Lesson 3: Balanced and orthogonal designs, fitting the model, and center points
IASSC Body of Knowledge 4.4, Balanced and Orthogonal Designs · 14 practice questions
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Reading
Lesson 2 built the 2ᵏ full factorial and showed how to read main effects
and interactions off it. This lesson asks why the standard 2ᵏ layout lets
you trust those estimates, and what to do once the model is fit. Two
structural properties, balance and orthogonality, make a factorial design
trustworthy. From there, fitting the model, reading its ANOVA table,
checking residuals, and adding center points to test for curvature turn a
table of run results into a defensible conclusion about which X's matter.
A 2x2 factorial with a center point. The sign table shows why the design is balanced and orthogonal: every column, and the AB product column, sums to zero.
Balance and orthogonality
A design is balanced when every level of every factor
appears the same number of times. In the 2x2 layout above, A is low twice
and high twice, and the same is true of B: no level is over-represented.
A design is orthogonal when the coded columns for its
factors and interactions are uncorrelated with each other, checked for a
two-level design by multiplying corresponding entries and summing: if that
dot product is zero for every pair of columns, the design is orthogonal. In
the sign table above, the A and B columns are orthogonal to each other, and
both are orthogonal to the AB interaction column formed by multiplying them
row by row.
Orthogonality lets each effect be estimated independently. In an orthogonal
design, A's estimated effect does not shift depending on whether B is in the
model, because A and B share no information. In a non-orthogonal design, the
estimates become entangled: part of what looks like A's effect may really
belong to B. The standard 2ᵏ full factorial is automatically balanced and
orthogonal.
Fitting the model and the ANOVA table
Once runs are complete, the data is fit to a model, most simply
Y = b0 + b1A + b2B + b12AB, with coefficients equal to one-half the
corresponding effect estimates from Lesson 2. Software then produces an
ANOVA table partitioning Y's total variation into a piece
explained by the model and a piece left as error, exactly as a one-way ANOVA
does:
Source
What it captures
Model (A, B, AB, ...)
Variation in Y explained by the factors and their interactions
Residual (error)
Variation left over after the model, including run-to-run noise
Total
All variation in the observed Y values
The table's F-ratio (F = MS-model / MS-residual, the same logic as
F = MSbetween/MSwithin from Green Belt ANOVA) tests whether the model, or a
given term, explains significantly more variation than noise alone would. A
small p-value on a term means: reject the null hypothesis that the term has
no effect on Y.
Pareto of effects, normal probability plot, and reducing the model
With several factors, a full ANOVA table gets long, so two graphical
shortcuts help spot which effects matter:
A Pareto chart of effects ranks every effect's magnitude
largest to smallest against a significance reference line, so the handful
that matter stand out from the rest.
A normal probability plot of effects plots the effect
estimates against a normal distribution. Noise-driven effects fall in a
straight line near zero; a real effect falls off that line, usually toward
one tail.
Once significant effects are identified, the model is reduced
by dropping insignificant terms and re-fitting. One rule constrains this: the
hierarchy principle. A main effect in a significant
interaction is kept even if its own p-value is not significant, because a model that keeps an interaction but drops its parent main effect is hard to interpret and changes if the factors are recoded.
Only the reduced, stable model becomes the working equation for Y.
Why this matters
The exam likes to ask "should this term stay in the model?" when its main
effect is small but it appears inside a significant interaction. The
hierarchy principle is the tie-breaker: keep it.
Residual diagnostics
A fitted model is only trustworthy if its residuals, the differences
between observed and predicted Y values, behave the way ordinary least
squares assumes. Four checks, mirroring Green Belt regression's residual
analysis, are standard:
Normality: a normal probability plot of the residuals
should fall roughly on a straight line.
Constant variance: residuals plotted against predicted
values or run order should scatter randomly, with no funnel shape or trend.
Independence: residuals plotted in run order should show
no pattern tied to run sequence, which would suggest an uncontrolled variable
drifted through the experiment.
No unusual points: a residual far larger than the rest
deserves investigation before the model is trusted.
For example, if a GCHS team runs a 2² experiment on punch-list rework rate
(factors: crew size and inspection timing) and residuals rise steadily
across the week the experiment ran, that signals something changed during
the week, such as a different inspector, not that the two factors failed to
matter.
