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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.

User Volvox
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1 Answer

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Answer:


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)

User Waylander
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