Solution. The differences \(d=\text {before}-\text {after}\) are \[10,\hspace {0.3cm}20,\hspace {0.3cm}-10,\hspace {0.3cm}25,\hspace {0.3cm}60, \hspace {0.3cm}10,\hspace {0.3cm}15,\hspace {0.3cm}-5 .\] A positive \(d\) means the behaviour fell, so the drug working corresponds to positive differences and the alternative is one-sided upward. None is zero, so \(n=8\) throughout.

(a) Sign test. Six differences are positive and two negative, so \(N_{+}=6\) with \(N_{+}\sim \operatorname {Bin}\left (8,\tfrac {1}{2}\right )\) under \(H_0\). Large \(N_{+}\) favours \(H_1\): \[p=P\left (N_{+}\geq 6\right ) =\dfrac {\binom {8}{6}+\binom {8}{7}+\binom {8}{8}}{2^{8}} =\dfrac {28+8+1}{256}=\dfrac {37}{256}=0.1445 .\] Since \(0.1445>0.05\) we do not reject \(H_0\). The sign test finds no evidence that the drug works.

(b) Wilcoxon signed-rank test. Rank the absolute differences, averaging ties:

The three \(\left |d\right |=10\) would take ranks \(2,3,4\) and so each receives \(\frac {2+3+4}{3}=3\). Then \[T_{+}=3+3+5+6+7+8=32,\hspace {0.8cm}T_{-}=1+3=4,\] and the check holds: \(32+4=36=\dfrac {8\times 9}{2}\).

The alternative says the differences run positive, which drives \(T_{-}\) down, so \(T_{-}\) is the statistic and small values are significant. For \(n=8\) one-sided at \(\alpha =0.05\) the critical value is \(T_0=5\), since \(P\left (T\leq 5\right )=0.0391\) while \(P\left (T\leq 6\right )=0.0547\). Here \[T_{-}=4\leq 5,\] so we reject \(H_0\). The exact \(p\)-value is \(P\left (T\leq 4\right )=\frac {7}{256}=0.0273\).

(c) Comment. The two tests disagree: \(p=0.1445\) against \(p=0.0273\), on identical data. The reason is that the two negative differences, \(-10\) and \(-5\), are among the smallest in the set, while the positive ones include \(60\), \(25\) and \(20\). The sign test counts six arrows up and two down and finds that unremarkable in eight children. The signed-rank test sees that the two children who worsened did so barely, and that between them they account for only \(4\) of the \(36\) available rank points.

Where both tests apply, the signed-rank test is the more powerful and is the one to report. It buys that power by assuming the differences are symmetric about their median, which the sign test does not require — so on markedly skewed differences the sign test’s verdict would be the one to trust.

Solution. Under \(H_0\) each of the ranks \(1,2,3,4\) independently carries a plus or a minus with probability \(\tfrac {1}{2}\), so there are \(2^{4}=16\) equally likely sign patterns and \(T_{+}\) is the sum of the ranks carrying a plus. Enumerating the subsets of \(\{1,2,3,4\}\) by their totals:

The two subsets summing to \(3\) are \(\{3\}\) and \(\{1,2\}\); those summing to \(4\) are \(\{4\}\) and \(\{1,3\}\), and so on. The distribution is symmetric about \(\dfrac {n(n+1)}{4}=5\) and the totals check: \(1+1+1+2+2+2+2+2+1+1+1=16\).

Accumulating the lower tail, \[P\left (T_{+}\leq 0\right )=\dfrac {1}{16}=0.0625,\hspace {0.6cm} P\left (T_{+}\leq 1\right )=\dfrac {2}{16}=0.1250,\hspace {0.6cm} P\left (T_{+}\leq 2\right )=\dfrac {3}{16}=0.1875 .\]

Critical values. There are none. Even the most extreme outcome possible — every one of the four differences pointing the same way — has one-sided probability \(0.0625\), which exceeds \(0.05\). No critical region of size at most \(0.05\) exists, one-sided or two-sided, so at \(n=4\) the test cannot reject \(H_0\) at the \(5\%\) level whatever the data.

Comment. This is a property of the design, not of the sample. With four paired observations the smallest attainable significance level is \(2/16=0.125\) two-sided, and no amount of favourable data will improve it. The practical consequence is that the sample size must be checked before the study is run: for a two-sided \(5\%\) signed-rank test \(n\geq 6\) is needed for any critical region to exist at all, and \(n=6\) only rejects when all six differences agree in sign.