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Effect Sizes and Power · Tutorial 106 of 595

Cohen's f for One-Way Effects

Learn how Cohen’s \(f\) measures one-way group-mean separation relative to within-group variation, and how it relates to eta-squared.

Intermediate 9 min read

What You'll Learn

  • Define Cohen’s \(f\) using weighted population group means and the common error standard deviation
  • Choose weights that match the group proportions of the population or design being described
  • Derive the relationship between population Cohen’s \(f\) and population eta-squared
  • Calculate \(f\) from specified population means and from a sample eta-squared
  • Interpret the magnitude and limitations of \(f\) without treating it as an outcome-unit difference

Why Standardize the Separation Among Group Means?

An omnibus ANOVA asks whether population group means are all equal. An effect size answers a different question: how far apart are the group means, relative to the variation of individual outcomes within groups? The answer depends on both the pattern of means and the scale of within-group variability. The same differences between means can be substantial when outcomes vary little within groups and modest when outcomes vary greatly.

Cohen’s \(f\) is a dimensionless measure of that standardized separation for a one-way effect. It uses the weighted spread of the population group means around their grand mean, divided by the common within-group standard deviation. The weights matter: they describe the group proportions for the population or design being considered.

Definition and Model

Let there be \(k\) groups. Let \(\mu_i\) be the population mean in group \(i\), and let \(w_i\) be that group’s weight, with \(w_i \geq 0\) and \(\sum_{i=1}^{k}w_i=1\). Define the weighted grand mean as \(\mu=\sum_{i=1}^{k}w_i\mu_i\). Suppose the outcome has the same within-group population standard deviation \(\sigma>0\) in every group.

Definition: Cohen’s \(f\) for a one-way effect is the square root of the weighted variance of the population group means around their weighted grand mean, divided by the common within-group standard deviation:
$$ f=\frac{\sqrt{\sum_{i=1}^{k} w_i(\mu_i-\mu)^2}}{\sigma}, \qquad \mu=\sum_{i=1}^{k}w_i\mu_i. $$

The numerator is the weighted standard deviation of the group means, measured in the outcome’s units. The denominator is the within-group standard deviation, in those same units. Dividing makes \(f\) unitless. If all group means are equal, the numerator and \(f\) are zero. As the means spread farther apart relative to \(\sigma\), \(f\) grows.

A standard one-way model for using this definition is \(Y_{ij}=\mu_i+\varepsilon_{ij}\), where \(Y_{ij}\) is the outcome for observation \(j\) in group \(i\). The usual model assumes independent errors with mean zero, a common variance \(\sigma^2\), and normally distributed errors for exact classical ANOVA inference. Thus observations are independent, outcomes within each group are normally distributed under the model, and the within-group variances are equal. Cohen’s \(f\) is defined from population means and a common population standard deviation; these assumptions describe the standard model in which the effect is used.

Worked Example: Equal Weights and Three Population Means

Suppose a hypothetical one-way setting has three equally weighted groups with population means of 10, 12, and 14 outcome units, and each group has a common population standard deviation of 4 outcome units. Assume independent observations and normally distributed outcomes with the stated common standard deviation in each group.

The weights are \(w_1=w_2=w_3=1/3\). The weighted grand mean is

$$ \mu=\frac{10+12+14}{3}=12. $$

The weighted variance of the group means is

$$ \sum_{i=1}^{3}w_i(\mu_i-\mu)^2 =\frac{1}{3}\left[(10-12)^2+(12-12)^2+(14-12)^2\right] =\frac{8}{3}. $$

Therefore, the weighted standard deviation of the means is \(\sqrt{8/3}\approx1.633\) outcome units, and

$$ f=\frac{\sqrt{8/3}}{4}\approx0.4082. $$

The group means spread by about 0.408 common within-group standard deviations according to this weighted measure. This is a standardized summary of all three means, not a difference between any particular pair.

Weights and Unequal Group Proportions

For equally represented groups, equal weights are appropriate. When groups have unequal proportions in the target population or design, use those proportions rather than treating the groups as equally weighted. The resulting grand mean and mean dispersion describe the group mixture represented by those weights.

