Question

The thickness of 81 randomly selected aluminum sheets were found to have a variance of 3.23. construct the 98% confidence interval for the population variance of the thickness of all aluminum sheets in this factory. Assume that the population is normally distributed. Round your answer to two decimal places.

I have been working on these problems for 8 hours and im still not getting it. They are due tomorrow. Thank you.

Answer 1

Confidence interval for population variance is given as below:

[(n – 1)*S² / χ²_{α/2, n– 1} ] < σ < [(n – 1)*S² / χ²_{1 -α/2, n– 1} ]

We are given

Confidence level = 98%

Sample size = n = 81

Degrees of freedom = n – 1 = 80

Sample Variance = S^2 = 3.23

χ²_{α/2, n – 1} = 112.3288

χ²_{1 -α/2, n– 1} = 53.5401

(By using chi square table)

[(n – 1)*S² / χ²_{α/2, n– 1} ] < σ^2 < [(n – 1)*S² / χ²_{1 -α/2, n– 1} ]

[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]

2.3004 < σ^2 < 4.8263

Lower limit = 2.30

Upper limit = 4.83