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

​$3.05.

Assume the sample is taken from a normally distributed population. Construct

90​%

confidence intervals for​ (a) the population variance

sigmaσsquared2

and​ (b) the population standard deviation

sigmaσ.

Interpret the results.

​(a) The confidence interval for the population variance is

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

(a)

s = $3.05

Point estimate = s2 = 9.3025

2R = 2/2,df = 22.362

2L = 21 - /2,df = 5.892

The 90% confidence interval for 2 is,

(n - 1)s2 / 2/2 < 2 < (n - 1)s2 / 21 - /2

(13)(9.3025) / 22.362 < 2 < (13)(9.3025) / 5.892

5.41 < 2 < 20.53

(5.41 , 20.53)

(b)

s = $3.05

s2 = 9.3025

2R = 2/2,df = 22.362

2L = 21 - /2,df = 5.892

The 90% confidence interval for is,

(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2

(13)(9.3025) / 22.362 < < (13)(9.3025) / 5.892

2.33 < < 4.53

(2.33 , 4.53)