The short answer
Ask whether the factor differences stay parallel.
A crossed two-way ANOVA models a continuous response using two categorical factors, their interaction and residual variation. The interaction null hypothesis says the systematic difference between supplement methods is the same at every dose—or, equivalently, the dose pattern is the same for both supplements.
The exact ToothGrowth design
R's built-in ToothGrowth dataset records odontoblast
length in 60 guinea pigs. Each animal received one of three
vitamin C doses—0.5, 1 or 2 mg/day—delivered as orange juice
(OJ) or ascorbic acid (VC). That is a balanced 3 × 2 crossed
design with ten independent observations in every cell.
| Dose | OJ mean (n = 10) | VC mean (n = 10) | OJ − VC |
|---|---|---|---|
| 0.5 mg | 13.23 | 7.98 | 5.25 |
| 1 mg | 22.70 | 16.77 | 5.93 |
| 2 mg | 26.06 | 26.14 | −0.08 |
The exact ANOVA result
| Source | df | F | P | Partial η² |
|---|---|---|---|---|
| Supplement × dose | 2, 54 | 4.107 | 0.02186 | 0.132 |
| Dose | 2, 54 | 92.000 | 4.05×10⁻¹⁸ | 0.773 |
| Supplement | 1, 54 | 15.572 | 0.000231 | 0.224 |
The interaction is the first result to interpret: F(2, 54) = 4.107, P = 0.02186. The mean OJ–VC difference is about 5.25 units at 0.5 mg and 5.93 units at 1 mg, then nearly disappears at 2 mg (−0.08). The overall supplement average of 3.70 units blends those different dose-specific comparisons.
What to examine after the interaction
- Plot cell means and raw observations. A line plot makes nonparallel profiles visible, but the raw values show overlap, spread and unusual observations.
- Name the simple effects in advance. In this example, the model-based OJ–VC tests give P = 0.00209 at 0.5 mg, P = 0.000590 at 1 mg and P = 0.961 at 2 mg.
- Control the intended family. Select the comparisons and multiplicity method from the scientific question; do not search every possible pair after seeing the interaction.
- Report effect sizes and intervals. A P value does not show how large or precise the interaction or cell differences are.
Design and assumption checks
- Each row is an independent experimental unit. Technical replicates do not create ten independent animals.
- The residual distribution and variance are plausible within the factorial model; inspect residuals rather than testing six cells in isolation.
- Every dose occurs with every supplement. Missing or unequal cells require explicit choices about the estimand and sums of squares.
- Dose is treated as a three-level categorical factor here. A continuous dose-response model answers a different question.
A compact reporting template
“Odontoblast length was analyzed with an ordinary two-way ANOVA including supplement, dose and their interaction (n = 10 independent animals per cell). The supplement-by-dose interaction was F(2,54) = 4.107, P = 0.02186, partial η² = 0.132. Cell means and pre-specified simple comparisons were reported because the OJ–VC difference varied by dose.”
What Plotwright checks
Plotwright's executable published-example suite submits all 60 measurements to the production two-way ANOVA engine. The gate checks the supplement, dose and interaction F tests against the R benchmark with explicit tolerances. Separate tests cover sums of squares, cell means, simple effects, multiple comparisons, effect sizes and residual diagnostics.
Those checks validate defined numerical routes. They cannot establish independence, justify the factor model, select a comparison family or turn an interaction into a causal claim.
Sources and reproducibility
- R datasets documentation: ToothGrowth—the 60-observation dataset, factors, doses and original references.
- NIST/SEMATECH: two-way crossed ANOVA—the factorial model, variance partition, tests and assumptions.
- Kim (2014), interaction in two-way ANOVA—an interaction-first explanation for clinical researchers.
- Plotwright statistical validation—the published expected values, tolerance and evidence boundaries.