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 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.

Answer 1

The 90% confidence interval would be given by (57.006;62.994)

Explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

`\bar X=60` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=10` represent the population standard deviation

n=30 represent the sample size

Assuming the X follows a normal distribution

`X \sim N(\mu, \sigma=10)`

The sample mean
`\bar X` is distributed on this way:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

The confidence interval on this case is given by:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

The next step would be find the value of
`\z_(\alpha/2)`,
`\alpha=1-0.90=0.1`,
`\alpha/2 =0.05` and
`z_\alpha/2=1.64`

Using the normal standard table, excel or a calculator we see that:

`z_(\alpha/2)=1.64`

Since we have all the values we can replace:

`60 - 1.64(10)/(√(30))=57.006`

`60 + 1.64(10)/(√(30))=62.994`

So on this case the 90% confidence interval would be given by (57.006;62.994)