Question

Suppose overtime of employees at Company XYZ are normally distributed and have a known population standard deviation of 4 hours per month and an unknown population mean. A random sample of 21 employees is taken and gives a sample mean of 61 hours per month. Find the confidence interval for the population mean with a 98% confidence level.

Round your answer to TWO decimal places.

Answer 1

Solution :

Given that,

sample mean = = 61

Population standard deviation =    = 4

Sample size = n =21

At 98% confidence level the z is ,

Z_/2 = Z_0.01 = 2.326
Margin of error = E = Z/2 * ( /n)

= 2.326 * ( 4/  21 )

= 2.03
At 98% confidence interval of the population mean
is,

- E < < + E

61 - 2.03 <   < 61+ 2.03

58.97<   < 63.03

( 58.97 ,63.03 )