13-3 Wilcoxon Signed-Ranks Test for Matched Pairs 661 Rationale: In Example 1, the unsigned ranks of 1 through 8 have a total of 36, so if there are no significant differences, each of the two signed-rank totals should be around 36 , 2 = 18. That is, the negative ranks and positive ranks should split up as 18–18 or something close, such as 17–19. Table A-8, the table of critical values, shows that at the 0.05 significance level with 8 pairs of data, the critical value is 4, so a split of 4–32 represents a significant departure from the null hypothesis, and any split that is farther apart will also represent a significant departure from the null hypothesis. Conversely, splits like 5–31 do not represent significant departures from an 18–18 split, and they would not justify rejecting the null hypothesis. The Wilcoxon signedranks test is based on the lower rank total, so instead of analyzing both numbers constituting the split, we consider only the lower number. The sum of all the ranks 1 + 2 + 3 + g+ n is equal to n1n + 12>2. If this rank sum is to be divided equally between two categories (positive and negative), each of the two totals should be near n1n + 12>4, which is half of n1n + 12>2. Recognition of this principle helps us understand the test statistic used when n 7 30. Claims About the Median of a Single Population The Wilcoxon signed-ranks test can also be used to test a claim that a single population has some claimed value of the median. The preceding procedures can be used with one simple adjustment: When testing a claim about the median of a single population, create matched pairs by pairing each sample value with the claimed value of the median. The procedure included in Example 1 can then be used. Statdisk Body Temperatures EXAMPLE 2 Data Set 5 “Body Temperatures” in Appendix B includes measured body temperatures of adults. Use the 106 temperatures listed for 12 AM on Day 2 with the Wilcoxon signed-ranks test to test the claim that the median is less than 98.6°F. Use a 0.05 significance level. YOUR TURN. Do Exercise 9 “Body Temperatures.” SOLUTION REQUIREMENT CHECK (1) The design of the experiment that led to the data in Data Set 5 justifies treating the sample as a simple random sample. (2) The requirement of an approximately symmetric distribution of differences is satisfied, because a histogram of those differences is approximately symmetric. By pairing each individual sample value with the median of 98.6°F, we are working with matched pairs. Shown in the margin is the Statdisk display showing the test statistic of T = 661, which converts to the test statistic z = -5.67. (The display is from a two-tailed test; for this left-tailed test, the critical value is -1.645.) The test statistic of z = -5.67 yields a P-value of 0.000, so we reject the null hypothesis that the population of differences between body temperatures and the claimed median of 98.6°F is zero. There is sufficient evidence to support the claim that the median body temperature is less than 98.6°F. This is the same conclusion that results from the sign test in Example 4 in Section 13-2.
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