Answer to Question #223401 in Statistics and Probability for Lovely

Question

According to official census figures, 8% of couples living together ate not married. A reacher took a random sample of 400 couples and found that 9.5% of them are not married. Test at the 15% significance level if the current percentage of unmarried couples is different from 8%

Answer 1

Because this sample meets the $np=400(0.08)=32>10$ and $n(1-p)=400(1-0.08)$ $=368>10$ conditions, we can assume it is approximately normal. 

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

$H_0: p=0.8$

$H_1: p\not=0.8$

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 $\alpha=0.15,$ and the critical value for a two-tailed test is $z_c=1.4395.$

The rejection region for this two-tailed test is $R=\{z:|z|>1.4395\}.$  

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.095-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{400}}}$$

$$\approx1.1058$$

Since it is observed that $|z|=1.1058<1.4395=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p=2P(Z>1.1058)=0.2688,$ and since $p=0.2688>0.15=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than $0.08,$ at the $\alpha=0.15$ significance level.