598 CHAPTER 11 Goodness-of-Fit and Contingency Tables McNemar’s test requires that for a table such as Table 11-9, the frequencies are such that b + c Ú 10. The test is a right-tailed chi-square test with the following test statistic: x2 = 1 b - c - 122 b + c P-values are typically provided by software, and critical values can be found in Table A-4 using 1 degree of freedom. Caution: When applying McNemar’s test, be careful to use only the two frequency counts from discordant (different) pairs, such as the frequency b in Table 11-9 (with different pairs of cured>not cured) and frequency c in Table 11-9 (with different pairs of not cured>cured). Are Hip Protectors Effective? EXAMPLE 5 A randomized controlled trial was designed to test the effectiveness of hip protectors in preventing hip fractures in the elderly. Nursing home residents each wore protection on one hip, but not the other. Results are summarized in Table 11-10 (based on data from “Efficacy of Hip Protector to Prevent Hip Fracture in Nursing Home Residents,” by Kiel et al., Journal of the American Medical Association, Vol. 298, No. 4). McNemar’s test can be used to test the null hypothesis that the following two proportions are the same: ■ The proportion of subjects with no hip fracture on the protected hip and a hip fracture on the unprotected hip. ■ The proportion of subjects with a hip fracture on the protected hip and no hip fracture on the unprotected hip. TABLE 11-10 Randomized Controlled Trial of Hip Protectors No Hip Protector Worn No Hip Fracture Hip Fracture Hip Protector Worn No Hip Fracture 309 10 Hip Fracture 15 2 Using the discordant (different) pairs with the general format from Table 11-9 we have b = 10 and c = 15, so the test statistic is calculated as follows: x2 = 1 b - c - 122 b + c = 1 10 - 15 - 122 10 + 15 = 0.640 With a 0.05 significance level and degrees of freedom given by df = 1, we refer to Table A-4 to find the critical value of x2 = 3.841 for this right-tailed test. The test statistic of x2 = 0.640 does not exceed the critical value of x2 = 3.841, so we fail to reject the null hypothesis. (Also, the P-value is 0.424, which is greater than 0.05, indicating that we fail to reject the null hypothesis.) The proportion of hip fractures with the protectors worn is not significantly different from the proportion of hip fractures without the protectors worn. The hip protectors do not appear to be effective in preventing hip fractures.
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