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 ​$3.56. Assume the sample is taken from a normally distributed population. Construct 95​% confidence intervals for​ (a) the population variance sigmaσsquared2 and​ (b) the population standard deviation sigmaσ.

Interpret the results.

Answer 1

Solution :

Given that,

s = 3.56

Point estimate = s² = 12.6736

n = 14

Degrees of freedom = df = n - 1 = 14 - 1 = 13

(a)

.At 95% confidence level the ² value is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

1 - / 2 = 1 - 0.025 = 0.975

²_L = ²_/2,df = 24.736

²_R = ²_{1 - /2,df} = 5.009

The 95% confidence interval for ² is,

(n - 1)s² / ²_/2 < ² < (n - 1)s² / ²_{1 - /2}

13 * 12.6736 / 24.736 < ² < 13 * 12.6736 / 5.009

6.66 < ² < 32.89

(6.66 , 32.89)

(b)

The 95% confidence interval for is,

2.58 < < 5.73

(2.58 , 5.73)