Question

The following statistics are computed by sampling from three normal populations whose variances are equal: (You may find it useful to reference the t table and the q table.)

x−1x−1 = 19.8, n₁ = 4; x−2x−2 = 24.0, n₂ = 10; x−3x−3 = 28.0, n₃ = 6; MSE = 27.8

a. Calculate 95% confidence intervals for μ₁ − μ₂, μ₁ − μ₃, and μ₂ − μ₃ to test for mean differences with Fisher’s LSD approach. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

Answer 1

MSE=			27.800
df(error)=			17
number of treatments =			3
pooled standard deviation=Sp =√MSE=			5.273

critical t with 0.05 level and 17 df=

2.110

Fisher's (LSD) =(t)*(sp*√(1/ni+1/nj)

			Confidence interval
			Lower bound	Upper bound	differ
	(xi-xj )	ME	(xi-xj)-ME	(xi-xj)+ME
μ1-μ2	-4.20	6.58	-10.78	2.38	not significant difference
μ1-μ3	-8.20	7.18	-15.38	-1.02	significant difference
μ2-μ3	-4.00	5.74	-9.74	1.74	not significant difference