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.

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".

represent the sample mean for the sample  

population mean (variable of interest)

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:

  (1)

Since the Confidence is 0.91 or 91%, the value of and , 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

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

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

the answer is b hope you pass