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.

a. Is this a test of one mean or proportion?

b. State the null and alternative hypotheses.

$H_0:$

$H_a:$

c. Is this a right-tailed, left-tailed, or two-tailed test?

d. What symbol represents the random variable for this test?

e. In words, define the random variable for this test.

f. Calculate the following:

i.$x=$

ii.role="math" localid="1649760873126" $n=$

iii.$p^{\text{'}}=$

g. Calculate role="math" localid="1649760901479" ${\sigma }_{\mathrm{x}}=$Show the formula set-up.

h. State the distribution to use for the hypothesis test.

i. Find the $p$-value.

j. At a pre-conceived $\alpha =0.05$, what is your:

i. Decision:

ii. Reason for the decision:

iii. Conclusion (write out in a complete sentence):

Short answer

a. This is a test on one proportion.

b. The null and alternative hypotheses are:

$$\begin{gathered} H_0:p\ge 0.095 \\ {H}_a:p<0.095 \end{gathered}$$

c. The test is left tailed test.

d. The random variable is $p_x$, which represents the percentage of persons in town who are depressed or have a depressive disease.

e.The random variable for this test can be defined as the proportion of people in town suffering from depression or depressive illness.

f. i. The value is $x=7$.

ii. The Sample size $n=100$.

iii. The value of $p^{\text{'}}$is $0.905$.

g. The standard deviation is $2.93$. The formula used is $\sigma =\sqrt{n\times p\times (1-p)}$.

h. Normal Distribution Should be used.

i. The required $p$value is $0.24196$.

j. i. The decision is to accept the null value.

ii. The $P$-value $P=0.1977$is greater than the significance level $\alpha =0.05$which means that the test statistic $z=-0.85$is not on the rejection region.

iii. There is not sufficient evidence at the $0.05$level of significance to conclude that the proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Step-by-step solution

Given Information

Let $p=9.5\%$be the true proportion of adults that suffer from depression i any one-year period.

Let $\hat{p}=\frac{7}{100}=7\%$be the proportion of adults that suffer from depression according to the survey.

Explanation Part a

The null value is $H_0:p=9.5\%$

The alternative hypothesis is:

$H_1:p<9.5\%$

The test statistic is:

$z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}$

This is a one-tailed test comparing a sample proportion to an established a population proportion.

Explanation part b

In the problem, the null hypothesis is:

$H_o:p\ge 0.095$

The alternative hypothesis is a claim about the population that is contradictory to$H_O$and what we conclude when we reject $H_O$.

In the problem, the alternative hypothesis is that the proportion of the people in town suffering from depressive illness is less than $9.5\%$.

In the problem, the alternative hypothesis is:

Explanation  part c

The goal of the test is to see if the genuine rate of depression in that town is lower than the rate in the overall population.

Thetestisleft-tailedsincetheclaimcontainstheterm"lowerthan."

The test is a left-tailed test.

Explanation part d

Let $x$be the percent of people in that town suffering from depression or depressive illness.

Let $p$be the mean proportion of people in town suffering from depression or depressive illness.

The random variable is the proportion of people in town suffering from depression or depressive illness.

The random variable is $p_x$.

Explanation part e

The random variable for this test can be defined as the proportion of people in town suffering from depression or depressive illness.

Explanation  part f.i

i.The number of people $x$, who respond in the positive, or those who suffered from depression is:

$x=7$

The required value is$x=7$

Explanation  part f.ii

Observe that the number of people surveyed are$100$ .

Therefore, it follows the sample size is:

$n=100$

Explanation part f.iii

It is mentioned that $9.5\%$of American adults suffer from depression or a depressive illness:

$p=9.5\%$

Convert the proportion into decimal:

$p=0.095$

The value of $p^{\text{'}}$is the complement of the value of $p$:

$p^{\text{'}}=1-p$

Substitute $p=0.095$into the equation:

$p^{\text{'}}=1-0.095$

$p^{\text{'}}=0.905$

Explanation for g

The standard deviation for a proportion distribution with proportion $\boldsymbol{p}$ and sample size $\boldsymbol{n}$ is expressed:

$\sigma =\sqrt{n\times p\times (1-p)}$

Substitute $n=100$and $p=9.5\%$to the equation above:

$\sigma =\sqrt{100\times 9.5\%\times (1-9.5\%)}$

Find the value of $\sigma$:

$\sigma \approx 2.93$

Explanation for h

Normal distribution can be used if $np\ge 5$and $nq\ge 5$.

Substitute $n=100,p=0.095$and $q=0.905$into the expressions to check if the inequalities are true:

$100\times 0.095=9.5\ge 5$

$100\times 0.905=90.5\ge 5$

Since both inequalities are true, normal distribution can be used for the hypothesis test.

Explanation for part i

Substitute $p=0.095,q=0.905,\hat{p}=0.07$and in the test statistic for population proportion:

$z=\frac{0.07-0.0905}{\sqrt{\frac{0.095·0.905}{100}}}$

Solve for $z$:

$z\approx -0.7$

In the z-table the p-value that corresponds to $-0.7$is $0.24196$.

Explanation for part j.i

The critical value with a $95\%$confidence level and the value of test statistic.

The value of test statistic falls in the acceptance region.

$-1.64<-0.896$

The decision is to accept the null hypothesis.

Explanation  for j.ii

The hypothesis test for the proportion reveals that the $P$-value is $P=0.1977$.

If the $P$-value is less than (or equal to) $\alpha$, reject the null hypothesis in favor of the alternative hypothesis. If the $P$-value is greater than $\alpha$, do not reject the null hypothesis.

Do not reject the null hypothesis.

Since the $P$-value $P=0.1977$is greater than the significance level $\alpha =0.05$, there is no sufficient evidence to reject the null hypothesis.

Explanation for j.iii

The significance value for this test is $\alpha =0.05$and the $p$- value $=0.1977$

$0.1977>0.05$

Since the $p$- value is greater than $\alpha$, the value of the test statistic is not in the rejection region so don't reject the null hypothesis.

There is not sufficient evidence at the $0.05$level of significance to conclude that the proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.