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.

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.

(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

Answer 1

To find the 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job, we can use the formula for the confidence interval for the difference between two proportions:

CI = (p1 - p2) +/- Z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))

Where:
p1 and p2 are the proportions of workers who work at more than one job in each city,
n1 and n2 are the sample sizes of each city,
Z is the critical value for a 99% confidence level.

First, calculate the proportions of workers who work at more than one job in each city:

p1 = 112 / 384 = 0.2917 (rounded to four decimal places)
p2 = 91 / 432 = 0.2106 (rounded to four decimal places)

Next, we need to find the critical value (Z) for a 99% confidence level. Looking up in the standard normal distribution table, the critical value for a 99% confidence level is approximately 2.576.

Now we can plug in the values into the formula:

CI = (0.2917 - 0.2106) +/- 2.576 * sqrt((0.2917 * (1 - 0.2917) / 384) + (0.2106 * (1 - 0.2106) / 432))

Calculating this expression, we get:

CI = 0.0811 +/- 2.576 * sqrt(0.000201 + 0.000181)

Simplifying:

CI = 0.0811 +/- 2.576 * sqrt(0.000382)

CI = 0.0811 +/- 2.576 * 0.019546

Calculating the bounds of the confidence interval:

Lower bound = 0.0811 - (2.576 * 0.019546) = 0.031055976

Upper bound = 0.0811 + (2.576 * 0.019546) = 0.131144024

Rounded to three decimal places:

Lower bound ≈ 0.031
Upper bound ≈ 0.131

Thus, the 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job is approximately (0.031, 0.131). Therefore, the correct answer is:
(0.031, 0.131)