Question

A study found that 68% of the population owns a home in a random sample of 150 households 92 owned a home at the a=0.01 level is there enough evidence to reject the claim

Answer 1

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

$H_0:p=0.68$

$H_a:p\not=0.68$

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 $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{92}{150}-0.68}{\sqrt{\dfrac{0.68(1-0.68)}{150}}}$$ $$\approx-1.75035$$

Since it is observed that $|z| = 1.75035<2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-1.75035)=0.080058,$ and since $p=0.080058>0.01=\alpha,$ it is concluded that the null hypothesis is notrejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than $0.68,$ at the $\alpha = 0.01$ significance level.