Question

1. Are college students who take a freshman orientation course more likely to stay in college than those who do not take such a course? An article reported that out of 100 randomly selected college students who participated in an orientation course 57 returned for a second year. Of the 95 randomly selected students who did not take the orientation course, 40 returned for a second year.

Test the claim that a larger proportion of college students return for a second year if they took an orientation course. Use an α = 0.05 level of significance.

1. State the null and alternate hypotheses.

2. Find the test statistic and p-value. Show Work.

3. What is the decision for this test? (circle or highlight the correct answer)

1. Reject Null Hypothesis
2. Fail to Reject Null Hypothesis
3. Reject Alternative Hypothesis
4. Accept Alternative Hypothesis

4. Is the orientation class successful at getting more students return for a second year? Explain.

Answer 1

The null and alternate hypothesis are:

H0: $P_1\leq P_2$

Ha: $P_1>P_2$

The test statistic is given by:

$z=\frac{(\widehat{p_{1}}-\widehat{p_{2}})}{\sqrt{\frac{\widehat{p_{1}}(1-\widehat{p_{1}})}{n_{1}}+\frac{\widehat{p_{2}}(1-\widehat{p_{2}})}{n_{2}}}} =\frac{(\frac{57}{100}-\frac{40}{95})}{\sqrt{\frac{\frac{57}{100}(1-\frac{57}{100 })}{100}+\frac{\frac{40}{95}(1-\frac{40}{95})}{95}}} =2.10$

Since this is a right-tailed test, so the p-value is given by:

$P(Z>2.10)=1-P(Z<2.10)=1-0.98214=0.01786$

Since p-value is less than 0.05, so we have sufficient evidence to reject the null hypothesis H0.

Thus, we reject the null hypothesis.

Hence 1st option.

Thus we can conclude that $P_1>P_2$ . So, the orientation class is successful at getting more students return for a second year.