Latin Squares are very efficient by including two blocking factors, however, the d.f. for error are often too small. In these situations, we consider replicating a Latin Square. Let's go back to the factory scenario again as an example and look at n = 3 repetitions of a 4 × 4 Latin square.
We labeled the row factor the machines, the column factor the operators and the Latin letters denoted the protocol used by the operators which were the treatment factor. We will replicate this Latin Square experiment n = 3 times. Now we have total observations equal to \(N = t^{2}n\).
You could use the same squares over again in each replicate, but we prefer to randomize these separately for each replicate. It might look like this:
Ok, with this scenario in mind, let's consider three cases that are relevant and each case requires a different model to analyze. The cases are determined by whether or not the blocking factors are the same or different across the replicated squares. The treatments are going to be the same but the question is whether the levels of the blocking factors remain the same.
Case 1 Section
Here we will have the same row and column levels. For instance, we might do this experiment all in the same factory using the same machines and the same operators for these machines. The first replicate would occur during the first week, the second replicate would occur during the second week, etc. Week one would be replication one, week two would be replication two and week three would be replication three.
We would write the model for this case as:
\(Y_{hijk}=\mu +\delta _{h}+\rho _{i}+\beta _{j}+\tau _{k}+e_{hijk}\)
where:
\(h = 1, \dots , n\)
\(i = 1, \dots , t\)
\(j = 1, \dots , t\)
\(k = d_{h}(i,j)\)  the Latin letters
This is a simple extension of the basic model that we had looked at earlier. We have added one more term to our model. The row and column and treatment all have the same parameters, the same effects that we had in the single Latin square. In a Latin square, the error is a combination of any interactions that might exist and experimental error. Remember, we can't estimate interactions in a Latin square.
Let's take a look at the analysis of variance table.
AOV

df

df for Case 1

SS

rep=week 
\(n  1\)

2


row=machine 
\(t  1\)

3


column=operator 
\(t  1\)

3


treatment=protocol 
\(t  1\)

3


error 
\((t  1) [n( t + 1)  3]\)

36


Total

\(nt^{2}  1\)

47


Case 2 Section
In this case, one of our blocking factors, either row or column, is going to be the same across replicates whereas the other will take on new values in each replicate. Back to the factory example e.g., we would have a situation where the machines are going to be different (you can say they are nested in each of the repetitions) but the operators will stay the same (crossed with replicates). In this scenario, perhaps, this factory has three locations and we want to include machines from each of these three different factories. To keep the experiment standardized, we will move our operators with us as we go from one factory location to the next. This might be laid out like this:
There is a subtle difference here between this experiment in a Case 2 and the experiment in Case 1  but it does affect how we analyze the data. Here the model is written as:
\(Y_{hijk}=\mu +\delta _{h}+\rho _{i(h)}+\beta _{j}+\tau _{k}+e_{hijk}\)
where:
\(h = 1, \dots , n\)
\(i = 1, \dots , t\)
\(j = 1, \dots , t\)
\(k = d_{h}(i,j)\) the Latin letters
and the 12 machines are distinguished by nesting the i index within the h replicates.
This affects our ANOVA table. Compare this to the previous case:
AOV

df

df for Case 2

SS

rep = factory 
\(n  1\)

2

See text p. 144. 
row (rep) = machine (factory) 
\(n(t  1)\)

9


column = operator 
\(t  1\)

3


treatment = protocol 
\(t  1\)

3


error 
\((t  1) (nt  2)\)

30


Total

\(nt^{2}  1\)

47

Note that Case 2 may also be flipped where you might have the same machines, but different operators.
Case 3 Section
In this case, we have different levels of both the row and the column factors. Again, in our factory scenario, we would have different machines and different operators in the three replicates. In other words, both of these factors would be nested within the replicates of the experiment.
We would write this model as:
\(Y_{hijk}=\mu +\delta _{h}+\rho _{i(h)}+\beta _{j(h)}+\tau _{k}+e_{hijk}\)
where:
h = 1, ... , n
i = 1, ... , t
j = 1, ... , t
\(k = d_{h}(i,j)\)  the Latin letters
Here we have used nested terms for both of the block factors representing the fact that the levels of these factors are not the same in each of the replicates.
The analysis of variance table would include:
AOV

df

df for Case 3

SS

rep = factory 
\(n  1\)

2

See text p. 144. 
row (rep) = machine (factory) 
\(n(t  1)\)

9


column (rep) = operator (factory) 
\(n(t  1)\)

9


treatment protocol 
\(t  1\)

3


error 
\((t  1) [n(t  1)  1]\)

24


Total

\(nt^{2}  1\)

47

Which case is best? Section
There really isn't a best here... the choice of case depends on how you need to conduct the experiment. If you are simply replicating the experiment with the same row and column levels, you are in Case 1. If you are changing one or the other of the row or column factors, using different machines or operators, then you are in Case 2. If both of the block factors have levels that differ across the replicates, then you are in Case 3. The third case, where the replicates are different factories, can also provide a comparison of the factories. The fact that you are replicating Latin Squares does allow you to estimate some interactions that you can't estimate from a single Latin Square. If we added a treatment by factory interaction term, for instance, this would be a meaningful term in the model, and would inform the researcher whether the same protocol is best (or not) for all the factories.
The degrees of freedom for error grows very rapidly when you replicate Latin squares. But usually if you are using a Latin Square then you are probably not worried too much about this error. The error is more dependent on the specific conditions that exist for performing the experiment. For instance, if the protocol is complicated and training the operators so they can conduct all four becomes an issue of resources then this might be a reason why you would bring these operators to three different factories. It depends on the conditions under which the experiment is going to be conducted.
Situations where you should use a Latin Square are where you have a single treatment factor and you have two blocking or nuisance factors to consider, which can have the same number of levels as the treatment factor.