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A study of the career paths of hotel general managers sent questionnaires to an SRS of 140 hotels belonging to major U.S. hotel chains. There were 101 responses. The average time these 101 general managers had spent with their current company was 11.1 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 4.5 years.)

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Answer:

CI = {9.947; 12.253}

Step-by-step explanation:

Average time (μ) = 11.1 years

Standard deviation (σ) = 4.5 years

Sample size (n) = 101 responses

Z-score for a 99% confidence interval (z) = 2.576

The confidence interval is defined by the following expression:

[tex]\mu \pm z*\frac{\sigma}{\sqrt{n}}[/tex]

The lower (L) and upper (U) limits of the 99% confidence interval is:

[tex]L= 11.1 - 2.576*\frac{4.5}{\sqrt{101}} \\L=9.947\\U= 11.1+ 2.576*\frac{4.5}{\sqrt{101}} \\U=12.253\\[/tex]

The 99% confidence interval  for the mean number of years general managers of major-chain hotels have spent with their current company is CI = {9.947; 12.253}

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