Question

Find the 99% confidence intervals for population variation, population standard deviation. (round to nearest integer)

Score: 0 of 1 pt 7 of 10 (10 complete) W Score: 75%, 7.5 of 10 X 6.4.20-T Question Help As part of a survey, a marketing representative asks a random sample of 29 business owners how much they would be willing to pay for a website for their company. She finds that the sample standard deviation is $3325. Assume the sample is taken from a normally distributed population. Construct 99% confidence intervals for (a) the population variance o2 and (b) the population standard deviation o. Interpret the results. (a) The confidence interval for the population variance is (Round to the nearest integer as needed.) Enter your answer in the edit fields and then click Chart A

Answer 1

Solution:

Confidence interval for population variance

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

We are given

Confidence level = 99%

Sample size = n = 29

Degrees of freedom = n – 1 = 28

Sample standard deviation = S = 3325

χ²_{α/2, n – 1} = 50.9934

χ²_{1 -α/2, n– 1} = 12.4613

(By using chi square table)

(29 – 1)* 3325^2 / 50.9934 < σ² < (29 – 1)* 3325^2 / 12.4613

6070543.3253 < σ² <24841437.5659

Lower limit = 6070543

Upper limit = 24841438

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(6070543.3253) < σ < Sqrt(24841437.5659)

2463.8473 < σ < 4984.1185

Lower limit = 2464

Upper limit = 4984