Question

Construct the confidence interval for the population standard deviation for the given values. Round your answers to one decimal place. n=21 , s=3.3, and c=0.9

Answer 1

The correct answer is "
`2.633< \sigma < 4.480`".

Explanation:

Given:

n = 21

s = 3.3

c = 0.9

now,

`df = n-1`

`=20`

⇒
`x^2_{(\alpha)/(2), n-1 }` =
`x^2_{(0.9)/(2), 21-1 }`

=
`31.410`

⇒
`x^2_{1-(\alpha)/(2), n-1 }` =
`10.851`

hence,

The 90% Confidence interval will be:

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

=
`\sqrt{((21-1)3.3^2)/(31.410) } < \sigma < \sqrt{((21.1)3.3^2)/(10.851) }`

=
`\sqrt{(20* 3.3^2)/(31.410) } < \sigma < \sqrt{(20* 3.3^2)/(10.851) }`

=
`2.633< \sigma < 4.480`