Use the sample data and confidence level given below to complete parts (a) through (d). A drug is used to help prevent blood clots in certain patients. In clinical trials, among 4399 patients treated with the drug, 131 developed the adverse reaction of nausea. Construct a $99 \%$ confidence interval for the proportion of adverse reactions. a) Find the best point estimate of the population proportion $p$. 0.03 (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 patients treated with the drug is 4399 and the number of patients who developed nausea is 131.

Step 2 ：The best point estimate of the population proportion \(p\) is the sample proportion, which is the number of successes (patients who developed nausea) divided by the total number of trials (total patients treated with the drug). This can be calculated as \(\frac{131}{4399} = 0.029779495339849967\).

Step 3 ：Rounding to three decimal places, the best point estimate of the population proportion \(p\) is \(\boxed{0.030}\).

Step 4 ：The margin of error \(E\) is the amount that is subtracted from and added to the point estimate to construct the confidence interval. It is calculated using the formula for the standard error of a proportion, multiplied by the z-score corresponding to the desired level of confidence.

Step 5 ：Given that the confidence level is 0.99, the corresponding z-score is 2.5758293035489004.

Step 6 ：Substituting the values into the formula, we get \(E = 2.5758293035489004 * \sqrt{\frac{0.029779495339849967 * (1 - 0.029779495339849967)}{4399}} = 0.006601368702287208\).

Step 7 ：Rounding to three decimal places, the margin of error \(E\) is \(\boxed{0.007}\).