Question

A sample of 29 boxes of cereal has a sample standard deviation of 0.81 ounces. Construct a 95% confidence interval to estimate the true standard deviation of the filling process for the boxes of cereal.

a. ( 0.427,1.144)
b. (0.515,1.105)
c. (0.726,0.726)

Answer 1

The 95% confidence interval is
`0.6428 < \sigma < 1.0954`

Explanation:

From the question we are told that

The sample size is n = 29

The sample standard deviation is
`s = 0.81`

Generally the 95% confidence interval to estimate the true standard deviation is mathematically represented as

`\sqrt{((n- 1)s^2 )/( X^2_( (\alpha )/(2 ) , n-1)) } < \sigma < \sqrt{((n- 1)s^2 )/( X^2_(1- (\alpha )/(2 ) , n-1)) }`

Generally from the chi -distribution table the critical value of
`(\alpha )/(2)` at a degree of freedom of
`df = 29 - 1 = 28` is

`X^2 _{(\alpha )/(2) , 28 } =44.46079184`

Generally from the chi -distribution table the critical value of
`1 - (\alpha )/(2)` at a degree of freedom of
`df = 29 - 1 = 28` is

`X^2 _{(1 - (\alpha )/(2)) , 28 } = 15.30786055`

So

`\sqrt{((29- 1)0.81^2 )/( 44.46079184 ) } < \sigma < \sqrt{((29- 1)0.81^2 )/(15.315.307860551 )`

=>
`0.6428 < \sigma < 1.0954`