Answer to Question #334779 in Statistics and Probability for Joy

Question

In a certain Barangay, 550 random residents are chosen in which 225 are males. At a=0.01, test the hypothesis that more than 35% of the Barangay population are male residents.

Answer 1

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

$H_0:p\le0.35$

$H_a:p>0.35$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{225}{550}-0.35}{\sqrt{\dfrac{0.35(1-0.353)}{550}}}\approx2.9054$$

The p-value is $p =P(Z>0.0970)= 0.001834$

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263.$

The rejection region for this right-tailed test is $R = \{z: z> 2.3263\}.$

Since it is observed that $z = 2.9054 \ge2.3263= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p =0.922726,$ and since $p = 0.001834 \le 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than 0.35, at the $\alpha = 0.01$ significance level.