9.1 - \(3^k\) Designs in \(3^p\) Blocks

Let's begin by taking the \(3^k\) designs and we will describe partitioning where you take one replicate of the design and put it into blocks. We will then take that structure and look at \(3^{k-p}\) factorials. These designs are not used for screening as the \(2^k\) designs were; rather with three levels we begin to think about response surface models. Also, \(3^{k}\) designs become very large as k gets large With just four factors a complete factorial is already 81 observations, i.e. \(N = 3^4\). In general, we won't consider these designs for very large k, but we will point out some very interesting connections that these designs reveal.

Reiterating what was said in the introduction, consider the two-factor design \(3^2\) with factors A and B, each at 3 levels. We denote the levels 0, 1, and 2. The \(A \times B\) interaction, with 4 degrees of freedom, can be split into two orthogonal components. One way to define the components is that AB component will be defined as a linear combination as follows:

\(L_{AB}=X_{1}+X_{2}\ (mod3)\)

and the \(AB^2\) component will be defined as:

\(L_{AB^2}=X_{1}+2X_{2}\ (mod3)\)

 \(A\)   \(B\)   \(AB\)  \(AB^{2}\)
0 0 0 0
1 0 1 1
2 0 2 2
0 1 1 2
1 1 2 0
2 1 0 1
0 2 2 1
1 2 0 2
2 2 1 0

In the table above for the \(AB\) and the \(AB^2\) components we have 3 0's, 3 1's and 3 2's, so this modular arithmetic gives us a balanced set of treatments for each component. Note that we could also find the \(A^{2}B\) and \(A^{2}B^{2}\) components but when you do the computation you discover that \(AB^{2}=A^{2}B\) and \(AB=A^{2}B^{2}\).

We will use this to construct the design as shown below.

We will take one replicate of this design and partition it into 3 blocks. Before we do, let’s consider the analysis of variance table for this single replicate of the design.

AOV df
A 3 - 1 = 2
B 3 - 1 = 2
A x B 2 * 2 = 4
Error 3 - 1 = 2
Total 3 - 1 = 2

We have partitioned the \(A \times B\) interaction into \(AB\) and \(AB^2\), the two components of the interaction, each with 2 degrees of freedom. So, by using modular arithmetic, we have partitioned the 4 degrees of freedom into two sets, and these are orthogonal to each other. If you create two dummy variables for each of these factors, \(A\), \(B\), \(AB\) and \(AB^{2}\) you would see that each of these sets of dummy variables are orthogonal to the other.

These pseudo components can also be manipulated using a symbolic notation. This is included here for completeness, but it is not something you need to know to use or understand confounding. Consider the interaction between \(AB\) and \(AB^{2}\). Thus \(AB \times AB^2\) which gives us \(A^2 B^3\) which using modular (3) arithmetic gives us \(A^2 B^0 = A^2 = (A^2)^2 = A\). Therefore, the interaction between these two terms gives us the main effect. If we wanted to look at a term such as \(A^2 B\) or \(A^2 B^2\), we would reduce it by squaring it which would give us: \((A^2 B)^2=AB^2\) and likewise \((A^2 B^2)^2 = AB\). We never include a component that has an exponent on the first letter because by squaring it we obtain an equivalent component. This is just a way of partitioning the treatment combinations and these labels are just an arbitrary identification of them.

Let's now look at the one replicate where we will confound the levels of the AB component with our blocks. We will label these 0, 1, and 2 and we will put our treatment pairs in blocks from the following table.

 \(A\)   \(B\)   \(AB\)  \(AB^{2}\)
0 0 0 0
1 0 1 1
2 0 2 2
0 1 1 2
1 1 2 0
2 1 0 1
0 2 2 1
1 2 0 2
2 2 1 0

Now we assign the treatment combinations to the blocks, where the pairs represent the levels of factors A and B.

