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.

a) What are the correct hypotheses?

Ho: p>0.093 H: p <0.093 Ho: p=0.093 H1 : p>0.093 Но : .:: O*: 07: Oui Oui p=0.111 H: p>0.111 Ho: u= 0.093 p>0.093 Ho: p<0.093 H1 : p20.093

b) The p-value is 0.0028. Choose the correct decision and summary.

Answer 1

, n=2000, a = 0.05 p=0.093 â= 0.) Te Hypothesis Hoi p=0.093 V Hii p>0.093 That statistic: PP 0.))) - 0.093 = 2.792 POP V 0.093 (1-0-093) 2000 = 2. 772 the For finding the p value is excel, use command = NORMIDIS (- 2.772) = 0.0028 Pvalue = 0.0028 Decision Ewe J p value 2 p value sa > & reject Ho Fail to reject Ho b) Reject Ho, there jg enough evidence to support the claim that the peopoction of people is that city suffeng from depression de a depressive iploegg is more than the 9.3%

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