8.5 Full Factorial Experiments and Design Selection

What a full factorial is, and why the BoK lists it separately

A full factorial experiment runs every combination of every level of every factor. For $k$ factors at two levels with $n$ replicates of each combination, the run count is:

$$N = n \times 2^k$$

Because every combination is actually run, a $2^k$ design estimates all $2^k - 1$ effects -- $k$ main effects, $\binom{k}{2}$ two-factor interactions, and every higher-order interaction -- with no aliasing whatever. That single property is what separates it from the fractional designs of section 8.6, and it is why the Body of Knowledge lists full factorial experiments (VII.A.6) as a topic distinct from two-level fractional factorials (VII.A.5).

Read the last column as the cost of completeness. At $k = 6$, 42 of the 63 estimable effects are three-factor interactions and above, which are almost never real. That is the economic argument for fractionating -- but only once $k$ is large. At $k \le 4$ the full factorial is usually the cheaper decision, because a fraction that later needs resolving costs more in total runs than the full design would have.

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The design matrix, standard order, and orthogonality

Write the design in standard (Yates) order: the first column alternates $-,+$; the second alternates in pairs; the third in fours. Interaction columns are formed by multiplying the parent columns row by row.

Two properties of this matrix carry the whole analysis. It is balanced -- each factor is run at its high level exactly as often as at its low level -- and it is orthogonal -- the element-by-element product of any two columns sums to zero. Orthogonality means each effect is estimated independently of the others: dropping a non-significant term from the model leaves every remaining estimate unchanged, and the sums of squares add exactly to the total. Standard order is also the order in which the design is written, never the order in which it is run; runs are randomized (section 8.2).

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From contrast to effect to sum of squares

For any column, the contrast is the sum of the run totals carrying a $+$ sign minus the sum of those carrying a $-$ sign. Three quantities follow directly:

$$\text{Effect} = \frac{\text{Contrast}}{n \, 2^{k-1}} \qquad SS = \frac{(\text{Contrast})^2}{n \, 2^{k}} \qquad \hat{\beta} = \frac{\text{Effect}}{2}$$

The effect is the change in the mean response moving from the low to the high setting, so it equals $\bar{y}_{+} - \bar{y}_{-}$. The regression coefficient for a coded model is half the effect, because the coded variable moves two units ($-1$ to $+1$) while the effect spans that same range. Confusing the two is a routine exam trap.

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Worked example: a $2^3$ with two replicates

A team investigates weld shear strength (MPa) against $A$ = weld current, $B$ = electrode force, and $C$ = surface cleaning. Sixteen runs, fully randomized, $n = 2$.

The grand mean is 56.00 MPa. Applying the formulas above:

Replication is what supplies those 8 error degrees of freedom: $2^k(n-1) = 8 \times 1$. The critical value $F_{0.05,\,1,\,8} = 5.32$, so $A$, $B$, and $AB$ are significant and $C$, $AC$, $BC$, and $ABC$ are not. The model explains $R^2 = 2206.52 / 2258.74 = 97.7\%$ of the variation.

Note what happens to $C$. Its effect of $+2.25$ MPa is not zero, and a manager looking only at the number would call cleaning worthwhile. With $F = 3.10$ against a critical value of 5.32, the design cannot distinguish it from run-to-run noise. Report it as undetected at this sample size, not as proven absent -- with $n = 2$ the experiment simply lacks the power to resolve an effect that small.

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Reading the interaction

$AB$ is significant, so the effect of electrode force depends on the current setting. Collapse the design to the four $A \times B$ cells:

Raising electrode force is worth 4.6 MPa at low current and 18.5 MPa at high current. The interaction effect is exactly half that difference: $(18.500 - 4.600)/2 = 6.950$, matching the ANOVA table. On an interaction plot the two lines are non-parallel; parallel lines mean no interaction.

The governing rule: when an interaction is significant, its parent main effects must not be interpreted in isolation. "Increase force by one setting to gain 11.55 MPa" is wrong at either individual current level -- 11.55 is only the average across both. Recommend factor settings as a combination.

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When there is no replication

An unreplicated $2^k$ ($n = 1$) has zero degrees of freedom for error, so no $F$ test exists. Two standard remedies apply:

1. Pool negligible terms. Drop the three-factor and higher interactions into the error term on the sparsity-of-effects argument. In the example above, pooling $AC$, $BC$, and $ABC$ gives $MS_E = 4.7664$ with 11 degrees of freedom. Pool only terms that are genuinely small; pooling a real effect deflates the error and manufactures significance elsewhere.
2. Plot the effects. On a normal probability plot, inert effects scatter along a straight line through the origin because they are noise from a common distribution; active effects fall off it. A half-normal plot does the same with absolute values, which makes the departures easier to see.

