Solution. \(H\) is computed from the ranks of the pooled sample, and ties leave the ranking undefined until a convention is imposed.

Step one: mid-ranks. Observations that are equal are given the average of the ranks they would jointly have occupied. If the fourth, fifth and sixth smallest values in the pooled sample are all equal, each receives \(\frac {4+5+6}{3}=5\). The rank sums \(R_j\) and the statistic \[H=\dfrac {12}{N(N+1)}\sum ^{k}_{j=1}\dfrac {R_j^{2}}{n_j}-3(N+1)\] are then formed from these mid-ranks in the usual way. Note that the total of all ranks is still \(N(N+1)/2\), so the arithmetic check is unaffected by ties.

Step two: correct the statistic. Averaging ranks reduces the variance of the rank sums, which makes \(H\) systematically too small. The corrected statistic divides by a factor computed from the tie pattern: \[H^{*}=\dfrac {H}{1-\dfrac {\sum _{p}\left (t_p^{3}-t_p\right )}{N^{3}-N}},\] the sum running over every group of tied values in the pooled sample and \(t_p\) being the number of observations in the \(p^{th}\) group. Untied observations form groups of size \(1\) and contribute nothing, since \(1^{3}-1=0\).

\(H^{*}\) is then referred to \(\chi ^{2}_{k-1}\) as before.

Direction and size of the correction. The divisor is at most \(1\), so \(H^{*}\geq H\): the correction always inflates the statistic and makes rejection easier. Omitting it is therefore conservative — the test loses power but does not overstate significance — which is why the omission so often goes unnoticed. The effect is negligible unless ties are numerous, but with heavily tied data, such as ordinal scores on a five-point scale, it can be substantial.

Solution. Two facts about the pooled ranking are needed. Every rank from \(1\) to \(N\) is used exactly once, so \[\sum ^{k}_{j=1}R_j=\sum ^{N}_{i=1}i=\dfrac {N(N+1)}{2}, \hspace {1cm}\text {and}\hspace {1cm}\sum ^{k}_{j=1}n_j=N .\]

Expand the square: \begin {align*} F &=A\sum ^{k}_{j=1}\dfrac {1}{n_j} \left [R_j^{2}-2R_j\cdot \dfrac {n_j(N+1)}{2}+\dfrac {n_j^{2}(N+1)^{2}}{4}\right ]\\[4pt] &=A\sum ^{k}_{j=1}\left [\dfrac {R_j^{2}}{n_j}-R_j(N+1) +\dfrac {n_j(N+1)^{2}}{4}\right ]\\[4pt] &=A\sum ^{k}_{j=1}\dfrac {R_j^{2}}{n_j} -A(N+1)\sum ^{k}_{j=1}R_j +\dfrac {A(N+1)^{2}}{4}\sum ^{k}_{j=1}n_j . \end {align*}

Substituting the two sums, \begin {align*} F &=A\sum ^{k}_{j=1}\dfrac {R_j^{2}}{n_j} -A(N+1)\cdot \dfrac {N(N+1)}{2} +\dfrac {A(N+1)^{2}}{4}\cdot N\\[4pt] &=A\sum ^{k}_{j=1}\dfrac {R_j^{2}}{n_j} -\dfrac {AN(N+1)^{2}}{2}+\dfrac {AN(N+1)^{2}}{4}\\[4pt] &=A\sum ^{k}_{j=1}\dfrac {R_j^{2}}{n_j}-\dfrac {AN(N+1)^{2}}{4} . \end {align*}

Finally, with \(A=\dfrac {12}{N(N+1)}\), \[\dfrac {AN(N+1)^{2}}{4} =\dfrac {12}{N(N+1)}\cdot \dfrac {N(N+1)^{2}}{4} =\dfrac {12(N+1)}{4}=3(N+1),\] so \[F=A\sum ^{k}_{j=1}\dfrac {R_j^{2}}{n_j}-3(N+1). \qquad \blacksquare \]

Solution. The procedure. A significant Friedman test says only that the \(k\) treatments are not all alike. To find which pairs differ, the Nemenyi test — the rank analogue of Tukey’s — compares every pair of mean ranks against a single critical difference \[CD=q_{\alpha }\sqrt {\dfrac {k(k+1)}{6n}}, \hspace {0.8cm}q_{\alpha }=\dfrac {q_{\infty ,k,\alpha }}{\sqrt {2}},\] declaring treatments \(i\) and \(j\) different when \(\left |\overline {R}_{\cdot i}-\overline {R}_{\cdot j}\right |\geq CD\), where \(\overline {R}_{\cdot j}=R_{\cdot j}/n\). Using one threshold for all \(\binom {k}{2}\) pairs is what keeps the family-wise error rate at \(\alpha \); testing each pair at \(\alpha \) separately would not.

If instead every treatment is compared only with a control, there are \(k-1\) comparisons rather than \(\binom {k}{2}\), and a Bonferroni adjustment — each comparison at \(\alpha /(k-1)\) — is available and more powerful than Nemenyi, which pays for comparisons that are not being made.

The omnibus test. First check the rank sums: \(34+18+32+16=100=\dfrac {nk(k+1)}{2}=\dfrac {10\times 4\times 5}{2}\). Then \begin {align*} F_R &=\dfrac {12}{nk(k+1)}\sum ^{k}_{j=1}R_{\cdot j}^{2}-3n(k+1)\\ &=\dfrac {12}{10\times 4\times 5}\left (34^{2}+18^{2}+32^{2}+16^{2}\right ) -3\times 10\times 5\\ &=0.06\left (1156+324+1024+256\right )-150\\ &=0.06(2760)-150=165.6-150=15.6 . \end {align*}

With \(k-1=3\) degrees of freedom, \(\chi ^{2}_{(0.05,3)}=7.815\), and \(15.6>7.815\), so \(H_0\) is rejected.

The comparisons. The mean ranks are \(3.4\), \(1.8\), \(3.2\), \(1.6\). With \(q_{0.05}=3.633/\sqrt {2}=2.569\), \[CD=2.569\sqrt {\dfrac {4\times 5}{6\times 10}}=2.569\times 0.577=1.483 .\]

Treatments \(1\) and \(3\) form one group and treatments \(2\) and \(4\) another: each of \(1\) and \(3\) differs from \(4\), and \(1\) differs from \(2\), while the members of each pair cannot be separated. Note that \(2\) against \(3\), at \(1.4\), falls just short of the threshold even though \(2\) against \(1\) at \(1.6\) clears it — with overlapping groups like this the honest report is the set of differences that reached \(CD\), not a clean ranking of all four.