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−1 = 20.2, n1 = 7; x−2 = 26.0, n2 = 9; x−3 = 30.4, n3 = 4; MSE = 32.7 a. Calculate 99% 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=			32.700
df(error)=			17
number of treatments =			3
pooled standard deviation=Sp =√MSE=			5.718

critical t with 0.01 level and 17 df=

2.898

Fisher's (LSD) =(t)*(sp*√(1/ni+1/nj) =

			Lower bound	Upper bound	differ
	(xi-xj )	LSD(ME)	(xi-xj)-ME	(xi-xj)+ME
μ1-μ2	-5.80	8.35	-14.15	2.55	not significant difference
μ1-μ3	-10.20	10.39	-20.59	0.19	not significant difference
μ2-μ3	-4.40	9.96	-14.36	5.56	not significant difference