QuestionFrom a random sample of 80 women, 48 say they have been sexually harassed. From another random sample of 65 men, 13 say they have been sexually harassed. Construct a $99 \%$ confidence interval for the difference between the proportions of women and men who say they have been sexually harassed.

Studdy Solution
Construct the confidence interval:
$(\hat{p}_w - \hat{p}_m) \pm ME$ $(0.6 - 0.2) \pm 0.2100$ $0.4 \pm 0.2100$
Lower bound: $0.4 - 0.2100 = 0.1900$ Upper bound: $0.4 + 0.2100 = 0.6100$
The 99% confidence interval for the difference between the proportions of women and men who say they have been sexually harassed is:
$$\boxed{(0.1900, 0.6100)}$$
This means we can be 99% confident that the true difference in proportions of women and men who have been sexually harassed is between 0.1900 and 0.6100.