In: Statistics and Probability
A study of arrival times for a number of international flights. The times were randomly recorded from a normally distributed population with the following results:
n = 19 and s = 12.1. Find the 95% confidence interval for σ.
Solution :
Given that,
s = 12.1
s2 = 3.4785
n = 19
Degrees of freedom = df = n - 1 = 19 - 1 = 18
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
2L = 2/2,df = 31.526
2R = 21 - /2,df = 8.231
The 95% confidence interval for is,
(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2
(18) (3.4785) / 31.526 < < (18)(3.4785) / 8.231
1.41 < < 2.76
(1.41 , 2.76 )