A magazine includes a report on the energy costs per year for 32-inch liquid cry

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 $3.07 . Assume the sample is taken from a normally distributed population. Construct 99 % confidence intervals for (a) the population variance sigma squared and (b) the population standard deviation sigma . Interpret the results.

(a) The confidence interval for the population variance is (----,---- ). (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 1 % confidence, you can say that the population variance is between ___and ___ .

B. With 99 % confidence, you can say that the population variance is less than ____ .

C. With 1 % confidence, you can say that the population variance is greater than ____.

D. With 99 % confidence, you can say that the population variance is between ___and___ .

(b) The confidence interval for the population standard deviation is (____,____). (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 1 % confidence, you can say that the population standard deviation is between ___and___ dollars per year.

B. With 99 % confidence, you can say that the population standard deviation is between ___and___ dollars per year.

C. With 1 % confidence, you can say that the population standard deviation is less than ___ dollars per year.

D. With 99 % confidence, you can say that the population standard deviation is greater than ___dollars per year.

Explanation / Answer

a.
Confidence Interval
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size
Since aplha =0.01
^2 right = (1 - Confidence Level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
^2 left = 1 - ^2 right = 1 - 0.005 = 0.995
the two critical values ^2 left, ^2 right at 13 df are 29.8195 , 3.565
S.D( S^2 )=3.07
Sample Size(n)=14
Confidence Interval for Variance = [ 13 * 9.4249/29.8195 < ^2 < 13 * 9.4249/3.565 ]
= [ 122.5237/29.8195 < ^2 < 122.5237/3.565 ]
= [ 4.1088 < ^2 < 34.3685 ]       

D. With 99 % confidence, you can say that the population variance is between 4.1088 and 34.3685

b.
Confidence Interval for standard deviation = [ Sqrt(4.1088) < < Sqrt(34.3685) ]   
= [ 2.02701 < < 5.86246 ]

With 99 % confidence, you can say that the population standard deviation is between 2.02701 and 5.86246
dollars per year.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.