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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.

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.

In the problem, the null hypothesis is:

$H_o:p鈮�0.095$

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:

The test is a left-tailed test.

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 $p_x$.

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

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

$x=7$

The required value is$x=7$

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

Therefore, it follows the sample size is:

$n=100$

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:

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

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

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

$蟽=\sqrt{n脳p脳(1-p)}$

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

$蟽=\sqrt{100脳9.5\%脳(1-9.5\%)}$

Find the value of $蟽$:

$蟽鈮�2.93$

Normal distribution can be used if $np鈮�5$and $nq鈮�5$.

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

$100脳0.095=9.5鈮�5$

$100脳0.905=90.5鈮�5$

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

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

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

Solve for $z$:

$z鈮�-0.7$

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

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.

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) $伪$, reject the null hypothesis in favor of the alternative hypothesis. If the $P$-value is greater than $伪$, 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 $伪=0.05$, there is no sufficient evidence to reject the null hypothesis.

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

$0.1977>0.05$

Since the $p$- value is greater than $伪$, 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.