For example, if the target proportions are 0.50, 0.30, and 0.20, the grand mean is \(0.50\mu_1+0.30\mu_2+0.20\mu_3\). It is not generally the unweighted average of the three means. A different choice of weights can produce a different \(f\), even if the means and common standard deviation stay the same.

Worked Example: Unequal Group Weights

Consider a hypothetical population with three groups whose proportions are 0.50, 0.30, and 0.20. Their population means are 20, 23, and 29 outcome units, respectively, and the common within-group standard deviation is 5 outcome units. Suppose observations are independent and group outcomes follow normal distributions with this common standard deviation.

First calculate the weighted grand mean:

$$ \mu=0.50(20)+0.30(23)+0.20(29) =10+6.9+5.8=22.7. $$

The weighted variance of the group means is

$$ \begin{aligned} \sum_{i=1}^{3}w_i(\mu_i-\mu)^2 &=0.50(20-22.7)^2+0.30(23-22.7)^2+0.20(29-22.7)^2\\ &=0.50(7.29)+0.30(0.09)+0.20(39.69)\\ &=3.645+0.027+7.938=11.61. \end{aligned} $$

Its square root is \(\sqrt{11.61}\approx3.407\) outcome units. Hence

$$ f=\frac{\sqrt{11.61}}{5}\approx0.6814. $$

The 20% group with mean 29 contributes to the weighted spread, but it does not count as much as it would under equal weighting. The result describes mean separation for this particular group mixture.

Connection to Eta-Squared

Population eta-squared, denoted here by \(\eta^2\), is the proportion of the total outcome variance in the weighted group mixture that is attributable to differences among the group means. Under the common-variance model, the total variance separates into the weighted variance of group means plus the common within-group variance. In symbols, if \(G\) denotes group membership, then

$$ \operatorname{Var}(Y) =\operatorname{Var}\bigl(E[Y\mid G]\bigr)+E\bigl[\operatorname{Var}(Y\mid G)\bigr] =\sum_{i=1}^{k}w_i(\mu_i-\mu)^2+\sigma^2. $$

The first term is the between-group component, and the second is the within-group component. By definition, population eta-squared is the first term divided by total variance. Let \(B=\sum_{i=1}^{k}w_i(\mu_i-\mu)^2\). Then \(\eta^2=B/(B+\sigma^2)\), while \(f^2=B/\sigma^2\). Substituting and rearranging gives the relationship.

Result: For a one-way population effect under the common within-group variance model,
$$ \eta^2=\frac{f^2}{1+f^2}, \qquad f=\sqrt{\frac{\eta^2}{1-\eta^2}}. $$
The second expression applies when \(0\leq\eta^2<1\). The first applies for finite \(f\geq0\).

This relationship can also be derived directly: from \(\eta^2=B/(B+\sigma^2)\), it follows that \(1-\eta^2=\sigma^2/(B+\sigma^2)\). Therefore \(\eta^2/(1-\eta^2)=B/\sigma^2=f^2\). Taking the nonnegative square root yields \(f=\sqrt{\eta^2/(1-\eta^2)}\). This is an exact population relationship, not a claim that the two measures use different information about the group-mean separation.

Worked Example: Converting Population Eta-Squared to Cohen’s f

Suppose a hypothetical one-way population effect has \(\eta^2=0.20\), with independent observations, normally distributed outcomes in each group, and equal within-group variances. The proportion of total variance attributable to differences among group means is 0.20. The corresponding Cohen’s \(f\) is

$$ f=\sqrt{\frac{\eta^2}{1-\eta^2}} =\sqrt{\frac{0.20}{1-0.20}} =\sqrt{\frac{0.20}{0.80}} =\sqrt{0.25}=0.500. $$

Thus the weighted standard deviation of the population group means is one-half of the common within-group standard deviation. The calculation converts a variance proportion into a standardized spread; it does not convert the effect into outcome units.

