In: Statistics and Probability
A random sample of 25 employees of a local utility firm showed that their monthly incomes had a sample standard deviation of $112. Provide a 90% confidence interval estimate for the standard deviation of the incomes for all the firm's employees.
Solution :
Given that,
s = 112
s2 = 10.5830
n = 25
Degrees of freedom = df = n - 1 = 25 - 1 = 24
At 90% confidence level the 2 value is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
1 - / 2 = 1 - 0.05 = 0.95
2L = 2/2,df = 36.415
2R = 21 - /2,df = 13.848
The 90% confidence interval for is,
(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2
(24)(10.5830) / 36.415 < < (24)(10.5830) / 13.848
2.64 < < 4.28
(2.64 , 4.28)