Question

Six measurements were made of the mineral content of spinach, with the following results. Assuming that the population is normally distributed, construct a 99% confidence interval for the population standard deviation of the mineral content of spinach, and write a sentence that interprets the interval.

19.1

20.8

20.8

21.4

20.5

19.7

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 = 0.837655

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)* 0.837655^2 / 16.7496] < σ < sqrt[(6 – 1)* 0.837655^2 / 0.4117]

Sqrt(0.2095) < σ < Sqrt(8.5207)

0.4577 < σ < 2.9190

Lower limit = 0.4577

Upper limit = 2.9190