Homework Answers
N = 40
Replications, r = 10
ΣX = 120
(ΣX)² =14400
ΣX² = 640
SSA = Σ((ΣXⱼ)²/nⱼ) - (ΣX)²/N = (50²/20 + 70²/20) - 14400/40 = 10
SSB = Σ((ΣXᵢ)²/nᵢ) - (ΣX)²/N = (90²/20 + 30²/20) - 14400/40 = 90
SSBN = Σ((ΣX)²/n) - (ΣX)²/N = 100
SSAxB = SSBN - SSA - SSB = 0
SSW = SST - SSA - SSB - SSAxB = 180
SST = ΣX² - (ΣX)²/N = 640 - 14400/40 = 280
dfA = a - 1 = 1
dfB = b-1 = 1
dfAxB = (a-1)*(b-1) = 1
dfW = ab(r-1) = 36
dfT = N-1 = 39
MSA = SSA/dfA = 10/1 = 10
MSB = SSB/dfB = 90/1 = 90
MSAxB = SSAxB/dfAxB = 0/1 = 0
MSW = SSW/dfW = 180/36 = 5
F for Factor A = MSA/MSW = 2
F for Factor B = MSB/MSW = 18
F for interaction = MSAxB/MSW = 0
| ANOVA | ||||
| Source of Variation | SS | df | MS | F |
| Between Treatment | 100 | 3 | ||
| Factor A | 10 | 1 | 10 | 2.00 |
| Factor B | 90 | 1 | 90 | 18.00 |
| Interaction | 0 | 1 | 0 | 0.00 |
| Within | 180 | 36 | 5 | |
| Total | 280 | 39 |
1)
Critical value for Factor A = F.INV.RT(0.05, 1, 36) = 4.113
2)
As F = 2 < 4.113, there is no main effect of Factor A.
3)
Critical value for Factor B = F.INV.RT(0.05, 1, 36) = 4.113
4)
As F = 18 > 4.113, there is main effect of Factor B.
5)
Critical value for Interaction = F.INV.RT(0.05, 1, 36) = 4.113
6)
As F = 0 < 4.113, there is no interaction.
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