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Consider a $2^4$ factorial experiment with two blocks within a replication, laid out in two replications with partial confounding (0 and 1 denote factor levels):
The interactions confounded are:
| Replication I - Block 1 | Replication I - Block 2 | Replication II - Block 1 | Replication II - Block 2 |
|---|---|---|---|
| 0000 | 0001 | 0000 | 0010 |
| 0011 | 0010 | 0001 | 0011 |
| 0101 | 0100 | 0110 | 0100 |
| 0110 | 0111 | 0111 | 0101 |
| 1001 | 1000 | 1010 | 1000 |
| 1010 | 1011 | 1011 | 1001 |
| 1100 | 1101 | 1100 | 1110 |
| 1111 | 1110 | 1101 | 1111 |
EXPLANATION
Factorial Experiments , Complete & Partial Co
- Identifying Confounded Factorial Interactions from Layout:
- In a $2^4$ factorial experiment with 2 blocks per replication, the interaction confounded corresponds to the defining contrast linear combination $L = \sum \alpha_i x_i \pmod 2$.
- Replication I Analysis:
- Check Block 1 treatments: 0000, 0011, 0101, 0110, 1001, 1010, 1100, 1111.
- Sum of all four coordinates: $L = x_1 + x_2 + x_3 + x_4 \pmod 2$.
- $0+0+0+0 = 0$
- $0+0+1+1 = 0$
- $0+1+0+1 = 0$
- $0+1+1+0 = 0$
- $1+0+0+1 = 0$
- $1+0+1+0 = 0$
- $1+1+0+0 = 0$
- $1+1+1+1 = 0$
- All units in Block 1 yield $L \equiv 0 \pmod 2$, whereas units in Block 2 yield $L \equiv 1 \pmod 2$. Thus, ABCD is confounded in Replication I.
- Replication II Analysis:
- Check Block 1 treatments: 0000, 0001, 0110, 0111, 1010, 1011, 1100, 1101.
- Sum of first three coordinates: $L = x_1 + x_2 + x_3 \pmod 2$.
- All units in Block 1 yield $x_1 + x_2 + x_3 \equiv 0 \pmod 2$, whereas units in Block 2 yield $x_1 + x_2 + x_3 \equiv 1 \pmod 2$.
- Thus, ABC is confounded in Replication II.
- Conclusion: ABCD in Replication I and ABC in Replication II (Option 4).
Correct option: (4) ABCD in Replication | and ABC in Replication II
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