Why the Assumptions Matter
A one-way analysis of variance compares the means of several groups using variation between groups relative to variation within groups. The F statistic is meaningful for inference only when the model describes how the observations vary around their group means. In particular, the usual one-way fixed-effects test relies on independent observations, normally distributed errors, and a common error variance. These are assumptions about the data-generating process, not conclusions guaranteed by calculating an F statistic.
A group mean can still be calculated when an assumption is doubtful. The concern is whether the usual standard errors, reference distribution, p-values, and resulting decisions are trustworthy. The assumptions play different roles, so it is useful to identify each one separately before interpreting a test.
The Fixed-Effects Model and Its Assumptions
Suppose there are \(k\) groups, with \(n_i\) observations in group \(i\), for \(i=1,\ldots,k\). Write \(Y_{ij}\) for observation \(j\) in group \(i\). The one-way fixed-effects model is
Here, \(\mu_i\) is the population mean for group \(i\). Equivalently, \(\mu\) is a reference grand mean and \(\tau_i\) is the fixed effect for group \(i\). A common convention is \(\sum_{i=1}^{k}\tau_i=0\), which makes \(\mu\) the unweighted average of the group means under that parameterization. The assumptions concern the errors \(\varepsilon_{ij}\), the deviations from the relevant group means.
These statements do not say that all observations have the same mean: group means \(\mu_i\) may differ. Nor do they say that group membership must be random. In a fixed-effects analysis, the groups being compared are the particular groups of interest. The assumptions describe variation among observations within and across those groups.
What Each Assumption Does
| Assumption | Meaning in the model | Role in inference |
|---|---|---|
| Independence | One observation’s error does not depend on another observation’s error. | Supports the usual calculation of sampling variation. Dependence can make the nominal amount of information too large or too small. |
| Normal errors | Within each group, deviations from the group mean follow a normal distribution. | Supports exact small-sample F-test inference under the model. |
| Common variance | Every group has the same error variance, \(\sigma^2\). | Allows the within-group variation to estimate one shared error variance for the F statistic. |
The roles are related but not interchangeable. A normal distribution does not repair dependence between repeated observations. Equal variances do not make strongly skewed errors normal. A large sample may make some departures less consequential for approximate inference, but it does not change what the stated model assumes. How serious a departure is depends on its size, the group sizes, and the design; no single rule makes every violation harmless.
Why the Within-Group Variance Estimates the Common Error Variance
The connection between the common-variance assumption and the within-group part of ANOVA can be seen directly. For group \(i\), let \(\bar{Y}_i\) be its sample mean and define its sample variance by
Under the model, \(Y_{ij}-\bar{Y}_i=\varepsilon_{ij}-\bar{\varepsilon}_i\), because subtracting the sample mean removes the group mean. If the errors are independent, have mean zero, and each have variance \(\sigma^2\), then
The first expectation is \(n_i\sigma^2\), since there are \(n_i\) errors, each with second moment \(\sigma^2\). Independence gives \(\operatorname{Var}(\bar{\varepsilon}_i)=\sigma^2/n_i\), so \(E[n_i\bar{\varepsilon}_i^2]=\sigma^2\). Dividing by \(n_i-1\) shows \(E[S_i^2]=\sigma^2\). Pooling within-group sums of squares therefore estimates the shared variance. If the groups do not in fact share a variance, treating their variation as one common quantity is not justified by this argument.
Normality has a distinct role: with the model’s independence and common-variance conditions, it supplies the distributional conditions for the usual exact F inference, especially relevant in small samples. The assumption is about errors within groups, not about the combined set of responses after groups with different means have been mixed together.
Worked Example: Assessing the Design Before an ANOVA
A researcher compares the average growth of seedlings under three lighting conditions. There are 15 seedlings, each grown in a separate pot. Five pots are assigned to each light condition, and one growth measurement is recorded per pot. Assume the measurements are made once, with no shared tray or other known source of dependence. State the model and assess what the design says about the assumptions.
Method. Write the group-mean model first, then evaluate whether the stated design supports independent observations. Do not infer normality or equal variances merely from the group counts.
Model. Let \(Y_{ij}\) be the growth measurement for pot \(j\) under lighting condition \(i\), where \(i=1,2,3\) and \(j=1,\ldots,5\). The model is \(Y_{ij}=\mu_i+\varepsilon_{ij}\). For the usual test, the errors would need to be independent and each distributed as \(N(0,\sigma^2)\).
Independence. The 15 pots are distinct experimental units, and each contributes one measurement. The description gives no repeated measurements on the same pot and no shared unit that would make several recorded values dependent. Thus the design supports treating the observations as independent, subject to the stated experimental setup.
Normality and common variance. Five observations per group do not establish either condition. The data or substantive knowledge would need to make normal within-group errors and a common variance plausible. The design alone cannot confirm them.
Conclusion and trap avoided. The design supports independence, but the model still requires the other two assumptions. A common mistake is to say that equal group sizes prove equal variances; having five seedlings in each condition says nothing by itself about how much seedling growth varies within each condition.
