Welch's ANOVA

A one-way ANOVA that does not assume equal group variances, weighting each group by n/s².

w_i = \frac{n_i}{s_i^2},\quad W = \sum w_i,\quad \tilde y = \frac{\sum w_i \bar y_i}{W},\quad A = \frac{\sum w_i(\bar y_i - \tilde y)^2}{k-1},\quad \Lambda = \sum\frac{(1 - w_i/W)^2}{n_i - 1}

$\bar y_i, s_i, n_i$: Mean, SD (n − 1) and size of group i.
$w_i$: Precision weight of group i.
$\tilde y$: Weighted grand mean.

F = \frac{A}{1 + \frac{2(k-2)\Lambda}{k^2-1}},\qquad df = \left(k - 1,\ \frac{k^2-1}{3\Lambda}\right)

The denominator df is usually fractional.

$k$: Number of groups.

When to use

Comparing several means when group SDs or sizes differ — a safe default, as Welch's t-test is for two groups. Each group needs at least 2 observations and a positive SD.

Assumptions

Independent observations; roughly normal data within groups (or reasonably large groups). Equal variances are not required.

How to interpret

Same as classic ANOVA: a small p-value is evidence that not all means are equal. For pairwise follow-ups with unequal variances, use Welch pairwise t-tests with a multiple-comparison adjustment (Games–Howell is a common alternative).

Common mistakes

Reporting the classic F's df (k − 1, N − k) with Welch's F.
Following Welch's ANOVA with Tukey HSD, which assumes equal variances.

Example

For the groups (4, 5, 6, 5), (7, 8, 6, 7), (5, 6, 7, 6), Welch's $F = 5.4$ on (2, ≈ 6.0) df, $p = 0.046$ (classic ANOVA: $F = 6.0$ , $p = 0.022$ ). With equal SDs the conclusions agree; Welch is a little less powerful here and safer when SDs differ.