Question

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)

Answer 1

Solution :

Given that,

s = 3.55

s² = 12.6025

²_R = ²_/2,df = 39.364

²_L = ²_{1 - /2,df} = 12.401

The 95% confidence interval for is,

(n - 1)s² / ²_/2 < < (n - 1)s² / ²_{1 - /2}

(24)(12.6025) / 39.364 < < (24)(12.6025) / 12.401

2.77 < < 4.94

(2.77 to 4.94)