Out of a sample of 600 adults aged 18 to 30,210 still lived with their parents. Based on this, construct a $90 \%$ confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents. Give your answers rounded to 4 decimal places. $<$ Select an answer $\vee<$ Submit Question

Step 1 ：The problem is asking for a 90% confidence interval for the population proportion. The formula for a confidence interval for a population 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 2 ：In this case, \(\hat{p} = \frac{210}{600}\), \(Z\) is the Z-score for a 90% confidence interval, and \(n = 600\).

Step 3 ：We can use a standard normal distribution table or a Z-score calculator to find the Z-score for a 90% confidence interval. The Z-score for a 90% confidence interval is approximately 1.645.

Step 4 ：Let's plug these values into the formula and calculate the confidence interval. \[p_{hat} = 0.35\] \[Z = 1.645\] \[n = 600\] \[margin_{of_{error}} = 0.03203177296321055\] \[lower_{bound} = 0.318\] \[upper_{bound} = 0.382\]

Step 5 ：Final Answer: The 90% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents is \(\boxed{(0.318, 0.382)}\).