650 CHAPTER 13 Nonparametric Tests Claims Involving Nominal Data with Two Categories In Chapter 1 we defined nominal data to be data that consist of names, labels, or categories only. The nature of nominal data limits the calculations that are possible, but we can identify the proportion of the sample data that belong to a particular category, and we can test claims about the corresponding population proportion p. The following example uses nominal data consisting of genders (girls>boys). The sign test is used by representing girls with positive 1+2 signs and boys with negative 1-2 signs. (Those signs are chosen arbitrarily—honest.) The null hypothesis is the claim of no difference between measured weights and reported weights of males, and the alternative hypothesis is the claim that there is a difference. H0: There is no difference. 1The median of the differences is equal to 0.2 H1: There is a difference. 1The median of the differences is not equal to 0.2 Following the sign test procedure summarized in Figure 13-1, we let n = 9 (the total number of positive and negative signs) and we let x = 3 (the number of the less frequent sign, or the smaller of 3 and 6). The sample data do not contradict H1, because there is a difference between the 6 positive signs and the 3 negative signs. The sample data show a difference, and we need to continue with the test to determine whether that difference is significant. Figure 13-1 shows that with n = 9, we should proceed to find the critical value from Table A-7. We refer to Table A-7, where the critical value of 1 is found for n = 9 and a = 0.05 in two tails. Since n … 25, the test statistic is x = 3 (and we do not convert x to a z score). With a test statistic of x = 3 and a critical x value of 1, we fail to reject the null hypothesis of no difference. (See Note 2 included with Table A-7: “Reject the null hypothesis if the number of the less frequent sign (x) is less than or equal to the value in the table.” Because x = 3 is not less than or equal to the critical value of 1, we fail to reject the null hypothesis.) There is not sufficient evidence to warrant rejection of the claim that for males, the median of the differences “measured weight – reported weight” is equal to 0. YOUR TURN. Do Exercise 5 “Measured and Reported Weights.” INTERPRETATION We conclude that there is not sufficient evidence to reject the claim that for males, there is no difference between measured weights and reported weights. Gender Selection EXAMPLE 3 The Genetics & IVF Institute conducted a clinical trial of its methods for gender selection for babies. Before the clinical trials were concluded, 879 of 945 babies born to parents using the XSORT method of gender selection were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection is effective in increasing the likelihood of a baby girl. SOLUTION REQUIREMENT CHECK The only requirement is that the sample is a simple random sample. Based on the design of this experiment, we can assume that the sample data are a simple random sample. Let p denote the population proportion of baby girls. The claim that girls are more likely with the XSORT method can be expressed as p 7 0.5, so the null and alternative hypotheses are as follows:
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