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 ​$3.37. Assume the sample is taken from a normally distributed population. Construct 95​% confidence intervals for​ (a) the population variance sigma squared and​ (b) the population standard deviation sigma. Interpret the results.

Answer 1

Solution :

Given that,

s = 3.37

Point estimate = s² = 11.3569

²_R = ²_/2,df = 24.736

²_L = 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 * 11.3569 / 24.736 < ² < 13 * 11.3569 / 5.009

5.9687 < ² < 29.4767

(5.9687 , 29.4767)

(b)

The 95% confidence interval for is,

2.4431 < < 5.4292

(2.4431 , 5.4292)