3.9 Ties in the Rank-Sum Test

As with the signed-rank test, tied observations are given average ranks and the variance must be reduced to match. Writing \(N = m + n\) for the combined sample size, \[\operatorname {var}(U) = \dfrac {mn}{12}\left [ (N + 1) - \dfrac {\sum _{p}\left (t_p^{3} - t_p\right )}{N(N-1)}\right ],\] where \(t_p\) is the size of the \(p\)-th group of tied values in the combined sample — both groups pooled, since ranking is done on the pooled data. With no ties the subtracted term is zero and the variance returns to \(\dfrac {mn(N+1)}{12}\).

A continuity correction of \(0.5\) may also be applied, replacing \(U\) by \(U \pm 0.5\) in the direction that moves \(Z\) towards zero.

Since \(U\) assumes only integer values, a continuity correction of \(0.5\) may be used alternatively the Wilcoxon-Mann-Whitney statistic is computed as follows.
We assume the 2 samples are drawn from a continuous distribution so that the probability of \(X_i=Y_j\) for some \(i\hspace {0.3cm} \text {and}\hspace {0.3cm} j\) need not to be considered. If \(mn\) indicators random variables are defined as \[D_{ij}= \begin {cases} 1, &\text {if}\hspace {0.3cm} Y_j<X_i\\ 0, &\text {if}\hspace {0.3cm} Y_j>X_i\\ \end {cases} \] for \(i=1,2,\dots ,n\) and \(j=1,2,\dots ,m\).

The \(U\) statistic is given by \(\displaystyle {U=\sum ^n_{i=1}\sum ^m_{j=1}D_{ij}}.\)

The logical rejection region \(R\), test statistic for 1-sided \(H_1\), that the \(Ys\) are stochastically lager than the \(Xs\). \[H_1:F_Y(x)\leq F_X(x)\] strict equality for some \(x\). Would clearly be small values of \(U\).

\begin {align*} U_1 &=R_1-\frac {n_1(n_1+1)}{2}, \hspace {1cm}\text {the } X\text {'s}\\ U_2 &=R_2-\frac {n_2(n_2+1)}{2}, \hspace {1cm}\text {the } Y\text {'s}\\ R_1+R_2 &=\frac {N(N+1)}{2}, \hspace {1.55cm} N=n_1+n_2\\ U_1+U_2 &=n_1n_2, \hspace {2.1cm}\text {so } U_1=n_1n_2-U_2 \end {align*}

The last two identities are the arithmetic checks used above: if either fails, the ranking is wrong.

Generalised
The general sample-problem \((K>2)\). The natural extension of the two sample problem is the \(K\) sample problem where observations are taken under a variate of independent conditions.
Assume that we have \(K\) independent observations, one from each of the \(K\) continuous population \(F_1(X_1), F_2(X_2),\dots , F_n(X_n)\) where the \(i^{th}\) random sample is of size \(n\), \(n_i=1,\dots ,K\) and there are a total of \(N=\sum \limits ^K_{i=1}n_i\) observations.
Note that we assume the independence extensions across samples to within samples. The extension of the two sample hypothesis to the \(K\) sample problems is that all \(K\) sample are drawn from the same population \[H_0:F_1(x)=F_2(x)=\dots =F_K(x)\hspace {0.4cm}\text {for all}\hspace {0.3cm} x\] then the general alternative \[H_a:F_1(x)\neq F(x)\] The location made for the \(K\) sample problem for the cdf are \[F_1(x-\theta _1), F_2(x-\theta _2),\dots , F_k(x-\theta _k)\] respectively where \(\theta \) denotes a location parameter for the \(i^{th}\) parameter usually interpreted as the median or treatment effect.
The null hypothesis are now written as \(H_0:\theta _1=\theta _2=\theta _3=\dots =\theta _k\hspace {0.2cm}\) and the general alternative is \(H_1:\theta _i\neq \theta _j\hspace {0.4cm}\text {for some}\hspace {0.4cm}i\neq j\)
Then non-parametric techniques which have been developed for this \(k\) sample problem requires no assumptions beyond continuous populations and there application to any situation and thus involve only simple calculations which cover here there extension of those sample problems, the Kruskal-Wallis Anova test including comparisons with a control of standardisation.