Question

The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance in the pounds of the impurities. The sample variance was found to be 6.62. Find a 95% confidence interval for the population variance in the pounds of impurities. State any assumption you need to make to be able to answer this problem.

Answer 1

Solution :

Given that,

s² = 6.62

²_R = ²_/2,df = 32.852

²_L = ²_{1 - /2,df} = 8.907

The 95% confidence interval for ² is,

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

  19 * 6.62 / 32.852 < ² < 19 * 6.62 / 8.907

3.83 < ² <  14.12  

(3.83 , 14.12)