Question

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.3% of American adults suffer from depression or a depressive illness. Suppose that in a survey of 2000 people in a certain city, 11.1% of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that city suffering from depression or a depressive illness is more than the 9.3% in the general adult American population. Test the relevant hypotheses using a 5% level of significance.

Answer 1

Answer:

There is enough evidence to support the claim that the true proportion of people in that city suffering from depression or a depressive illness is more than the 9.3% in the general adult American population.  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 2000

p = 9.3% = 0.093

Alpha, α = 0.05

$\hat{p} = 11.1\% = 0.111$

First, we design the null and the alternate hypothesis  

$H_{0}: p = 0.093\\H_A: p > 0.093$

This is a one-tailed(right) test.

Formula:

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.111-0.093}{\sqrt{\frac{0.093(1-0.093)}{2000}}} = 2.7716$

Now, we calculate the p-value from the table.

P-value = 0.002789

Since the p-value is less than the significance level, we fail to accept the null hypothesis and accept the alternate hypothesis.

Thus, there is enough evidence to support the claim that the true proportion of people in that city suffering from depression or a depressive illness is more than the 9.3% in the general adult American population.