652 CHAPTER 13 Nonparametric Tests Claims About the Median of a Single Population The next example illustrates the procedure for using the sign test in testing a claim about the median of a single population. See how the negative and positive signs are based on the claimed value of the median. Body Temperatures EXAMPLE 4 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 sign test to test the claim that the median is less than 98.6°F. Use a 0.05 significance level. Of the 106 subjects, 68 had temperatures below 98.6°F, 23 had temperatures above 98.6°F, and 15 had temperatures equal to 98.6°F. SOLUTION REQUIREMENT CHECK The only requirement is that the sample is a simple random sample. Based on the design of this experiment, we assume that the sample data are a simple random sample. The claim that the median is less than 98.6°F is the alternative hypothesis, while the null hypothesis is the claim that the median is equal to 98.6°F. H0: Median is equal to 98.6°F. 1median = 98.6°F2 H1: Median is less than 98.6°F. 1median 6 98.6°F2 Following the procedure outlined in Figure 13-1, we use a negative sign to represent each temperature below 98.6°F, and we use a positive sign for each temperature above 98.6°F. We discard the 15 data values of 98.6, since they result in differences of zero. We have 68 negative signs and 23 positive signs, so n = 91 and x = 23 (the number of the less frequent sign). The sample data do not contradict the alternative hypothesis, because most of the 91 temperatures are below 98.6 °F. The value of n exceeds 25, so we convert the test statistic x to the test statistic z: z = 1 x + 0.52 - a n 2b 2 n 2 = 123 + 0.52 - a 91 2 b 2 91 2 = -4.61 P-Value In this left-tailed test, the test statistic of z = -4.61 yields a P-value of 0.0000 (Table: 0.0001). Because that P-value is so small, we reject the null hypothesis. Critical Value In this left-tailed test with a = 0.05, use Table A-2 to get the critical z value of -1.645. From Figure 13-3 on the next page we see that the test statistic of z = -4.61 is within the critical region, so reject the null hypothesis.
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