Discussion
Suppose we wish to have a confounding arrangement in \(2^{r}\) blocks of a \(2^{n}\) factorial experiment. Then, how many interactions have to be confounded ?
Answer: D
The number of confounded interactions is \(2^{r} - 1\).
Dividing the \(2^{n}\) treatment combinations into \(2^{r}\) blocks uses up \(2^{r} - 1\) degrees of freedom between blocks. In a \(2^{n}\) factorial every effect carries exactly one degree of freedom, so \(2^{r} - 1\) effects are confounded with blocks. Of these, \(r\) are chosen independently and the remaining \(2^{r} - r - 1\) are their generalised interactions.
For example, a \(2^{3}\) experiment in \(2^{2} = 4\) blocks confounds 3 effects, such as AB, BC and their generalised interaction AC.
\(r - 1\) and \(2^{r-1}\) undercount, and \(n - r + 1\) does not count block degrees of freedom at all.
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