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. $2.2819<\sigma<4.3323$

Step-by-step explanation:

The confidence interval for population variance is given by :-

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

, 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 :

$\dfrac{(3.08)^2(14-1)}{23.6845}<\sigma^2<\dfrac{(3.08)^2(14-1)}{6.5706}\\\\=5.20691591547\sigma^2<18.7689404316\\\\\approx5.2069<\sigma^2<18.7689$

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

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

$\sqrt{5.2069}<\sigma<\sqrt{18.7689}$

$\approx2.2819<\sigma<4.3323$