Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

Answer to Problem 14E

Null hypothesis:

$H_0:p=0.15$

That is, $15\%$ of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

$H_1:p<0.15$

That is less than $15\%$ of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random $1325$ adults, $149$ adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than $15\%$ of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than $15\%$ of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

$H_0:p=0.15$

That is, $15\%$ of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

$H_1:p<0.15$

That is less than $15\%$ of U.S. adults have a great deal of confidence in banks.

(b)

To determine

The test statistic $z$.

Answer to Problem 14E

$z=-3.83$.

Explanation of Solution

Given information:

In a survey of random $1325$ adults, $149$ adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than $15\%$ of US adults have confidence in banks.

Formula used:

$\begin{array}{l}\widehat{p}=\frac{x}{n}\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\end{array}$

Calculation:

In a survey of random $1325$ adults, $149$ adults have a great deal of confidence in banks and other financial institution. i.e. $n=1,325\text{ and }x=149$

Based on the survey an economist claims that less than $15\%$ of US adults have confidence in banks. i.e. $p_0=0.15$.

So the sample proportion $\widehat{p}$.

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{149}{1,325}\\ =0.1125\end{array}$

Compute the test statistic $z$ by using the formula as given below.

$\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.1125-0.15}{\sqrt{\frac{0.15\left(1-0.15\right)}{1,325}}}\\ =\frac{-0.0375}{0.0098}\\ =-3.83\end{array}$

Thus, the value of test statistic is $z=-3.83$.

(c)

To determine

Whether the claim made by economist is true for $\alpha =0.05$.

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random $1325$ adults, $149$ adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than $15\%$ of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, $\alpha =0.05$

From Table $9.1$, the critical value for the level of significance $\alpha =0.05\text{ is }-1.645$.

Rejection rule for left-tailed test:

If $z\le -z_{\alpha }\left(=-1.645\right)$ then reject the null hypothesis

Conclusion for $\alpha =0.05$

The value of test statistic is $z=-3.83$ and the critical value is $z_{\alpha }=-1.645$.

Here, the value of test statistic is lesser than the critical value.

That is $z\left(=-3.83\right)\le z_{\alpha }\left(=-1.645\right)$

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than $15\%$ of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

(d)

To determine

Whether the claim made by economist is true for $\alpha =0.01$.

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random $1325$ adults, $149$ adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than $15\%$ of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, $\alpha =0.01$

From Table $9.1$, the critical value for the level of significance $\alpha =0.01\text{ is }-2.326$.

Rejection rule for left-tailed test:

If $z\le -z_{\alpha }\left(=-2.326\right)$ then reject the null hypothesis

Conclusion for $\alpha =0.01$

The value of test statistic is $-3.83$ and the critical value is $z_{\alpha }=-2.326$.

Here, the value of test statistic is lesser than the critical value.

That is $z\left(=-3.83\right)\le z_{\alpha }\left(=-2.326\right)$

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than $15\%$ of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.