Question

Problem 2 (15 marks) Consider a multinomial experiment involving n = 200 students of a large high school. The attendance department recorded the number of students who were absent during the weekdays. The null hypothesis to be tested is: Ho: p1 = .10, p2 = 25, p3 = .30, p4 = 20, p5 = 15. The following data regarding absenteeism have been collected during the last week. Day of the Week Mon. Tues. Wed. Thurs. Fri. Number Absent 16 44 56 48 36 At a significance level a=0.05, test the null hypothesis.

Answer 1

Categories Observed Expected (fo-fe)?/fe Category 1 16 200*0.1=20 (16-20)2/20 = 0.8 Category 2 44 200*0.25=50 (44-50)/50 = 0.72 Category 3 56 200*0.3=60 (56-60)2/60 = 0.267 Category 4 48 200*0.2=40 (48-40)2/40 = 1.6 Category 5 36 200*0.15=30 (36-30)/30 = 1.2 Sum = 200 200 4.587 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho : p1 = 0.1, p2 = 0.25, p3 = 0.3, P4 = 0.2, p5 = 0.15 He: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 5-1 = 4, so then the rejection region for this test is R= {x? : x² > 9.488). (3) Test Statistics The Chi-Squared statistic is computed as follows: (O; - – E;) = 0.8 +0.72 +0.267 +1.6 +1.2 = 4.587 Ei 2=1 (4) Decision about the null hypothesis Since it is observed that x = 4.587 < xã = 9.488, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a = 0.05 significance level.

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