Why the Experimental Unit Matters
A field trial might measure many plants, but the number of measurements is not automatically the number of independent experimental units. If irrigation is assigned to whole plots, the treatment is applied to plots. Plants sampled within the same plot share that assignment and much of their growing environment. Treating every plant as an independently randomized replicate can make the evidence for treatment differences appear stronger than it is.
This tutorial analyzes a hypothetical trial comparing four irrigation regimes. The main questions are what counts as a replicate, how to fit the one-way ANOVA model, what the omnibus test says, and how to report treatment differences and an effect size.
From the Field Design to a One-Way Model
Suppose a researcher divides a field into separate plots and randomly assigns each plot to one of four irrigation regimes. The response is harvested yield, measured in kilograms per plot. If there are \(k\) regimes and \(n_i\) independently randomized plots under regime \(i\), let \(Y_{ij}\) be the yield from plot \(j\) assigned to regime \(i\). The one-way treatment model is
Here \(\mu_i\) is the mean yield per plot under regime \(i\), and \(\varepsilon_{ij}\) is the deviation of that plot's yield from its regime mean. In the classical one-way ANOVA, the errors are independent, have mean zero and common variance \(\sigma^2\), and follow a normal distribution. Equivalently, \(Y_{ij}\sim N(\mu_i,\sigma^2)\) independently. Independence is supported by the design when plots are separately randomized and do not interfere with one another; normality and similar error spreads are assessed using the residuals, with particular caution when the number of plots is small.
A useful equivalent model writes \(\mu_i=\mu+\tau_i\), where \(\mu\) is a common reference level and \(\tau_i\) is the effect of regime \(i\). The hypotheses concern equality of the \(\mu_i\)'s, so they are unchanged by the choice of reference or by a constraint such as \(\sum_i\tau_i=0\).
Worked Example: Finding the Replicates in a Plot Trial
In a hypothetical trial, four irrigation regimes are each assigned to three plots. A technician measures five plants within every plot and records the mean plant yield for that plot. What are the experimental units, and how many replicates are there per regime?
Irrigation was assigned to plots, so each plot—not each plant—is an experimental unit. There are three independently assigned plots per regime, hence \(n_i=3\) for each regime and \(N=4(3)=12\) experimental units in total. The five plant measurements contribute to a plot's response summary; they do not turn that plot into five independent treatment replicates.
For instance, if the five plant yields in one plot are 4.0, 4.4, 4.6, 5.0, and 5.0 kilograms, the plot's mean plant yield is
If the response of interest is mean plant yield, that 4.60 is one plot-level response. The analysis uses the 12 plot responses, with their actual assignment structure, rather than treating all 60 plants as if irrigation had been independently assigned to each one.
How the Omnibus F Test Is Built
For each regime, the sample mean is \(\bar{x}_i\), and the grand mean across all plots is \(\bar{x}\). The treatment sum of squares measures variation among the regime means, weighted by the number of plots in each regime. The error sum of squares measures variation of plots around their own regime mean.
The total sum of squares is \(SS_{\text{total}}=\sum_i\sum_j(Y_{ij}-\bar{x})^2\). The degrees of freedom are \(k-1\) for treatment and \(N-k\) for error. Dividing each sum of squares by its degrees of freedom gives the corresponding mean square. The test statistic is the ratio of the treatment mean square to the error mean square.
Under \(H_0\) and the model assumptions, this statistic follows an \(F\) distribution with \(k-1\) and \(N-k\) degrees of freedom. A large \(F\) means the observed variation among treatment means is large relative to the within-treatment variation. The p-value is the upper-tail probability under this null model: \(P(F_{k-1,N-k}\geq F_{\text{observed}})\). It is not the probability that \(H_0\) is true.
Worked Example: Testing Four Irrigation Regimes
Consider hypothetical yields, in kilograms per plot, from four plots assigned to each regime. The observations are:
| Regime | Plot yields (kg per plot) | Mean (kg per plot) |
|---|---|---|
| A | 18, 19, 21, 22 | 20 |
| B | 22, 23, 25, 26 | 24 |
| C | 26, 27, 29, 30 | 28 |
| D | 30, 31, 33, 34 | 32 |
There are \(k=4\) regimes and \(N=16\) independently randomized plots. The grand mean is
The treatment sum of squares is
Within each regime, the deviations from its mean are \(-2,-1,1,2\) kilograms, whose squared values sum to \(4+1+1+4=10\). Thus
The degrees of freedom are \(df_{\text{treatment}}=4-1=3\) and \(df_{\text{error}}=16-4=12\). Therefore,
The upper-tail probability for \(F=32.0\) with 3 and 12 degrees of freedom is less than 0.001. At \(\alpha=0.05\), reject \(H_0\): under the one-way model, the data provide evidence that the mean yield is not the same for all four irrigation regimes. The result does not establish that every pair of regimes differs.
