91Ó°ÊÓ

Out of 5000 e-mails distributed, 717 surveys were returned.

The null hypothesis is written as follows:

The proportion of surveys that were returned equals15%.

$H_0:p=0.15$

The alternative hypothesis is written as follows:

The proportion of surveys that were returned is less than 15%.

$H_1:p<0.15$

The test is left-tailed.

The sample size is n=5000.

The sample proportion of surveys that were returnedis as follows:

$$\begin{gathered} \hat{p}=\frac{717}{5000} \\ =0.143 \end{gathered}$$

The population proportion of surveys that were returned is 0.15.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.143-0.15}{\sqrt{\frac{0.15\left(1-0.15\right)}{5000}}} \\ =-1.307 \end{gathered}$$

Thus, z=-1.307.

Referring to the standard normal table, the critical value of z at $\mathrm{Î\pm }=0.01$ for a left-tailed test is -2.3263.

Referring to the standard normal table, the p-value for the test statistic value of-1.307 equals0.0956.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to support the claim that the proportion of surveys that were returned is less than 0.15.