Question

An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s2 = 0.01532 (fluid ounce). If the variance of fill volume is too large, an unacceptable proportion of bottles will be under- or overfilled. We will assume that the fill volume is approximately normally distributed. Determine the 95% upper confidence bound of the variance.

Answer 1

Solution :

Given that,

c = 0.95

s ²= 0.01532

n = 20

At 95% confidence level the is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

_/2,df = _0.025,18 = 32.85

and

_{1- /2,df} = _0.975,18 = 8.91

²_L = ²_/2,df = 32.85

²_R = ²_{1 - /2,df} = 8.91

The 95% confidence interval for ² is,

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

( 19 *0.01532) / 32.85 < ² < ( 19 * 0.01532 ) / 8.91

0.0089< ² < 0.0327

(0.0089 , 0.0327)

The 95% upper confidence bound of the variance is 0.0327

The 95% confidence interval for is,

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

₃₈( 19 - 1 ) / 34.81 < < ₃₈( 19 - 1 ) / 7.01

27.33 < < 97.57

( 27.33 , 97.57 )