Question

95% conf, n=41 , X bar=$67,600, s=18,484 use chi-square critical values Finding confidence interval for population standard deviation. Assume simple random sample has a normal distribution.

Answer 1

Solution :

Given that,

s = 18484

s² = 341658256

n = 41

Degrees of freedom = df = n - 1 = 41 - 1 = 40

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

²_L = ²_/2,df = 59.342

²_R = ²_{1 - /2,df} = 24.433

The 95% confidence interval for is,

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

40 * 18484 / 59.342 < < 40 * 18484 / 24.433

15175.60 < < 23650.30

( , )