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the national institute of mental health published an article stating th…

Question

the national institute of mental health published an article stating that in any one - year period, approximately 9.1% of american adults suffer from depression or a depressive illness. suppose that in a survey of 2000 people in a certain city, 10.4% 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 9.1% in the general adult american population. test the relevant hypotheses using a 5% level of significance. give answer to at least 4 decimal places.
a. what are the correct hypotheses? (select the correct symbols and use decimal values not percentages.)

b. based on the hypotheses, find the following:

c. test statistic =

d. p - value =

e. based on the above we choose to select an answer

f. the correct summary would be: select an answer that the true proportion of people in that city suffering from depression or a depressive illness is more than the percent in the general adult american population of 9.1%.

question help: video message instructor

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0\) is that the proportion \(p\) is equal to the general - population proportion. The alternative hypothesis \(H_1\) is that the proportion \(p\) is greater than the general - population proportion.
\(H_0:p = 0.091\)
\(H_1:p>0.091\)

Step2: Calculate the sample proportion \(\hat{p}\)

The sample size \(n = 2000\), and the sample proportion \(\hat{p}=0.104\)

Step3: Calculate the test statistic \(z\)

The formula for the test statistic in a one - sample proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(p = 0.091\), \(\hat{p}=0.104\), and \(n = 2000\) into the formula:

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Step4: Calculate the p - value

Since this is a right - tailed test (\(H_1:p>0.091\)), the p - value is \(P(Z>z)\) where \(Z\sim N(0,1)\) and \(z = 2.0218\)
\(p - value=P(Z>2.0218)=1 - P(Z\leq2.0218)\)
Using a standard normal table or a calculator, \(P(Z\leq2.0218)\approx0.9784\)
\(p - value=1 - 0.9784=0.0216\)

Step5: Make a decision

Since the p - value (\(0.0216\)) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis \(H_0\)

Answer:

a. \(H_0:p = 0.091\); \(H_1:p>0.091\)
b.
c. Test Statistic \(z\approx2.0218\)
d. \(p - value\approx0.0216\)
e. Reject \(H_0\)
f. There is sufficient evidence