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06. Comparing 3+ Groups (ANOVA & Kruskal-Wallis)7 min readAdvanced Lesson

The Multiplicity Problem & The Omnibus F-Ratio

How repeated pairwise t-tests escalate the Familywise Error Rate and how One-Way ANOVA keeps overall alpha at 5%.

In this lesson you will learn

  • 1Compute the Familywise Error Rate (1 − (1 − α)^c) across k study groups to explain why unadjusted multiple t-tests are invalid.
  • 2Decompose total sum of squares into Between-Group signal (MS_between) and Within-Group noise (MS_within) to form F.
  • 3Interpret Eta-Squared (η² = SS_between / SS_total) as the proportion of outcome variance explained by group membership.
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01

Multiplicity & Variance Partitioning

What this section is about

This section defines Familywise Error Rate (FWER), the Omnibus F-ratio, Between-Group vs Within-Group Mean Squares, and Eta-Squared (η²).

Why it matters for your study

Reviewers immediately reject manuscripts that compare 3 or more study arms using separate unadjusted t-tests.

When a clinical trial includes k ≥ 3 arms (for example: Placebo, Low-Dose, and High-Dose), comparing every pair requires c = k(k − 1) / 2 tests. Because each independent test carries a 5% false-positive risk, the cumulative probability of making at least one Type I error across the family of tests is:

FWER = 1 − (1 − α)^c   |   F = MS_between (Treatment Signal) / MS_within (Within-Arm Noise)

Sir Ronald Fisher developed Analysis of Variance (ANOVA) to solve this problem. Instead of running multiple t-tests, ANOVA compares the variance between group means against the average variance within groups in a single ratio F. Under the null hypothesis of equal population means, F ≈ 1.0. When treatments diverge significantly, F rises well above 1.0.

ANOVA Protected Limit (5.0%)5.0%k=2 (1 tests)14.3%k=3 (3 tests)26.5%k=4 (6 tests)40.1%k=5 (10 tests)53.7%k=6 (15 tests)
Figure:Unadjusted pairwise testing causes false-positive risk to escalate rapidly as the number of study arms increases.

Multiplicity Error Rate & Omnibus F-Ratio Calculator

Watch false-positive probability escalate with multiple t-tests and see how within-group noise shapes F

Number of Study Arms (k):4
Within-Group Noise (SD):6
Pairwise Tests (c)
6
k(k − 1) / 2
True False-Alarm Risk
26.49%
1 − (0.95)^6
Bonferroni Threshold
.0083
0.05 / 6
Omnibus F-Ratio
F = 44.44
η² = 55.3%
With 4 groups, running 6 separate t-tests inflates your familywise Type I error rate to about 26.49% (assuming independent tests). One-Way ANOVA tests all arms simultaneously while holding alpha strictly at 5.0%.

Clinical Example

A 4-arm knee osteoarthritis trial (N = 160; 40 per arm) compares 8-week VAS pain reduction across Placebo, Celecoxib 100 mg, Celecoxib 200 mg, and a Biologic regimen.

SS_between = 14,210.4 (df = 3, MS = 4,736.8) | SS_within = 3,804.8 (df = 156, MS = 24.39) → F(3, 156) = 194.21, p < .001, η² = .789

Interpretation: Running 6 separate t-tests would have inflated false-positive risk to 26.5%. The single Omnibus ANOVA confirms a highly significant difference across the 4 regimens (F(3, 156) = 194.21, p < .001) explaining 78.9% of pain variance.

Common Pitfall

Running post-hoc pairwise comparisons and reporting a 'significant difference between Group A and Group C' when the overall Omnibus ANOVA F-test was non-significant (p > .05).

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The Multiplicity Problem & The Omnibus F-Ratio | Biostatistics Handbook | Analysez