Question

A study of the career paths of hotel general managers sent questionnaires to an SRS of 250 hotels belonging to major U.S. hotel chains. There were 127 responses. The average time these 127 general managers had spent with their current company was 8.92 years. (Take it as known that the standard deviation of time with the company for all general managers is 2.8 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. (Enter: ''DECREASES'', ''DOES NOT CHANGE'' or ''INCREASES'', without the quotes.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8.92

Population standard deviation =    = 2.8

Sample size = n = 127

a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z/2 * ( /n)

E = 1.645 * ( 2.8 /  127 )

E = 0.41 years

b) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

Margin of error = E = Z/2 * ( /n)

E = 2.576 * ( 2.8 /  127 )

E = 0.64 years

(c) In general, increasing the confidence level  ''INCREASES the margin of error (width) of the confidence interval