A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Calculate a 95% confidence interval of the population mean number of days it takes to find a job.
A) [40.4000, 79.6000]
B) [55.9085, 64.0915]
C) [56.4216, 63.5784]
D) [56.9966, 63.0034]

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for a population mean:

CI = x̄ ± z*(σ/√n)

where:

x̄ = sample mean = 60 days
σ = population standard deviation = 10 days
n = sample size = 30
z = z-score for the desired confidence level. For a 95% confidence level, z = 1.96.

Plugging in the values, we get:

CI = 60 ± 1.96*(10/√30)
CI = 60 ± 3.5784
CI = [56.4216, 63.5784]

So the best choice is C, which matches our calculated interval.

A Website Advertises Job Openings on Its Website, but Job