For a one-factorial layout with non-normally distributed residuals
the Fligner-Wolfe test can be used.
Let there be \(k-1\)-treatment groups and one control group, then
the null hypothesis, H\(_0: \theta_i - \theta_c = 0 ~ (1 \le i \le k-1)\)
is tested against the alternative (greater),
A\(_1: \theta_i - \theta_c > 0 ~ (1 \le i \le k-1)\),
with at least one inequality being strict.
Let \(n_c\) denote the sample size of the control group,
\(N^t = \sum_{i=1}^{k-1} n_i\) the sum of all treatment
sample sizes and \(N = N^t + n_c\). The test statistic without taken
ties into account is
$$
W = \sum_{j=1}^{k-1} \sum_{i=1}^{n_i} r_{ij} -
\frac{N^t \left(N^t + 1 \right) }{2}
$$
with \(r_{ij}\) the rank of variable \(x_{ij}\).
The null hypothesis is rejected,
if \(W > W_{\alpha,m,n}\) with
\(m = N^t\) and \(n = n_c\).
In the presence of ties, the statistic is
$$
\hat{z} = \frac{W - n_c N^t / 2}{s_W},
$$
where
$$ s_W =
\frac{n_c N^t}{12 N \left(N - 1 \right)}
\sum_{j=1}^g t_j \left(t_j^2 - 1\right),
$$
with \(g\) the number of tied groups and \(t_j\)
the number of tied values in the \(j\)th group. The null hypothesis
is rejected, if \(\hat{z} > z_\alpha\) (as cited in EPA 2006).
If dist = Wilcoxon, then the \(p\)-values are estimated from the Wilcoxon
distribution, else the Normal distribution is used. The latter can be used,
if ties are present.