A study of the career paths of hotel general managers sent questionnaires to an SRS of 150 hotels belonging to major U.S. hotel chains. There were 104 responses. The average time these 104 general managers had spent with their current company was 10.7 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.)

Given that a study of career paths of hotel general managers sent questionnaires to an SRS of 150 hotels belonging to major U.S. hotel chains. There were 104 responses.

The average time these 104 general managers had spent with their current company was 10.7 years.

Here we have sample size [tex]n=104\\\bar x= 10.7\\\sigma = 3.2[/tex]

Since population standard deviation is known and also sample size is greater than 30, we can use Z critical value for finding 99% confidence interval.

99% Z critical value = 2.58

Hence confidence interval 99%

=[tex](10.7-2.58*\frac{3.2}{\sqrt{104} }, 10.7+2.58*\frac{3.2}{\sqrt{104} })[/tex]

=[tex](9.8904,11.5096)[/tex]