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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 -

    Let's look at your data:
    n1 = 384
    n2 = 432
    p1 = 112/384
    p2 = 91/432

    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 -

    The answer is A, Math guru know what he be doin!

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