Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 30 / 5 / Solution

Code the two levels of each factor as $\pm 1$. For a factor subset $A$, its factorial contrast column is the character ${\chi }_A(x)={\prod }_{j\in A}{x}_j$. Products of columns correspond to symmetric differences of subsets, because $x_j^2=1$.

Let $G_1,\dots ,G_k$ be defining words whose incidence vectors are linearly independent over ${\mathbb{F}}_2$. Write $x_j=(-1{)}^{u_j}$, with $u_j\in {\mathbb{F}}_2$. Specifying signs ${\chi }_{G_i}(x)=s_i$ is then a rank-$k$ system of binary linear equations. Every one of its $2^k$ right-hand sides has $2^{n-k}$ solutions. Thus $k$ independent defining contrasts split the full $2^n$ design into $2^k$ equal fractions. Independence is necessary: dependent defining words do not yield $2^k$ distinct sets.

The defining contrast subgroup has $2^k$ words. On a selected fraction its word $G$ has a constant sign $s_G$, soHence the coefficients of $A$ and $AG$ cannot be estimated separately without assumptions or extra runs: this is aliasing in a fractional factorial design. Conversely, averaging a character over the affine binary solution space vanishes unless its word is in the defining subgroup, so these signed subgroup cosets give all aliases. Here a lowercase treatment label records the factors at their high levels, and $(1)$ puts all five at their low levels; signs must be retained when listing aliases.