Nonparametric Methods: Analysis of Ranked Data Chapter 18

GOALS Conduct the sign test for dependent samples using the binomial and standard normal distributions as the test statistics. Conduct a test of hypothesis for dependent samples using the Wilcoxon signed-rank test. Conduct and interpret the Wilcoxon rank-sum test for independent samples. Conduct and interpret the Kruskal-Wallis test for several independent samples. Compute and interpret Spearman’s coefficient of rank correlation. Conduct a test of hypothesis to determine whether the correlation among the ranks in the population is different from zero.

The Sign Test EXAMPLE The director of information systems at Samuelson Chemicals recommended that an in-plant training program be instituted for managers. The objective is to improve the knowledge of database usage in accounting, procurement, production, and so on. A sample of 15 managers was selected at random. A panel of database experts determined the general level of competence of each manager with respect to using the database. Their competence and understanding were rated as being either outstanding, excellent, good, fair, or poor. After the three-month training program, the same panel of information systems experts rated each manager again. The two ratings (before and after) are shown along with the sign of the difference. A “+” sign indicates improvement, and a “-” sign indicates that the manager’s competence using databases had declined after the training program. Did the in-plant training program effectively increase the competence of the managers using the company’s database? The Sign Test is based on the sign of a difference between two related observations. Assumption regarding the shape of the population of differences is NOT necessary. The binomial distribution is the test statistic for small samples and the standard normal (z) for large samples. The test requires dependent (related) samples. Procedure to conduct the test: Determine the sign (+ or -) of the difference between pairs. Determine the number of usable pairs. Compare the number of positive (or negative) differences to the critical value. n is the number of usable pairs (without ties), X is the number of pluses or minuses, and the binomial probability π = .5

The Sign Test – Example Step 1: State the Null and Alternative Hypotheses H0: π ≤.5 (There is no increase in competence as a result of the in-plant training program.) H1: π >.5 (There is an increase in competence as a result of the in-plant training program.) Step 2: Select a level of significance. We chose the .10 level. Step 3: Decide on the test statistic. It is the number of plus signs resulting from the experiment. Step 4: Formulate a decision rule. In this example α is .10. The probability of 3 or fewer successes is .029, found by .000 + .001 + .006 + .022. The probability of 11 or more successes is also .029. Adding the two probabilities gives .058. This is the closest we can come to .10 without exceeding it. Hence, the decision rule for a two-tailed test would be to reject the null hypothesis if there are 3 or fewer plus signs, or 11 or more plus signs. . Step 5: Make a decision regarding the null hypothesis. Eleven out of the 14 managers in the training course increased their database competency. The number 11 is in the rejection region, which starts at 10, so the null hypothesis is rejected. We conclude that the three-month training course was effective. It increased the database competency of the managers.

Normal Approximation If the number of observations in the sample is larger than 10, the normal distribution can be used to approximate the binomial. EXAMPLE The market research department of Cola, Inc., has been given the assignment of testing a new soft drink. Two versions of the drink are considered—a rather sweet drink and a somewhat bitter one. A preference test is to be conducted consisting of a sample of 64 consumers. Each consumer will taste both the sweet cola (labeled A) and the bitter one (labeled B) and indicate a preference. Conduct a test of hypothesis to determine if there is a difference in the preference for the sweet and bitter tastes. Use the .05 significance level. Step 1: State the null hypothesis and the alternate hypothesis. H0: π = .50 There is no preference H1: π ≠ .50 There is a preference Step 2: Select the level of significance. α = 0.05 as stated in the problem Step 3: Select the test statistic. Use Z-distribution where µ=.50n and σ=.50 Step 4: Formulate the decision rule. Referring to Appendix B.1, Areas under the Normal Curve, for a two-tailed test (because states that π ≠ .50), at the .05 significance level, the critical values are -1.96 and +1.96. Step 5: Compute z, compare the computed value with the critical value, and make a decision regarding H0 Preference for cola A was given a “+”sign and preference for B a “-” sign. Out of the 64 in the sample, 42 preferred the sweet taste, which is cola A. Therefore, there are 42 pluses. Since 42 is more than n/2 =64/2=32, we use: The computed z of 2.38 is beyond the critical value of 1.96. Conclusion: The null hypothesis of no difference is rejected at the .05 significance level. There is evidence of a difference in consumer preference. That is, we conclude consumers prefer one cola over another.

Wilcoxon Signed-Rank Test for Dependent Samples If the assumption of normality is violated for the paired-t test, use the Wilcoxon Signed-rank test. The test requires the ordinal scale of measurement. The observations must be related or dependent. The steps for the test are: Compute the differences between related observations. Drop observations with 0 difference from the sample. Rank the absolute differences from low to high. Return the signs to the ranks and sum positive and negative ranks. Compare the smaller of the two rank sums with the T value, obtained from Appendix B.7. EXAMPLE Fricker’s is a family restaurant chain located primarily in the southeastern part of the United States. It offers a full dinner menu, but its specialty is chicken. Recently, Fricker, the owner and founder, developed a new spicy flavor for the batter in which the chicken is cooked. Before replacing the current flavor, he wants to conduct some tests to be sure that patrons will like the spicy flavor better. Bernie selects a random sample of 15 customers, each customer is given a small piece of the current chicken and asked to rate its overall taste on a scale of 1 to 20. A value near 20 indicates the participant liked the flavor, whereas a score near 0 indicates they did not like the flavor. Next, the same 15 participants are given a sample of the new chicken with the spicier flavor and again asked to rate its taste on a scale of 1 to 20. The results are reported in the table below. Is it reasonable to conclude that the spicy flavor is preferred? Use the .05 significance level.

