Correct
Consider two independent random samples with the following results:
\[
\begin{array}{ll}
n_{1}=573 & n_{2}=604 \\
\hat{p}_{1}=0.63 & \hat{p}_{2}=0.49
\end{array}
\]

Use this data to find the $95 \%$ confidence interval for the true difference between the population proportions.

Step 2 of 3: Find the margin of error. Round your answer to six decimal places.

Answer
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Step 1 ：Consider two independent random samples with the following results: \(n_{1}=573, n_{2}=604, \hat{p}_{1}=0.63, \hat{p}_{2}=0.49\). We are asked to find the 95% confidence interval for the true difference between the population proportions.

Step 2 ：The margin of error for a confidence interval for the difference between two population proportions can be calculated using the formula: \(E = Z \sqrt{ \left(\frac{p1(1-p1)}{n1}\right) + \left(\frac{p2(1-p2)}{n2}\right) }\) where: E is the margin of error, Z is the Z-score (for a 95% confidence interval, the Z-score is approximately 1.96), p1 and p2 are the sample proportions, and n1 and n2 are the sample sizes.

Step 3 ：Substitute the given values into the formula to find the margin of error: \(n1=573, n2=604, p1=0.63, p2=0.49, Z=1.96\).

Step 4 ：The margin of error for the 95% confidence interval for the difference between the two population proportions is approximately 0.05614460984264031 when rounded to six decimal places.

Step 5 ：Final Answer: The margin of error for the 95% confidence interval for the difference between the two population proportions is approximately \(\boxed{0.056145}\).