Question

The director of admissions at a large university says that 15% of high school juniors to whom she sends university literature eventually apply for admission. In a sample of 300 persons to whom materials were sent, 30 students applied for admission. In a two-tail test at the 0.05 level of significance, should we reject the director’s claim?

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p=0.15$

$H_1:p\not=0.15$

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

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is$z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.1-0.15}{\sqrt{\dfrac{0.15(1-0.15)}{300}}}$$

$$\approx-2.4254$$

Since it is observed that $|z| = 2.425 > 1.96=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(z<-2.4254)= 0.0152934,$ and since $p= 0.0152934<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is different than 0.15, at the $\alpha = 0.05$  significance level.