Question

The thicknesses of 55 randomly selected ceramic tiles were found to have a variance of 3.18 Construct the 98% confidence interval for the population variance of the thicknesses of all ceramic tiles in this factory. Round your answers to two decimal places.

Answer 1

df = n - 1 = 55 - 1 = 54

Chi square critical values are 0.02 level = 32.793 , 81.069

98% confidence interval for $\sigma$² is

(n-1) S² / $\chi ^{2}$$\alpha$/2 <  $\sigma$² < (n-1) S² / $\chi ^{2}$1-$\alpha$/2

54 * 3.18 / 81.069 < $\sigma$² < 54 * 3.18 / 32.793

2.12 < $\sigma$² < 5.24

98% CI is ( 2.12 , 5.24 )