Question

A hotel chain wanted to learn about the level of experience of its general managers. A random sample of 14 general managers was taken, and these managers had a mean of 11.72 years of experience. Suppose that the standard deviation of years of experience for all general managers in the chain is known to be 3.2 years. What is the lower limit of a 95% confidence interval for the mean experience of all general managers in this hotel chain?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 11.72

Population standard deviation =    = 3.2

Sample size = n = 14

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z0.025 = 1.960

Z/2 = 1.960  

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

= 1.960 * ( 3.2 /   14)

=1.676

At 95% confidence interval estimate of the population mean is,

- E < < + E

11.72 - 1.676 < < + 11.72 +1.676

10.04 <   < 13.40
( 10.04 , 13.40)

Lower limit is =10.04.