A college claims that the proportion, $p$, of students who commute more than fifteen miles to school is less than $15 \%$. A researcher wants to test this. A random sample of 265 students at this college is selected, and it is found that 23 commute more than fifteen miles to school. Is there enough evidence to support the college's claim at the 0.05 level of significance?
Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)
(a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$.
\[
\begin{array}{l}
H_{0}: \square \\
H_{1}: \square
\end{array}
\]
(b) Determine the type of test statistic to use.
(Choose one) $\boldsymbol{\nabla}$
(c) Find the value of the test statistic. (Round to three or more decimal places.)
(d) Find the critical value. (Round to three or more decimal places.)
(e) Is there enough evidence to support the claim that the proportion of students who commute more than fifteen miles to school is less than $15 \%$ ?
Yes
No
\begin{tabular}{ccc}
$\mu$ & $\sigma$ & $p$ \\
$\bar{x}$ & $s$ & $\hat{p}$ \\
$\square^{\square}$ & $\square \square$ & $\frac{\square}{\square}$ \\
$\square=\square$ & $\square \leq \square$ & $\square> \square$ \\
$\square \neq \square$ & $\square< \square$ & $\square> \square$ \\
$\times$ & & $S$
\end{tabular}
Answer
Conclude that there is enough evidence to support the claim that the proportion of students who commute more than fifteen miles to school is less than 15%, \(\boxed{H_{1}: p < 0.15}\)
Step 1 ：State the null hypothesis $H_{0}: p = 0.15$ and the alternative hypothesis $H_{1}: p < 0.15$
Step 2 ：Identify that a z-test for proportions will be used
Step 3 ：Calculate the sample proportion $\hat{p} = \frac{23}{265} = 0.087$
Step 4 ：Calculate the test statistic using the formula $z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$, which gives $z = \frac{0.087 - 0.15}{\sqrt{\frac{0.15(1-0.15)}{265}}} = -2.875$
Step 5 ：Identify the critical value for a one-tailed test at the 0.05 level of significance as -1.645
Step 6 ：Since the test statistic $-2.875$ is less than the critical value $-1.645$, reject the null hypothesis
Step 7 ：Conclude that there is enough evidence to support the claim that the proportion of students who commute more than fifteen miles to school is less than 15%, \(\boxed{H_{1}: p < 0.15}\)
