We again start out with a \(3^3\) design which has 27 treatment combinations and assign them to 3 blocks. What we want to do in this lesson, going beyond the \(3^2\) design, is to describe the AOV for this \(3^3\) design. Then we also want to look at the connection between confounding in blocks and \(3^{kp}\)fractional factorials. This story will be very similar to what we did in the \(2^{kp}\)designs previously. There is a direct analogue here that you will see.
From the previous section we had the following design, \(3^3\) treatments in 3 blocks with the \(ABC\) pseudo factor confounded with blocks, i.e.,
\(L_{ABC}\)  

0  1  2 
0, 0, 0  1, 0, 0  2, 0, 0 
2, 1, 0  0, 1, 0  1, 1, 0 
1, 2, 0  2, 2, 0  0, 2, 0 
2, 0, 1  0, 0, 1  1, 0, 1 
1, 1, 1  2, 1, 1  0, 1, 1 
0, 2, 1  1, 2, 1  2, 2, 1 
1, 0, 2  2, 0, 2  0, 0, 2 
0, 1, 2  1, 1, 2  2, 1, 2 
2, 2, 2  0, 2, 2  1, 2, 2 
The three (color coded) blocks are determined by the levels of the \(ABC\) component of the threeway interaction which is confounded with blocks. If we only had one replicate of this design we would have 26 degrees of freedom. So, let's pretend that this design is Rep 1 and we will add Reps 2, 3, 4, just as we did with the twofactor case. This would result in a total of 12 blocks.
If we did this as our basic design and replicate it three more times our AOV would look like the following:
AOV  df 

\(Reps\)  3 
\(Blocks(Rep)\)  4 × (3  1) = 8 
\(ABC\)

2 
\(Rep \times ABC\)

6 
\(A\)  2 
\(B\)  2 
\(C\)  2 
\(A \times B\)  4 
\(A \times C\)  4 
\(B \times C\)  4 
\(A \times B \times C\)  6 
\(Error\)  72 
Total  107 
We would have Reps with \(3\ df\), blocks nested in Reps with \(2\ df\times 4\ Reps = 8\ df\), then we would have all of the unconfounded effects as shown above. The \(A\times B\times C\) would only have \(6\ df\) because one component (\(ABC\)) is confounded with blocks. Error is \(24\times 3=72\ df\) and our total is \((3^3\times 4)1=107\ df\).
Now, we have written Blocks(Rep) with \(8\ df\) equivalently (in the blue font above) as \(ABC\) with \(2\ df\), and \(Rep\times ABC\) with \(6\ df\), but now we are considering the 0, 1, and 2 as levels of the ABC factor. In this case, \(ABC\) is one component of the interaction and still has meaning in terms of the levels of \(ABC\), just not very interesting since it is part of the threeway interaction. Had we confounded the main effect with blocks, we certainly would have wanted to analyze it, as seen above where the main effect was confounded with blocks. Then it had important meaning and you certainly would want to pull this out and be able to test it.
Now we have a total of \(3\times 4=12\) blocks and the \(11\ df\) among them are the interblock part of the analysis. If we averaged the nine observations in each block and got a single number, we could analyze those 12 numbers and this would be the interblock part of this analysis.
How do we accomplish this in Minitab? If you have a set of data labeled by rep, blocks, and A, B, and C, then you would have everything you need and you can fit a general linear model:
\(Y = Rep\ Blocks(Rep)\ A\ \ B\ \ C\)
This would generate the analysis since \(A\ \ B\ \ C\) expands to all main effects and all interactions in GLM of Minitab.
An Alternate Design  Partial Confounding Section
In thinking about how this design should be implemented a good idea would be to followed this first Rep with a second Rep that confounds \(L_{AB^2 C}\), confound \(L_{ABC^2}\) in Rep three, and finally confound \(L_{AB^2 C^2}\) in fourth Rep. Now we could estimate all four components of the threeway interactions because in three of the Reps they would be unconfounded. There is no information available in the way we had approached it previously. There is lots of information available using this partial confounding strategy of the threeway interactions.
\(3^{kp}\) designs  Fractional Factorial 3level Designs Section
The whole point of looking at this structure is because sometimes we want to only conduct a fractional factorial. We sometimes can't afford 27 runs, certainly not 108 runs. Often we can only afford a fraction of the design. So, let's construct a \(3^{31}\) design which is a \(\frac{1}{3}\) fraction of a \(3^3\) design. In this case, \(N = 3^{31} = 3^2 = 9\), the total number of runs. This is a small, compact design. For the case where we use the \(L_{ABC}\) pseudo factor to create the design, we would use just one block of the design above, and below here is the alias structure:
\(I = ABC\)
\(A = A\times ABC = (A^2 BC) = AB^2 C^2\)
\(A = A\times (ABC)^2 = A^3 B^2 C^2 = (B^2 C^2)^2 = BC\)
\(B = B\times ABC = AB^2 C\)
\(B = B\times (ABC)^2 = A^2 B^3 C^2 = AC\)
\(C = C\times ABC = ABC^2\)
\(C = C\times (ABC)^2 = A^2 B^2 C^3 = (A^2 B^2)^2 = AB\)
Here A is confounded with part of the 3way and part of the 2way interaction, likewise for B and for C. This design only has 9 observations. It has A, B and C main effects estimable and if we look at the AOV we only have nine observations so we can only include the main effects:
AOV  df 

