# 7.6 - Example 1

7.6 - Example 1

## Example 7.1

Let's take another example where $$k = 4$$, and $$p = 2$$. This is one step up in the number of treatment factors. And now we have block size $$= 2^{4 - 2}$$ or 4. Again, we have to choose two effects to confound. We will show three cases to illustrate this.

1. Let's try ABCD and a 3-way, ABC. This implies $$ABCD \times ABC = A^{2}B^{2}C^{2}D = D$$ is also confounded. We usually do not want to confound a main effect. It seems that if you reach too far then you fall short. So, the question is: what is the right compromise?
2. We could try ABCD and just AB. In this case we get $$ABCD \times AB = A^{2}B^{2}CD = CD$$. Here we have the 4-way interaction and just two of the 2-way interactions confounded. Can we do better than this? Do you know that one or more of your 2-way interaction effects are not important? This is something you probably don't know, but you might. In this case you could pick this interaction and very carefully assign treatments based on this knowledge.
3. One more try. How about confounding two 3-way interactions? What if we use ABC and BCD. This would give us the interactions of those, $$ABC \times BCD = AB^{2}C^{2}D \text{which} = AD$$.

Which of these three attempts is better? The first try (a) is definitely not good because it confounds the main effect. So, which of the second or third do you prefer? The third (c) is probably the best because it has the fewest lower order interactions confounded. Generally, it is assumed that the higher order interactions are less important, so this makes the (c) case the best choice. Both cases (b) and (c) confound 2-way interactions but the (b) case confounds two of them and the (c) case only one.

If we look at Minitab the program defaults are always set to choose the best of these options. Use this short viewlet to see how Minitab v.17 selects these:

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