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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!

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  • $\begingroup$ If you can present your table, even empty, it will be helpful to understand your question. $\endgroup$
    – user158565
    Nov 9, 2018 at 3:02
  • $\begingroup$ @a_statistician I've edited the question by adding the given table. If the "Replications" row isn't included I know how to solve this problem exactly, but with it , I'm not sure how to account for that exactly. More importantly, I don't understand why it is there in the first place; it seems unnecessary to me because I don't see why we would want to see if replications is significant or not. $\endgroup$
    – WaveX
    Nov 9, 2018 at 3:13
  • $\begingroup$ Without your table, I though the replication is ERROR or ABC. With your table, I do not think that Replication should be there. Without replication, you already separate the total SS into 8 parts. It is the finest separation and you cannot separate SS further, because you have 8 cells in total (assume $2^3$ design). $\endgroup$
    – user158565
    Nov 9, 2018 at 3:29
  • $\begingroup$ This is what I thought as well, which is why when I received the question it puzzled me. For this particular question do you think it's supposed to be the SSE broken down into two parts: true error and error due to replication? And then in practice, you say that it probably shouldn't be included at all, correct? $\endgroup$
    – WaveX
    Nov 9, 2018 at 3:34
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    $\begingroup$ I have no way to break down SSE. I think Replication should not be in the table at all. $\endgroup$
    – user158565
    Nov 9, 2018 at 3:40

1 Answer 1

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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.

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  • $\begingroup$ Thank you for your answer. So, in this case, you would include "replication" in the ANOVA table? Would you then compare the F-statistic of, say, effect "A" against the error term - where it now has less SS than if replication had not been included? I think the source of my confusion is why in some instances replication is not included in the ANOVA table - even though the experiment is replicated - and sometimes it is not. Is this something you could elaborate on? $\endgroup$
    – Bergson
    May 12, 2023 at 9:21
  • $\begingroup$ @Bergson part of the problem is that the word "replication" is overloaded, meaning different things in different contexts. I interpreted this scenario as something like doing the same study in two different laboratories, with each laboratory restricted to just one observation for each of the 27 treatment combinations. Then the "replication" includes any systematic mean difference between laboratories, which should be taken into account. That's different from doing two "replicates" of each treatment combination in the same laboratory, where the "replicates" would only contribute to the Error. $\endgroup$
    – EdM
    May 12, 2023 at 14:33
  • $\begingroup$ @Bergson this Penn State course page nicely summarizes the potential advantage of separate "replications" or "blocks" in the first sense of my comment above: "the statistical cost of blocking is really the loss of one degree of freedom for error, and the potential gain if the block explains significant variation would be to reduce the size of the MSE and thereby increase the power of the tests." $\endgroup$
    – EdM
    May 12, 2023 at 14:43

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