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The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.8% 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 9.8% 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.)H0: H1:  b. Based on the hypotheses, find the following:c. Test Statistic =  d. p-value =    e. Based on the above we choose to f. The correct summary would be: 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 9.8%.

Question

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.8% 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 9.8% 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.)H0: H1:  b. Based on the hypotheses, find the following:c. Test Statistic =  d. p-value =    e. Based on the above we choose to f. The correct summary would be: 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 9.8%.

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Solution

a. The correct hypotheses are:

H0: p = 0.098 (Null hypothesis: The proportion of people in the city suffering from depression or a depressive illness is equal to the proportion in the general adult American population.)

H1: p > 0.098 (Alternative hypothesis: The proportion of people in the city suffering from depression or a depressive illness is greater than the proportion in the general adult American population.)

b. To find the test statistic and p-value, we first need to calculate the standard error (SE). The formula for SE is sqrt[p(1-p)/n], where p is the proportion in the general population and n is the sample size.

SE = sqrt[0.098(1-0.098)/2000] = 0.0067

c. The test statistic (z) is calculated as (p_sample - p_population) / SE. Here, p_sample is the proportion in the sample, which is 10.4% or 0.104.

z = (0.104 - 0.098) / 0.0067 = 0.8955

d. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. We find this value using a z-table or a statistical software. For a z-value of 0.8955, the p-value is approximately 0.1854.

e. Since the p-value (0.1854) is greater than the significance level (0.05), we fail to reject the null hypothesis.

f. The correct summary would be: Based on the hypothesis test, we do not have sufficient evidence to conclude 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 9.8%.

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Similar Questions

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