Question

14. The following table shows the frequency distribution of births by days of the week for all registered births in the United States in the year 1971. Test the null hypothesis that all seven days of the week are equally likely for childbirth, using the 1% significance Number of Births by Day of the Week, United States, 1971 Monday Tuesday Wednesday Thursday Friday Saturday Sunday Number of Births 520,900 544,600 536,800 516,800 538,300 472,100 443,600 Source: R. Rindfuss, et al., "Convenience and the Occurrence of Births - Induction of Labor in the United States and Canada," International Journal of Health Services, 9:3, (1979), p.441, Table 1. (A) Reject Ho since 42> 18.475 (B) Reject Ho since 42 > 18.548 (C) Reject Ho since 42>16.812 7, 4 341S1. 1? (E) Reject Ho since 17094 > 18.548

Answer 1

We test this with chi - square goodness of fit test. Null hypothesis: True proportions in all 7 categories is equal to their claimes value. Alternative hypothesis: Atleast one proportions is different from its claimes value. H_0: p_1 = 0.142857, p_2 = 0.142857, p_3 = 0.142857, p_4 = 0.142857, p_5 = 0.142857, p_6 = 0.142857, p_7 = 0.142857 H_a: Atleast on p_i do not match the claimes value \r\n Test stastistic: chi^2 = 17093.661 Significance level = 0.01 Degrees of freedom, df = k - 1 = 6 chi^2_0.01 has area of 0.01 to its right. To find chi^2_0.01, open any Chi-square table, go across from row df = v = 6 and down from column 0.02 and get 16.812 Critical value: chi^2_0.01 = 16.812 Decision Rule: Reject H_0 if chi^2 > 16.812 We can also use excel to find exact value: CHISQ. Inv. RT (0.01, 6) = 16.8118938297709 To get p - value, we can use excel function CHISQ. Dist.