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

`46816-1.70(12557)/(√(651))=45979.35`

`46816+1.70(12557)/(√(651))=47652.65`

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

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=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)(\sigma)/(√(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(12557)/(√(651))=45979.35`

`46816+1.70(12557)/(√(651))=47652.65`

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