Submit Question Question 7 $0 / 2$ pts $100 \rightleftarrows 2$ Details We wish to estimate what percent of adult residents in a certain county are parents. Out of 300 adult residents sampled, 183 had kids. Based on this, construct a $99 \%$ confidence interval for the proportion of adult residents who are parents in this county. Give your answers as decimals to three places. Submit Question Question 8 $4 / 4$ pts $\bigcirc 99 \rightleftarrows 2$ Details Score on last try: 4 of 4 pts. See Details for more.

Step 1 ：The question is asking for a 99% confidence interval for the proportion of adult residents who are parents in a certain county. The sample size is 300 and 183 of them are parents.

Step 2 ：To calculate a confidence interval for a proportion, we can use the formula: \(CI = \hat{p} \pm Z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score corresponding to the desired confidence level (for a 99% confidence level, \(Z\) is approximately 2.576), and \(n\) is the sample size.

Step 3 ：First, we need to calculate the sample proportion \(\hat{p}\), which is the number of successes (parents) divided by the sample size. Then we can substitute \(\hat{p}\), \(Z\), and \(n\) into the formula to get the confidence interval.

Step 4 ：Given: \(n = 300\), \(x = 183\), \(Z = 2.576\), \(\hat{p} = 0.61\), \(se = 0.028160255680657446\), \(ci_{lower} = 0.5374591813666264\), \(ci_{upper} = 0.6825408186333736\)

Step 5 ：The 99% confidence interval for the proportion of adult residents who are parents in this county is approximately \([0.537, 0.683]\). This means we are 99% confident that the true proportion of adult residents who are parents in this county is between 53.7% and 68.3%.

Step 6 ：So, the final answer is \(\boxed{[0.537, 0.683]}\).