FInd upper and lower bound of 2 samples

Question

The sample of 20 female servers worked an average of 25 hours per week, with a standard deviation of 3. A sample of 26 male servers worked an average of 35 hours per week, with a standard deviation of 2. Construct a 99% confidence interval for the mean difference in hours worked for male and female servers. You may assume that the populations are normal.

What is the Lower Bound and Upper Bound?

Answer 1

CI = mean male - mean female +/- z-critical value for 99% CI * SE(mean male - mean female)

mean male - mean female = (35-25) = 10

z-critical value for 99% CI is z-score for 0.995 probability = 2.58

SE = sqrt (s male **2/n1 + s female**2 /n2) where s male = 2 and s female = 3 and n1 = 26 and n2 = 20