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Full factorials — Examples

1 · Prerequisites

2 · Summary

The examples compute a two-factor interaction on the raw and logarithmic response scales, then add one unequal replicate to show how the weighted design matrix loses orthogonality while still sending only the zero coefficient vector to zero. These calculations distinguish the balanced sign-design identities from assumptions about response additivity or error distributions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Unequal replication removes factorial-column orthogonality

Example

Take the two-factor run set ${(- 1, - 1), (+ 1, - 1), (- 1, + 1), (+ 1, + 1)}$ and replicate the $(- 1, - 1)$ run twice, with each other run observed once. All four treatment combinations are present, but the intercept and first-factor columns have inner product $- 1$ in the five-row design matrix. Thus unequal replication loses orthogonality even though it does not make the four coefficient directions linearly dependent.

Facts & Assumptions

[F1]

Two-level full factorial designs and coded effects: for two factors the full run set contains all four sign pairs, and the columns are $1$ , $x_{1}$ , $x_{2}$ , and $x_{1} x_{2}$ .

[F2]

The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products: the real coordinate inner product is the sum of coordinatewise products.

[F3]

Orthogonal vectors and subspaces, orthogonal and orthonormal sets, and orthonormal bases: two columns are orthogonal exactly when their inner product is zero.

[F4]

Orthogonality of the full two-level factorial contrasts: with one copy of each run, the four columns satisfy $X^{T} X = 4 I_{4}$ .

Verification

Given: The four run combinations in [F1], with multiplicities $2, 1, 1, 1$ in the order shown in the Example. Verify loss of orthogonality and retained full rank.

1.1givenF1F2algebra

The intercept column is $1$ in each row, while the first-factor column takes values $- 1, - 1, + 1, - 1, + 1$ in the five-row design. By [F2], their inner product is $- 1 - 1 + 1 - 1 + 1 = - 1$ .

2.1F3step 1.1

By [F3], orthogonality requires a zero inner product, so step 1.1 shows these columns are not orthogonal. The same four distinct rows, before the extra replicate, still occur, so any vector in the kernel of the five-row matrix lies in the kernel of the original four-row matrix.

3.1F4step 2.1algebra∎

If $b$ is in that kernel, the original four-row matrix $X$ satisfies $X b = 0$ . Then $0 = b^{T} X^{T} X b = 4 b^{T} b$ by [F4], so $b = 0$ . Thus the five-row matrix has full column rank: unequal replication removes orthogonality here but not estimability of these four coefficients.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A two-factor interaction depends on response scale

Example

In a two-factor full factorial, let the four positive responses be $y_{- -} = 1$ , $y_{+ -} = 2$ , $y_{- +} = 3$ , and $y_{+ +} = 6$ . The coded $A B$ effect is $1$ on the raw response scale and $0$ on the logarithmic scale. The same design therefore shows interaction on one response scale and no interaction on another.

Facts & Assumptions

[F1]

Two-level full factorial designs and coded effects: for a nonempty index subset $S$ , its factorial effect is the mean response among runs with $x_{S} = + 1$ minus the mean among runs with $x_{S} = - 1$ .

Verification

Given: The four positive responses in the Example, with factor levels coded $- 1, + 1$ .

1.1givenF1algebra

Here $x_{{1, 2}} = x_{1} x_{2}$ is $+ 1$ on runs $- -$ and $+ +$ , and $- 1$ on runs $+ -$ and $- +$ . By [F1], the raw-scale $A B$ effect is $\frac{1 + 6}{2} - \frac{2 + 3}{2} = 1$ .

1.2givenF1algebra

The log-transformed responses remain indexed by these same four runs, so [F1] uses the same two sign groups. The log-scale $A B$ effect is $\frac{\log 1 + \log 6}{2} - \frac{\log 2 + \log 3}{2} = 0$ , using $\log 1 = 0$ and $\log 6 = \log 2 + \log 3$ .

2.1step 1.1step 1.2algebra∎

The raw-scale interaction effect is nonzero while the log-scale interaction effect is zero, so this table's interaction depends on the declared response scale. The run set alone specifies no response additivity or physical mechanism.