A local school district is considering changing the food that is offered in the

Question

A local school district is considering changing the food that is offered in the cafeteria in the hope that students will eat healthier lunches. They plan to offer broccoli, asparagus, spinach, peas, or green beans with every lunch. However, they are concerned that students will simply choose not to eat what is offered if they don't like it. Rather than send a survey out to every family in the district, which they fear will not accurately reflect the students' views, they ask a statistics class to conduct a study to determine what percent of students would eat one of the green vegetables at least three times per week. The class decides to interview people during lunchtime to determine how students will respond to the change in menu. a. What considerations are important for the class to consider in conducting the study? b. The districts want to determine if more than 25% of the students would likely eat the green vegetables at least three times per week. Do you think the statistics class should set the alpha level at 0.05 or 0.10? Explain. The class finds that 26 of 90 students say they would eat the vegetables at least three times per week. Can they conclude that more than one-fourth of the student body would accept the change and eat the green vegetables? Explain. Construct and interpret a confidence interval for the percent of students who would eat the green vegetables at least three times per week. Based on the results, what should the school district do? Explain. c. d. e.

Explanation / Answer

a. Students should be sampled randomly, and sample size should be large enough so that normal sampling distribution is followed.

b. The researcher would like to establish their tentative proposition by rejecting null hypothesis [H0:p=0.25 (25% of students would like to eat vegetables at least three times per week). Setting a lower alpha, increases the chance of rejecting null hypothesis. Therefore, Statistics class should set alpha=0.05.

c. Compute 1-proportion z test.

Z=(phat-p)/sqrt[p(1-p)/N], where, phat is sample proportion, p is population proportion, N is sample size.

=(26/90-0.25)/sqrt[0.25(1-0.25)/90]

=0.85

p value: 0.197

Per rejection rule based on p value, reject null hypothesis if p value is less than alpha=0.05. Here, 0.197>0.05, therefore, fail to reject null hypothesis. There is insuffiicent sample evidence to suggest that more than 25% of students would eat vegetables at least three times per week.

d. The 95% c.i=phat+-zalpha/2sqrt[phat(1-phat)/n], where, phat is sample proportion, n is sample size.

=26/90+-1.96sqrt[{26/90(1-26/90)}/90]

=(0.1952,0.3825)

Thus, one can be 95% confident that the interval captures the true population proportion of students who eat vegetables at least three times per week.

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