Question

Random samples of 400 men and 600 women were asked whether they would like

to have flyover near their residence. 200 men and 325 women were in favor of the

proposal. Test the hypothesis that proportions of men and women in favor of the

proposal are same against that they are not, at 5% l.o.s.

Answer 1

Sample Proportion 1 $\hat{p_1}=0.5$

Favorable Cases 1 $X_1=200$

Sample Size 1 $n_1=400$

Sample Proportion 2 $\hat{p_2}=\dfrac{13}{24}\approx0.5417$

Favorable Cases 2 $X_2=325$

Sample Size 2 $n_2=600$

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{200+325}{400+600}=0.525$$

Significance Level $\alpha=0.05$

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

$H_0:p_1=p_2$

$H_1: p_1\not=p_2$

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

(2) Rejection Region

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:

​

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

$$\approx\dfrac{0.5-0.5417}{\sqrt{0.525(1-0.525)(1/400+1/600)}}\approx-1.2926$$

Since it is observed that $|z|=1.2926<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $\hat{p}_1$ is different than $\hat{p}_2,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=2P(Z<-1.2926)\approx0.19615,$ and since $p=0.19615>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $\hat{p}_1$ is different than $\hat{p}_2,$ at the $\alpha=0.05$ significance level.