Discussion
A \(2^{3}\) design with 4 replicates are under consideration. Two of the replicates are shown below :
| Replicate 1 | Replicate 2 | ||
|---|---|---|---|
| (1) c ab abc | a b ac bc | (1) a bc abc | b c ab ac |
Identify which of the treatment combinations are partially confounded in each replicate ?
Answer: B
The confounded interactions are AB and BC.
In each replicate, the confounded effect is the one whose plus and minus treatment combinations form the two blocks. The sign of a treatment in AB is + when it contains both or neither of \(a\) and \(b\).
Replicate 1: the block {(1), \(c\), \(ab\), \(abc\)} contains exactly the treatments with an even number of letters from \(\{a, b\}\), the + set of AB, and the other block is the - set. So AB is confounded.
Replicate 2: the block {(1), \(a\), \(bc\), \(abc\)} has an even number of letters from \(\{b, c\}\) in each treatment, the + set of BC. So BC is confounded. Each is lost only in its own replicate, which is partial confounding.
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