Question

The thicknesses 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 thicknesses of all aluminum sheets in this factory. Round your answers to two decimal places

Answer 1

The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.

Explanation:

The confidence interval for population variance is given as below:

`[(n - 1)* S^(2) / X^(2) \alpha/2, n-1 ] < \alpha < [(n- 1)* S^(2) / X^(2) 1- \alpha/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

`X^(2)_([\alpha/2, n - 1]) = 112.3288\\\X^(2) _(1 -\alpha/2,n- 1) = 53.5401`

(By using chi-square table)

[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 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.