0673chapter13 - Chapter 13 Nonparametric Statistics 13-2 Sign Test 1 The sign test is nonparametric or distribution-free because it does not require the

0673chapter13 - Chapter 13 Nonparametric Statistics 13-2...

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Chapter 13 Nonparametric Statistics13-2 Sign Test 1. The sign test is “nonparametric” or “distribution-free” because it does not require the data to come from a particular distribution defined in terms of certain parameters. 2. The procedure in this section is called the “sign” test because it reduces the data to plus and minus signs in order to perform the analysis. 3. The alternative hypothesis is that the proportion of girls is greater than 0.5, but the sample proportion of 39/(172+39) = 39/211 is less than 0.5. Without doing any mathematics, it is apparent that the conclusion must be that there is not enough evidence to support the claim that the method increases the likelihood of girl births. In order for the data to support the claim the number of girl births would have to be significantly greater than (½)·(211) = 105.5. 4. The efficiency of the sign test is 0.63. This means that for normal populations, the sign test requires a sample of size n=100 to identify departures from the null hypothesis that the appropriate corresponding parametric test could identify with a sample of size n=63. 5. 13 +’s and 1 –’s n = 14 25; use C.V. = 2 from Table A-7 Since min(13,1) = 1 2 = C.V., reject the hypothesis of no difference. Since there were more +’s, conclude that the first variable has the larger scores. 6. 5 +’s and 7 –’s n = 12 25; use C.V. = 2 from Table A-7 Since min(5,7) = 5 > 2 = C.V., do not reject the hypothesis of no difference. There is not sufficient evidence to reject the hypothesis of no difference. 7. 360 +’s and 374 –’s n = 734 > 25; use C.V. = -1.96 from the z table z = [(x+0.5) – n/2]/[n/2]= [360.5 – 367]/[734/2]= -6.5/13.546 = -0.480 Since -0.480 > -1.96, do not reject the hypothesis of no difference. There is not sufficient evidence to reject the hypothesis of no difference. 8. 512 +’s and 327 –’s n = 839 > 25; use C.V. = -1.96 from the z table z = [(x+0.5) – n/2]/[n/2]= [327.5 – 419.5]/[839/2]= -92/14.483 = -6.35 Since -6.35 < -1.96, reject the hypothesis of no difference. Since there were more +’s, conclude that the first variable has the larger scores.
442 CHAPTER 13 Nonparametric Statistics NOTES FOR THE REMAINING EXERCISES IN THIS SECTION. (1) FORn 25. Table A-7 gives only xL, the lowercritical value for the sign test. And so the text lets x be the smallerof the number of +’s or the number of –’s, and warns the reader to use common sense to avoid concluding the reverse of what the data indicates. But the problem’s symmetry means that the upper critical value is xU= n – xLand that μx= n/2, the natural expected value of x when Hois true. For completeness, this manual indicates those values whenever using the sign test – and uses a normal curve for illustration, even though the distribution of x is discrete. Letting x always be the number of +’s is an alternative approach that maintains the natural agreement between the alternative hypothesis and the critical region – and is consistent with the logic and notation of parametric tests. Many sign test software programs use this approach.

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