**This answer has been revised after being accepted, as I did not adequately appreciate Wilcoxon's critique of the sign test to extend the null hypothesis. I address the difference between the revised and previous answer at the end**

The Wilcoxon sign rank test has these null and alternative hypotheses (see Snedecor, G. W. and Cochran, W. G. (1989) *Statistical Methods*, 8th edition. Iowa State University Press: Ames, IA.):

$\text{H}_{0}$: The magnitude of paired differences are symmetrically distributed about zero.

$\text{H}_{\text{A}}$: The magnitude of paired differences are either not symmetrically distributed, or are not distributed about zero, or both.

(The symmetric distribution about zero rapidly approaches a normal distribution as the sample size increases. See Belera, 2010.)

Many introductory texts motivate the signed rank test as a test of median difference, or more rarely in my experience, mean difference, without mentioning that two fairly strict assumptions are required for this interpretation:

The distribution of both groups must have the same shape.

The variance of both groups must be equal.

If both these assumptions are true, then the signed rank test can validly be interpreted as having a null hypothesis of equal medians (or equal means).

**References**

Bellera, C. A., Julien, M., and Hanley, J. A. (2010). Normal approximations to the distributions of the Wilcoxon statistics: Accurate to what $n$? Graphical insights. *Journal of Statistics Education*, 18(2):1–17.

Wilcoxon, F. (1945). Individual comparisons by ranking methods. *Biometrics Bulletin*, 1(6):80–83.

**Motivation for my revised answer**

My previously accepted answer was that the null and alternative hypotheses were in paired observations:

$$H_0:P(X_A>X_B)=0.5;H_A:P(X_A>X_B)≠0.5$$

These are null and alternative hypotheses about *relative stochastic size* (sometimes *zeroth-order stochastic dominance*). In plain language the null hypothesis is that the probability of a random observation from group $A$ exceeding the paired observation drawn from group $B$ is one half (i.e. a random observation in group $A$ has just as much probability of being greater than, as being less than its paired observation in group $B$).

In plain language the alternative hypothesis is that this probability is not one half (i.e. one of the groups is more likely to be greater than the other than less than the other).

But the sign rank can be false due to asymmetry (because of the magnitude of rank differences are larger in one direction than the other), so that even when $P(X_A>X_B)=0.5$ we reject the null if the magnitudes when $X_{A}>X_{B}$ are, say, much larger than when $X_{A}<X_{B}$. Magnitude of rank difference is therefore why the sign rank test must incorporate symmetry into the null as in my revision. My thanks to @SalMangiafico for his patient tutelage.

## Best Answer

BBR Section 7.2 and especially 7.2.1 cover this. It shows how to scale the signed rank statistic to [0,1] representing the probability that a randomly chosen pair of observations sum to a positive number.

Don't state P < 0.05. Give the P-value, and a proper conclusion is not reject/accept $H_0$ but rather, when P is small, that there is evidence against the supposition of no difference.