Question

Q. An experiment was set up to measure the yield provided by each of three catalysts in a certain reaction. The experiment was repeated three times for each catalyst. The reactor yields, in grams, are below. Catalyst 1: 84.33 90.25 85.62 Catalyst 2: 88.44 89.81 86.53 Catalyst 3: 94.71 91.19 92.81 (a) Construct the ANOVA table for this problem. (b) Determine if differences exist between the catalysts at alpha = 0.05. If there are differences in the catalysts, perform the appropriate multiple comparison procedure to determine which catalyst(s) results in the highest yield.

Answer 1

	Group 1	Group 2	Group 3	Total
Sum	260.2	264.78	278.71	803.69
Count	3	3	3	9
Mean, Sum/n	86.7333	88.2600	92.9033
Sum of square, Ʃ(xᵢ-x̅)²	19.3825	5.4278	6.2083

a)

Null and Alternative Hypothesis:  

Ho: µ1 = µ2 = µ3  

H1: At least one mean is different.  

Number of treatment, k =    3

Total sample Size, N =    9

df(between) = k-1 =    2

df(within) = N-k =    6

df(total) = N-1 =    8

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N =    61.9602

SS(within) = SS1 + SS2 + SS3 =    31.0185

SS(total) = SS(between) + SS(within) =    92.9787

MS(between) = SS(between)/df(between) =    30.9801

MS(within) = SS(within)/df(within) =    5.1698

F = MS(between)/MS(within) =    5.9926

p-value = F.DIST.RT(5.9926, 2, 6) =    0.0371

Decision:  

P-value < α, Reject the null hypothesis.  

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	61.9602	2	30.9801	5.9926	0.0371
Within Groups	31.0185	6	5.1698
Total	92.9787	8

b)

At α = 0.05, N-K = 6, t critical value, t_c =T.INV.2T(0.05, 6) = 2.447

LSD = t_c* √(2*MSW*/n) = 2.447√(2*5.1698/3) = 4.54

Comparison	Diff. = (xi - xj)	LSD	Results
x̅1-x̅2	-1.53	4.54	Means are not different
x̅1-x̅3	-6.17	4.54	Means are different
x̅2-x̅3	-4.64	4.54	Means are different