A survey revealed that a substantial proportion of university students were not engaging in regularexercise. In response to a question on regular exercise, 60% of all graduates reported getting noregular exercise, 25% reported exercising sporadically and 15% reported exercising regularly asundergraduates. The next year the University launched a health promotion campaign on campus inan attempt to increase healthy behaviors among undergraduates. To evaluate the impact of theprogram, the University again surveyed graduates and asked the same questions. The survey wascompleted by 470 graduates and the following data were collected on the exercise question:No Regular ExerciseSporadic ExerciseRegular ExerciseTotalNumber of Students25512590470Based on the data, is there evidence of a shift in the distribution of responses to the exercisequestion following the implementation of the health promotion campaign on campus? Run the testat a 5% level of significance.

The test hypotheses are given below: H0: The distribution of responses to the exercise question follows the implementation of the health promotion campaign on campus (H0 : pNo reg=0.60; pspor=0.25; preg=0.15). H1: The distribution of responses to the exercise question does not follow the implementation of the health promotion campaign on campus. The test statistic is calculated by the formula, χ2=∑(O−E)2E Where, O represents the observed frequency and E represents the expected frequency. Methods for finding the expected frequency: Method 1: If the all the expected frequencies are equal, then E=nk Method 2: If the expected frequencies are not equal, then E=np for each category. Where, n is the total number of frequencies, k is the categories and p is the probability of occurrence.Here, the total number of frequencies is 470 and there are 3 outcomes then k is 3. Also, the probabilities are different. Thus, by using the second method the expected frequencies are calculated as follows. Digits E.frequency E1 121.5 (=(470)(0.6)) E2 36.45 (=(470)(0.25) E3 461.7 (=(470)(0.15)) Observed Expected Residual (O-E)2E Frequency Frequency No regular exercise 255 282 2.5851 Sporadic exercise 125 117.5 0.4787 Regular exercise 90 70.5 5.3936 Thus, the test-statistic value, χ2 is 8.4574 (=2.5851+0.4787+5.3936). P-value: The level of significance is 0.05. The P-value of the sample statistic is obtained using Excel formula. From EXCEL formula “=CHISQ.DIST.RT(8.4574, 2)”, the P-value is 0.0146. Decision rule: If the P-value is less that the significance level, reject the null hypothesis. Otherwise fail to reject the null hypothesis. Conclusion: Here, the P-value is less that the significance level (=5%%). Therefore, by the decision rule, reject the null hypothesis. There is no evidence of a shift in the distribution of responses to the exercise question following the implementation of the health promotion campaign on campus.

Math

Statistics

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? Run the test at a 5% level of significance.

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? Run the test at a 5% level of significance.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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