Statistics
posted by Huh .
A random sample of 384 people in a midsized city (city one) revealed 112 individuals who worked at more than one job. A second random sample of 432 workers from another midsized city (city two) found 91 people who work at more than one job. Find a 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job.
(0.003, 0.159)
(0.021, 0.141)
(0.159, 0.004)
(0.031, 0.131)
Sample sizes aren't large enough to justify using zprocedures

Let's look at your data:
n1 = 384
n2 = 432
p1 = 112/384
p2 = 91/432
Formula:
CI99 = (p1  p2) ± 2.58 √(p1(1p1)/n1 + p2(1p2)/n2)
Substitute the values into the formula and calculate. (Convert all fractions to decimals.)
You should be able to select your answer once you have determined the interval. 
The answer is A, Math guru know what he be doin!