Question

As part of a​ survey, a marketing representative asks a random sample of 3030 business owners how much they would be willing to pay for a website for their company. She finds that the sample standard deviation is ​$34043404. Assume the sample is taken from a normally distributed population. Construct 9090​% 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 the nearest integer as​ needed.)

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 = 90%

Sample size = n = 30

Degrees of freedom = n – 1 = 30 – 1 = 29

Sample standard deviation = S = 3404

χ²_{α/2, n – 1} = 42.5570

χ²_{1 -α/2, n– 1} = 17.7084

(By using chi square table)

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

(30 – 1)*3404^2/42.5570 < σ² <(30 – 1)*3404^2/17.7084

7895986.9935 < σ² < 18975734.8388

Lower limit = 7895986.9935

Upper limit = 18975734.8388

We are 90% confident that the population variance will lies within $7895987 and $18975735.

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(7895986.9935) < σ < Sqrt(18975734.8388)

2809.9799 < σ < 4356.1146

Lower limit = 2809.9799

Upper limit = 4356.1146

We are 90% confident that the population standard deviation will lies within $2810 and $4356.