Question

In a survey of 200 students, 78 of the 120 females in the sample passed math 17 on their first take while this figure is 60 among the males. Will you agree that the proportion of males who passed math 17 on their first take is higher than the proportion of females who passed math 17 on their first take?test at a 0.05

Answer 1

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{78+60}{120+80}=\dfrac{69}{100}=0.69$$

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

$H_0:p_1\ge p_2$

$H_a:p_1<p_2$

This corresponds to a left-tailed test, and a z-test for two population proportions will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is$z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z:z< -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}$$

$$=\dfrac{\dfrac{78}{120}-\dfrac{60}{80}}{\sqrt{0.69(1-0.69)(1/120+1/80)}}$$

$$=-1.4980$$

Since it is observed that $z=-1.4980> -1.6449=z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z<-1.4980)=0.067067,$ and since $p=0.067067>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p_2$ is greater than $p_1,$ at the $\alpha = 0.05$ significance level.