Question

According to a polling​ organization, 22​% of adults in a large region consider themselves to be liberal. A survey asked 200 respondents to disclose their political​ philosophy: Conservative,​ Liberal, Moderate. Treat the results of the survey as a random sample of adults in this region. Do the survey results suggest the proportion is higher than that reported by the polling​ organization? Use an alphaequals0.01 level of significance.

Answer 1

No. There is not enough evidence to support the claim that the proportion of liberals is higher than that reported by the polling​ organization (P-value = 0.0366).

Explanation:

The question is incomplete: there is no information about the results of the survey. We will assume that 55 of the subjects answer "liberal", and test the claim.

This is a hypothesis test for a proportion.

The claim is that the proportion of liberals is higher than that reported by the polling​ organization.

Then, the null and alternative hypothesis are:

`H_0: \pi=0.22\\\\H_a:\pi>0.22`

The significance level is 0.01.

The sample has a size n=200.

The sample proportion is p=0.275.

`p=X/n=55/200=0.275`

The standard error of the proportion is:

`\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.22*0.78)/(200)}\\\\\\ \sigma_p=√(0.000858)=0.029`

Then, we can calculate the z-statistic as:

`z=(p-\pi-0.5/n)/(\sigma_p)=(0.275-0.22-0.5/200)/(0.029)=(0.053)/(0.029)=1.792`

This test is a right-tailed test, so the P-value for this test is calculated as:

`\text{P-value}=P(z>1.792)=0.0366`

As the P-value (0.0366) is greater than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of liberals is higher than that reported by the polling​ organization.