Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Question Part 1 The chair of a department at a university claims that at least 6

ID: 3326367 • Letter: Q

Question

Question

Part 1

The chair of a department at a university claims that at least 60 percent of its students participated in a community service project in the previous semester. In order to test the claim, you collect a sample of 40 students and find that 20 of them did participate in a community service project in the previous semester. Using a level of significance of 0.05, carry out a hypothesis test for the department chair’s claim.

Part 2

The chair of a different department at the same university also claims that at least 60 percent of its students participated in a community service project in the previous semester. In order to test the claim of that chair, you collect a sample of 80 students and find that 40 of them did participate in a community service project in the previous semester. Using a level of significance of 0.05, carry out a hypothesis test for that department chair’s claim.

Part 3

Explain why we reject H0 in one of the two above parts of this question and not in the other, even though the two sample proportions are the same.

Explanation / Answer

Part 1:
Below are the null and alternate hypothesis
H0: p = 0.6
H1: p < 0.6

pcap = 20/40 = 0.5
SE = sqrt(p*(1-p)/n) = sqrt(0.6*0.4/40) = 0.0775

test statistics, z = (0.5 - 0.6)/0.0775 = -1.2903

p-value = 0.0985

Fail to reject null hypothesis as p-value is greater than the significance level.

Part 2:
Below are the null and alternate hypothesis
H0: p = 0.6
H1: p < 0.6

pcap = 40/80 = 0.5
SE = sqrt(p*(1-p)/n) = sqrt(0.6*0.4/80) = 0.0548

test statistics, z = (0.5 - 0.6)/0.0548 = -1.8248

p-value = 0.034

Reject null hypothesis as p-value is less than the significance level.

Part 3:
This is due to sample size. In part 1 sample size used is 40 which resulted into higher value of SE which in turn caused z value = -1.2903 compared to z = -1.8248 in part 2. Due to this p-value is less in part 2 and it is less than significance level of 0.05 hence null hypothesis was rejected.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote