in heart patients. The cholesterol levels before and after drug treatment were obtained for a random sample of 25 heart patients with the following results: 15.78 Worker Fatigue To investigate methods of reducing fatigue among employees EX1578 whose jobs involve a monotonous assembly procedure, 12 randomly selected employees were asked to perform their usual job under each of three trial conditions. As a measure of fatigue, the experimenter used the number of assembly line stoppages during a 4-hour period for each trial condition. Patient Before After Patient Before After Employee 1 2 3 4 5 6 7 8 9 10 11 12 13 257 222 177 258 294 244 390 247 409 214 217 340 364 243 217 174 260 295 236 383 233 410 216 210 335 343 14 15 16 17 18 19 20 21 22 23 24 25 210 263 214 392 370 310 255 281 294 257 227 385 217 243 198 388
357 299 258 276 295 227 231 374 1 2 3 4 5 6 7 8 9 10 11 12 Conditions 1 31 20 26 31 12 22 28 15 41 19 31 18 2 22 15 21 22 16 29 17 9 31 19 34 11 3 26 23 18 32 18 34 26 12 46 25 41 21 a. Use the sign test to determine whether or not this drug reduces the cholesterol levels of heart patients. Use a.01. b. Use the Wilcoxon signed-rank test to test the hypothesis in part a at the 1% level of signiﬁcance. Are your conclusions the same as those in part a? 15.77 Legos® The time required for kindergarten children to assemble a speciﬁc Lego EX1577 creation was measured for children who had been instructed for four different lengths of time. Four children were randomly assigned to each instructional group, but two were eliminated during the experiment because of sickness. The length of time (in minutes) to assemble the Lego creation was recorded for each child in the experiment. Training Period (hours).5 8 14 9 12 1.0 9 7 5 1.5 4 6 7 8 2.0 4 7 5 Use the Kruskal–Wallis H-Test to determine whether there is a differnce in the distribution of times for the four different lengths of instructional time. Use a.01. a. What type of experimental design has been used in this experiment? b. Use the appropriate nonparametric test to determine whether the distribution of assembly line stoppages (and consequently worker fatigue) differs for these three conditions. Test at the 5% level of signiﬁcance. 15.79 Ranking Quarterbacks A ranking of the quarterbacks in the top eight teams of the National Football League was made by polling a number of professional football coaches and sportswriters. This “true ranking” is shown below, together with “my ranking.” Quarterback True Ranking My Ranking. Calculate rs. b. Do the data indicate a positive correlation between my ranking and that of the experts? Test at the 5% level of signiﬁcance. CASE STUDY Eggs CASE STUDY ❍ 677 How’s Your Cholesterol Level? As consumers become more and more interested in eating healthy foods, many “light,” “fat-free,” and “cholesterol-free” products are appearing in the marketplace. One such product is
the frozen egg substitute, a cholesterol-free product that can be used in cooking and baking in many of the same ways that regular eggs can—though not all. Some consumers even use egg substitutes for Caesar salad dressings and other recipes calling for raw eggs because these products are pasteurized and thus eliminate worries about bacterial contamination. Unfortunately, the products currently on the market exhibit strong differences in both ﬂavor and texture when tasted in their primary preparation as scrambled eggs. Five panelists, all experts in nutrition and food preparation, were asked to rate each of three egg substitutes on the basis of taste, appearance, texture, and whether they would buy the product.8 The judges tasted the three egg substitutes and rated them on a scale of 0 to 20. The results, shown in the table, indicate that the highest rating, by 23 points, went to ConAgra’s Healthy Choice Egg Product, which the tasters unanimously agreed most closely resembled eggs as they come from the hen. The secondplace product, Morningstar Farms’ Scramblers, struck several tasters as having an “oddly sweet ﬂavor... similar to carrots.” Finally, none of the tasters indicated that they would be willing to buy Fleishmann’s Egg Beaters, which was described by the testers as “watery,” “slippery,” and “unpleasant.” Oddly enough, these results are contrary to a similar taste test done 4 years earlier, in which Egg Beaters were considered better than competing egg substitutes. Taster Healthy Choice Scramblers Egg Beaters Dan Bowe John Carroll Donna Katzl Rick O’Connell Roland Passot Totals 16 16 14 15 13 74 9 7 8 16 11 51 7 8 4 9 2 30 Source: Data from “Eggs Substitutes Range in Quality,” by K. Sakekel, The San Francisco Chronicle, February 10, 1993, p. 8. Copyright © 1993 San Francisco Chronicle. 