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

Answer:

$95082.787-1.782\frac{1857.716}{\sqrt{13}}=94164.485$  

$95082.787+1.782\frac{1857.716}{\sqrt{13}}=96001.089$  

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

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:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (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):  

$95082.787-1.782\frac{1857.716}{\sqrt{13}}=94164.485$  

$95082.787+1.782\frac{1857.716}{\sqrt{13}}=96001.089$  

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