Question

(1 point) A study of the career paths of hotel general managers sent questionnaires to an SRS of 260 hotels belonging to major U.S. hotel chains. There were 148 responses. The average time these 148 general managers had spent with their current company was 9.64 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

a) Margin of error is given by

For 90% confidence interval , zc = 1.645 (from z table )

Thus we get , margin of error for 90% confidence interval

b) For 99% confidence , zc = 2.58 ( from z table )

Thus we get , margin of error for 99% confidence interval

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

Note : C% Confidence interval for population mean is given by

As we increase level of confidence interval , the width of interval increases

Note : For 90% confidence , from z table we get , zc = 1.645

that is P( -1.645 < z < 1.645) = 0.90

For 99% confidence , zc =2.58

that is P( -2.58 < z < 2.58) = 0.99