Question

The amount of cereal in fifteen boxes of Brand A breakfast cereal were found to have a mean of 14 ounces and a standard deviation of 0.76 ounces. Construct a 90% confidence interval for the true standard deviation of the amount of cereal in Brand A boxes of breakfast cereal. Interpret your results

Answer 1

Solution :

Given that,

c = 0.90

s = 0.76

n = 15

At 90% confidence level the is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

_/2,df = _0.05,14 = 23.68

and

_{1- /2,df} = _0.95,14 = 6.57

²_L = ²_/2,df = 23.68

²_R = ²_{1 - /2,df} = 6.57

The 90% confidence interval for is,

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

0.76 ( 15 - 1 ) / 23.68 < < 0.76 ( 15 - 1 ) / 6.57

0.58 < < 1.11

( 0.58 , 1.11)

Therefore based on the data provided, the 90% confidence interval for the true standard deviation of the amount of cereal in Brand A boxes of breakfast cereal is 0.58 < < 1.11