Question

An economist wants to estimate the variance of employee test scores. A random sample of 38 scores had a sample standard deviation of 10.4. Find a 95% confidence interval for the population variance.

Answer 1

Solution :

Given that,

s = 10.4

Point estimate = s² = 3.2249

n = 38

Degrees of freedom = df = n - 1 = 38 - 1 = 37

At 95% confidence level the ² value is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

1 - / 2 = 1 - 0.025 = 0.975

²_L = ²_/2,df = 55.67

²_R = ²_{1 - /2,df} = 22.11

The 95% confidence interval for ² is,

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

(37)(3.2249) / 55.67 < ² < (37)(3.2249) / 22.11

2.14 < ² < 5.40

(2.14 , 5.40)