# 7.7 - Example 2

7.7 - Example 2

## Example 7.2

Let's try an example where $$k = 5$$, and $$p = 2$$.

a. If we choose to confound two 4-way interactions ABCD and BCDE, this would give us $$ABCD \times BCDE = AB^{2}C^{2}D^{2}E = AE$$, confounded as well, which is a 2-way interaction. Not so good.

If we choose ABC and CDE, this would give us $$ABC \times CDE = ABC^{2}DE = ABDE$$. So, with this choice we are confounding the higher level 4-way interaction and two 3-way interactions instead of the 2-way interaction as above.

Let's see what Minitab chooses...

If you were planning to replicate one of these designs, you would not need to use the same three factors for blocking in each replicate of the design, but instead could choose a different set of effects to use for each replicate of the experiment. More on that later.

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