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Here, the number of levels of Factor A is, k1 = 6.
The number of levels of Factor B is, k2 = 4.
The number of replications per cell is, r = 6.
Thus, the degrees of freedom (df) for Factor A is, k1 – 1 = 5; the df for Factor B is, k2 – 1 = 3.
Since Interaction = Factor A × Factor B, df for Interaction is 5 × 3 = 15.
Since there are 6 levels of Factor A and 4 levels of Factor B, there are a total of 6 × 4 = 24 cells. Each cell has 6 replicates. So, total number of observations, n = 24 × 6 = 144. Thus, total df is, n – 1 = 143.
As a result, error df is, 143 – 5 – 3 – 15 = 120.
The MS or mean of squares is the SS or sum of squares, divided by the corresponding df.
The F for a particular factor is the MS for that factor, divided by the error MS; same for the Interaction.
Using this information, the ANOVA table is completed as follows:
Parts 2 and 3:
The F-test is used for testing the significant effects of Factor A, Factor B, and Interaction.
The numerator and denominator df’s for testing the effect of Factor A are, 5, 120.
The significance level is, α = 0.05.
Thus, the critical value is, F0.05; 5, 120.
Using Excel formula, =F.INV.RT(0.05,5,120), the critical value for Factor A is 2.29.
The numerator and denominator df’s for testing the effect of Factor B are, 3, 120.
Thus, the critical value is, F0.05; 3, 120.
Using Excel formula, =F.INV.RT(0.05,3,120), the critical value for Factor A is 2.68.
The numerator and denominator df&r...
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