\begin{tabular}{l} Working at home: According to the U.S. Census Bureau, \( 34 \% \) of men who worked at home were college graduates. In a sample of 500 women who worked at \\ home, 170 were college graduates. \\ Part: \( 0 / 3 \) \\ Part 1 of 3 \\ (a) Find a point estimate for the proportion of college graduates among women who work at home. Round the answer to at least three decimal places. \\ The point estimate for the proportion of college graduates among women who work at home is 0.340 , \\ Part: \( 1 / 3 \) \\ Part 2 of 3 \\ (b) Construct a \( 90 \% \) confidence interval for the proportion of women who work at home who are college graduates. Round the answer to at least three \\ decimal places. \\ A \( 90 \% \) conficence interval for the proportion of women who work at home who are college graduates is \( \square p<\square \) \\ \hline\end{tabular}

To find a \( 90\% \) confidence interval for the proportion of college graduates among women who work at home, we start by calculating the point estimate, which in this case is \( \hat{p} = \frac{170}{500} = 0.340 \). Next, we need to calculate the standard error (SE) of the proportion: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.340(1 - 0.340)}{500}} \approx 0.0223 \] For a \( 90\% \) confidence level, the z-score is approximately \( 1.645 \). Now we can calculate the margin of error (ME): \[ ME = z \cdot SE \approx 1.645 \cdot 0.0223 \approx 0.0367 \] Finally, we construct the confidence interval: \[ CI = \hat{p} \pm ME = 0.340 \pm 0.0367 \] This gives us: \[ \text{Lower limit} = 0.340 - 0.0367 \approx 0.3033 \] \[ \text{Upper limit} = 0.340 + 0.0367 \approx 0.3767 \] Thus, rounding to three decimal places, we have: \[ 0.303 < p < 0.377 \] So, a \( 90\% \) confidence interval for the proportion of women who work at home who are college graduates is \( 0.303 < p < 0.377 \).