Question

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 31 customers an average of 40 days to find a job. Assume the population standard deviation is 10 days. Calculate a 99% confidence interval of the population mean number of days it takes to find a job.

Answer 1

Using the z-distribution, the 99% confidence interval of the population mean number of days it takes to find a job is (35.38, 44.62).

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

In this problem, we have a 99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so the critical value is z = 2.575.

The other parameters are:

$\overline{x} = 40, \sigma = 10, n = 31$

Hence the bounds of the interval are:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 40 - 2.575\frac{10}{\sqrt{31}} = 35.38$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 40 + 2.575\frac{10}{\sqrt{31}} = 44.62$

More can be learned about the z-distribution at brainly.com/question/25890103

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