In a sample of 390 adults, 355 had children. Construct a $90 \%$ confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places \[ <\mathrm{p}< \] Question Help: Message instructor Submit Question

Step 1 ：Given a sample size \(n = 390\) and the number of successes \(x = 355\), where success is defined as an adult having children.

Step 2 ：Calculate the sample proportion \(\hat{p} = \frac{x}{n} = \frac{355}{390} = 0.910\).

Step 3 ：Use the Z-score for a 90% confidence level, which is \(Z = 1.645\).

Step 4 ：Calculate the standard error using the formula \(SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.910(1 - 0.910)}{390}} = 0.014\).

Step 5 ：Calculate the lower and upper bounds of the confidence interval using the formulas \(\hat{p} - Z \cdot SE\) and \(\hat{p} + Z \cdot SE\) respectively. This gives us \(0.910 - 1.645 \cdot 0.014 = 0.886\) and \(0.910 + 1.645 \cdot 0.014 = 0.934\).

Step 6 ：The 90% confidence interval for the true population proportion of adults with children is \(\boxed{0.886}\) to \(\boxed{0.934}\).