Question

one of the homework questions in our text had a set of 13 observations of the time in seconds for a specific sprinkler system type to activate. in that problem, the sample data resulted in x-bar = 27.92 and S² = 5.62. if we assume that the population of sprinkler activation times is normal, then find a 90% confidence interval for the population variance sigma²

Answer 1

Solution :

Given that,

Point estimate = s² = 5.62

n = 13

Degrees of freedom = df = n - 1 = 12

²_L = ²_/2,df = 21.026

²_R = ²_{1 - /2,df} = 5.226

The 90% confidence interval for ² is,

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

12 * 5.62 / 21.026 < ² < 12 * 5.62 / 5.226

3.21 < ² < 12.90

(3.21 , 12.90)