Question

Use technology to construct the confidence intervals for the population variance sigma squared and the population standard deviation sigma. Assume the sample is taken from a normally distributed population, C= 0.90, S^2=10.24, N= 30

The confidence interval for the population variance is:

The confidence interval for the population standard deviation is:

Answer 1

Solution :

Given that,

c = 0.90

s² = 10.24

n = 30

At 90% confidence level the is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

_/2,df = _0.05,29 = 42.56

and

_{1- /2,df} = _0.95,29 = 17.71

Point estimate = s² = 10.24

²_L = ²_/2,df = 42.56

²_R = ²_{1 - /2,df} = 17.71

The 90% confidence interval for ² is,

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

( 29 *10.24 ) / 42.56 < ² < ( 29 * 10.24) / 17.71

6.98 < ² < 16.77

(6.98 , 16.77)

The 90% confidence interval for is,

s (n-1) / _/2,df < < s (n-1) / _{1- /2,df}

3.2 ( 30 - 1 ) / 42.56 < < 3.2( 30 - 1 ) / 17.71

2.64 < < 4.10

( 2.64 , 4.10)