Why Measure the Between-Group Share of Variation?
A one-way ANOVA separates observed variation into a between-group part and a within-group part. The F statistic compares mean squares to assess evidence against the equal-means null hypothesis. Eta-squared answers a different question: what proportion of the total sample variation is represented by differences between the observed group means? It is a descriptive measure of the size of the between-group component in the data.
This distinction matters in reporting results. A test statistic is used for inference under a statistical model; eta-squared summarizes the observed partition of variation. A small or large value does not, by itself, establish whether a result is statistically significant, practically important, or causal.
For group \(i\), write the observations as \(x_{ij}\), where \(i=1,\ldots,k\) and \(j=1,\ldots,n_i\). Let \(N=\sum_{i=1}^k n_i\), \(\bar{x}_i\) be the mean of group \(i\), and \(\bar{x}\) be the grand mean of all \(N\) observations. The sums of squares are
In a one-way ANOVA, these quantities satisfy \(SS_T=SS_B+SS_W\). Consequently, eta-squared can also be calculated as \(1-SS_W/SS_T\), if the total and within-group sums of squares are known. The sums of squares use squared deviations, not sample variances: their common scale is what makes their ratio a proportion.
Why the ANOVA Partition Gives a Proportion
The partition follows from splitting each observation's deviation from the grand mean into two pieces: its deviation from its own group mean and the deviation of that group mean from the grand mean. The cross-product terms vanish when summed within each group.
For every observation, \(x_{ij}-\bar{x}=(x_{ij}-\bar{x}_i)+(\bar{x}_i-\bar{x})\).
Expand the square for each observation. The expansion contains the squared within-group deviation, the squared group-mean deviation, and a cross-product.
In group \(i\), \(\sum_{j=1}^{n_i}(x_{ij}-\bar{x}_i)=0\), so the cross-product summed over that group is zero.
The within-group squared deviations sum to \(SS_W\). The group-mean term occurs \(n_i\) times for group \(i\), giving \(SS_B\). Thus \(SS_T=SS_W+SS_B\).
In algebra, the cross-product for group \(i\) is \(2(\bar{x}_i-\bar{x})\sum_j(x_{ij}-\bar{x}_i)\), which equals zero. The remaining group-mean term is \(\sum_j(\bar{x}_i-\bar{x})^2=n_i(\bar{x}_i-\bar{x})^2\). Summing over groups yields the stated partition. Dividing \(SS_B\) by \(SS_T=SS_B+SS_W\) therefore gives the share of total squared deviation assigned to between-group differences.
Calculating Eta-Squared From Raw Observations
When the raw observations are given, calculate group means and the grand mean before forming the sums of squares. A useful check is to calculate \(SS_T\) directly and confirm it equals \(SS_B+SS_W\). The examples below use complete observations from a single one-way grouping; the calculation requires a positive total sum of squares.
Worked Example: Two Groups With Raw Data
A hypothetical analyst compares the processing times, in minutes, for two independent teams completing the same type of task. The recorded times are Team A: 2, 4, 6 and Team B: 5, 7, 9. Calculate eta-squared and interpret it within these six observations.
Method and group means. Team A has mean \((2+4+6)/3=4\); Team B has mean \((5+7+9)/3=7\). The grand mean is \((2+4+6+5+7+9)/6=33/6=5.5\).
Within-group sum of squares. For Team A, the squared deviations from 4 are \((2-4)^2=4\), \((4-4)^2=0\), and \((6-4)^2=4\), giving 8. For Team B, the squared deviations from 7 are \((5-7)^2=4\), \((7-7)^2=0\), and \((9-7)^2=4\), also giving 8. Thus \(SS_W=8+8=16\).
Between-group sum of squares. Each group has three observations, so \(SS_B=3(4-5.5)^2+3(7-5.5)^2=3(2.25)+3(2.25)=13.5\).
