3.3 The Wilcoxon Signed-Rank Test

The sign test discards a great deal. It records that a difference was positive but not that it was large, so a subject who improved enormously counts exactly as much as one who improved by a hair. When the differences are measured on a scale where their sizes mean something, that information is worth keeping, and the Wilcoxon signed-rank test keeps it.

Proposed by Wilcoxon in the same 1945 paper as the rank-sum test, it applies to the same situations as the sign test — one sample, or paired data reduced to differences — and is almost always the better choice there.

For a sample \(X_1,\dots ,X_n\) and a hypothesised median \(\theta _0\), form the differences \(d_i=X_i-\theta _0\); for paired data \(\left (Y_i,Z_i\right )\) take \(d_i=Y_i-Z_i\). Then

(i).

discard any \(d_i=0\) and reduce \(n\) accordingly, as in the sign test;

(ii).

rank the absolute differences \(\left |d_i\right |\) from \(1\) to \(n\), smallest first, using mid-ranks for ties;

(iii).

attach to each rank the sign of its \(d_i\);

(iv).

let \(T_{+}\) be the sum of the ranks carrying a plus and \(T_{-}\) the sum of those carrying a minus.

Since every rank goes into one sum or the other, \[T_{+}+T_{-}=\sum ^{n}_{i=1}i=\dfrac {n(n+1)}{2},\] which is the arithmetic check: if the two sums do not add to \(n(n+1)/2\), the ranking is wrong. For a two-sided test the statistic is \[T=\min \left (T_{+},T_{-}\right ).\]

\[H_0:\text {the differences are symmetrically distributed about } 0 .\]

Symmetry is doing essential work here, and it is an assumption the sign test does not need. Under \(H_0\) a difference of any given magnitude is as likely to be negative as positive, independently across observations, so each rank \(1,2,\dots ,n\) carries a plus or a minus with probability \(\tfrac {1}{2}\) each. That is what makes the null distribution of \(T_{+}\) computable without knowing anything else about the population — and it is why the test is not valid for strongly skewed differences, where the sign test remains available.

Enumerating all \(2^{n}\) sign patterns gives the exact distribution. From it, \[E\left (T_{+}\right )=\dfrac {n(n+1)}{4},\hspace {0.8cm} \operatorname {var}\left (T_{+}\right )=\dfrac {n(n+1)(2n+1)}{24},\] the mean being half the total \(n(n+1)/2\) by symmetry. For \(n>20\), \[Z=\dfrac {T_{+}-\dfrac {n(n+1)}{4}}{\sqrt {\dfrac {n(n+1)(2n+1)}{24}}}\ \thicksim \ N(0,1);\] below that, exact tables of \(T\) are used, and \(H_0\) is rejected for small \(T\).

Hypotheses and the matching statistic. Write \(D_1\), \(D_2\) for the two distributions. The statistic to use is the one that becomes small under the alternative:

The test on the example. For a two-sided alternative at \(\alpha = 0.05\) with \(n' = 6\), the critical value is \(T_0 = 0\); for a one-sided test at \(\alpha = 0.05\) it is \(T_0 = 2\). Here \[T = \min (12, 9) = 9 \ \not \leq \ T_0 ,\] so we fail to reject \(H_0\): there is no evidence of a shift. Where the critical values come from. With \(n' = 6\) there are \(2^{6} = 64\) equally likely sign patterns under \(H_0\), so the null distribution of \(T_{+}\) can be written down exactly: \[P(T_{+} \leq 0) = \tfrac {1}{64} = 0.0156,\quad P(T_{+} \leq 1) = \tfrac {2}{64} = 0.0312,\quad P(T_{+} \leq 2) = \tfrac {3}{64} = 0.0469,\] \[P(T_{+} \leq 3) = \tfrac {5}{64} = 0.0781 .\] For a one-sided test the whole of \(\alpha \) sits in one tail, so the largest usable cut-off is \(T_0 = 2\), not \(3\): a cut-off of \(3\) would give a \(7.8\%\) test, not a \(5\%\) one. For a two-sided test each tail may carry only \(0.025\), and since \(P(T_{+}\leq 1) = 0.0312\) already exceeds that, the cut-off drops to \(T_0 = 0\) — a two-sided test at \(n' = 6\) can reject only when every single difference points the same way.

At such small \(n'\) the attainable levels are coarse, and it is worth quoting the exact tail rather than the nominal \(\alpha \). The two-sided test above is really a \(3.1\%\) test, and there is no way to make it a \(5\%\) one.