1. What type of design has been used in this taste-testing experiment? 2. Do the data satisfy the assumptions required for a parametric analysis of variance? Explain. 3. Use the appropriate nonparametric technique to determine whether there is a signiﬁcant difference between the average scores for the three brands of egg substitutes. This page intentionally left blank Appendix I Tables CONTENTS Table 1 Table 2 Table 3 Table 4 Table 5 Table 6 Table 7 Table 8
Table 9 Table 10 Table 11 Cumulative Binomial Probabilities 680 Cumulative Poisson Probabilities 686 Areas under the Normal Curve 688 Critical Values of t 691 Critical Values of Chi-Square 692 Percentage Points of the F Distribution 694 Critical Values of T for the Wilcoxon Rank Sum Test, n1 n2 702 Critical Values of T for the Wilcoxon Signed-Rank Test, n 5(1)50 704 Critical Values of Spearman’s Rank Correlation Coefficient for a One-Tailed Test 705 Random Numbers 706 Percentage Points of the Studentized Range, q(k, df) 708 679 680 ❍ APPENDIX I TABLES p(x) 0 k x TABLE 1 Cumulative Binomial Probabilities Tabulated values are P(x k) p(0) p(1) p(k). (Computations are rounded at the third decimal place.) n 2 p k 0 1 2.01.05.10.20.30.40.50.60.70.80.90.95.99.980 1.000 1.000.902.998 1.000.810.990 1.000.640.960 1.000.490.910 1.000.360.840 1.000.250.750 1.000.160.640 1.000.090.510 1.000.040.360 1.000.010.190 1.000.002.098 1.000.000.020 1.000 n 3 p k 0 1 2 3.01.05.10.20.30.40.50.60.70.80.90.95.99.970 1.000 1.000 1.000.857.993 1.000 1.000.729.972.999 1.000.512.896.992 1.000.343.784.973 1.000.216.648.936 1.000.125.500.875 1.000.064.352.784 1.000.027.216.657 1.000.008.104.488 1.000.001.028.271 1.000.000.007.143 1.000.000.000.030 1.000 01.05.10.20.30.40.50.60.70.80.90.95.99.961.999 1.000
1.000 1.000.815.986 1.000 1.000 1.000.656.948.996 1.000 1.000.410.819.973.998 1.000.240.652.916.992 1.000.130.475.821.974 1.000.062.312.688.938 1.000.026.179.525.870 1.000.008.084.348.760 1.000.002.027.181.590 1.000.000.004.052.344 1.000.000.000.014.185 1.000.000.000.001.039 1.000 APPENDIX I TABLES ❍ 681 TABLE 1 (continued01.05.10.20.30.40.50.60.70.80.90.95.99.951.999 1.000 1.000 1.000 1.000.774.977.999 1.000 1.000 1.000.590.919.991 1.000 1.000 1.000.328.737.942.993 1.000 1.000.168.528.837.969.998 1.000.078.337.683.913.990 1.000.031.188.500.812.969 1.000.010.087.317.663.922 1.000.002.031.163.472.832 1.000.000.007.058.263.672 1.000.000.000.009.081.410 1.000.000.000.001.023.226 1.000.000.000.000.001.049 1.000 n 6 p.01.05.10.20.30.40.50.60.70.80.90.95.99 941.999 1.000 1.000 1.000 1.000 1.000.735.967.998 1.000 1.000 1.000 1.000.531.886.984.999 1.000 1.000 1.000.262.655.901.983.998 1.000 1.000.118.420.744.930.989.999 1.000.047.233.544.8
21.959.996 1.000.932.998 1.000 1.000 1.000 1.000 1.000 1.000.698.956.996 1.000 1.000 1.000 1.000 1.000.478.850.974.997 1.000 1.000 1.000 1.000.210.577.852.967.995 1.000 1.000 1.000.082.329.647.874.971.996 1.000 1.000.028.159.420.710.904.981.998 1.000.004.041.179.456.767.953 1.000.001.011.070.256.580.882 1.000.000.002.017.099.345.738 1.000.000.000.001.016.114.469 1.000.000.000.000.002.033.265 1.000.000.000.000.000.001.059 1.000.016.109.344.656.891.984 1.000 p.002.019.096.290.580.841.972 1.000.000.004.029.126.353.671.918 1.000.000.000.005.033.148.423.790 1.000.000.000.000.003.026.150.522 1.000.000.000.000.000.004.044.302 1.000.000.000.000.000.000.002.068 1.000.008.062.227.500.773.938.992 1.000 p.01.05.10.20.30.40.50.60.70.80.90.95.99 01.05.10.20.30.40.50.60.70.80.90.95.99.923.997 1.000 1.000 1.000 1.000 1.000 1.000 1.000.663.943.994 1.000 1.000 1.000 1.000 1.000 1.000.430.813.962.995 1.000 1.000 1.000 1.000 1.000.168.503.797.944.990.
