The Rank-Sum Test Section 15.2

Objectives Assign ranks to a sample Test hypotheses using the rank-sum test

Assign ranks to a sample Objective 1 Assign ranks to a sample

Ranks To rank the values in a sample, we order them from smallest to largest. The smallest value is assigned a rank of 1, the next smallest is assigned a rank of 2, and so forth. For values in the data set whose ranks are tied, each value is assigned the average of the ranks. Example: The heights of ten fourth-graders, in inches, are 48, 46, 45, 54, 46, 52, 50, 46, 52, 49. Assign ranks to these values. Solution: We arrange the values in increasing order. We notice there are three values tied at 46, and two tied at 52. We assign the ranks 1 through 10 as preliminary ranks. The three values of 46 have preliminary ranks 2, 3, 4. We compute the average of these preliminary ranks: 2 + 3 + 4 3 = 9 3 =3. The rank assigned to each of these values is therefore 3. The two values of 52 have preliminary ranks 8 and 9. We assign each the rank 8 + 9 2 = 17 2 =8.5. Height 45 46 48 49 50 52 54 Preliminary Ranks 1 2 3 4 5 6 7 8 9 10 Ranks 8.5

Test hypotheses using the rank-sum test Objective 2 Test hypotheses using the rank-sum test

The Rank-Sum Test Wilcoxon’s rank-sum test, also known as the Mann-Whitney test, is a nonparametric test to determine whether two population medians are equal. It is a nonparametric alternative to the two-sample 𝑡-test. The test requires that the populations have approximately the same shape. When two populations have the same shape and their medians are equal, they have the same distribution. For this reason the rank-sum test is sometimes described as a test to determine whether two populations differ. The main advantage of the rank-sum test is that it does not require that the populations be approximately normal. The main disadvantage is that the null hypothesis can be falsely rejected when the population medians are equal, if the shapes of the two populations differ substantially.

The Rank-Sum Test The rank-sum test is used to test the null hypothesis that two population medians are equal. Independent samples are drawn from each population. The combined samples are then ranked. The smallest value in the two samples combined is assigned a rank of 1, the second smallest is assigned a rank of 2, and so forth. If the two populations are the same, the ranks for the values assigned to one sample should not be substantially larger or smaller than the ranks assigned to the other. Thus the rank-sum test rejects the null hypothesis when the sum of the ranks for one of the samples is exceptionally large or small. Null hypothesis: 𝐻 0 : 𝑚 1 = 𝑚 2 Alternate hypothesis: Left-tailed: 𝐻 1 : 𝑚 1 < 𝑚 2 Right-tailed: 𝐻 1 : 𝑚 1 > 𝑚 2 Two-tailed: 𝐻 1 : 𝑚 1 ≠ 𝑚 2

Test Statistic To construct the test statistic, let 𝑛 1 be the smaller of the two sample sizes and let 𝑛 2 be the larger. Let 𝑆 be the sum of the ranks for the smaller sample. If both samples are the same size, then 𝑛 1 and 𝑛 2 are both equal to that size, and 𝑆 may be the sum of the ranks of either sample. When 𝐻 0 is true, it can be shown that 𝑆 has mean 𝜇 𝑆 and standard deviation 𝜎 𝑆 given by 𝜇 𝑆 = 𝑛 1 𝑛 1 +𝑛+1 2 𝜎 𝑆 = 𝑛 1 𝑛 2 ( 𝑛 1 + 𝑛 2 +1) 12 We construct the test statistic as 𝑧= 𝑆− 𝜇 𝑆 𝜎 𝑆 . When both samples sizes are 10 or more, the test statistic has an approximately standard normal distribution, so P-values and critical values can be found using the normal curve (Table A.2 or technology). When one or both of the sample sizes is less than 10, special tables can be used. However, since most samples in practice consist of at least 10 values, we do not consider this situation here.

Example – Rank-Sum Test A sample of 14 students took a statistics class online, and another sample of 12 students took an equivalent class in a traditional classroom. Both classes were given the same final exam at the end of the course. Can you conclude that there is a difference between the median scores from the two classes? Use the 𝛼=0.05 level of significance. Solution: The null hypothesis is that there is no difference in the scores between the two teaching methods. The alternate hypothesis is that there is a difference. 𝐻 0 : 𝑚 1 = 𝑚 2 𝐻 1 : 𝑚 1 ≠ 𝑚 2 Online 78 82 83 87 75 63 60 94 62 98 90 97 81 Traditional 73 72 92 100 74 64 84 77 89 70

Example – Rank-Sum Test Solution (continued): We arrange the combined samples in increasing order and assign ranks. Score 60 62 63 64 70 72 73 74 75 77 78 Sample O T Rank 1 2 3 4.5 6 7 8 9 10 11 12.5 81 82 83 84 87 89 90 92 94 97 98 100 14 15 16 17 18 19 20.5 22 23 24 25 26 The sample sizes are 𝑛 1 =12 and 𝑛 2 =14. The smaller sample belongs to the traditional class, so 𝑆 is the sum of the ranks corresponding to that sample. We compute 𝑆=154.5, 𝜇 𝑆 =162, and 𝜎 𝑆 =19.442. The value of the test statistic is 𝑧= 𝑆− 𝜇 𝑆 𝜎 𝑆 = 154.5−162 19.442 =−0.39.

Example – Rank-Sum Test Solution (continued): Since this is a two-tailed test, we may find the P-value by summing the area to the left of 𝑧=−0.39 and to the right of 𝑧=0.39. From Table A.2, we find the P-value to be 0.6965. Technology yields a P-value of 0.6997. Since P > 0.05, we do not reject 𝐻 0 . We cannot conclude that there are differences in test scores between the two methods of instruction.

You Should Know… The primary advantage and disadvantage of the rank-sum test How to assign ranks How to perform the rank-sum test to test the median of a population