Posted by **Huh** on Sunday, November 4, 2012 at 12:03pm.

A random sample of 384 people in a mid-sized city (city one) revealed 112 individuals who worked at more than one job. A second random sample of 432 workers from another mid-sized 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 z-procedures

- Statistics -
**MathGuru**, Monday, November 5, 2012 at 7:46pm
Let's look at your data:

n1 = 384

n2 = 432

p1 = 112/384

p2 = 91/432

Formula:

CI99 = (p1 - p2) ± 2.58 √(p1(1-p1)/n1 + p2(1-p2)/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.

- Statistics -
**Arihan**, Monday, March 3, 2014 at 2:36pm
The answer is A, Math guru know what he be doin!

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