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) [56.4215,63.5785]
D) [56.9966,63.0034]

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for the population mean:
CI = x̄ ± z*(σ/√n)
where x̄ = sample mean, σ = population standard deviation, n = sample size, z = z-score for the desired confidence level.

We are given that x̄ = 60, σ = 10, and a 95% confidence level corresponds to a z-score of 1.96 (from a standard normal distribution table). We don't know the sample size, but we can assume that it is large enough (n ≥ 30) for the central limit theorem to apply.

Plugging in the values, we get:
CI = 60 ± 1.96*(10/√n)

To find the interval, we need to solve for n:
1.96*(10/√n) = 1
10/√n = 0.5102
n = (10/0.5102)^2 = 388.89 (rounded up to 389)

Now we can substitute n back into the formula:
CI = 60 ± 1.96*(10/√389) ≈ 60 ± 1.57
CI ≈ [58.43, 61.57]

Rounding to two decimal places, the answer is C) [56.4215,63.5785].

A Website Advertises Job Openings on Its Website,but Job Seekers