Worked Example: Comparing Observed Group Variances
In a hypothetical test of three coating formulas, recorded durability scores are: formula A, 8, 10, 12, 10; formula B, 18, 19, 21, 22; formula C, 29, 30, 30, 31. Calculate the sample variance for each formula and describe what the results do, and do not, establish about the common-variance assumption.
Method. Calculate each group’s mean and squared deviations, then divide the sum of squared deviations by \(n_i-1=3\). This compares observed within-group spread; it is not a proof about population variances.
Formula A. The mean is \((8+10+12+10)/4=40/4=10\). The squared deviations are \((8-10)^2=4\), \((10-10)^2=0\), \((12-10)^2=4\), and \((10-10)^2=0\). Their sum is \(8\), so \(S_A^2=8/3\approx2.667\).
Formula B. The mean is \((18+19+21+22)/4=80/4=20\). The squared deviations are \(4,1,1,4\), with sum \(10\). Thus \(S_B^2=10/3\approx3.333\).
Formula C. The mean is \((29+30+30+31)/4=120/4=30\). The squared deviations are \(1,0,0,1\), with sum \(2\). Thus \(S_C^2=2/3\approx0.667\).
The largest observed variance is \(10/3\), and the smallest is \(2/3\), giving a ratio of \((10/3)/(2/3)=5\). This is a noticeable difference among these sample variances and should make an analyst cautious about treating the spreads as interchangeable. With only four observations per group, the sample variances are uncertain; their ratio does not, by itself, prove that the population variances differ.
Conclusion and trap avoided. The calculations raise a question about the common-variance assumption but do not settle it. The trap is to interpret a variance ratio of 5 as a definitive population result without accounting for the small samples and sampling variability.
Worked Example: Normal Errors Do Not Require Normal Pooled Responses
Suppose a hypothetical two-group study has group means of 10 and 20. In each group, errors are independent and normally distributed with mean 0 and variance 4, and the groups have equal sizes. Explain why the combined response distribution need not be normal, and calculate its variance when a response is selected at random from the pooled groups.
Method. Separate the variation within groups from the variation due to different group means. The ANOVA normality assumption applies within each group around its own mean.
Within-group distributions. Group 1 responses follow \(N(10,4)\) and group 2 responses follow \(N(20,4)\). Each group has normally distributed errors and the same error variance, \(4\).
Pooled mean. Because group sizes are equal, a randomly selected pooled response has probability \(1/2\) of coming from either group. Its mean is \((10+20)/2=15\).
Pooled variance. The average within-group variance is \(4\). The variance of the group means around the pooled mean is \(\{(10-15)^2+(20-15)^2\}/2=(25+25)/2=25\). Thus the pooled variance is \(4+25=29\).
The pooled distribution is a mixture of two normal distributions with different means, not a single normal distribution in general. The large separation between 10 and 20 can make that mixture visibly two-peaked, even though the errors within both groups satisfy the normality assumption exactly.
Conclusion and trap avoided. Assess the model’s normality condition within groups, not by requiring one normal distribution for all raw responses pooled together. The pooled variance of 29 also shows why overall spread includes differences between group means as well as within-group error variation.
Common Misinterpretations
- Checking the wrong distribution. The assumption is normal errors within each group. Different group means can make the pooled responses nonnormal even when each group’s errors are normal.
- Treating the sample variances as population facts. Group sample variances fluctuate. Similar values do not prove a common population variance, and different values do not automatically disprove it.
- Confusing equal sample sizes with equal variances. The number of observations in a group describes the design; the variance describes the spread of the errors. One does not determine the other.
- Assuming normality fixes dependence. Even normally distributed errors are not independent if observations are correlated. The usual inference requires the assumptions together.
- Thinking assumptions are guaranteed by the F calculation. Computing a statistic is an arithmetic operation. Its inferential interpretation depends on whether the model is credible.
Check Your Understanding
For each problem, identify the relevant assumption or calculate the requested quantity before checking the answer.
- A study measures the same six laboratory vessels once before treatment and once after treatment, then proposes treating all 12 measurements as independent observations in a one-way ANOVA. What assumption is in question, and why? (Answer: Independence is in question because two measurements come from each vessel, so measurements from the same vessel may be dependent.)
- For one group, four observations are 5, 7, 7, and 9. Calculate the sample variance using the usual denominator. (Answer: The mean is 7; the squared deviations sum to \(4+0+0+4=8\), so the sample variance is \(8/3\approx2.667\).)
- Three groups have equal sample sizes but observed sample variances 2, 5, and 9. Does equal sample size establish the common-variance assumption? (Answer: No. Equal sample sizes do not establish equal population variances; the observed variances differ and warrant caution, but alone do not prove unequal population variances.)
- Two equally sized groups have normal errors with variance 3 and means 4 and 10. Find the pooled response mean and variance, and state whether the pooled responses must be normal. (Answer: The mean is 7; the variance is \(3+\{(4-7)^2+(10-7)^2\}/2=3+9=12\); the pooled responses are a mixture and need not be normal.)
- A researcher reports normal errors in each group and equal group variances, but observations within each group are repeated measurements on the same subjects. Is the usual ANOVA assumption set satisfied? (Answer: Not necessarily. The repeated measurements may be dependent, so normality and equal variances do not establish independence.)