Question

A random sample of 51 charge sales showed a sample standard deviation of $47. A 90% confidence interval estimate of the population standard deviation is (Round your answers to 2 decimal places.)

Answer 1

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} ]

We are given

Confidence level = 90%

Sample size = n = 51

Degrees of freedom = n – 1 = 50

Sample standard deviation = S = 47

χ²_{α/2, n – 1} = 67.5048

χ²_{1 -α/2, n– 1} = 34.7643

(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[(51 – 1)*47^2 / 67.5048] < σ < sqrt[(51 – 1)*47^2 / 34.7643]

40.45 < σ < 56.37

Lower limit = 40.45

Upper limit = 56.37