Question

STATISTICS. CONFIDENCE REGIONS.

The daily number of pieces manufactured by an "A" machine in 5 days has been : 50,48, 53, 60, 37 ; while, in those same days, a "B" machine has manufactured :40, 51, 62, 55, 64.

Assume that the two populations studied, number of pieces manufactured by the machines A and B, follow a normal distribution not necessarily with the same variance.

Construct a confidence interval of level 0.90 for the quotient of variances.

Thank you for your explanations.

Answer 1

for machine A

= 49.6, s1 = 8.3845, n1 = 5

for machine B

= 54.4, s2 = 9.6073, n2 = 5

Df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

= ((8.3845)^2/5 + (9.6073)^2/5)^2/(((8.3845)^2/5)^2/4 + ((9.7073)^2/5)^2/4) = 8

At 90% confidence interval the critical value is t^* = 1.860

The 90% confidence interval the critical value is

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (49.6 - 54.4) +/- 1.86 * sqrt(((8.3845)^2/5 + (9.6073)^2/5)

= -4.8 +/- 10.61

= -15.41, 5.81