In a simple random sample of 1600 people age 20 and over in a certain country the proportion with a certain disease was found to be 0.135 (or $13.5 \%$ ). Complete parts (a) through (d) below.
a. What is the standard error of the estimate of the proportion of all people in the country age 20 and over with the disease?
\[
S E_{\text {est }}=0085
\]
(Round to four decimal places as needed.)
b. Find the margin of error, using a $95 \%$ confidence level, for estimating this proportion.
\[
m=
\]
(Round to three decimal places as needed)

Step 1 ：Given that the proportion (p) is 0.135 and the sample size (n) is 1600, we can calculate the standard error of the estimate of the proportion using the formula: \(SE = \sqrt{ \frac{p(1-p)}{n} }\).

Step 2 ：Substituting the given values into the formula, we get \(SE = \sqrt{ \frac{0.135(1-0.135)}{1600} }\) which simplifies to approximately 0.0085.

Step 3 ：Next, we can calculate the margin of error using the formula: \(m = Z * SE\), where Z is the Z-score corresponding to the desired confidence level. For a 95% confidence level, Z is approximately 1.96.

Step 4 ：Substituting the values into the formula, we get \(m = 1.96 * 0.0085\) which simplifies to approximately 0.017.

Step 5 ：\(\boxed{\text{The standard error of the estimate of the proportion of all people in the country age 20 and over with the disease is approximately 0.0085.}}\)

Step 6 ：\(\boxed{\text{The margin of error, using a 95% confidence level, for estimating this proportion is approximately 0.017.}}\)