R-squared, adjusted R-squared, and predicted R-squared
Three statistics summarize model fit. R-squared (R² = SSR/SST, model
sum of squares over total sum of squares) is the fraction of variation in Y
explained, but it mechanically rises with every added term, even a useless
one. Adjusted R-squared corrects for that:
Adjusted R² = 1 - (1 - R²) x (n - 1) / (n - p - 1)
where n is the number of runs and p is the number of model terms (excluding
the intercept). Predicted R-squared goes further: it leaves
out each run in turn, refits without it, and checks how well the model
predicts the left-out value (using the PRESS statistic, prediction error sum
of squares). If predicted R-squared is much lower than adjusted R-squared,
the model fits these particular runs, not the underlying process, and will
predict new runs poorly.
Worked example. A GCHS 2² rework experiment, replicated twice, has n = 8 runs
and a model with p = 3 terms (two main effects, one interaction). The ANOVA
table gives SST = 100 and model sum of squares SSR = 82, so:
The drop from 0.82 to 0.685 reflects the penalty for 3 terms on 8 runs; a
predicted R² near 0.68 would mean the model is stable, but one near 0.2
would suggest over-fitting.
Center points: testing for curvature and estimating pure error
A 2ᵏ design only runs each factor at two levels, so alone it cannot tell
whether X and Y are truly linear or curve between them. Adding center
points, runs with every factor halfway between low and high, coded
(0, 0, ...), solves two problems at once:
It gives a formal test for curvature: if the center
points' average is close to the corner runs' average, a straight line
adequately fits the range studied. If the center points average meaningfully
higher or lower, that is significant curvature, and the team should move to
a response surface design with quadratic terms.
It gives an estimate of pure error, run-to-run noise
with no factor changed, since every center point runs under identical
conditions, without replicating the whole factorial.
Worked example. A GCHS team adds 3 center points to a 2²
design (4 corner runs) on inspection turnaround time. The 4 corner runs
average y-bar-F = 45.0 days; the 3 center points average y-bar-C = 51.0 days,
with a center-point standard deviation of 1.0 day (2 degrees of freedom).
The curvature sum of squares is:
SS-curvature = (nF x nC x (y-bar-F - y-bar-C)²) / (nF + nC) = (4 x 3 x 6.0²) / 7 = 432 / 7 = 61.7
With 1 degree of freedom for curvature and a pure-error mean square of
1.0² = 1.0 (2 degrees of freedom), F = 61.7 / 1.0 = 61.7, far above the
F critical value of 18.51 at alpha = 0.05. Curvature is significant: the
straight-line model does not fit, and the team should move to a response
surface design.
Practice questions
1. A design is described as "balanced." What does that mean?
The residuals are normally distributed
Every run produced the same output value
Every level of every factor appears the same number of times
The design has an equal number of main effects and interactions
Answer: c. Balance means every level of every factor appears equally often in the design, so no level is over-represented in the data.
2. How is orthogonality between two coded design columns checked?
Their ranges must be equal, meaning both columns span from -1 to +1 in every row
Their row-by-row products must sum to zero (zero dot product)
They must contain the same number of runs, with each factor's runs listed in the same order
Their column averages must be equal to the response average
Answer: b. Two columns are orthogonal when the sum of their corresponding products, their dot product, is zero.
3. Why does orthogonality matter for interpreting a factorial's results?
It removes the need for an ANOVA table, because effects can simply be read off the signs
It doubles the number of runs required, because every column needs a mirror-image run
Each factor's effect can be estimated independently of every other factor's effect
It guarantees the process is in statistical control during the whole experiment
Answer: c. Orthogonal columns carry no shared information, so each effect estimate stands on its own instead of being entangled with another factor.
4. In a factorial ANOVA table, what does a small p-value on a term's F-test indicate?
Reject the null hypothesis that this term has no effect on the response Y
The term should be removed from the model regardless of the hierarchy principle
The design is not orthogonal, so the effect estimates cannot be trusted
Residuals are non-normal, so the ANOVA F-test must be repeated
Answer: a. A small p-value means the term's F-test rejects the null hypothesis that it has no effect on Y.
5. What does a Pareto chart of effects show?
The residuals plotted against predicted values, to check for constant variance
The center-point average versus the corner-point average, to test for curvature
The order the runs were performed in, plotted to look for time trends
Every effect's magnitude, ranked against a significance reference line
Answer: d. A Pareto of effects ranks all effects by magnitude with a reference line for significance, so the few effects that matter stand out.
