Question

A commercial farm uses a machine that packages strawberries in six ounce portions. A sample of 19 19 packages of strawberries has a variance of 0.41 0.41 . Construct the 80% 80 % confidence interval to estimate the variance of the weights of the packages prepared by the machine. Round your answers to two 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 = 80%

Sample size = n = 19

Degrees of freedom = n – 1 = 18

Sample standard deviation = S = 0.640312

χ2 α/2, n – 1 = 25.9894

χ2 1 - α/2, n – 1 = 10.8649

(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[(19 – 1)*0.41 / 25.9894] < σ < sqrt[(19 – 1)*0.41 / 10.8649]

0.5329 < σ < 0.8242

Lower limit = 0.53

Upper limit = 0.82