Total and effect size. The partition gives \(SS_T=16+13.5=29.5\). As a direct check, the six squared deviations from 5.5 are \(12.25, 2.25, 0.25, 0.25, 2.25, 12.25\); their sum is \(29.5\). Therefore \(\eta^2=13.5/29.5=0.457627\ldots\), or 0.4576 to four decimals (45.76% to two decimal places).
Interpretation and trap avoided. In these observations, 45.76% of the total sum of squared deviations from the grand mean is represented by the difference between the two observed team means. This is not a claim that team membership caused that share of variation. Do not divide \(SS_B\) by \(SS_W\): \(13.5/16=0.84375\) is not eta-squared because the denominator must be total variation, \(SS_T=29.5\).
Using Group Summaries or an ANOVA Table
Raw observations are not necessary when group sizes, group means, and within-group sums of squares are available. The between-group sum of squares uses the group sizes as weights, so unequal group sizes must be handled explicitly. If an ANOVA table is provided, use its sums of squares; if some entries are missing, first recover them from the one-way partition and the degrees of freedom or mean-square definitions supplied.
Worked Example: Unequal Group Sizes From Summaries
A hypothetical study records a numerical score for three independent training formats. The summaries are: Format A, \(n_1=2\), mean 10, within-group sum of squares 2; Format B, \(n_2=3\), mean 12, within-group sum of squares 8; Format C, \(n_3=1\), mean 16, within-group sum of squares 0. Calculate eta-squared.
Method and grand mean. The total number of observations is \(N=2+3+1=6\). The weighted total of the observations is \(2(10)+3(12)+1(16)=20+36+16=72\), so \(\bar{x}=72/6=12\). Weighting matters because the group sizes differ.
Between-group variation. \(SS_B=2(10-12)^2+3(12-12)^2+1(16-12)^2=2(4)+3(0)+1(16)=24\).
Within-group and total variation. The within-group sum is \(SS_W=2+8+0=10\). Hence \(SS_T=24+10=34\), and \(\eta^2=24/34=12/17=0.705882\ldots\), or 0.7059 to four decimals (70.59% to two decimal places).
Interpretation and trap avoided. For this observed sample, 70.59% of total squared variation is between the three group means. An unweighted average of the means, \((10+12+16)/3\), would not be the grand mean here; using it would give an incorrect between-group sum of squares. Use the observation-count-weighted grand mean.
Worked Example: Recovering Eta-Squared From an F Statistic and Partial Table
A one-way ANOVA compares three groups with \(N=12\) observations. A partial table gives \(SS_W=36\) and observed \(F=2.25\). Find eta-squared. Treat the supplied \(F\) as the observed ANOVA statistic, not as a critical value.
Method and degrees of freedom. The between-group degrees of freedom are \(k-1=3-1=2\); the within-group degrees of freedom are \(N-k=12-3=9\). Therefore \(MS_W=SS_W/9=36/9=4\).
Recover the between-group mean square. By definition, \(F=MS_B/MS_W\). Thus \(MS_B=F(MS_W)=2.25(4)=9\). Since \(MS_B=SS_B/2\), it follows that \(SS_B=2(9)=18\).
Find total variation and eta-squared. The ANOVA partition gives \(SS_T=SS_B+SS_W=18+36=54\). Therefore \(\eta^2=18/54=1/3=0.3333\) to four decimals, or 33.33% to two decimal places.
Check and interpretation. The reconstructed mean squares give \(MS_B/MS_W=9/4=2.25\), agreeing with the stated F statistic. In this sample, one-third of the total sum of squares is between groups. Do not report \(F=2.25\) as the proportion: the F statistic is a ratio of mean squares, whereas eta-squared is a ratio of sums of squares.
Interpreting the Measure Carefully
Eta-squared is bounded between 0 and 1 when \(SS_T>0\), because \(SS_B\) and \(SS_W\) are nonnegative and sum to \(SS_T\). A value of 0 means all group means equal the grand mean, so \(SS_B=0\). A value of 1 means \(SS_W=0\): every observation equals its own group mean, leaving no within-group variation. Intermediate values describe the observed partition. They do not automatically imply a particular strength of evidence or a practical consequence.
