One-way ANOVA calculator: the F statistic, p-value, sums of squares, mean squares and η² for two or more groups, with the full ANOVA table.
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Reject H₀ at α = 0.05. If all population group means were equal, a result at least this extreme would turn up with probability 0.0032. The p-value is not the probability that H₀ is true, and a significant result says nothing about how large or important the effect is. Group membership accounts for 61.5% of the total variation (η²). ANOVA does not say which groups differ: follow up with a post-hoc comparison such as Tukey’s HSD.
| Source | SS | df | MS | F | p-value |
|---|---|---|---|---|---|
| Between groups | 27.8973 | 2 | 13.9487 | 9.5911 | 0.0032 |
| Within groups | 17.452 | 12 | 1.4543 | — | — |
| Total | 45.3493 | 14 | — | — | — |
H₁: at least one population mean differs.
Critical F at α = 0.05 is 3.8853.
One-way analysis of variance (ANOVA) tests whether two or more independent groups share the same mean. It splits the total variation into a between-group sum of squares and a within-group sum of squares, divides each by its degrees of freedom (k − 1 and N − k) to get mean squares, and takes their ratio as the F statistic. The p-value is the area of the F distribution beyond that F.
Typical uses are comparing crop yields under three fertilisers, exam scores across teaching methods or response times across product versions. The default data, the NIST e-Handbook example with three groups of five, give F = 9.59 on 2 and 12 degrees of freedom and p = 0.0032, so at α = 0.05 the three means are not all equal.
The test assumes independent observations, roughly normal data in each group and similar group variances. A significant F says that at least one mean differs, not which one.
One-way ANOVA tests whether group means are equal with F = MSB ÷ MSW, the mean square between groups over the mean square within them. Three groups of five with means 5.34, 7.72 and 8.56 give F = 9.59 on 2 and 12 df and p = 0.0032: the means are not all equal.
The data are the NIST e-Handbook example: 6.9, 5.4, 5.8, 4.6, 4.0; 8.3, 6.8, 7.8, 9.2, 6.5; and 8.0, 10.5, 8.1, 6.9, 9.3. Worked by hand:
For the example, the table reads:
The two sums of squares add to the total variation around the grand mean, and so do the degrees of freedom. The total row has no mean square or F, and the within row has no F of its own.
If the population means are equal, F tends to land near 1. A large F means the group means are further apart than the scatter inside the groups would explain.
How large is large enough depends on both degrees of freedom. With 2 and 12 df, the 5% critical value is 3.885 and the 1% value is 6.927, so F = 9.59 is significant at either level.
The same F can carry different evidence. F = 13.5 gives p = 0.0020 with 2 and 9 df, but p = 0.0213 with 1 and 4 df.
A p-value says whether the means differ; η² = SSB ÷ SST says how much of the variation group membership explains. For the example, η² = 27.897 ÷ 45.349 = 0.615.
Cohen (1988) proposed 0.01, 0.06 and 0.14 as small, medium and large effects, so 0.615 is very large. η² overstates the population effect in small samples, and ω² corrects for that bias.
The F test assumes independent observations, a roughly normal distribution in each group and equal population variances. It tolerates moderate non-normality when the groups are of similar size.
Moore and McCabe's rule of thumb accepts equal variances when the largest group SD is less than twice the smallest. The example passes: its SDs are 1.117, 1.103 and 1.378.
When the rule fails, Welch's ANOVA is the usual alternative, and the Kruskal–Wallis test suits clearly non-normal data. Repeated measurements on the same subjects need a repeated-measures model, not independent groups.
A significant F says at least one mean differs, not which. A post-hoc test answers that. Tukey's HSD compares every pair while holding the family-wise error rate at α, and Bonferroni tests each of m pairs at α/m, which is 0.05/3 = 0.0167 for three groups.
Separate t-tests at 0.05 on every pair raise the chance of at least one false positive toward 1 − 0.95³ = 14% for three comparisons.
With exactly two groups, one-way ANOVA matches the pooled two-sample t-test: F = t². Groups 1, 2, 3 and 4, 5, 6 give t = −3.674 and F = 13.5, with the same p-value, 0.0213. Welch's t-test makes no equal-variance assumption, so its result can differ.
| Scenario | F statistic | p-value |
|---|---|---|
| Default | 9.591107 | 0.003248 |
| Unequal groups | 13.500000 | 0.001953 |
| Equal means | 0.000000 | 1.000000 |
| Two groups | 13.500000 | 0.021312 |
"Default" is the NIST example. "Unequal groups" is 23, 25, 21, 22; 28, 30, 27, 26, 29; and 24, 26, 25, with η² = 0.75. "Two groups" is 1, 2, 3 and 4, 5, 6.
"Equal means" is 1, 2, 3; 0, 2, 4; and −1, 2, 5. The spreads differ, but every mean is 2, so SSB, F and η² are all 0 and p = 1.
To check the F tail for other degrees of freedom, use the t, chi-square and F distribution calculator. For two groups without the equal-variance assumption, use Welch's test in the t-test calculator.
Checked against: NIST/SEMATECH e-Handbook §7.4.3.3 ANOVA table (SS 27.897 / 17.452, F = 9.59)
Checked against: Sums of squares with exact fractions (SSB 62.25, SSW 20.75); A&S 26.6.4 for d₁ = 2 gives p = (9/(9 + 2F))^4.5 = 4^−4.5 = 1/512
Checked against: F = t² for two groups; pooled t from with exact fractions, two-sided p from the A&S 26.7.4 closed form with ν = 10
Checked against: Every group mean is 2, so SS between = 0 and F = 0
It is the probability of an F statistic at least as large as the one observed if every group mean were equal. With the default data p = 0.0032: equal means would produce F ≥ 9.59 in about 3 samples out of 1,000. It is not the probability that the null hypothesis is true, and a small p-value does not say how large the differences are; η² measures that.
F is the between-group mean square divided by the within-group mean square. Equal population means give F values near 1; larger values point to real differences. How large is large enough depends on the degrees of freedom: with 2 and 12 df the 5% critical value is 3.885, so F = 9.59 is significant at α = 0.05 and F = 3 would not be.
Independent observations, a roughly normal distribution in each group, and equal population variances. The F test tolerates moderate non-normality when groups are of similar size. Moore and McCabe's rule of thumb accepts the equal-variance assumption if the largest group standard deviation is less than twice the smallest; otherwise use Welch's ANOVA, and for clearly non-normal data the Kruskal–Wallis test.
Run a post-hoc test. Tukey's honestly significant difference (HSD) compares every pair while holding the family-wise error rate at α; the Bonferroni method tests each of the m pairs at α/m, which is 0.05/3 ≈ 0.0167 for three groups. Separate t-tests at 0.05 on every pair push the chance of at least one false positive well above 5%, towards 1 − 0.95³ ≈ 14% for three comparisons.
η² is the between-group sum of squares divided by the total sum of squares: the share of variation explained by group membership. Cohen (1988) proposed 0.01, 0.06 and 0.14 as small, medium and large effects. The default data give 27.897/45.349 = 0.615, a very large effect. η² overstates the population effect in small samples; ω² corrects for that bias.
This calculator includes 4 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.