Question

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5% of American adults suffer from depression or a depressive illness. Suppose that in a survey of hundred 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 percentage in the general adult American population

Answer 1

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Answer:

The null hypothesiscan be stated as shown below:

That mean, the proportion of people suffering from depression in the town is at least 9.5% of American adults.

And, the alternative hypothesiscan be stated as shown below:

That mean, the proportion of people suffering from depression in the town is less than 9.5% of American adults.

Here, x is the number of successes from total number in the sample. Therefore, x is.

iHere, n is the total number in the sample. Therefore, n is.

Point estimatecan be calculated as shown below:

The calculation ofis shown below:

The distribution foris shown below:

Therefore,

To find p-value in excel, enter value of p andas shown below:

Now, calculate z-statistics in cell B4 by inserting formula.

The screenshot is shown below:

Finally, calculate p-value as shown below:

Hence, p-value is.

Decision: Do not reject the null hypothesis.

Reason for the decision: Reject because. But here, p-value is greater than alpha level 0.05. That is,

Conclusion: It is given that the p-value is 0.196. Therefore, for 5% of established alpha level it can say that do not reject the null hypothesis because p-value is greater than alpha level 0.05. Hence, there is no sufficient evidence to insure that the proportion of people suffering from depression in the town is less than 9.5% of American adults.