The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: SST= 248, SSA= 22, SSB= 21, SSAB= 155.
Set up the ANOVA table and test for significance using alpha= .05 . Show entries to 2 decimals, if necessary. If the answer is zero enter “0”.
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value |
Factor A | |||||
Factor B | |||||
Interaction | |||||
Error | |||||
Total |
Level of factor A, a = 4
Level of factor B,b = 3
Replications, r = 3
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value |
Factor A | 22 | a-1 = 3 |
22/3 = 7.3333 |
7.333/2.0833 = 3.52 |
F.DIST.RT(3.52, 3, 24) = 0.0303 |
Factor B | 21 | b-1 = 2 |
21/2 = 10.5 |
10.5/2.0833 = 5.04 |
F.DIST.RT(5.04, 2, 24) = 0.0149 |
Interaction | 155 | (a-1)*(b-1) = 6 |
155/6 = 25.8333 |
25.8333/2.0833 = 12.4 |
F.DIST.RT(12.4, 6, 24) = 0.0000 |
Error |
248 -22-21-155 = 50 |
ab(r-1) = 24 |
50/24 = 2.0833 |
||
Total | 248 | abr -1 = 35 |
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