The National Institute of Mental Health published an article stating that In any

Question

The National Institute of Mental Health published an article stating that In any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive Illness. Suppose that in a survey of 100 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 percent In the general adult American population. Is this a test of one mean or proportion? State the null and alternative hypotheses. Is this a right-tailed, left-tailed, or two-tailed test? What symbol represents the random variable for this test? In words, define the random variable for this test. Calculate the following: Calculate sigma x = Show the formula set-up. State the distribution to use for the hypothesis test. Find the p-value. At a pre-conceived alpha = 0.05, what is your Decision: Reason for the decision Conclusion (write out in a complete sentence):

Explanation / Answer

3. a. The variable in question is not an interval-ratio measurement. Therefore, sample proportion is suitable than sample mean.

b. H0: Pu=0.095 (sample proportion of American suffering from depression or depressive illness is same as that of population proportion).

H1: Pu < 0.095 (sample proportion of American suffering from depression or depressive illness is lower than that of population proportion).

c. Since alternative hypothesis tests whether true proportion of people suffering from depression or depressive illness is lower than that of American adults in general, the test is one-tailed (left) one.

d. The symbol N denotes the random variable, people in a certain town for this test.

e. 100 people in a certain town was surveyed and the number of people suffering from depression or depressive illness was noted.

g. Sqrt Pu(1-PU)/N= Sqrt [0.07(1-0.07)/100]=0.0255

f. x=7, N=100. Therefore, sample proportion p'=7/100=0.07.

h. The distribution to use for thi stest is 1-sample z distribution.

i. The z statistic is as follows:

z=(Ps-Pu)/Sqrt Pu(1-PU)/N=(0.095-0.07)/0.0255=0.980

The p value is 0.1635. [0.5-Area corresponding to mean and 0.980]=0.5-0.3365

j. The p value is not less than 0.05. Therefore, fail to reject null hypothesis.

There is not sufficient sample evidence to conclude than true proportion of adults suffering from depression or depressiveillness is lowwer than American population of adults in general.

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