In: Statistics and Probability
1. Suppose a sample of 30 students take an IQ test. If the sample has a standard deviation of 12.23, find a 90% confidence
interval for the population standard deviation.
Solution :
Given that,
s = 12.23
s2 = 149.5729
n = 30
Degrees of freedom = df = n - 1 = 30-1 = 29
At 90% confidence level the 2 value is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
1 - / 2 = 1 - 0.05 =0.95
2L = 2/2,df = 42.557
2R = 21 - /2,df = 17.7084
The 95% confidence interval for is,
(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2
(29)149.5729 /42.557 < < (29)149.5729/17.7084
10.0958< < 15.6508
( 10.0958 , 15.6508 )