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 2310 subjects randomly selected from an online group involved with ears. 1150 surveys were returned. Construct a $95 \%$ 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 $p$.
0.498
(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.)

Step 1 ：Given that the total number of surveys sent is 2310 and the number of returned surveys is 1150.

Step 2 ：The best point estimate of the population proportion p is the sample proportion, which is the number of successes (returned surveys) divided by the number of trials (total surveys sent). This can be calculated as \(p = \frac{x}{n} = \frac{1150}{2310} = 0.498\).

Step 3 ：The margin of error E can be calculated using the formula for the confidence interval for a proportion, which is \(E = z \sqrt{\frac{p(1-p)}{n}}\), where z is the z-score corresponding to the desired confidence level, p is the sample proportion, and n is the sample size.

Step 4 ：Substituting the given values into the formula, we get \(E = 1.96 \sqrt{\frac{0.498(1-0.498)}{2310}} = 0.020\).

Step 5 ：\(\boxed{\text{Final Answer: }}\) a) The best point estimate of the population proportion p is \(\boxed{0.498}\). b) The margin of error E is \(\boxed{0.020}\).