In: Statistics and Probability
Make a 95% confidence interval for sigma squared
if:
s²=83.72
n=36
Solution :
Given that,
s2 = 83.72
Degrees of freedom = df = n - 1 = 35
2L = 2/2,df = 53.203
2R = 21 - /2,df = 20.569
The 95% confidence interval for is,
(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2
35 * 83.72 / 53.203 < < 35 * 83.72 / 20.569
7.42 < < 11.54
(7.42 , 11.54)