\(L_{AB}\)
0 1 2
0, 0 1, 0 2, 0
2, 1 0, 1 1, 1
1, 2 2, 2 0, 2

This is how we get these three blocks confounded with the levels of the \(L_{AB}\) component of interaction.

Now, let's assume that we have four reps of this experiment - all the same - with AB confounding with blocks using the \(L_{AB}\). (each replicate is assigned to 3 blocks with AB confounded with blocks). We have defined one rep by confounding the AB component, and then we will do the same with 3 more reps.

Let's take a look at the AOV resulting from this experiment:

AOV df
Rep 4 - 1 = 3
Blk = AB 3 - 1 = 2
Rep x AB 3 * 2 = 6
Inter-block Total 11
A 3 - 1 = 2
B 3 - 1 = 2
A x B = AB2 3 - 1 = 2
Error (2 + 2 + 2) * (4 - 1) = 18
Total 3 * 3 * 4 - 1 = 35

Note that Rep as an overall block has 3 df. Within reps we have variation among the 3 blocks, which are the AB levels - this has 2 df. Then we have Rep by blk or Rep by AB which has 6 df. This is the inter-block part of the analysis. These 11 degrees of freedom represents the variation among the 12 blocks (3*4).

Next we consider the intra-block part: A with 2 df, B with 2 df and the \(A \times B\) or \(AB^{2}\) component that also has 2 df. Finally we have error, which we can get by subtraction, (36 observations = 35 total df, 35 - 17 = 18 df). Another way to think about the Error is the interaction between the treatments and reps which is \(6 \times 3 = 18\), which is the same logic as in a randomized block design, where the SSE is (a-1)(b-1). A possible confusion here is using the terminology of blocks at two levels, the reps are at an overall level, and then within each rep we have the smaller blocks which are confounded with the AB component.

We now examine another experiment, this time confounding the AB2 factor. We can construct another design using this component as our generator to confound with blocks.

 \(A\)   \(B\)   \(AB\)  \(AB^{2}\)
0 0 0 0
1 0 1 1
2 0 2 2
0 1 1 2
1 1 2 0
2 1 0 1
0 2 2 1
1 2 0 2
2 2 1 0

Using the AB2 then gives us the following treatment pairs (A,B) assigned to 3 blocks:

\(L_{AB^{2}}\)
0 1 2
0, 0 1, 0 2, 0
1, 1 2, 1 0, 1
2, 2 0, 2 1, 2

This partitions all nine of the treatment combinations into the three blocks.

Partial Confounding (optional section) Section

We now consider a combination of these experiments, in which we have 2 reps confounding AB and 2 reps confounding \(AB^{2}\). We again will have 4 reps but our AOV will look a little different:

  AOV df  
  \(Rep\) 4 - 1 = 3  
  \(Blk = AB\) 3 - 1 = 2  
  \(Blk = AB^2\) 3 - 1 = 2  
Inter-block Error \(Rep \times AB\) (2 - 1) * 2 = 2 4
\(Rep \times AB^2\) (2 - 1) * 2 = 2
Inter-block Total   11  
  \(A\) 3 - 1 = 2  
  \(B\) 3 - 1 = 2  
  \(A \times B\) 2 * 2 = 4  
  \(AB\) 3 - 1 = 2  4
  \(AB^2\) 3 - 1 = 2
  \(Error\) 2 * (4 - 1) +
2 * (4 - 1) +
2 * (2 - 1) +
2 * (2 - 1) = 16
  Total 3 * 3 * 4 - 1 = 35

There are only two reps with AB confounded, so \(Rep \times AB = (2-1) * (3-1) = 2 df\) . The same is true for the \(AB^2\) component. This gives us the same 11 df among the 12 blocks. In the intra-block section, we can estimate A and B, so they will have 2 df. \(A \times B\) will have 4 df now, and if we look at what this is in terms of the \(AB\) and the \(AB^2\) component each accounts for 2 df. Then we have Error with 16 df and the total stays the same. The 16 df comes from the unconfounded effects - \(\left( A \colon 2 \times 3 = 6 \text{ and } B \colon 2 \times 3 = 6 \right) \) - that's 12 of these df, plus each of the \(AB\) and the \(AB^{2}\) components which are confounded in two reps, and unconfounded in the other two reps - \( \left(2 \times \left(2-1 \right) = 2 \text { for AB and } 2 \times \left( 2-1 \right) = 2 \text{ for } AB^{2}\right)\) - which accounts for the remaining 4 of the total 16 df for error.

