Question

2. One‐Sample Univariate Hypothesis Testing with Proportions

For this question, show the results “by hand”, but you can use R to check your work. Suppose that the 4‐year graduation rate at a large, public university is 70 percent (this is the population proportion of successes). In an effort to increase graduation rates, the university randomly selected 200 incoming freshman to participate in a peer‐advising program. After 4 years, 154 of these students graduated. What are the null and alternative hypotheses? Can you conclude that this program was a success at the 5‐percent level of significance? Can you conclude that the program is a success at the 1‐percent level of significance? Show your work and explain. Since “success” is an increase in graduation rates, this is a one‐tailed test.

Answer 1

Given that , n=200, X=154,

and Population proportion is

(1) The following null and alternative hypotheses need to be tested:

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

(2) The significance level is .

(3) The z-statistic is computed as follows:

(4) Decision :The p-value is p = 0.015, and since p = 0.0154 < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion:  there is enough evidence to claim that the population proportion p is greater than 0.70​ , at the significance level. i.e The program is a success at the 5‐percent level of significance

(5)  The p-value is p = 0.0154, and since p = 0.0154 > 0.01, it is concluded that the null hypothesis is not rejected at  the 1‐percent level of significance.

Conclusion: there is not enough evidence to claim that the population proportion p is greater than 0.70​ , at the significance level.

i.e The program is not a success at the 1‐percent level of significance