In: Statistics and Probability
US department of labor wanted to estimate the variability in unemployment rate. They took a sample of size 25 and found that sample mean 20.32 and sample standard deviation is 3.55. Estimate the 95% confidence interval for true standard deviation.
Question 3 options:
(2.77 to 7.54) |
|
(2.77 to 4.94) |
|
(7.68 to 32.27) |
|
(1.45 to 7.68) |
|
(7.68 to 24.39) |
Solution :
Given that,
s = 3.55
s2 = 12.6025
2R = 2/2,df = 39.364
2L = 21 - /2,df = 12.401
The 95% confidence interval for is,
(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2
(24)(12.6025) / 39.364 < < (24)(12.6025) / 12.401
2.77 < < 4.94
(2.77 to 4.94)