For this hypothetical example, the observed residuals are balanced around each regime mean and have the same spread in all four groups. Those features are consistent with similar variances and do not show obvious skewness or outliers. They cannot prove that the errors are normally distributed. The design supplies the basis for treating plots as independent units, provided plots do not influence one another in ways that violate the trial's randomization and response model.
Why the Sum-of-Squares Decomposition Works
The F ratio compares between-regime variation with within-regime variation. The decomposition behind it follows by adding and subtracting each regime mean inside every deviation from the grand mean:
Squaring and summing over all plots gives a within-regime squared term, a between-regime squared term, and a cross-product. For each regime, the cross-product sums to zero because \(\sum_j(Y_{ij}-\bar{x}_i)=0\). Consequently,
This identity shows precisely how total variation is partitioned. Under the equal-means null hypothesis, both mean squares estimate the common error variance \(\sigma^2\); a large ratio is evidence against equal means. The reference \(F\) distribution calibrates how unusual that ratio would be if the null and model assumptions held.
Reporting Treatment Differences and Effect Size
An omnibus test addresses whether all means are equal, not the size or direction of each difference. A useful report includes the treatment means with units, selected mean differences, and a measure of the fraction of observed variation associated with treatment. One common measure for a one-way ANOVA is eta-squared:
Eta-squared describes the sample's variance partition. It is not a measure of the probability that a treatment caused a change, and its value alone does not show which regimes differ.
Worked Example: Describing Differences and Eta-Squared
Use the four-regime trial above, with means 20, 24, 28, and 32 kilograms per plot and sums of squares \(SS_{\text{treatment}}=320\) and \(SS_{\text{total}}=360\).
The estimated difference between regime D and regime A is
In this sample, plots assigned to D yielded an average of 12 kilograms per plot more than plots assigned to A. The adjacent sample-mean differences are each \(4\) kilograms per plot: \(24-20=4\), \(28-24=4\), and \(32-28=4\). These are descriptions of the observed means; the omnibus test by itself does not provide a separate significance test for any one contrast.
The effect size is
Thus, about 88.9% of the observed variation in plot yields in this hypothetical data set is between the four treatment means, with the remaining 11.1% within regimes. This is a sample description, not a claim that irrigation explains that percentage of yield variation in every field or population.
Assumptions, Interpretation, and a Common Pitfall
Before relying on the classical F test, connect each assumption to the actual trial. The plots must be the independently assigned units; separate plants within a plot do not supply independent treatment assignments. The error distribution should be reasonably consistent with normality, and the regime variances should be sufficiently similar for the pooled error mean square to be appropriate. Residual plots can reveal conspicuous outliers, skew, or differences in spread, but small samples limit how much diagnostics can establish. The randomization should also be carried out as described, and treatment application to one plot should not materially alter another plot's response.
- Independence: supported by separately randomized plots and a design without important plot-to-plot interference.
- Similar variances: assessed by comparing residual spreads across regimes, not just by comparing treatment means.
- Normal errors: assessed from residual behavior; the assumption concerns errors within regimes, not that all treatment means are equal.
- Correct unit and response: the analysis must use one appropriate response per independently randomized plot.
Common mistake: counting subsamples as replicates. A student might say, “There are five plants in every plot, so there are 20 observations per regime,” and enter all plant measurements into an ordinary one-way ANOVA as independent data. That reasoning confuses measurement count with treatment-assignment count. If treatment was assigned to plots, the independent replicates are plots. Analyze plot-level responses or use a model that represents the nested plant-within-plot structure; do not claim 80 independent treatment assignments when only 16 plots were randomized.
A second interpretive caution is just as important: a small omnibus p-value says that the observed means are not all alike under the model. It does not say that every treatment pair differs, that the largest sample mean is certainly the best in future fields, or that irrigation alone determines yield. Report the means and differences in context, state the test result and its degrees of freedom, and keep the scope of the conclusion aligned with the randomized plots studied.
Check Your Understanding
Answer these questions using the design and calculations in this tutorial.
- Irrigation is assigned to 10 plots per regime, and six plants are measured per plot. How many independently randomized replicates are there per regime, and what is the experimental unit?
- What does the alternative hypothesis in the four-regime omnibus test assert? What does it leave unspecified?
- For the hypothetical data, verify that \(SS_{\text{total}}=SS_{\text{treatment}}+SS_{\text{error}}\) and state the treatment and error degrees of freedom.
- Explain why \(F=32.0\) with 3 and 12 degrees of freedom supports rejecting equal treatment means at \(\alpha=0.05\), without concluding that every pair differs.
- Interpret \(\eta^2=0.8889\) in the context of the observed plot yields. What does it not establish?