Question

A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​ (LCD) televisions. The article states that

1414

randomly selected​ 32-inch LCD televisions have a sample standard deviation of

​$3.773.77.

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

9898​%

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

​(nothing​,nothing​).

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

Interpret the results. Select the correct choice below and fill in the answer​ box(es) to complete your choice.

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

A.With

9898​%

​confidence, you can say that the population variance is between

nothing

and

nothing.

B.With

22​%

​confidence, you can say that the population variance is between

nothing

and

nothing.

C.With

9898​%

​confidence, you can say that the population variance is

lessless

than

nothing.

D.With

22​%

​confidence, you can say that the population variance is

greatergreater

than

nothing.

​(b) The confidence interval for the population standard deviation is

​(nothing​,nothing​).

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

Interpret the results. Select the correct choice below and fill in the answer​ box(es) to complete your choice.

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

A.With

9898​%

​confidence, you can say that the population standard deviation is

greatergreater

than

nothing

dollars per year.

B.With

22​%

​confidence, you can say that the population standard deviation is between

nothing

and

nothing

dollars per year.

C.With

9898​%

​confidence, you can say that the population standard deviation is between

nothing

and

nothing

dollars per year.

D.With

22​%

​confidence, you can say that the population standard deviation is

lessless

than

nothing

dollars per year.

Answer 1

Solution:

Part a

Confidence interval for population variance is given as below:

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

We are given

Confidence level = 98%

Sample size = n = 14

Degrees of freedom = n – 1 = 13

Sample standard deviation = S = 3.77

χ²_{α/2, n – 1} = 27.6882

χ²_{1 -α/2, n– 1} = 4.1069

(By using chi square table)

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

(14 – 1)*3.77^2/27.6882 < σ² < (14 – 1)*3.77^2/4.1069

6.6731 < σ² < 44.9894

Lower limit = 6.6731

Upper limit = 44.9894

The confidence interval for the population variance is (6.67, 44.99).

Interpret the results.

Correct Answer: A. With 98​% confidence, you can say that the population variance is between 6.67 and 44.99.

Part b

Confidence interval for population standard deviation is given as below:

Sqrt[(n – 1)*S² / χ²_{α/2, n– 1} ] < σ < sqrt[(n – 1)*S² / χ²_{1 -α/2, n– 1} ]

Sqrt(6.6731) < σ < sqrt(44.9894)

2.5832 < σ <6.7074

The confidence interval for the population standard deviation is (2.58, 6.71)

Interpret the results.

Correct Answer: C. With 98​% confidence, you can say that the population standard deviation is between 2.58 and 6.71 dollars per year.