Question

a random sample of 85 group leaders, supervisors and similar personnel revealed that a person spent an average of 6.5 years on the job before being promoted. The standard deviation was 1.7 years. Using the 99% degree of confidence what is the confidence interval for the population mean. Please show all steps

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 6.5

Population standard deviation =    = 1.7

Sample size = n = 85

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)

= 2.576 * ( 1.7 /  85 )

= 0.47

At 99% confidence interval estimate of the population mean is,

- E < < + E

6.5 - 0.47 <   < 6.5 + 0.47

6.03 <   < 6.97

( 6.03 , 6.97 )

The 99% confidence interval for the population mean is : ( 6.03 , 6.97 )