 Solution

The experimental unit for comparing the treatments is the pot: its two plants form a pair. Let $D_i=C_i-S_i$ be cross-fertilised minus self-fertilised height in pot $i$, for $i=1,\dots ,15$. The paired Wilcoxon signed-rank test assumes independent differences with a continuous distribution symmetric about a location $\delta$. Its null is $\delta =0$, with the one-sided formulation $\delta \le 0$, against $\delta >0$. Under this location model it asks whether cross-fertilisation produces a positive shift within a pot. The location is a median, and also a mean if it exists; a hypothesis merely about a median without symmetry does not justify the signed-rank null distribution.

Test 1 uses this paired test and its exact distribution. Form the fifteen differences, omit any zero differences, rank their absolute values, and sum the statistical ranks belonging to positive differences. That positive-rank sum is the reported $V=96$. Tied absolute differences receive average statistical ranks. Under the symmetric zero-location null, conditional on their magnitudes, the signs are independent fair signs. The exact upper-tail probability is obtained from all sign allocations to the observed statistical ranks: with fifteen nonzero differences there are $2^{15}$ equiprobable allocations. For untied absolute differences the rank weights are $1,\dots ,15$; ties would require the corresponding conditional rank weights. The reported probability of a positive-rank sum at least as large as observed is $0.0206$. No normal approximation is used.

Test 2 makes exactly the same paired comparison and has the same null and alternative hypotheses. exact=F replaces the exact distribution by a normal approximation with a continuity correction. Under an untied null the positive-rank sum hasMore generally, conditional on the nonzero absolute statistical ranks $r_i$, its variance is $\frac{1}{4}{\sum }_i{r}_i^2$. The upper-tail correction places the boundary half a lattice unit below the observed statistic. Standardizing gives the reported $Z=2.0163$, and its approximate $p$-value is $1-\Phi (Z)=0.0219$. Thus the slight difference from Test 1 reflects approximation, not a different scientific question.

Test 3 omits paired=T, so it is the unpaired Wilcoxon rank sum test. Its usual distribution-free null is that two independent samples have the same continuous distribution. In a common-shape location model this is a zero location difference, against a positive cross-minus-self shift. More generally “greater” directs the test towards the cross distribution being shifted upwards; it does not turn the test into a general comparison of means or medians for arbitrary unequal shapes.

To form its statistic, pool the thirty heights, rank them together with averaged statistical ranks for ties, and add the cross-group statistical ranks to get $R_C$. Equivalently use $U=R_C-n_C(n_C+1)/2$. Under independent untied samples,If the pooled ties have sizes $t_h$ and $N=n_C+n_S$, the variance becomesSubtract the null mean and the upper-tail continuity correction, then divide by this null standard deviation. This procedure gives the reported rank-sum $Z=3.0105$. Its approximate one-sided probability is $1-\Phi (Z)=0.0013$. This describes how the statistic and probability are obtained without reconstructing the unprinted heights.

For the statistician, use the paired analysis, preferably the exact Test 1 for these fifteen pairs. The independent units are the pots, while the two heights within a pot share growing conditions and cannot be treated as independent samples. Test 2 is a reasonable approximation to the same paired test and agrees closely. Test 3 discards the matching and uses a null variance that assumes independence between the samples, so its smaller $p$-value is not a reason to prefer it. The appropriate paired result gives evidence at $5\%$, but not at $1\%$, that cross-fertilised plants have a positive height shift relative to their paired self-fertilised plants, subject to the symmetry assumption for differences.