We could determine the Error df simply by subtracting from the Total df, but, if it is helpful to think about randomized block designs where you have blocks and treatments and the error is the interaction between them. Note that here we use the term replicates instead of blocks, so actually we consider replicates as sort of super-blocks. In this case, the error would be the interaction between replicates and unconfounded treatments. This RCBD framework is a foundational structure that we use again and again in experimental design.

This is a good example of the benefit of partial confounding because the interaction of the pseudo factors are confounded in only half of the design, so we can estimate the interaction A*B from the other half. You get overall exactly half the information on the interaction from this partially confounded design.

Confounding a main effect (an important idea) Section

Now let’s think further outside of the box. What if we confound the main effect A? What would this do to our design? What kind of experimental design would this be?

Now we define or construct our blocks by using levels of A from the table above. A single replicate of the design would look like this.

A
0 1 2
0, 0 1, 0 2, 0
0, 1 1, 1 2, 1
0, 2 1, 2 2, 2

Then we could replicate this design four times. Let's consider an agricultural application and say that A = irrigation method, B = crop variety, and the Blocks = whole plots of land to which we apply the irrigation type. By confounding a main effect we're going to get a split-plot design in which the analysis will look like this:

  AOV df
  \(Reps\) 3  
  \(A\) 2  
  \(Rep times A\) 6  
Inter-block Total   11  
  \(B\) 2  
  \(A \times B\) 4  
  \(Error\) 18  
  Total 35  

In this design, there are four reps (3 df), and the blocks within reps are actually the levels of A which has 2 df, \(Rep \times A\) has 6 df. The interblock part of the analysis here is just a randomized complete block analysis of four reps, three treatments, and their interactions. The intra-block part contains B which has 2 df, and the \(A \times B\) interaction which has 4 df. Therefore this is another way to understand a split-plot design, where you confound one of the main effects.

More examples of confounding Section

Let's look at the \(k = 3\) case - an increase in the number of treatments by one. Here we will look at a \(3^3\) design confounded in \(3^1\) blocks, or we could look at a \(3^{3}\) design confounded in \(3^2\) blocks. In a \(3^3\) design confounded in three blocks, each block would have nine observations now instead of three.

To create the design shown in Figure 9-7 below, follow the following commands:

Stat > DOE > Factorial > Create Factorial Design

  • click on General full factorial design,
  • set Number of factors to 3
  • set Number of levels of each factor to 3
  • under options, deselect the randomize button
  • Then use Calc menu and subtract 1 from each of column A, B, and C (We could have initially made levels 0, 1 and 2).

Now the levels of the three factors are coded with (0, 1, 2). We are ready to calculate the pseudo factor, \(AB^{2}C^{2}\), which we will abbreviate as \(AB2C2\).

Label the next blank column, \(AB2C2\). Again, using the Calc menu, let \(AB2C2 = Mod(A + 2 \times B + 2 \times C, 3)\), which creates the levels of the pseudo factor \(L_{AB^{2}C^{2}}\) described on the page 371.

Here is a link to a Minitab project file that implements this: Figure-9-7.MPJ

Let's look at the \(k = 3\) case - a \(3^3\) design confounded in \(3^1\) blocks. In a \(3^3\) design confounded in three blocks, each block would have nine observations now.