Lenth's method (Lenth, 1989) makes that judgment formal without replication. From the $m$ estimated effects, compute:

$$s_0 = 1.5 \times \text{median}|\hat{\theta}_i| \qquad PSE = 1.5 \times \text{median}\{|\hat{\theta}_i| : |\hat{\theta}_i| < 2.5 s_0\}$$

Trimming at $2.5 s_0$ removes the active effects so the remainder estimates noise. For an unreplicated $2^4$ with the 15 absolute effects $0.3,\ 0.4,\ 0.5,\ 0.6,\ 0.7,\ 0.8,\ 0.9,\ 1.0,\ 1.1,\ 1.2,\ 1.5,\ 3.1,\ 4.8,\ 8.6,\ 12.4$: the median is 1.0, so $s_0 = 1.5$ and the trim point is 3.75. The twelve surviving values have median 0.85, so $PSE = 1.275$. The margin of error is $ME = t_{0.975,\,d} \times PSE$ with $d = m/3 = 5$, giving $ME = 2.571 \times 1.275 = 3.28$. Only the three effects of magnitude 4.8, 8.6, and 12.4 clear it; the effect of 3.1 does not.

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Center points and the test for curvature

A two-level design fits a plane. It cannot detect curvature, because with only two levels per factor there is nothing between them to bend. Adding $n_C$ center point runs at the mid-setting of every factor fixes this cheaply and supplies pure-error degrees of freedom at the same time:

$$SS_{\text{Curvature}} = \frac{n_F \, n_C \, (\bar{y}_F - \bar{y}_C)^2}{n_F + n_C}$$

If the true surface is planar, the centre response should equal the average of the factorial corners. A gap between them is curvature.

Worked case. Eight factorial runs average $\bar{y}_F = 52.6$; four center runs (53.4, 56.1, 54.8, 55.3) average $\bar{y}_C = 54.9$ with $s^2 = 1.2867$ on 3 degrees of freedom.

$$SS_{\text{Curvature}} = \frac{8 \times 4 \times (52.6 - 54.9)^2}{12} = \frac{32 \times 5.29}{12} = 14.107$$ $$F = \frac{14.107}{1.2867} = 10.96 \quad \text{versus} \quad F_{0.05,\,1,\,3} = 10.13$$

Curvature is detected, but only just. With four center points the test carries 3 error degrees of freedom and an enormous critical value, so it will miss moderate curvature entirely -- budget more center runs when curvature is a live question. A significant result means the two-level model is inadequate and the design should be augmented with axial points into a central composite design (section 8.7). Center points also cost nothing in balance: they are orthogonal to every factorial column, so adding them changes no effect estimate.

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The bridge to fractional designs

When the full design will not fit the budget, assign an extra factor to a high-order interaction column. Assigning $D = ABC$ in the $2^3$ above produces a $2^{4-1}$ in 8 runs with defining relation $I = ABCD$, and every effect acquires an alias found by multiplying it by that relation. Resolution equals the length of the shortest word in the defining relation -- here 4, so main effects are clear of two-factor interactions but two-factor interactions are aliased in pairs ($AB = CD$, $AC = BD$, $AD = BC$). Section 8.6 develops this fully.

The design-selection consequence is worth stating plainly. Suppose that $2^{4-1}$ returns a significant $AB + CD$ string. A full fold-over -- reversing every sign -- de-aliases main effects from two-factor interactions but leaves $AB$ still aliased with $CD$. Separating them needs a single-factor fold-over on $A$: 8 more runs reversing only column $A$, which frees $A$ and all of its two-factor interactions. Total spend: 16 runs, exactly the $2^4$ full factorial that was avoided in the first place, now collected in two sequential blocks. Fractionate when $k$ is large or the factors are likely inert; run the full factorial when $k \le 4$ and interactions are the actual question.

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Choosing the design

Three-level full factorials ($3^k$) do estimate quadratic effects directly, but 27 runs for three factors and 81 for four make them a poor trade against a $2^k$ with center points followed by a response surface augmentation, which reaches the same model for far fewer runs. Reach for $3^k$ mainly when a factor is genuinely categorical with three states.