Question

A company seeks to employ a new public relations manager. The hiring committee surveys 13 public relations managers and finds the average salary is $95082.787 with standard deviation of $1857.716. What is the 90% confidence interval for the true average public relations manager salary?

Answer 1

`95082.787-1.782(1857.716)/(√(13))=94164.485`

`95082.787+1.782(1857.716)/(√(13))=96001.089`

So on this case the 90% confidence interval would be given by (94164.485;96001.089)

Step-by-step explanation:

Previous concepts

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".

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".

`\bar X=95082.787` represent the sample mean

`\mu` population mean (variable of interest)

s=1857.716 represent the sample standard deviation

n=13 represent the sample size

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=13-1=12`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,12)".And we see that
`t_(\alpha/2)=1.782`

Now we have everything in order to replace into formula (1):

`95082.787-1.782(1857.716)/(√(13))=94164.485`

`95082.787+1.782(1857.716)/(√(13))=96001.089`

So on this case the 90% confidence interval would be given by (94164.485;96001.089)