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.

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 $C{I}_{lower}=0.34-1.645\ast 0.04737=0.2621$ and $C{I}_{upper}=0.34+1.645\ast 0.04737=0.4179$ 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 $\boxed{{\displaystyle [0.2621,0.4179]}}$.