999 1.000 1.000 1.000.058.255.552.806.942.989.999 1.000 1.000.017.106.315.594.826.950.991.999 1.000.004.035.145.363.637.855.965.996 1.000.001.009.050.174.406.685.894.983 1.000.000.001.011.058.194.448.745.942 1.000.000.000.001.010.056.203.497.832 1.000.000.000.000.000.005.038.187.570 1.000.000.000.000.000.000.006.057.337 1.000.000.000.000.000.000.000.003.077 1.000 682 ❍ APPENDIX I TABLES TABLE 1 (continued) n 9.01.05.10.20.30.40.50.60.70.80.90.95.99 p k.01.05.10.20.30.40.50.60.70.80.90.95.99.914.997 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.630.929.992.999 1.000 1.000 1.000 1.000 1.000 1.000.387.775.947.992.999 1.000 1.000 1.000 1.000 1.000.134.436.738.914.980.997 1.000 1.000 1.000 1.000.040.196.463.730.901.975.996 1.000 1.000 1.000.010.071.232.483.733.901.975.996 1.000 1.000 10.904.996 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.599.914.988.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000.349.736.930.987.998 1.000 1.000 1.000 1.000 1.
000 1.000.107.376.678.879.967.994.999 1.000 1.000 1.000 1.000.028.149.383.650.850.953.989.998 1.000 1.000 1.000.006.046.167.382.633.834.945.988.998 1.000 1.000.000.004.025.099.267.517.768.929.990 1.000.000.000.004.025.099.270.537.804.960 1.000.000.000.000.003.020.086.262.564.866 1.000.000.000.000.000.001.008.053.225.613 1.000.000.000.000.000.000.001.008.071.370 1.000.000.000.000.000.000.000.000.003.086 1.000.002.020.090.254.500.746.910.980.998 1.000 p.000.002.012.055.166.367.618.833.954.994 1.000.000.000.002.011.047.150.350.617.851.972 1.000.000.000.000.001.006.033.121.322.624.893 1.000.000.000.000.000.000.002.013.070.264.651 1.000.000.000.000.000.000.000.001.012.086.401 1.000.000.000.000.000.000.000.000.000.004.096 1.000.001.011.055.172.377.623.828.945.989.999 1.000 p.01.05.10.20.30.40.50.60.70.80.90.95.99.895.995 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.569.898.985.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.314.697.
