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Chi Square Test for Qualitative Data Student name University name 19.9. (a) Hypothesis formulated: Null Hypothesis H0 = Crimes are not equally likely to be committed on any day of the week. Alternative Hypothesis H1: Crimes are equally likely to be committed on any day of the week. P Sunday = P Monday = P Tuesday = P Wednesday = P Thursday = P Friday = P Saturday = H1 = H0 is false The null hypothesis is rejected at 0.01 level of significance i.e. x2 > 16.81. Freq. Monday Tuesday Wednesday Thursday Friday Saturday Sunday Total f0 17 21 22 18 23 24 15 140 fe 20 20 20 20 20 20 20 140 X2 = = = = = 3.4 The observed x2 is 3.4 and is less than the critical x2 of 16.81 i.e. 3.4<16.81. So we reject the null hypothesis formulated at 0.01 level of H = = 0.01 (2304.9+2310.4+2560) - 930.01 [2304.9+2310.4+2560] – 93 = -20.89 H = 20.89 We reject the null hypothesis because observed H value is greater than critical H of 5.99. 20.10 We formulate the hypothesis: Null hypothesis H0: All three programs have equal distribution Alternative Hypothesis H1: The three programs are not equally distributed. If observed H value is greater than the critical H of 5.99 at a significance level of 0.05 w2 reject the null hypothesis H0. H = µ1 = 81.5 µ2 = 53.5 µ3 = 141 n= 23 H = = = 0.02 [3516.32] – 72 = -1.67 H = 1.67 We do not have sufficient evidence to reject the null at the significance level - 0.05. It is because the calculated H is less than critical H value i.e. 1.67< 5.99. Hence we accept the Null hypothesis H0: All three programs have equal distribution. [...]
Order Description:
Show all relevant work; use the equation editor in Microsoft Word when necessary. Chapter 19, numbers 19.9, 19.10, 19.13, 19.14, and 19.16 Chapter 20, numbers 20.5, 20.6, 20.7, and 20.10
Subject Area: Statistics
Document Type: Paraphrasing
