Constructing and Interpreting Confidence Intervals. In Exercises 13–16, use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion p; (b) identify the value of the margin of error E; (c) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval.

Survey Return Rate In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys.

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 $z_{\frac{\alpha }{2}}$for is equal to 1.645.

The margin of error is calculated below:

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

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.