Question

A study of the career paths of hotel general managers sent questionnaires to an SRS of 280 hotels belonging to major U.S. hotel chains. There were 177 responses. The average time these 177 general managers had spent with their current company was 12.39 years. (Take it as known that the standard deviation of time with the company for all general managers is 3.7 years.)
(a) Find the margin of error for a 90% confidence interval to estimate the mean time a general manager had spent with their current company: years
(b) Find the margin of error for a 99% confidence interval to estimate the mean time a general manager had spent with their current company: years
(c) In general, increasing the confidence level the margin of error (width) of the confidence interval.

Answer 1

Answer and Step-by-step explanation:

we have the following data:

Point estimate = sample mean = \ bar x = 12.39

Population standard deviation = \ sigma = 3.7

Sample size = n = 177

a) the margin of error with a 90% confidence interval

α = 1 - 90%

alpha = 1 - 0.90 = 0.10

alpha / 2 = 0.05

Z \ alpha / 2 = Z0.05 = 1,645

Margin of error = E = Z \ alpha / 2 * (\ sigma / \ sqrtn)

we replace:

E = 1.645 * (3.7 / \ sqrt177)

Outcome:

E = 0.46

b) margin of error with a 99% confidence interval

α = 1-99%

alpha = 1 - 0.99 = 0.01

alpha / 2 = 0.005

Z \ alpha / 2 = Z0.005 = 2,576

Margin of error = E = Z \ alpha / 2 * (\ sigma / \ sqrtn)

we replace:

E = 2,576 * (3.7 / \ sqrt177)

Outcome:

E = 0.72

c) A larger confidence interval value will increase the margin of error.