Question

10. 0/10 points Previous Answers YMSStat2 10..013. My A study of the career paths of otel general managers sent questionnaires to an SRS of 160 hotels belonging to major U.S. hotel chains. There were 109 responses. The average time these 109 general managers had spent with their current company was 11.5 years. How large a sample of hotel managers would be needed to estimate the mean within 1.08 year with 80% confidence? (Take it as known that the standard deviation of time with the company for all general managers is 4.23 years.) 25.13 X

Answer 1

Solution :

Given that,

Population standard deviation = $\sigma$ = 4.23

Margin of error = E = 1.08

At 80% confidence level the z is,

$\alpha$ = 1 - 80%

$\alpha$ = 1 - 0.80 = 0.20

$\alpha$/2 = 0.10

Z_$\alpha$/2 = 1.282

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [1.282 * 4.23 / 1.08 ]²

n = 25.21

Sample size = n = 26