Question

A study regarding pizza eaters showed that in a random sample of 21 individuals, the standard deviation for the number of slices of pizza an individual can eat is 1.3 slices. Find the 90% confidence interval of the variance and the standard deviation. Show your work to receive credit.

Answer 1

Solution :

Given that,

c = 0.90

s = 1.3

n = 21

At 90% confidence level the is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

_/2,df = _0.05,20 = 31.41

and

_{1- /2,df} = _0.95,20 = 10.85

Point estimate = s² = 1.69

²_L = ²_/2,df = 31.41

²_R = ²_{1 - /2,df} = 10.85

The 90% confidence interval for ² is,

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

( 20 *1.69) / 31.41 < ² < ( 20 * 1.69) / 10.85

1.08 < ² < 3.12

(1.08 , 3.12)

The 90% confidence interval for is,

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

1.3 ( 21 - 1 ) / 31.41 < < 1.3 ( 21 - 1 ) / 10.85

1.04 < < 1.76

( 1.04 , 1.76)