Calculating a Sample-Based Value

In practice, population means and \(\sigma\) are usually unknown. A one-way ANOVA table provides the between-groups sum of squares, \(SS_{\text{between}}\), and the total sum of squares, \(SS_{\text{total}}\). The sample eta-squared is \(\eta^2=SS_{\text{between}}/SS_{\text{total}}\). A descriptive sample-based value of \(f\) can be obtained by applying the population conversion to this sample eta-squared:

$$ \hat{f}=\sqrt{\frac{\eta^2}{1-\eta^2}} =\sqrt{\frac{SS_{\text{between}}}{SS_{\text{total}}-SS_{\text{between}}}} =\sqrt{\frac{SS_{\text{between}}}{SS_{\text{within}}}}. $$

Here \(SS_{\text{within}}=SS_{\text{total}}-SS_{\text{between}}\). This is a plug-in effect-size calculation from the observed sample, not an unbiased estimate of population \(f\). Sampling variability can make sample effect sizes differ from the population effect. In particular, a sample eta-squared and the corresponding \(\hat f\) should not be described as known population values.

Worked Example: Obtaining a Sample-Based f From Group Data

Suppose a hypothetical sample has four outcome measurements in each of three groups:

GroupObserved outcomesSample mean
18, 8, 12, 1210
210, 10, 14, 1412
312, 12, 16, 1614

The grand mean is 12. The between-groups sum of squares is

$$ SS_{\text{between}} =\sum_{i=1}^{3}n_i(\bar{x}_i-\bar{x})^2 =4[(10-12)^2+(12-12)^2+(14-12)^2] =32. $$

Within each group, the four observations are two units below or above their group mean. Each group therefore contributes \(4(2^2)=16\) to the within-groups sum of squares, so \(SS_{\text{within}}=3(16)=48\). Thus \(SS_{\text{total}}=32+48=80\), and the sample eta-squared is

$$ \eta^2=\frac{32}{80}=0.400. $$

The sample-based Cohen’s \(f\) is

$$ \hat{f}=\sqrt{\frac{0.400}{1-0.400}} =\sqrt{\frac{0.400}{0.600}} =\sqrt{\frac{2}{3}}\approx0.8165. $$

Equivalently, \(\hat{f}=\sqrt{32/48}\approx0.8165\). These are descriptive calculations from the observed sums of squares. They do not establish that the population effect is exactly 0.8165. For classical ANOVA interpretation, the design should support independent observations, and the common-variance and normal-error assumptions should be plausible; this small sample by itself cannot verify those population assumptions.

Interpretation and Common Pitfalls

Cohen’s \(f\) is nonnegative and dimensionless. It summarizes the overall spread of weighted group means relative to within-group variability; it does not show which groups differ, the direction of any difference, or the size of a difference in the outcome’s original units. Two very different patterns of group means can yield the same \(f\). Use the means and suitable comparisons when those details matter.

Cohen’s conventional reference values of \(f=0.10\), \(0.25\), and \(0.40\) are sometimes described as small, medium, and large. They are rough guides, not universal boundaries. Whether an effect matters depends on the outcome, context, and consequences. A value should not be called important solely because it crosses one of these reference points.

A common mistake is to say, “The sample eta-squared is 0.20, so Cohen’s \(f\) is also 0.20.” That confuses a variance proportion with a standardized standard deviation. The correct conversion is \(f=\sqrt{\eta^2/(1-\eta^2)}\); for \(\eta^2=0.20\), it gives \(f=0.500\). Another mistake is to use equal weights automatically when the group proportions are unequal. The weights must match the population or design being summarized.

Key takeaway: Cohen’s \(f\) is the weighted standard deviation of population group means divided by the common within-group standard deviation. Under the one-way common-variance model, it and population eta-squared are related by \(\eta^2=f^2/(1+f^2)\).

Check Your Understanding

Use the definitions and relationships above to answer these questions.

  1. Three equally weighted population groups have means 5, 5, and 9, with a common within-group standard deviation of 4. Find the weighted grand mean and Cohen’s \(f\).
  2. Why can unequal group weights change Cohen’s \(f\), even when the group means and common standard deviation do not change?
  3. If population \(\eta^2=0.36\), calculate the corresponding population Cohen’s \(f\) and interpret it in terms of standard deviations.
  4. In a one-way ANOVA, \(SS_{\text{between}}=18\) and \(SS_{\text{within}}=50\). Calculate the sample eta-squared and sample-based \(\hat f\).
  5. Why does a single Cohen’s \(f\) value not tell you which pair of groups has the largest mean difference?
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