Question

Problem 9 Owing to the variability of trade-in allowance, the profit per new car sold by an automobile dealer varies from car to car. The profits per sale in hundreds of dollars) tabulated for the past week were 2.1, 3.0, 1.2, 6.2, 4.5, and 5.1. Assume the profits are normally distributed. a) Find a 95% confidence interval for u, the mean profit of a new car sold by the automobile dealer. b) Find a 95% confidence interval for o, the standard deviation of the profit of a new car sold by the automobile dealer.

Answer 1

Part a

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 3.683333333

S = 1.905168409

n = 6

df = n – 1 = 6 – 1 = 5

Confidence level = 95%

Critical t value = 2.5706

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 3.683333333 ± 2.5706*1.905168409/sqrt(6)

Confidence interval = 3.683333333 ± 2.5706*0.777781746

Confidence interval = 3.683333333 ± 1.9994

Lower limit = 3.683333333 - 1.9994 = 1.6840

Lower limit = 1.6840

Upper limit = 3.683333333 + 1.9994 = 5.6827

Upper limit = 5.6827

Part b

The confidence interval for the 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} ]

We are given

Confidence level = 95%

Sample size = n = 6

Degrees of freedom = n – 1 = 5

Sample standard deviation = S = 1.9052

χ²_{α/2, n – 1} = 12.8325

χ²_{1 -α/2, n– 1} = 0.8312

(By using chi square table)

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

Sqrt[(6 – 1)* 1.9052^2/ 12.8325] < σ < Sqrt[(6 – 1)* 1.9052^2/ 0.8312]

Sqrt(1.4143) < σ < Sqrt(21.8343)

1.1892 < σ < 4.6727

Lower limit = 1.1892

Upper limit = 4.6727