Question

In a random sample of 651 computer scientists who subscribed to a web-based daily news update, it was found that the average salary was $46,816 with a population standard deviation of $12,557. Calculate a 91 percent confidence interval for the mean salary of computer scientists.

Answer 1

Answer:

$46816-1.70\frac{12557}{\sqrt{651}}=45979.35$    

$46816+1.70\frac{12557}{\sqrt{651}}=47652.65$    

So on this case the 91% confidence interval would be given by (45979.35;47652.65)

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".

$\bar X=46816$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=12557$ represent the population standard deviation

n=651 represent the sample size  

Solution to the problem

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.91 or 91%, the value of $\alpha=0.09$ and $\alpha/2 =0.045$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.045,0,1)".And we see that $z_{\alpha/2}=1.70$

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

$46816-1.70\frac{12557}{\sqrt{651}}=45979.35$    

$46816+1.70\frac{12557}{\sqrt{651}}=47652.65$    

So on this case the 91% confidence interval would be given by (45979.35;47652.65)