Question

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. Is this a test of one mean or proportion

Answer 1

Answer:

This is a hypothesis test for a proportion.

There is not enough evidence to support the claim that the true proportion of people in this town suffering from depression or a depressive illness is lower than the percent in the general adult American population (P-value=0.248).

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the true proportion of people in this town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.095\\\\H_a:\pi<0.095$

The significance level is 0.05.

The sample has a size n=100.

The sample proportion is p=0.07.

$p=X/n=7/100=0.07$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.095*0.905}{100}}\\\\\\ \sigma_p=\sqrt{0.001}=0.029$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.07-0.095+0.5/100}{0.029}=\dfrac{-0.02}{0.029}=-0.682$

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

$P-value=P(z<-0.682)=0.248$

As the P-value (0.248) is bigger than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the true proportion of people in this town suffering from depression or a depressive illness is lower than the percent in the general adult American population.