Question 12
We wish to estimate what percent of adult residents in a certain county are parents. Out of 300 adult residents sampled, 234 had kids. Based on this, construct a $95 \%$ confidence interval for the proportion $\pi$ of adult residents who are parents in this county.
Give your answers as decimals, to three places.
\[
< \pi<
\]
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Step 1 ：The question is asking for a 95% confidence interval for the proportion of adult residents who are parents in a certain county. The sample size is 300 and the number of successes (adults who are parents) is 234.

Step 2 ：The formula for a confidence interval for a proportion is given by: \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\hat{p}\) is the sample proportion, Z is the Z-score corresponding to the desired level of confidence, and n is the sample size.

Step 3 ：In this case, \(\hat{p} = \frac{234}{300}\), Z = 1.96 (for a 95% confidence interval), and n = 300.

Step 4 ：We can plug these values into the formula to find the confidence interval.

Step 5 ：Calculate the standard error (SE) using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) which gives SE = 0.0239.

Step 6 ：Calculate the lower and upper bounds of the confidence interval using the formula \(\hat{p} \pm Z \times SE\) which gives CI_lower = 0.733 and CI_upper = 0.827.

Step 7 ：The 95% confidence interval for the proportion of adult residents who are parents in this county is \(\boxed{[0.733, 0.827]}\).