Question

A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​ (LCD) televisions. The article states that 14 randomly selected​ 32-inch LCD televisions have a sample standard deviation of ​$3.08. Assume the sample is taken from a normally distributed population. Construct 90​% confidence intervals for​:

a. the population variance sigma squared.
b. the population standard deviation sigma.

Answer 1

Answer: a.
`5.2069<\sigma^2<18.7689`

b.
`2.2819<\sigma<4.3323`

Explanation:

The confidence interval for population variance is given by :-

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

, where n= sample size

s = sample standard deviation.

As per given , we have

n= 14

s=​$3.08

For 90% confidence , the significance level =
`\alpha=1-0.90=0.10`

Critical values using chi-square distribution table ,

`\chi^2_(\alpha/2, n-1)=\chi^2_(0.05,13)=23.6845\\\\\chi^2_(1-\alpha/2, n-1)=\chi^2_(0.95,13)=6.5706`

Now, the required confidence interval for population variance :

`((3.08)^2(14-1))/(23.6845)<\sigma^2<((3.08)^2(14-1))/(6.5706)\\\\=5.20691591547\sigma^2<18.7689404316\\\\\approx5.2069<\sigma^2<18.7689`

i.e. 90​% confidence intervals for the population variance:
`5.2069<\sigma^2<18.7689`

Then, 90​% confidence intervals for population standard deviation :

`√(5.2069)<\sigma<√(18.7689)`

`\approx2.2819<\sigma<4.3323`