Question

The following data were randomly drawn from an approximately normal population. 37,38, 40, 44, 47,50 Based on these data, find a 99% confidence interval for the population standard deviation. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval? x 5 ?

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 = 99%

Sample size = n = 6

Degrees of freedom = n – 1 = 5

Sample standard deviation = S = 5.2026

χ2 α/2, n – 1 = 16.7496

χ2 1 - α/2, n – 1 = 0.4117

(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)* 5.2026^2 / 16.7496] < σ < sqrt[(6 – 1)* 5.2026^2/ 0.4117]

2.8425 < σ < 18.1298

Lower limit = 2.8425

Upper limit = 18.1298