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:
[tex]H_0: \pi=0.095\\\\H_a:\pi<0.095[/tex]
The significance level is 0.05.
The sample has a size n=100.
The sample proportion is p=0.07.
[tex]p=X/n=7/100=0.07[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.095*0.905}{100}}\\\\\\ \sigma_p=\sqrt{0.001}=0.029[/tex]
Then, we can calculate the z-statistic as:
[tex]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[/tex]
This test is a left-tailed test, so the P-value for this test is calculated as:
[tex]P-value=P(z<-0.682)=0.248[/tex]
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.