A  2 
B  2 
C  2 
Error  2 
Total  8 
Below is the \(3^3\) design where we partitioned the treatment combinations for one Rep of the experiment using the levels of \(L_{ABC}\). It is of interest to notice that a \(3^{31}\) fractional factorial design is also a design we previously discussed. Can you guess what it is?
If we look at the first light blue column, we can call A the row effect, B the column effect and C the Latin letters, or in this case 0, 1, 2. We would use this procedure to assign the treatments to the square. This is how we get a 3 × 3 Latin square. So, a one third fraction of a \(3^3\) design is the same as a 3 × 3 Latin square design that we saw earlier in this course. Watch the video below to see how this works.
It is important to see the connection here. We have three factors, A, B, C, and before when we talked about Latin squares, two of these were blocking factors and the third was the treatment factor. We could estimate all three main effects and we could not estimate any of the interactions. And now you should be able to see why. The interactions are all aliased with the main affects.
Let's look at another component \(L_{AB^2 C}\) of the three factor interaction: \(A\times B\times C\):
\([)L_{AB^2 C}=X_{1}+2X_{2}+X_{3}\ (mod 3)\)
We can now fill out the table by first plugging in the levels of \(X_{1}\), \(X_{2}\) and \(X_{3}\) from the levels of A, B and C to generate the column \(L_{AB^2 C}\). When you assign treatments to the level of \(L_{AB^2 C}=0\) you get an arrangement that follows (only the principle block filled in):
\(L_{AB^{2}C}\)  

0  1  2 
0, 0, 0  
1, 1, 0  
2, 2, 0  
2, 0, 1  
0, 1, 1  
1, 2, 1  
1, 0, 2  
2, 1, 2  
0, 2, 2 
Then it also generates its own Latin square using the same process that we used above. You should be able to follow how this Latin square was assigned to the nine treatment combinations from the table above.
B  
C  0  1  2  
0  0  1  2  
A  1  2  0  1 
2  1  2  0 
The benefit of doing this is to see that this one third fraction is also a Latin square. This is a Resolution III design, (it has a three letter word generator), and so it has the same properties that we saw at the two level designs, i.e. the main effects are clear of each other and estimable and aliased with higher order interactions including twoway. In fact, since the \(ABC\) and the \(AB^2 C\) are orthogonal to each other  they partition the \(A\times B\times C\) interaction  the two Latin squares we constructed are orthogonal Latin Squares.
Example: Four Factors Section
Now let's take a look at the \(3^{42}\) design. How do we create this design? In this case we would have to pick 2 generators. We have four factors, A, B, C and D. So, let's say we will begin (trial and error) by selecting \(I = ABC = BCD\) as our generators then we will also have the generalized interactions between those generators which are also included. Thus we will also confound:
\(ABC\times BCD = AB^2 C^2 D\), and
\(ABC\times (BCD)^2 = AD^2\)
This is a Resolution II design  there are only two letters in the second component and we should be able to do better.
Let's try again, how about
\(I = ABC = BC^2 D\) as our generators. This confounds:
\(I = ABC = BC^2 D = AB^2 D\)
\(I = ABC = (BC^2 D)^2 = AC^2 D^2\)
This is much better because there is nothing smaller than a three letter word in the generator set so this is a Resolution III design. Now, how do we generate the design? It is a design with four factors but how many observations are there? Nine. It is still a design with only nine observations, or a \(\frac{1}{9}^{th}\) fraction of a \(3^4\) design or 81 observations. If we can write out the basic design with nine observations, which we can do with A and B, it gives us the basic design, and then we use our generators to give us C and D. We can use \(ABC\) such that:
\(L_{ABC} = 0\) this principle fraction implies that \(X_{3}=2X_{1} + 2X_{2}\ (mod 3)\).
\(L_{BC^2 D} = 0\) this implies that \(X_{4} = 2X_{2} + X_{3}\ (mod 3)\)
If we were confounding this in blocks we will want a principal block where these two defining relationships are both zero. You will see that by defining \(X_{3}\) and \(X_{4}\) in this way results in \(ABC\) being equal to zero. Take a look and make sure that you understand how column C was generated by the function \(X_{3}=2X_{1}+2X_{2}\ (mod\ 3)\) yet still preserves the principle implied where \(L_{ABC}=0\). Also, by the same process column D was generated using the function \(X_{4} = 2X_{2} + X_{3}\ (mod 3)\) in such a way that it preserves the principle implied where in \(L_{BC^2 D}=0\).
A  B  C  D 

0  0  0  0 
1  0  2  2 
2  0  1  1 
0  1  2  1 
1  1  1  0 
2  1  0  2 
0  2  1  2 
1  2  0  1 
2  2  2  0 
Tip: Hover over the values in the C and D columns to see the corresponding substitutions.
And so, the \(3^{42}\) design is equivalent to the GraecoLatin square. There are two Latin squares, one for each component, C and D, superimposed as shown below:
So we can see that the GraecoLatin Square with three treatments is simply a fractional factorial of this \(3^4\) design!