Question

A random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 2.22.2 inches. Construct a​ 99% confidence interval for the population standard​ deviation, sigmaσ. Assume the sample is from a normally distributed population.

Answer 1

Answer:
`\approx1.49<\sigma<3.97`

Explanation:

We know that the confidence interval for population standard deviation is given by :-

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

, where n= sample size

s = sample standard deviation.

`\chi^2_(\alpha/2)` and
`\chi^2_(1-\alpha/2)`= Chi-square critical value for degree of freedom (n-1) and significance level (
`\alpha`).

Given :
`\alpha=1-0.99=0.01`

n= 16

Critical ch-square values for degree of freedom 15 and
`\alpha=0.01`will be :

`\chi_(\alpha/2)=\chi_(0.005)=32.80132`

`\chi_(1-\alpha/2)=\chi_(0.995)=4.6009`

Then , the required 99% confidence interval for the population standard​ deviation will be :

`\sqrt{((15)(2.2)^2)/(32.80132)}<\sigma<\sqrt{((15)(2.2)^2)/(4.6009)}`

`√(2.21332556129)<\sigma<√(15.779521398)`

`1.48772496157<\sigma<3.97234457191`

`\approx1.49<\sigma<3.97`

Hence, the a​ 99% confidence interval for the population standard​ deviation :