910.981.997 1.000 1.000 1.000 1.000 1.000 1.000 1.000.086.322.617.839.950.988.998 1.000 1.000 1.000 1.000 1.000.020.113.313.570.790.922.978.996.999 1.000 1.000 1.000.004.030.119.296.533.754.901.971.994.999 1.000 1.000.000.006.033.113.274.500.726.887.967.994 1.000 1.000.000.001.006.029.099.246.467.704.881.970.996 1.000.000.000.001.004.022.078.210.430.687.887.980 1.000.000.000.000.000.002.012.050.161.383.678.914 1.000.000.000.000.000.000.000.003.019.090.303.686 1.000.000.000.000.000.000.000.000.002.015.102.431 1.000.000.000.000.000.000.000.000.000.000.005.105 1.000 10 n 11 10 11 10 10 11 APPENDIX I TABLES ❍ 683 TABLE 1 (continued) n 12 10 11 12 n 15 10 11 12 13 14 15.01.05.10.20.30.40.50.60.70.80.90.95.99 p.886.994 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.540.882.980.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.282.659.889.974.996.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000.069.275.558.795.927.981.996.999 1.000 1.000 1.000 1.000 1.000.014.085.253.493
.724.882.961.991.998 1.000 1.000 1.000 1.000.002.020.083.225.438.665.842.943.985.997 1.000 1.000 1.000.000.000.003.015.057.158.335.562.775.917.980.998 1.000.000.000.000.002.009.039.118.276.507.747.915.986 1.000.000.000.000.000.001.004.019.073.205.442.725.931 1.000.000.000.000.000.000.000.001.004.026.111.341.718 1.000.000.000.000.000.000.000.000.000.002.020.118.460 1.000.000.000.000.000.000.000.000.000.000.000.006.114 1.000.000.003.019.073.194.387.613.806.927.981.997 1.000 1.000 p.01.05.10.20.30.40.50.60.70.80.90.95.99.860.990 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.463.829.964.995.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.206.549.816.944.987.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.035.167.398.648.836.939.982.996.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000.005.035.127.297.515.722.869.950.985.996.999 1.000 1.000 1.000 1.000 1.000.000.005.027.091.217.403.610
.787.905.966.991.998 1.000 1.000 1.000 1.000.000.000.004.018.059.151.304.500.696.849.941.982.996 1.000 1.000 1.000.000.000.000.002.009.034.095.213.390.597.783.909.973.995 1.000 1.000.000.000.000.000.001.004.015.050.131.278.485.703.873.965.995 1.000.000.000.000.000.000.000.001.004.018.061.164.352.602.833.965 1.000.000.000.000.000.000.000.000.000.000.002.013.056.184.451.794 1.000.000.000.000.000.000.000.000.000.000.000.001.005.036.171.537 1.000.000.000.000.000.000.000.000.000.000.000.000.000.000.010.140 1.000 10 11 12 10 11 12 13 14 15 684 ❍ APPENDIX I TABLES TABLE 1 (continued) n 20 10 11 12 13 14 15 16 17 18 19 20.01.05.10.20.30.40.50.60.70.80.90.95.99.818.983.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.358.736.925.984.997 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.122.392.677.867.957.989.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.
000.012.069.206.411.630.804.913.968.990.997.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.001.008.035.107.238.416.608.772.887.952.983.995.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.000.001.004.016.051.126.250.416.596.755.872.943.979.994.998 1.000 1.000 1.000 1.000 1.000 1.000.000.000.000.001.006.021.058.132.252.412.588.748.868.942.979.994.999 1.000 1.000 1.000 1.000.000.000.000.000.000.002.006.021.057.128.245.404.584.750.874.949.984.996.999 1.000 1.000.000.000.000.000.000.000.000.001.005.017.048.113.228.392.584.762.893.965.992.999 1.000.000.000.000.000.000.000.000.000.000.001.003.010.032.087.196.370.589.794.931.988 1.000.000.000.000.000.000.000.000.000.000.000.000.000.000.002.011.043.133.323.608.878 1.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.003.016.075.264.642 1.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.001.017.182 1.000 10 11 12 13 14 15 16 17 18 19 20 APPENDIX I TABLES ❍ 685 TABLE 1 (continued) n 25 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24 25.01.05.10.20.30.40.50.60.70.80.90.95.99.778.974.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.277.642.873.966.993.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.072.271.537.764.902.967.991.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.004.027.098.234.421.617.780.891.953.983.994.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.000.002.009.033.090.193.341.512.677.811.902.956.983.994.998 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.000.000.000.002.009.029.074.154.274.425.586.732.846.922.966.987.996.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.000.000.000.000.000.002.007.022.054.115.212.345.500.655.788.885.946.978.993.998 1.000 1.000 1.

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