Question 14 5 pts 01 Details Out of a sample of 100 adults aged 18 to 30,34 still lived with their parents. Based on this, construct a 905 confidence interval for the true population proportion of adults ages 18 to 30 that stilt live with their parents. Give your answers rounded to 4 decimal places.
Solution
Step 1 ：Given a sample of 100 adults aged 18 to 30, where 34 still lived with their parents, we are asked to construct a 90% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents.
Step 2 ：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 confidence level, and n is the sample size.
Step 3 ：In this case, \(\hat{p} = \frac{34}{100} = 0.34\), n = 100, and the Z-score for a 90% confidence interval is approximately 1.645.
Step 4 ：Substituting these values into the formula, we get the standard error (SE) as \(SE = \sqrt{\frac{0.34(1-0.34)}{100}} = 0.04737\).
Step 5 ：Then, we calculate the lower and upper bounds of the confidence interval as \(CI_{lower} = 0.34 - 1.645*0.04737 = 0.2621\) and \(CI_{upper} = 0.34 + 1.645*0.04737 = 0.4179\) respectively.
Step 6 ：Final Answer: The 90% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents is \(\boxed{[0.2621, 0.4179]}\).
