660 CHAPTER 13 Nonparametric Tests Step 3: Attach to each rank the sign of the difference from which it came. That is, insert the signs that were ignored in Step 2. EXAMPLE: The bottom row of Table 13-4 lists the same ranks found in the fourth row, but the signs of the differences shown in the third row are inserted. Step 4: Find the sum of the ranks that are positive. Also find the absolute value of the sum of the negative ranks. EXAMPLE: The bottom row of Table 13-4 lists the signed ranks. The sum of the positive ranks is 5 + 4 + 6 + 2 + 8 + 1 + 3 = 29. The sum of the negative ranks is -7, and the absolute value of this sum is 7. The two rank sums are 29 and 7. Step 5: Let T be the smaller of the two sums found in Step 4. Either sum could be used, but for a simplified procedure we arbitrarily select the smaller of the two sums. EXAMPLE: The data in Table 13-4 result in the rank sums of 29 and 7, so the smaller of those two sums is 7. Step 6: Let n be the number of pairs of data for which the difference d is not 0. EXAMPLE: The data in Table 13-4 have 8 differences that are not 0, so n = 8. Step 7: Determine the test statistic and critical values based on the sample size, as shown in the preceding Key Elements box. EXAMPLE: For the data in Table 13-4 the test statistic is T = 7. The sample size is n = 8, so the critical value is found in Table A-8. Using a 0.05 significance level with a two-tailed test, the critical value from Table A-8 is 4. Step 8: Reject the null hypothesis if the sample data lead to a test statistic that is in the critical region—that is, the test statistic is less than or equal to the critical value(s). Otherwise, fail to reject the null hypothesis. EXAMPLE: For the sample of matched pairs in the first two rows of Table 13-4, the test statistic is T = 7 and the critical value is 4, so the test statistic is not less than or equal to the critical value. Consequently, we fail to reject the null hypothesis that the matched pairs are from a population of matched pairs with differences having a median equal to zero. Conclusion Table A-8 includes a note stating that we should reject the null hypothesis if the test statistic T is less than or equal to the critical value. Because the test statistic of T = 7 is not less than or equal to the critical value of 4, we fail to reject the null hypothesis. YOUR TURN. Do Exercise 5 “Measured and Reported Weights.” INTERPRETATION We conclude that there is not sufficient evidence to support the claim that for males, there is a significant difference between measured weights and reported weights. Based on the very small sample, it appears that males do not tend to report weights that are much different from their actual weights. It is possible that a much larger sample would lead to a different conclusion.
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