9.1 Confounding in two blocks
Definition 9.1.1. An effect is confounded with blocks when its estimate is also an estimate of a difference between blocks, so that the two cannot be separated.
Consider a \(2^3\) design in two blocks of four runs. Place in one block the four runs with a \(-\) sign in the \(ABC\) column of the table of signs, and in the other the four with a \(+\):
| block \(1\) (\(ABC = -\)) | block \(2\) (\(ABC = +\)) |
| \((1)\), \(ab\), \(ac\), \(bc\) | \(a\), \(b\), \(c\), \(abc\) |
The \(ABC\) contrast is then (block \(2\) total) \(-\) (block \(1\) total), and any difference between the blocks enters it in full: \(ABC\) is confounded with blocks. Every other effect is unaffected.
Proposition 9.1.2. If a \(2^k\) design is divided into two blocks according to the sign of the column of an effect \(E\), and the block effects are additive, then \(E\) is confounded with blocks and the estimate of every other effect is unchanged by the block effects.
Proof. Let the block effects be \(\delta _1\) and \(\delta _2\). They add \(\delta _1\) to every run in block \(1\) and \(\delta _2\) to every run in block \(2\). For the confounded effect, the \(+\) runs are exactly block \(2\), so its contrast gains \(2^{k-1}(\delta _2 - \delta _1)\). For any other effect \(F\), the column of \(F\) is orthogonal to the column of \(E\), so within each block (each sign of \(E\)) the column of \(F\) has as many \(+\) as \(-\) signs: the product \(EF\) is a column, with as many \(+\) as \(-\) signs in each half where \(E\) is constant. Hence the contrast of \(F\) gains \(\delta _1(\text {\#}+ - \text {\#}-)_{\text {block }1} + \delta _2(\text {\#}+ - \text {\#}-)_{\text {block }2} = 0\). □
A convenient way to form the blocks uses the defining contrast \[ L = \alpha _1x_1 + \alpha _2x_2 + \dots + \alpha _kx_k \pmod 2 , \] where now \(x_i = 1\) if factor \(i\) is at its high level and \(0\) if low, and \(\alpha _i = 1\) if factor \(i\) appears in the effect to be confounded and \(0\) otherwise. Runs with \(L = 0\) go in one block and runs with \(L = 1\) in the other. For \(ABC\) in the \(2^3\), \(L = x_1 + x_2 + x_3\), which is \(0\) for \((1)\), \(ab\), \(ac\), \(bc\) and \(1\) for the others, reproducing the blocks above. (\(L\) is the parity of the number of letters a run shares with the confounded effect, and the sign of the effect column is \((-1)^{k_E - L}\) with \(k_E\) the number of letters in \(E\), so the two descriptions agree.)
The block containing \((1)\) is called the principal block. It has a useful structure.
Proposition 9.1.3. Multiply treatment combinations by writing their letters together and deleting any squared letter (so \(ab \cdot bc = ab^2c = ac\)). Then the principal block is closed under multiplication: it is a group, with identity \((1)\). Every other block is obtained by multiplying each run of the principal block by any one run not in it.
Proof. In the \(0/1\) coding, multiplying two runs adds their vectors modulo \(2\), and \(L\) is linear modulo \(2\). If \(L(\mathbf x) = L(\mathbf y) = 0\), then \(L(\mathbf x + \mathbf y) = 0\), so the principal block is closed. If \(\mathbf z\) has \(L(\mathbf z) = 1\), then \(L(\mathbf x + \mathbf z) = 1\) for every \(\mathbf x\) in the principal block, and \(\mathbf x \mapsto \mathbf x + \mathbf z\) is one-to-one, so it gives the whole of the other block. □
In practice one finds the principal block, then generates the others by multiplication. The run order within each block, and the assignment of the blocks to the physical blocks, are randomised.
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