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. Choose the closest value from your table.

Answer 1

`46816-1.695(12557)/(√(651))=45981.81`

`46816+ 1.695(12557)/(√(651))=46650.19`

And the confidence interval would be given by: (45981.81; 46650.19)

Explanation:

Data given

`\bar X=46816` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=12557` represent the population standard deviation

n=651 represent the sample size

Confidence interval

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 significacne is
`\alpha=0.09` and
`\alpha/2 =0.045`, and the critical value is
`z_(\alpha/2)=1.695`

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

`46816-1.695(12557)/(√(651))=45981.81`

`46816+ 1.695(12557)/(√(651))=46650.19`

And the confidence interval would be given by: (45981.81; 46650.19)