Question

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.Select one:a. (0.003, 0.159)b. (0.021, 0.141)c. (-0.159, 0.004)d. (0.031, 0.131)e. Sample sizes aren't large enough to justify using z-procedures

Answer 1

99% confidence interval is:

(0.00278 < P1 - P2< 0.15921)

Explanation:

For calculating a confidence intervale for the difference between the proportions of workers in the two cities, we calculate the following:

`[(p_(1) - p_(2)) \pm z_(\alpha/2) \sqrt{(p_(1)(1-p_(1)))/(n_(1)) + (p_(2)(1-p_(2)))/(n_(2)) }`

Where
`p_(1)` : proportion sample of individuals who worked

at more than one job in the city one

`n_(1)`: Number of respondents in the city one

`p_(1)` : proportion sample of individuals who worked

at more than one job in the city two

`n_(1)`: Number of respondents in the city two

Then

α = 0.01 and α/2 = 0.005

and
`z_(\alpha/2) = 2.575`

`p_(1) = (112)/(384) = 0.2916`

`p_(2) = (91)/(432) = 0.2106`

`n_(1)= 384` and
`n_(2)= 432`

The confidence interval is:

`[(0.2916 - 0.2106) \pm 2.575 \sqrt{(0.2916(1-0.2916))/(384) + (0.2106(1-0.2106))/(432) }`

(0.00278 < P1 - P2< 0.15921)