91影视

In a sample of emails of surveys sent to 5000 subjects, 717 surveys were returned.

(a)

The best point estimate of the proportion of surveys that were returned is computed below:

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

Thus, the sample proportion of surveys that were returned and equal to 0.143 is the best point estimate of the proportion of surveys that were returned.

(b)

The confidence level is equal to 90%. Thus, the corresponding level of significance is equal to 0.10.

From the standard normal distribution table, the right-tailed value of $z_{\frac{伪}{2}}$for is equal to 1.645.

The margin of error is calculated below:

$$\begin{gathered} E=1.645脳\sqrt{\frac{0.143脳0.857}{5000}} \\ =0.0082 \end{gathered}$$

Thus, the margin of error is equal to 0.0082.

(c)

The formula for computing the confidence interval estimate of the population proportion is written below:

$CI=\left(\hat{p}-E,\hat{p}+E\right)$

The 90% confidence interval becomes equal to:

$$\begin{gathered} CI=\left(0.143-0.0082,0.143+0.0082\right) \\ =\left(0.135,0.152\right) \end{gathered}$$

Therefore, the 90% confidence interval estimate of the population proportion of surveys that were returned is equal to (0.135, 0.152).

(d)

There is 90% confidence that the true proportion of surveys that were returned will lie between the values 0.135 and 0.152.