Question

use the sample data and confidence level given below to complete parts (a) through (d).
in a study of cell phone use and brain hemispheric dominance, an internet survey was e - mailed to 2595 subjects randomly selected from an online group involved with ears. 931 surveys were returned. construct a 90% confidence interval for the proportion of returned surveys.
click the icon to view a table of z scores.
a) find the best point estimate of the population proportion p.
0.359
(round to three decimal places as needed.)
b) identify the value of the margin of error e.
e =
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for margin of error

The formula for the margin of error \(E\) for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\).
For a \(90\%\) confidence level, \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\). From the standard normal table, \(z_{\alpha/2}=z_{0.05} = 1.645\).
We know that \(\hat{p}=0.359\) (from part a) and \(n = 2595\).

Step2: Calculate \(\hat{p}(1-\hat{p})\)

\(\hat{p}(1 - \hat{p})=0.359\times(1 - 0.359)=0.359\times0.641 = 0.230119\)

Step3: Calculate \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)

\(\sqrt{\frac{0.230119}{2595}}=\sqrt{0.00008868}=0.009416\)

Step4: Calculate the margin of error \(E\)

\(E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.645\times0.009416\approx0.015\)

Answer:

\(E = 0.015\)