Question

The Interstate Conference of Employment Security Agencies says the average workweek in the United States is down to only 35 hours, largely because of a rise in part-time workers. Suppose this figure was obtained from a random sample of 20 workers and that the standard deviation of the sample was 4.2 hours. Assume hours worked per week are normally distributed in the population. Use this sample information to develop a 98% confidence interval for the population variance of the number of hours worked per week for a worker. What is the point estimate?

Answer 1

Solution :

Given that,

c = 0.98

s = 4.2

n = 20

At 98% confidence level the is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

_/2,df = _0.01,19 = 36.19

and

_{1- /2,df} = _0.99,19 = 7.63

Point estimate = s² = 17.64

²_L = ²_/2,df = 36.19

²_R = ²_{1 - /2,df} = 7.63

The 98% confidence interval for ² is,

(n - 1)s² / ²_/2 < ² < (n - 1)s² / ²_{1 - /2}

( 19*17.64) / 36.19< ² < ( 19* 17.64) / 7.63

9.26< ² < 43.91

(9.26, 43.91)

The point estimate is 17.64