To construct a 99% confidence interval for the proportion of college graduates among women who work at home, we first need to calculate the sample proportion \( \hat{p} \):
\[ \hat{p} = \frac{x}{n} \] where:
- \( x \) is the number of college graduates (162),
- \( n \) is the total sample size (510).
\[ \hat{p} = \frac{162}{510} \approx 0.318 \]
Next, we calculate the standard error (SE) for the sample proportion:
\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.318(1 - 0.318)}{510}} \]
Calculating \( 1 - \hat{p} \):
\[ 1 - \hat{p} = 1 - 0.318 = 0.682 \]
Now calculating the numerator:
\[ \hat{p}(1 - \hat{p}) = 0.318 \times 0.682 \approx 0.2176 \]
Now the standard error:
\[ SE = \sqrt{\frac{0.2176}{510}} \approx \sqrt{0.000426} \approx 0.02066 \]
Now, for a 99% confidence interval, we need the z-value for 99% confidence. The z-value that corresponds to 99% confidence is approximately 2.576 (for a two-tailed test).
Now we can calculate the margin of error (ME):
\[ ME = z \cdot SE \approx 2.576 \cdot 0.02066 \approx 0.0533 \]
Now, we can determine the confidence interval:
\[ (\hat{p} - ME, \hat{p} + ME) = (0.318 - 0.0533, 0.318 + 0.0533) \approx (0.2647, 0.3713) \]
Thus, rounding to three decimal places, the 99% confidence interval for the proportion of college graduates among women who work at home is:
\[ (0.265, 0.371) \]