6. On a normal probability plot of effects, how does a real (significant) effect typically appear?
It cannot be distinguished from noise on this plot without replicating the design
It falls exactly on the straight line along with the noise effects near zero
It only appears if the design is not orthogonal or is unbalanced
It falls off the straight line formed by the noise effects
Answer: d. Noise-driven effects line up near zero on the straight line; a real effect deviates from that line, usually toward one tail.
7. A model has a significant AB interaction, but the main effect of A alone is not significant. Per the hierarchy principle, what should the team do?
Remove A immediately, since its own p-value is not significant and it adds nothing
Remove AB instead, since A is not significant on its own and interactions are optional
Ignore the hierarchy principle for two-level designs
Keep A, because it appears in the significant AB interaction
Answer: d. The hierarchy principle keeps a main effect in the model when it appears in a significant interaction, even if its own effect is small.
8. A residual-versus-run-order plot shows residuals steadily rising across the week an experiment was run. What does this most likely indicate?
R-squared cannot be calculated until the run-order trend has been removed from the data
The center points were mislabeled, so the corner averages are shifted by the same amount
An uncontrolled change over the week is affecting results (independence violated)
The design is not balanced, because later runs used more high-level settings
Answer: c. A trend in residuals over run order suggests something changed over time that the model does not account for, violating the independence assumption.
9. What is the key difference between R-squared and adjusted R-squared?
Adjusted R-squared penalizes the number of model terms, so it does not automatically rise when a useless term is added
R-squared applies only to designs with center points
Adjusted R-squared is calculated only from residual plots
They are always numerically identical
Answer: a. R-squared mechanically increases with every added term; adjusted R-squared corrects for the number of terms and the sample size.
10. Predicted R-squared is much lower than adjusted R-squared. What does that suggest?
The design is not orthogonal, so the coefficient estimates are entangled with each other
The residuals are perfectly normal
The model is over-fit to these runs and will predict new runs poorly
Center points must be removed from the analysis before predicted R-squared is reported
Answer: c. A large gap between predicted and adjusted R-squared is a classic sign of an over-fit model that will not generalize to new data.
11. A model has SST = 100 and a model sum of squares SSR = 82. What is R-squared?
0.18
0.82
1.22
8.2
Answer: b. R-squared = SSR/SST = 82/100 = 0.82.
12. Using n = 8 runs, p = 3 model terms, and R-squared = 0.82, what is the adjusted R-squared?
0.90
0.685
0.82
0.315
Answer: b. Adjusted R-squared = 1 - (1 - 0.82) x (8-1)/(8-3-1) = 1 - 0.18 x 1.75 = 1 - 0.315 = 0.685.
13. What does adding center points to a 2ᵏ factorial primarily let a team do?
Test for curvature and estimate pure error without fully replicating the design
Eliminate the need for orthogonality
Increase the number of factors in the design
Replace the need for an ANOVA table
Answer: a. Center points give a formal curvature test and a pure-error estimate, both without doubling the run count through full replication.
14. A 2² design has 4 corner runs (nF = 4) averaging 45.0 and 3 center points (nC = 3) averaging 51.0. The curvature sum of squares works out to SS-curvature = 61.7 with 1 degree of freedom, against a pure-error mean square of 1.0 with 2 degrees of freedom (F-critical = 18.51 at alpha = 0.05). What should the team conclude?
The design is not orthogonal, so the result is invalid
Curvature is not significant since F is a large number
The center points should be discarded and the test re-run
F = 61.7 exceeds 18.51, so curvature is significant and a straight-line model is not adequate over this range
Answer: d. F = 61.7 far exceeds the critical value of 18.51, so curvature is statistically significant and the team should consider a response surface design.
Glossary
Balanced design
A design in which every level of every factor appears the same number of times.
Orthogonal design
A design whose coded columns have a zero dot product with each other, letting each effect be estimated independently.
ANOVA table (factorial)
The variance-partitioning table (model, residual, total) with an F-test used to judge which terms in a factorial model are significant.
Pareto of effects
A chart ranking every effect estimate by magnitude against a significance reference line.
Hierarchy principle
The rule that a main effect is kept in a reduced model if it appears in a significant interaction, even if its own effect is small.
Residual
The difference between an observed value of Y and the value the fitted model predicts for that run.
Adjusted R-squared
R-squared corrected for the number of model terms and the sample size, so it does not automatically rise as terms are added.
Center point
A run with every factor set halfway between its low and high level, used to test for curvature and estimate pure error.