Because eta-squared is calculated from the sample, it describes the groups and observations actually analyzed. Generalizing beyond the sample requires an appropriate sampling design and inferential reasoning. The usual one-way ANOVA assumptions—including independent observations and, for the classical test, normal errors with a common variance—matter when using ANOVA for inference. They are not extra arithmetic conditions for computing this descriptive ratio from an observed partition. If all observations are identical, then \(SS_T=0\), and eta-squared is undefined because its denominator is zero.
| Quantity | Definition | What it describes |
|---|---|---|
| Eta-squared | \(SS_B/SS_T\) | Between-group share of total observed sum of squares |
| F statistic | \(MS_B/MS_W\) | Ratio of mean squares used for the ANOVA test |
| Within-group share | \(SS_W/SS_T\) | Share of total observed sum of squares within groups |
The partition also implies \(\eta^2+SS_W/SS_T=1\). This is a useful arithmetic check, but it does not turn eta-squared into a universal measure of importance. Interpretation should stay tied to the outcome, grouping, and data under study. A numerical effect-size convention cannot replace context.
Common Calculation and Interpretation Errors
- Using the wrong denominator. Eta-squared is \(SS_B/SS_T\), not \(SS_B/SS_W\). The total includes both between- and within-group variation.
- Forgetting the group-size weights. The between-group sum of squares is \(\sum_i n_i(\bar{x}_i-\bar{x})^2\), not merely the sum of squared differences between group means.
- Confusing the F statistic with eta-squared. F uses mean squares and accounts for their degrees of freedom; eta-squared uses sums of squares and expresses a proportion.
- Overstating the conclusion. Eta-squared describes the observed sample partition. It does not by itself prove statistical significance, establish causation, or measure the probability that the null hypothesis is true.
- Ignoring a zero total. If \(SS_T=0\), the ratio has no defined value. Do not report zero merely because the observations show no variation.
Check Your Understanding
For each problem, calculate eta-squared and state the requested interpretation. Answers are rounded to four decimals when needed.
- A one-way analysis has \(SS_B=12\) and \(SS_W=28\). Calculate eta-squared. (Answer: \(SS_T=40\), so \(\eta^2=12/40=0.3000\).)
- An ANOVA summary gives \(SS_T=30\) and \(SS_W=18\). Find \(SS_B\) and eta-squared. (Answer: \(SS_B=12\) and \(\eta^2=12/30=0.4000\).)
- Two groups each have two observations. Their means are 3 and 5, and each group has within-group sum of squares 2. Calculate eta-squared. (Answer: Grand mean \(=4\); \(SS_B=2(3-4)^2+2(5-4)^2=4\), \(SS_W=4\), and \(\eta^2=4/8=0.5000\).)
- Two training cohorts have summaries: Cohort A has \(n=3\), mean 4, and within-group sum of squares 4; Cohort B has \(n=2\), mean 7, and within-group sum of squares 2. Calculate eta-squared and interpret the proportion. (Answer: Grand mean \(=(3(4)+2(7))/5=5.2\); \(SS_B=3(4-5.2)^2+2(7-5.2)^2=4.32+6.48=10.8\); \(SS_W=6\), \(SS_T=16.8\), and \(\eta^2=10.8/16.8=9/14=0.6429\), or 64.29%.)
- An ANOVA with four groups and \(N=20\) has \(SS_W=48\) and observed \(F=1.5\). Recover \(SS_B\), \(SS_T\), and eta-squared. (Answer: \(df_W=20-4=16\), \(MS_W=48/16=3\), \(MS_B=1.5(3)=4.5\), \(df_B=3\), so \(SS_B=3(4.5)=13.5\); \(SS_T=61.5\) and \(\eta^2=13.5/61.5=0.2195\).)