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?

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".

[tex]\bar X=95082.787[/tex] represent the sample mean

[tex]\mu[/tex] 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:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=13-1=12[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], 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 [tex]t_{\alpha/2}=1.782[/tex]

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

[tex]95082.787-1.782\frac{1857.716}{\sqrt{13}}=94164.485[/tex]

[tex]95082.787+1.782\frac{1857.716}{\sqrt{13}}=96001.089[/tex]

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