When should we include the "Replicate" in an ANOVA table for Factorial Design?

Question

When I have been learning how to do Factorial Designs ($2^k$ mainly) the ANOVA table would usually include the main effects, all the interaction effects if necessary, and the error. However, on one of my exams it not only included the factors and their interactions, and the error, but it also included "replication".

First off, I was never taught that we could have replication as one of our sources of variation. So I don't know how that could effect the Sum of Squares of Error. I'm just guessing here, but I guess the degrees of freedom should be $n-1$ if we have $n$ replicates, but I am only guessing that because everything else appears to be of the form $\theta - 1$ for some parameter $\theta$.

Up to this point I have never included replication in my ANOVA table. Should replication be included and if it is included how does it affect the Sums of Squares?

The exact example we had was that we had a $3$ factor experiment with two replicates and it gave us part of the ANOVA table, and we had to fill in the rest. If it didn't include Replication, I would know exactly how to fill in the rest of the table, but I don't know exactly why Replication would need to be checked whether or not it is significant or not.

$$\begin{array}{c|c|c|c|c|c|} \text{Source of Variation} & \text{df} & \text{SS} & \text{MS} & \text{F-Value} & \text{p-value} \\ \hline \text{Replication} & & 300 \\ \hline \text{A} & & & 120\\ \hline \text{B} & & 320 \\ \hline \text{C} & & & 60 \\ \hline \text{AB} & & &55 \\ \hline \text{AC} & &160 \\ \hline \text{BC} & &100 \\ \hline \text{ABC} & & \\ \hline \text{Error} & & \\ \hline \text{Total} & & 2000\\ \hline \end{array}$$

We are given that each factor is three levels, two replicates, and that the treatment sum of squares is $1200$

If this table doesn't include the row "Replicates" I know how to fill in all the missing entries. But since it is there, I don't really know how to account for that correctly. Nor do I understand why we would ever want to have that included in our ANOVA

Any help on my confusion would be greatly appreciated!

Answer 1

We are given that each factor is three levels, two replicates, and that the treatment sum of squares is 1200.

The interpretation of this old question might be a bit ambiguous. I'll take it to mean that there were 2 separate experiments (replications), with 1 observation made for each of the 27 combinations of A, B and C (each a factor with 3 levels) within each replication. Thus there were 54 observations, for 53 overall degrees of freedom.

The replication would then be thought of as a 2-level fixed effect that isn't involved in interactions with A, B or C. It could account for an overall difference in means between the two replications.

Then work backwards. In that interpretation, there is 1 degree of freedom associated with replication.

Each 3-level factor uses up 2 degrees of freedom. Thus the missing SS for A is 240, the missing MS for B is 160, and the missing SS for C is 120.

Each 2-way interaction between two 3-level factors uses up $2 \times 2=4$ degrees of freedom. The missing SS for AB is 220. The missing MS values for AC and BC are 40 and 25, respectively.

By my count, that's a total SS of 1160 for A, B, C and their 2-way interactions. If the total "treatment" SS for all combinations is 1200, that leaves 40 for the SS of the 3-way interactions. The three-way interactions among all 3 factors use up $2 \times 2 \times 2=8$ degrees of freedom, for a MS of only 5 for the 3-way interactions.

What's left for Error? For SS, you have 2000 total minus 1200 for treatments, and also minus 300 for replication: 500.

For df, you have 53 total minus 1 for replication, minus 6 for the individual A, B, C combined (2 df each), minus 12 for their two-way interactions (4 df each), and minus 8 for the 3-way interaction: 26 df are left for Error. MS Error is thus 19.2, approximately.