One-Way ANOVA - The Statistics Notebook

Compare several treatment means by separating between-treatment variation from within-treatment error. This page covers the global test, equivalent model parameterizations, diagnostics, and the alternative framework for binary outcomes. Numerical examples use illustrative practice data.

At a Glance¶

Global question

Focused question

Are all treatment means equal? ANOVA provides one global test.

One-Way ANOVA¶

ANOVA compares one factor with $m$ treatment levels. The standard normal-response model is

Y_{ij}=\mu_i+\varepsilon_{ij},\qquad \varepsilon_{ij}\overset{\mathrm{iid}}{\sim}N(0,\sigma^2),\qquad N=\sum_{i=1}^m n_i.

(1)

H_0:\mu_1=\cdots=\mu_m,\qquad H_A:\text{not all treatment means are equal}.

(2)

The alternative does not say that every pair differs.

Decomposing Variation¶

Define the group means $\bar Y_{i\cdot}$ and the observation-weighted grand mean

\bar Y_{\cdot\cdot}=\frac1N\sum_i n_i\bar Y_{i\cdot}.

(3)

\begin{aligned} SS_T&=\sum_i\sum_j(Y_{ij}-\bar Y_{\cdot\cdot})^2,\\ SS_{\mathrm{Tr}}&=\sum_i n_i(\bar Y_{i\cdot}-\bar Y_{\cdot\cdot})^2,\\ SS_E&=\sum_i\sum_j(Y_{ij}-\bar Y_{i\cdot})^2. \end{aligned}

(4)

\boxed{SS_T=SS_{\mathrm{Tr}}+SS_E}

(5)

Treatment variation measures separation among group means. Error variation measures spread within each group. Some notation calls $SS_{\mathrm{Tr}}$ “condition SS” or $SS_C$ .

ANOVA Table¶

Source	Sum of squares	df	Mean square	Test
Treatment	$SS_{\mathrm{Tr}}$	$m-1$	$MS_{\mathrm{Tr}}=SS_{\mathrm{Tr}}/(m-1)$	$F=MS_{\mathrm{Tr}}/MSE$
Error	$SS_E$	$N-m$	$MSE=SS_E/(N-m)$
Total	$SS_T$	$N-1$

Under $H_0$ and the model assumptions,

F\sim F_{m-1,N-m},\qquad p=P(F_{m-1,N-m}\geq F_{\mathrm{obs}}).

(8)

Both mean squares estimate $\sigma^2$ under the null. Separated treatment means increase the numerator, so rejection is in the right tail. This is a test about means, unlike the two-sample variance-ratio test.

For $m=2$ , this ANOVA test equals the two-sided pooled $t$ test, with $F=t^2$ . It is not generally equivalent to Welch’s test.

Worked ANOVA Example

Treatment	Observations	Mean
A	5, 7, 6, 6	6
B	8, 9, 7, 8	8
C	10, 12, 11, 11	11

Here $N=12$ , $m=3$ , and $\bar y_{\cdot\cdot}=25/3=8.3333$ .

SS_{\mathrm{Tr}}=4[(6-25/3)^2+(8-25/3)^2+(11-25/3)^2]=50.6667.

(9)

Each group contributes 2 to within-group SS, so $SS_E=6$ .

Source	SS	df	MS	$F$
Treatment	50.6667	2	25.3333	38
Error	6	9	0.6667
Total	56.6667	11

p=P(F_{2,9}\geq38)\approx0.00004090.

(10)

There is evidence that the population means are not all equal under the model. The global test does not identify which comparisons drive the result.

Equivalent Model Parameterizations¶

The following models have the same fitted group means, residuals, and overall ANOVA test.

Model	Form	Interpretation
Means	$Y_{ij}=\mu_i+\varepsilon_{ij}$	A parameter for each group mean.
Effects	$Y_{ij}=\mu+\tau_i+\varepsilon_{ij}$ , $\sum_i\tau_i=0$	Mean of treatment means plus deviations.
Reference-group regression	$Y=\beta_0+\sum_{k=2}^m\beta_{k-1}I(\text{group}=k)+\varepsilon$	Reference mean plus differences from it.

In the effects model with an unweighted sum-to-zero constraint,

\mu=\frac1m\sum_i\mu_i,\qquad \widehat\mu=\frac1m\sum_i\bar Y_{i\cdot},\qquad \widehat\tau_i=\bar Y_{i\cdot}-\widehat\mu.

(11)

For unequal group sizes, this average of group means is not necessarily the observation-weighted grand mean used in the ANOVA decomposition. With balanced groups, they coincide.

For A, B, C with A as the reference,

Y=\beta_0+\beta_1I(B)+\beta_2I(C)+\varepsilon.

(12)

The worked example gives $\widehat\beta_0=6$ , $\widehat\beta_1=2$ , and $\widehat\beta_2=5$ . The global null is $\beta_1=\beta_2=0$ , with no requirement that the intercept be zero.

Avoid the dummy-variable trap: with an intercept, include only $m-1$ group indicators. Alternatively, use all $m$ indicators without an intercept.

Residual Diagnostics & Other Response Types¶

For one-way ANOVA, fitted values are group means and residuals are

\widehat Y_{ij}=\bar Y_{i\cdot},\qquad e_{ij}=Y_{ij}-\bar Y_{i\cdot}.

(13)

Check	What to look for	Implication
Assignment and measurement design	Repeated units, clusters, or matched observations.	Independence must come from the design, not a normality plot.
Residual Q-Q plot	Strong curvature or extreme tails.	Normal-theory inference may be unreliable, especially with small samples.
Residuals vs. fitted values	Changing spread across fitted groups.	Common variance may be unsuitable.
Residuals vs. collection order	Drift, cycles, long runs.	Time structure or omitted nuisance effects.
Unusual observations	Recording mistakes or exceptional conditions.	Investigate context; do not remove points just to improve significance.

Normality tests can supplement plots but do not certify a model. With unequal variances, consider a heteroscedastic analysis such as Welch’s ANOVA. For strongly skewed outcomes, a meaningful transformation or a suitable response model may be preferable. Count outcomes may need a count model; binary outcomes need a proportions or binary-response analysis.

Binary Outcomes: Comparing Several Proportions¶

For treatment $i$ , let $O_{1i}$ be failures and $O_{2i}$ successes, with $n_i=O_{1i}+O_{2i}$ . Test

H_0:\pi_1=\cdots=\pi_m.

(14)

Estimate the common success probability under the null:

\widehat\pi=\frac{\sum_iO_{2i}}N,\qquad E_{1i}=n_i(1-\widehat\pi),\quad E_{2i}=n_i\widehat\pi.

(15)

X^2=\sum_{i=1}^m\sum_{r=1}^2\frac{(O_{ri}-E_{ri})^2}{E_{ri}} \ \overset{\mathrm{approx}}{\sim}\ \chi^2_{m-1}.

(16)

Use the right tail. This approximation requires independent observations and sufficiently large expected counts; sparse tables call for an appropriate exact or simulation-based method. The global test does not identify which proportions differ.