Question

A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​...

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

​$3.843.84.

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.)

Homework Answers

Answer #1

Given that sample size n = 14

alpha level = 1- confidence level = 1-0.98 = 0.02

sample standard deviation s = 3.84

(A) confidence interval for variance

(B) Confidence interval for standard deviation is given as

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​...
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 ​$3.29. Assume the sample is taken from a normally distributed population. Construct 99​% confidence intervals for​ (a) the population variance sigmasquared and​ (b) the population standard deviation sigma. Interpret the results.
The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is...
The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is $200 with a standard deviation of $30. If 9 LCD computer monitors are randomly selected, then the distribution of the sample mean price is
The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is...
The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is $200 with a standard deviation of $30. If 9 LCD computer monitors are randomly selected, then the distribution of the sample mean price is Exactly normal with mean $200 and standard deviation $30. Exactly normal with mean $200 and standard deviation $10. Approximately normal with mean $200 and standard deviation $10. Can't be determined without more information
Use technology to construct the confidence intervals for the population variance sigmaσsquared2 and the population standard...
Use technology to construct the confidence intervals for the population variance sigmaσsquared2 and the population standard deviation sigmaσ. Assume the sample is taken from a normally distributed population. cequals=0.990.99 , ssquared2equals=12.2512.25 , nequals=2929 The confidence interval for the population variance is (nothing ,nothing ). (Round to two decimal places as needed.)
The number of hours of reserve capacity of 10 randomly selected automotive batteries is shown. 1.72...
The number of hours of reserve capacity of 10 randomly selected automotive batteries is shown. 1.72 1.84 1.59 1.62 1.76 1.94 1.31 1.56 1.41 2.08 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. ​(a) The confidence interval for the population variance is left parenthesis nothing comma nothing right parenthesis .,. ​(Round to three decimal places as​ needed.)
1) 100,709 79,677 47,982 92,189 74,995 80,964 73,768 98,821 84,535 61,753 78,061 44,544 80,986 79,359 82,841...
1) 100,709 79,677 47,982 92,189 74,995 80,964 73,768 98,821 84,535 61,753 78,061 44,544 80,986 79,359 82,841 67,361 86,766 92,379 63,708 51,565 61,170 63,654 73,598 67,210 57,866 74,535 56,526 94,385 54,897 50,392 46,724 65,956 78,528 60,835 89,116 (a) Find the sample mean. ​(b) Find the sample standard deviation. The sample standard deviation is defined as is equals StartRoot StartFraction Upper Sigma left parenthesis x minus x overbar right parenthesis squared Over n minus 1 EndFraction EndRoots=Σx−x2n−1. (c) Construct a​ 98% confidence...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT