Question

A butcher uses a machine that packages chicken livers in ten ounce portions. A sample of 9090 packages of chicken livers has a standard deviation of 0.380.38. Construct the 80%80% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine. Round your answers to two decimal places.

Lower endpoint: ??? and Upper endpoint : ???

Answer 1

Solution :

Given that,

s = 0.38

s² = 0.1444

n = 90

Degrees of freedom = df = n - 1 = 90 - 1 = 89

At 80% confidence level the ² value is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

1 - / 2 = 1 - 0.10 = 0.90

²_L = ²_/2,df = 106.469

²_R = ²_{1 - /2,df} = 72.387

The 80% confidence interval for is,

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

89 * 0.1444 / 106.469 < < 89 * 0.1444 / 72.387

0.35 < < 0.42

( 0.35 , 0.42)