Question

4. (18 points) A University conducted a survey of its recent graduates to collect demo- graphic and health information for future planning purposes as well as to assess stu- dents' satisfaction with their undergraduate experiences. The survey revealed that a substantial proportion of students were not engaging in regular exercise, many felt their nutrition was poor and a substantial number were smoking. In response to a question on regular exercise, 60% of all graduates reported getting no regular exercise, 25% reported exercising sporadically and 15% reported exercising regularly as under- graduates. The next year the University launched a health promotion campaign on campus in an attempt to increase health behaviors among undergraduates. The pro- gram included modules on exercise, nutrition and smoking cessation. To evaluate the impact of the program, the University again surveyed graduates and asked the same questions. The survey was completed by 470 graduates and the following data were collected on the exercise question: No Regular Exercise Sporadic Exercise Regular Exercise 255 125 90 Students Based on the data, is there evidence of a shift in the distribution of responses to the exercise question following the implementation of the health promotion campaign on campus? Perform a 4-step hypothesis test to assess the question using a critical value of 5.99.

Answer 1

Solutions:-

The following table is obtained: Categories Observed Expected (fo-fe)</fe No Regular Exercise 255 470*0.60=282 (255-282)2/282 = 2.585 Sporadic Exercise 125 470*0.25=117.5 (125-117.5)2/117.5 = 0.47 Regular Exercise 90 470*0.15=70.5 (90-70.5)2/70.5 = 5.394 Sum = 470 470 8.457 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p1 = 0.60, p2 = 0.25, p3 = 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 = 0.0599, the number of degrees of freedom is df = 3-1 = 2, so then the rejection region for this test is R= {x? : x2 > 5.63} (3) Test Statistics The Chi-Squared statistic is computed as follows: n (O; χ? = Σ E;) Ei = 2.585 +0.479 + 5.394 = 8.457 i=1 (4) Decision about the null hypothesis Since it is observed that x² = 8.457 > xã = 5.63, it is then concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a = 0.0599 significance level.

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