We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 208 had kids. Based on this, construct a $95 \%$ confidence interval for the proportion p of adult residents who are parents in this county.
Express your answer in tri-inequality form. Give your answers as decimals, to three places.
$< p< $ Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.
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Step 1 ：We are given that out of 400 adult residents sampled, 208 had kids. We are asked to construct a 95% confidence interval for the proportion p of adult residents who are parents in this county.

Step 2 ：We start by calculating the sample proportion, denoted as \(\hat{p}\), which is the number of successes (in this case, the number of parents) divided by the sample size. So, \(\hat{p} = \frac{208}{400} = 0.52\).

Step 3 ：We know that for a 95% confidence interval, the Z-score is approximately 1.96.

Step 4 ：We use the formula for the confidence interval for a proportion, which is given by \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where Z is the Z-score and n is the sample size.

Step 5 ：Substituting the values into the formula, we get \(0.52 \pm 1.96 \sqrt{\frac{0.52(1-0.52)}{400}}\).

Step 6 ：Solving the above expression, we get the confidence interval as \(0.471 < p < 0.569\).

Step 7 ：The point estimate is \(\hat{p} = 0.52\) and the margin of error is \(1.96 \sqrt{\frac{0.52(1-0.52)}{400}} = 0.049\).

Step 8 ：Thus, the 95% confidence interval for the proportion p of adult residents who are parents in this county is \(\boxed{0.471 < p < 0.569}\). The point estimate is \(\boxed{0.520}\) and the margin of error is \(\boxed{0.049}\).