Problem
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 2412 subjects randomly selected from an online group involved with ears. 1018 surveys were returned. Construct a $99 \%$ 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 $p$.
0.422
(Round to three decimal places as needed.)
b) Identify the value of the margin of error $E$.
\[
E=0.026
\]
(Round to three decimal places as needed.)
c) Construct the confidence interval.
$
Solution
Step 1 ：Given that the sample size (n) is 1018 and the sample proportion (p_hat) is 0.422.
Step 2 ：The z-score for a 99% confidence level (Z_{α/2}) is approximately 2.576.
Step 3 ：The margin of error (E) is calculated using the formula: \(E = Z_{α/2} * \sqrt{\frac{p_hat(1-p_hat)}{n}}\), which gives E = 0.026.
Step 4 ：The confidence interval is then calculated using the formula: \(CI = p_hat ± E\).
Step 5 ：Substituting the given values into the formula, we get the lower limit of the confidence interval as 0.396 and the upper limit as 0.448.
Step 6 ：\(\boxed{[0.396, 0.448]}\) is the 99% confidence interval for the proportion of returned surveys.
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Solution
Step 1 ：Given that the sample size (n) is 1018 and the sample proportion (p_hat) is 0.422.
Step 2 ：The z-score for a 99% confidence level (Z_{α/2}) is approximately 2.576.
Step 3 ：The margin of error (E) is calculated using the formula: \(E = Z_{α/2} * \sqrt{\frac{p_hat(1-p_hat)}{n}}\), which gives E = 0.026.
Step 4 ：The confidence interval is then calculated using the formula: \(CI = p_hat ± E\).
Step 5 ：Substituting the given values into the formula, we get the lower limit of the confidence interval as 0.396 and the upper limit as 0.448.
Step 6 ：\(\boxed{[0.396, 0.448]}\) is the 99% confidence interval for the proportion of returned surveys.
