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 ​$3.87.  Assume the sample is taken from a normally distributed population. Construct 90​% confidence intervals for​ (a) the population variance and​ (b) the population standard deviation

Interpret the results.

Answer 1

df = n - 1 = 14 - 1 = 13

chi square critical values at 13 df with 0.10 significance level = L = 5.892 , U = 22.362

a)

90% confidence interval for is

(n - 1) S² / U < < (n - 1) S² / L

13 * 3.87² / 22.362 < < 13 * 3.87² / 5.892

8.71 < < 33.04

90% CI is ( 8.71 , 33.04)

Interpretation - We are 90% confident that population variance is between 8.71 and 33.04

b)

90% confidence interval for is

sqrt [ (n - 1) S² / U]  < < sqrt [ (n - 1) S² / L]

sqrt [ 13 * 3.87² / 22.362] < < sqrt [ 13 * 3.87² / 5.892 ]

2.95 < < 5.75

90% CI is ( 2.95 , 5.75 )

Interpretation - We are 90% confident that population standard deviation is between 2.95 and 5.75