A B C
0 0 0
1 0 0
2 0 0
0 1 0
1 1 0
2 1 0
0 2 0
1 2 0
2 2 0
0 0 1
1 0 1
2 0 1
0 1 1
1 1 1
2 1 1
0 2 1
1 2 1
2 2 1
0 0 2
1 0 2
2 0 2
0 1 2
1 1 2
2 1 2
0 2 2
1 2 2
2 2 2

With 27 possible combinations, without even replicating, we have 26 df. These can be broken down in the following manner:

AOV df
\(A\) 2
\(B\) 2
\(C\) 2
\(A \times B\) 4
\(A \times C\) 4
\(B \times C\) 4
\(A \times B \times C\) 8
Total 26

The main effects all have 2 df, the three two-way interactions all have 4 df, and the three-way interaction has 8 df. If we think about what we might confound with blocks to construct a design we typically want to pick a higher order interaction.

The three-way interaction \(A × B × C\) can be partitioned into four orthogonal components labeled, \(ABC, AB^{2}C, ABC^{2} \text{ and } AB^{2}C^{2}\). These are the only possibilities where the first letter has exponent = 1. When the first letter has an exponent higher than one, for instance, \(A^{2}BC\), to reduce it we can first square it, \(A^{4}B^{2}C^{2}\), and then using mod 3 arithmetic on the exponent get \(AB^{2}C^{2}\), i.e. a component we already have in our set. These four components partition the 8 degrees of freedom and we can define them just as we have before. For instance:

\(L_{ABC}=X_{1}+X_{2}+X_{3}\ (mod 3)\)

This column has been filled out in the table below in two steps, the first column carries out the arithmetic (sum) and the next column applies the mod 3 arithmetic:

\(A\) \(B\) \(C\) \(A + B + C\) \(L_{ABC}\)
0 0 0 0 0
1 0 0 1 1
2 0 0 2 2
0 1 0 1 1
1 1 0 2 2
2 1 0 3 0
0 2 0 2 2
1 2 0 3 0
2 2 0 4 1
0 0 1 1 1
1 0 1 2 2
2 0 1 3 0
0 1 1 2 2
1 1 1 3 0
2 1 1 4 1
0 2 1 3 0
1 2 1 4 1
2 2 1 5 2
0 0 2 2 2
1 0 2 3 0
2 0 2 4 1
0 1 2 3 0
1 1 2 4 1
2 1 2 5 2
0 2 2 4 1
1 2 2 5 2
2 2 2 6 0

Using the \(L_{ABC}\)component to assign treatments to blocks we could write out the following treatment combinations for one of the reps:

\(L_{ABC}\)
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

This partitions the 27 treatment combinations into three blocks. The ABC component of the three-way interaction is confounded with blocks.

If we performed one block of this design perhaps because we could not complete 27 runs in one day - we might be able to accommodate nine runs per day. So perhaps on day one we use the first column of treatment combinations, on day two we used the second column of treatment combinations and on day three we use the third column of treatment combinations. This would conclude one complete replicate of the experiment. We can then continue a similar approach in the next three days to complete the second replicate. So, in twelve days four reps would have been performed.

How would we analyze this? We would use the same structure.

AOV df
\(Rep\) 4 - 1 = 3
\(ABC = Blk\) 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
\(AB^{2}C\) 2
\(ABC^{2}\) 2
\(AB^{2}C^{2}\) 2
Error 72
Total 108 - 1 = 107

We have (4 - 1) or 3 df for Rep, ABC is confounded with blocks so the ABC component of blocks has 2 df, the Rep by ABC (3*2) has 6 df. In summary to this point we have twelve of these blocks in our 4 reps so there are 11 df in our inter-block section of the analysis. Everything else follows below. The main effects have 2 df, the two-way interactions have 4 df, and the \(A\times B\times C\)would have 8 df, but it only has 6 df because the ABC component is gone, leaving the other three components with 2 df each.

Error will be the unconfounded terms times the number of reps -1, or 24 × (4 - 1) = 72.

Likewise, \(L_{AB^2 C}=X_{1}+2X_{2}+X_{3}\ (mod 3)\) can also be defined as another pseudo component in a similar fashion.