Question

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 103 responses. The average time these 103 general managers had spent with their current company was 11.7 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 3.6 years.)

Answer 1

(10.78483, 12.61517)

Explanation:

It is given that the genera mangers of few hotels were sent some questionnaires for conducting a study for the career paths in the major hotel chains of the United States.

Number of hotels = 160

Number of response received = 103

The average number of years these general mangers was in their current hotels,
`$\bar x$` = 11.7 years

Confidence Interval, CI = 0.99

Therefore,

a = 0.01, |Z(0.005)| (from standard normal table)

∴ 99% of CI =
`$\bar x \pm Z * (s)/(\sqrt n)$`

`$= 11.7 \pm 2.58 * \frac{3.6}{\sqrt {103}}$`

`$(10.78483, 12.61517)$`