Brown-Forsythe test

Overview
In statistics, when a usual one-way ANOVA is performed, it is assumed that the group variances are statistically equal. If this assumption is not valid, then the resulting F-test is invalid. The Brown-Forsythe test is a statistical test for the equality of group variances based on performing an ANOVA on a transformation of the response variable.

Transformation
The transformed response variable is constructed to measure the spread in each group. Let



z_{ij}=\left\vert y_{ij} - \tilde{y}_j \right\vert $$

where $$\tilde{y}_j$$ is the median of group j. In order to correct for the artificial zeros that come about with odd numbers of observations in a group, any zij that equals zero is replaced by the next smallest zij in group j. The Brown-Forsythe test statistic is the model F statistic from a one way ANOVA on zij:


 * $$F = \frac{(N-p)}{(p-1)} \frac{\sum_{j=1}^{p} n_j (z_{\cdot j}-z_{\cdot\cdot})^2} {\sum_{j=1}^{p}\sum_{i=1}^{N} (z_{ij}-z_{\cdot j})^2}$$

where p is the number of groups, nj is the number of observations in group j, and N is the total number of observations.

If the variances are indeed heterogenous, techniques that allow for this (such as the Welch one-way ANOVA) may be used instead of the usual ANOVA.

Comparison with Levene's test
Levene's test uses the mean instead of the median. Although the optimal choice depends on the underlying distribution, the definition based on the median is recommended as the choice that provides good robustness against many types of non-normal data while retaining good statistical power. If one has knowledge of the underlying distribution of the data, this may indicate using one of the other choices. Brown and Forsythe performed Monte Carlo studies that indicated that using the trimmed mean performed best when the underlying data followed a Cauchy distribution (a heavy-tailed distribution) and the median performed best when the underlying data followed a Chi-square distribution with four degrees of freedom (a heavily skewed distribution). Using the mean provided the best power for symmetric, moderate-tailed, distributions.