Discussion
In a \(2^{4}\) factorial design with two blocks of 8 plots each the treatment combinations are allotted as below :
| Block I | Block II |
|---|---|
| a | 1 |
| b | ab |
| ac | c |
| bc | abc |
| d | ad |
| abd | bd |
| cd | acd |
| abcd | bcd |
Hence the confounded treatment combination is :
Answer: A
The confounded interaction is ABD.
Block II contains the control (1), so it is the principal block: every treatment in it has an even number of letters in common with the confounded interaction. Test ABD against Block II:
- ab: a, b (2); c: none (0); abc: a, b (2)
- ad: a, d (2); bd: b, d (2)
- acd: a, d (2); bcd: b, d (2)
All even, while every Block I treatment (a, b, ac, bc, d, abd, cd, abcd) shares an odd number of letters with ABD. ACD fails at once on ab in Block II (one letter, a), BCD fails on ab (one letter, b), and ABCD fails on c (one letter).
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