Brown-Forsythe Test Calculator
Robust test for homogeneity of variances using deviations from group medians.
Group Data
Brown-Forsythe F Statistic
1.4561
Variances Are Homogeneous
Group Statistics
ANOVA of Deviations
What Is the Brown-Forsythe Test?
The Brown-Forsythe test is a robust statistical test for the equality of group variances. It is a modification of Levene's test that uses deviations from group medians rather than group means, making it more resistant to outliers and skewed data. This robustness is critical because the test for variance homogeneity is a prerequisite for ANOVA, and a variance test that is itself sensitive to non-normality undermines the entire assumption-checking process.
The test works by computing the absolute deviation of each observation from its group median, then performing a one-way ANOVA on these deviation scores. A significant F-statistic indicates that at least one group has a different variance from the others. Because medians are more stable than means in the presence of outliers, the Brown-Forsythe test maintains its Type I error rate better than Bartlett's test when data deviates from normality.
This calculator computes the Brown-Forsythe F statistic, the ANOVA of absolute deviations, group medians and variances, and a significance verdict at your chosen alpha level.
Brown-Forsythe Test Formula
The Brown-Forsythe test transforms each observation into an absolute deviation from the group median, then runs ANOVA on the transformed values. The F-statistic tests whether the mean absolute deviations differ across groups.
Brown-Forsythe F Statistic
Where:
- Z_ij= Absolute deviation of observation j in group i from the group i median
- MSB= Mean square between groups for the Z scores = SSB / (k−1)
- MSW= Mean square within groups for the Z scores = SSW / (N−k)
- k= Number of groups being compared
- N= Total number of observations across all groups
Interpreting Brown-Forsythe Results
The null hypothesis H₀ is that all groups have equal population variances. A large F-statistic (exceeding the critical value) or a small p-value indicates that the absolute deviations differ significantly across groups, meaning the variances are not equal.
| Verdict | Condition | What to Do |
|---|---|---|
| Variances Homogeneous | F ≤ critical value | Proceed with standard ANOVA — the equal-variance assumption is met |
| Variances Differ | F > critical value | Use Welch's ANOVA or consider data transformation |
The calculator also displays the ANOVA of deviations table (SSB, SSW, MSB, MSW), which lets you see how the variation in absolute deviations is partitioned. Group medians and variances are shown for reference — large differences in variance often correspond to visibly different spreads in the raw data.
How to Use This Calculator
Running the Brown-Forsythe test requires three group inputs:
- Enter Group 1 data: Provide comma-separated numeric values. Each group should have at least 2 observations for meaningful variance estimation.
- Enter Groups 2 and 3: Populate the remaining groups. The calculator supports three groups by default.
- Set significance level (α): Choose your threshold — commonly 0.05. The calculator compares the F-statistic against an approximate critical value.
- Read the results: The calculator displays the F-statistic, degrees of freedom, p-value, critical value, significance verdict, group statistics (median and variance per group), and the ANOVA of deviations table.
Real-World Applications
The Brown-Forsythe test is the recommended variance homogeneity test when data may deviate from normality — which is most real-world data. In clinical trials, patient outcomes are often skewed or contain outliers; applying Bartlett's test (which assumes normality) may yield misleading p-values. The Brown-Forsythe test provides more reliable variance comparisons for non-normal biomedical data.
In psychology and social sciences, Likert-scale responses, reaction times, and behavioral measures frequently exhibit non-normal distributions. Before comparing group means with ANOVA, researchers use the Brown-Forsythe test to verify variance homogeneity without the normality assumption that Bartlett's test requires. In environmental science, pollutant measurements are often log-normal or contain detection-limit censored observations, making robust variance tests essential.
Worked Examples
Non-Normal Data Variance Check
Problem:
Three groups of reaction time data (ms). Group 1: 230,250,280,300,320,350. Group 2: 200,220,350,380,420,450. Group 3: 180,280,300,350,400,480. Test variance equality at α = 0.05.
Solution Steps:
- 1Step 1: Enter the three groups. Set α = 0.05.
- 2Step 2: Group medians: G1=290, G2=365, G3=325. Absolute deviations from medians are computed for each observation.
- 3Step 3: A one-way ANOVA is performed on the absolute deviations. SSB and SSW are computed from the Z scores.
- 4Step 4: F = MSB/MSW is compared to the F-critical value. A significant F indicates variance differences.
Result:
Group 2 shows the largest spread (variance approximately 11,000) compared to Group 1 (variance ~2,000). The F-statistic should exceed the critical value, indicating that group variances are not homogeneous. Welch's ANOVA is the recommended next step.
Homogeneous Variances Confirmed
Problem:
Three groups with similar spread: Group 1: 45,48,50,52,55. Group 2: 47,49,51,53,54. Group 3: 46,50,51,52,56. Test at α = 0.05.
Solution Steps:
- 1Step 1: Enter groups with α = 0.05.
- 2Step 2: All group medians are near 50-51. Absolute deviations are small and similar across groups.
- 3Step 3: The ANOVA on deviations produces a small F-statistic — between-group variation in deviations is negligible.
- 4Step 4: F is well below the critical value. Fail to reject H₀ — variances are homogeneous.
Result:
The Brown-Forsythe test confirms that all three groups have similar variances. Since the data is also roughly symmetric, the researcher can proceed with standard one-way ANOVA without concerns about the equal-variance assumption.
Outlier-Robust Comparison
Problem:
A dataset with one extreme outlier in Group 2: Group 1: 10,12,11,13,12. Group 2: 10,11,12,50,11. Group 3: 9,11,12,10,13. Brown-Forsythe is robust to the outlier; Bartlett's test would not be.
Solution Steps:
- 1Step 1: Enter groups. Group 2 has one outlier (50) that dramatically inflates its variance.
- 2Step 2: Group 2 median = 11 (unaffected by the outlier). Absolute deviation for the outlier is |50−11|=39.
- 3Step 3: The ANOVA on absolute deviations still detects that Group 2's deviations are larger on average, but the median-based transformation is less distorted than a mean-based one.
- 4Step 4: The F-statistic may be significant, correctly flagging the variance disparity. Bartlett's test would have been distorted by the outlier.
Result:
The Brown-Forsythe test correctly identifies that Group 2 has larger spread, but the F-statistic is less inflated than it would be under Bartlett's test. The conclusion is robust — variances likely differ, but the test is not unduly influenced by the single outlier.
Tips & Best Practices
- ✓Brown-Forsythe is the default variance test when data is skewed or contains outliers — it should be your first choice.
- ✓Compare the group medians and variances side by side — large variance differences often correspond to uneven spreads.
- ✓The ANOVA of deviations table shows how variation in absolute deviations is partitioned between and within groups.
- ✓If the test is significant, Welch's ANOVA is the most common follow-up — it adjusts for unequal variances.
- ✓With extremely small samples (n < 5), interpret results cautiously — the test may lack power.
Frequently Asked Questions
Sources & References
Last updated: 2026-06-06
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Editorial Note
MyCalcBuddy Editorial Team
This page is maintained as an educational calculator reference.
Formula Source: Standard Mathematical References
by Various