Question

20- The thicknesses of 73randomly selected linoleum tiles were found to have a variance of 3.14 . Construct the 90% confidence interval for the population variance of the thicknesses of all linoleum tiles in this factory. Round your answers to two decimal places.

Step 1 of 1:

Answer 1

20)

Solution :

Given that,

Point estimate = s² = 3.14

n = 73

Degrees of freedom = df = n - 1 = 73 - 1 = 72

At 95% confidence level the ² value is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

1 - / 2 = 1 - 0.05 = 0.95

²_L = ²_/2,df = 92.808

²_R = ²_{1 - /2,df} = 53.462

The 90% confidence interval for ² is,

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

72 * 3.14 / 92.808 < ² < 72 * 3.14 / 53.462

2.44 < ² < 4.23

( 2.44 , 4.23)