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A study of the career paths of hotel general managers sent questionnaires to an SRS of 160 hotels belonging to major U.S. hotel chains. There were 114 responses. The average time these 114 general managers had spent with their current company was 11.78 years. Give a 99%; confidence interval for the mean number of years general managers of major-chain hotels have spent with their current company. (Take it as known that the standard deviation of time with the company for all general managers is 3.2 years.)

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Answer:

The 99% confidence interval for the mean number of years general managers of major-chain hotels have spent with their current company is between 11.01 years and 12.55 years.

Explanation:

We have that to find our
\alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:


\alpha = (1-0.99)/(2) = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of
1-\alpha.

So it is z with a pvalue of
1-0.005 = 0.995, so
z = 2.575

Now, find the margin of error M as such


M = z*(\sigma)/(โˆš(n))

In which
\sigma is the standard deviation of the population and n is the size of the sample.


M = 2.575*(3.2)/(โˆš(114)) = 0.77

The lower end of the interval is the sample mean subtracted by M. So it is 11.78 - 0.77 = 11.01 years

The upper end of the interval is the sample mean added to M. So it is 11.78 + 0.77 = 12.55 years

The 99% confidence interval for the mean number of years general managers of major-chain hotels have spent with their current company is between 11.01 years and 12.55 years.

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