Wilcoxon Signed-Rank Test for Dependent Samples - Example Hypothesis: H0: There is no difference in the ratings of the two flavors. H1: The spicy ratings are higher. Decision Rule: Reject H0 if the smaller of the rank sums is 25 or less. Computed T = 30 Critical T = 25 Decision is not to reject the null hypothesis. We cannot conclude there is a difference in the flavor ratings between the current and the spicy. The smaller of the two rank sums is used as the test statistic and referred to as T.

Wilcoxon Signed-Rank Test for Dependent Samples - Example The critical values for the Wilcoxon signed-rank test are located in Appendix B.7. A portion of that table is shown on the table below.

Wilcoxon Rank-Sum Test for Independent Samples The Wilcoxon Rank-Sum Test is used to determine if two independent samples came from the same or equal populations. No assumption about the shape of the population is required. The data must be at least ordinal scale. Each sample must contain at least eight observations. To determine the value of the test statistic W, all data values are ranked from low to high as if they were from a single population. The sum of ranks for each of the two samples is determined. The data are ranked as if the observations were part of a single sample. The sum of ranks for each of the two samples is determined If the null hypothesis is true, then the ranks will be about evenly distributed between the two samples, and the sum of the ranks for the two samples will be about the same.

Wilcoxon Rank-Sum Test for Independent Samples - Example Dan Thompson, the president of CEO Airlines, recently noted an increase in the number of no-shows for flights out of Atlanta. He is particularly interested in determining whether there are more no-shows for flights that originate from Atlanta compared with flights leaving Chicago. A sample of nine flights from Atlanta and eight from Chicago are reported on table. At the .05 significance level, can we conclude that there are more no-shows for the flights originating in Atlanta? Set up Hypothesis and Decision Rule: Hypothesis: H0: The population distribution of no-shows is the same or less for Atlanta and Chicago. H1: The population distribution of no-shows is larger for Atlanta than for Chicago. Decision Rule: Reject H0 if: computed Z > critical Z .05 level of significance = 1.65 critical Z Rank the observations from both samples as if they were a single group. The Chicago flight with only 8 no-shows had the fewest, so it is assigned a rank of 1. The Chicago flight with 9 no-shows is ranked 2, and so on.

Wilcoxon Rank-Sum Test for Independent Samples - Example The value of W is calculated for the Atlanta group and is found to be 96.5, which is the sum of the ranks for the no-shows for the Atlanta flights. The computed z value (1.49) is less than 1.65, the null hypothesis is not rejected. It appears that the number of no-shows is the same in Atlanta as in Chicago.

Kruskal-Wallis Test: Analysis of Variance by Ranks EXAMPLE A management seminar consists of executives from manufacturing, finance, and engineering. Before scheduling the seminar sessions, the seminar leader is interested in whether the three groups are equally knowledgeable about management principles. Plans are to take samples of the executives in manufacturing, in finance, and in engineering and to administer a test to each executive. If there is no difference in the scores for the three distributions, the seminar leader will conduct just one session. However, if there is a difference in the scores, separate sessions will be given. We will use the Kruskal-Wallis test instead of ANOVA because the seminar leader is unwilling to assume that (1) the populations of management scores follow the normal distribution or (2) the population standard deviations are the same. Step 1: Set up the Null and Alternate Hypotheses H0: The population distributions of the management scores for the populations of executives in manufacturing, finance, and engineering are the same. H1: The population distributions of the management scores for the populations of executives in manufacturing, finance, and engineering are NOT the same. Step 2: State the Decision Rule H0 is rejected if the computed H statistic is greater than critical χ2 value of 5.991 (There are 2 degrees of freedom at the .05 significance level.) This is used to compare three or more samples to determine if they came from equal populations. The ordinal scale of measurement is required. It is an alternative to the one-way ANOVA. The chi-square distribution is the test statistic. Each sample should have at least five observations. The sample data is ranked from low to high as if it were a single group.

Kruskal-Wallis Test: Analysis of Variance by Ranks - Example Step 3: Collect Data and Compute the Chisquare Statistic Considering the scores as a single population, the engineering executive with a score of 35 is the lowest, so it is ranked 1. There are two scores of 38. To resolve this tie, each score is given a rank of 2.5, found by (2+3)/2. This process is continued for all scores. The scores, the ranks, and the sum of the ranks for each of the three samples are given in the table below. Because the computed value of H (5.736) is less than the critical value of 5.991, the null hypothesis is not rejected. There is not enough evidence to conclude there is a difference among the executives from manufacturing, finance, and engineering with respect to their typical knowledge of management principles.

Rank-Order Correlation EXAMPLE Lorrenger Plastics, Inc., recruits management trainees at colleges and universities throughout the United States. Each trainee is given a rating by the recruiter during the on-campus interview. This rating is an expression of future potential and may range from 0 to 15, with the higher score indicating more potential. The recent college graduate then enters an in-plant training program and is given another composite rating based on tests, opinions of group leaders, training officers, and so on. The on-campus rating and the in-plant training ratings are given in the table on the right. Spearman’s coefficient of rank correlation reports the association between two sets of ranked observations. The features are: It can range from –1.00 up to 1.00. It is similar to Pearson’s coefficient of correlation, but is based on ranked data. It computed using the formula:

Rank-Order Correlation - Example Conclusion: The value of .726 indicates a strong positive association between the ratings of the on-campus recruiter and the ratings of the training staff. The graduates that received high ratings from the on-campus recruiter also tended to be the ones that received high ratings from the training staff.