It has to do with the assumptions of the test for which the distribution of the test statistic under the null is derived.
The variables are assumed to be continuous.
The probability of a tie is therefore 0 ... and this makes it possible to compute the permutation distribution of the test statistic under the null for given sample size.
Without that assumption being true, you could still do a test, but if you're going to get the null distribution of the test statistic, you'll have to try to compute it conditional on the pattern of tied values (or more easily, simulate).
The easier alternative is to only consider untied values.
Note further that observing ties is not 'evidence in favor of the null', it only contains a lack of evidence against it. With discrete distributions, a range of non-null alternatives are likely to produce ties, not just the null itself.
The 'correct' thing to do is not use a test that assumes continuous distributions on data that don't satisfy the assumptions. If you don't have that, you have to do something to deal with that failure.
I believe that conditioning on the untied data preserves the required properties for the significance level in a way that including ties in some way would not. We might check by simulation.
Brief sketch of ARE for one-sample $t$-test, signed test and the signed-rank test
I expect the long version of @Glen_b's answer includes detailed analysis for two-sample signed rank test along with the intuitive explanation of the ARE. So I'll skip most of the derivation. (one-sample case, you can find the missing details in Lehmann TSH).
Testing Problem: Let $X_1,\ldots,X_n$ be a random sample from location model $f(x-\theta)$, symmetric about zero. We are to compute ARE of signed test, signed rank test for the hypothesis $H_0: \theta=0$ relative to t-test.
To assess the relative efficiency of tests, only local alternatives are considered because consistent tests have power tending to 1 against fixed alternative.
Local alternatives that give rise to nontrivial asymptotic power is often of the form $\theta_n=h/\sqrt{n}$ for fixed $h$, which is called Pitman drift in some literature.
Our task ahead is
- find the limit distribution of each test statistic under the null
- find the limit distribution of each test statistic under the alternative
- compute the local asymptotic power of each test
Test statisics and asymptotics
- t-test (given the existence of $\sigma$) $$t_n=\sqrt{n}\frac{\bar{X}}{\hat{\sigma}}\to_dN(0,1)\quad \text{under the null}$$
$$t_n=\sqrt{n}\frac{\bar{X}}{\hat{\sigma}}\to_dN(h/\sigma,1)\quad \text{under the alternative }\theta=h/\sqrt{n}$$
- so the test that rejects if $t_n>z_\alpha$ has asymptotic power function
$$1-\Phi\left(z_\alpha-h\frac{1}{\sigma}\right)$$
- signed test $S_n=\frac{1}{n}\sum_{i=1}^{n}1\{X_i>0\}$
$$\sqrt{n}\left(S_n-\frac{1}{2}\right)\to_dN\left(0,\frac{1}{4}\right)\quad \text{under the null }$$
$$\sqrt{n}\left(S_n-\frac{1}{2}\right)\to_dN\left(hf(0),\frac{1}{4}\right)\quad \text{under the alternative }$$ and has local asymptotic power
$$1-\Phi\left(z_\alpha-2hf(0)\right)$$
- signed-rank test $$W_n=n^{-2/3}\sum_{i=1}^{n}R_i1\{X_i>0\}\to_dN\left(0,\frac{1}{3}\right)\quad \text{under the null }$$
$$W_n\to_dN\left(2h\int f^2,\frac{1}{3}\right)\quad \text{under the alternative }$$
and has local asymptotic power
$$1-\Phi\left(z_\alpha-\sqrt{12}h\int f^2\right)$$
Therefore, $$ARE(S_n)=(2f(0)\sigma)^2$$
$$ARE(W_n)=(\sqrt{12}\int f^2\sigma)^2$$
If $f$ is standard normal density, $ARE(S_n)=2/\pi$, $ARE(W_n)=3/\pi$
If $f$ is uniform on [-1,1], $ARE(S_n)=1/3$, $ARE(W_n)=1/3$
Remark on the derivation of distribution under the alternative
There are of course many ways to derive the limiting distribution under the alternative. One general approach is to use Le Cam's third lemma. Simplified version of it states
Let $\Delta_n$ be the log of the likelihood ratio. For some statistic
$W_n$, if
$$ (W_n,\Delta_n)\to_d N\left[\left(\begin{array}{c}
\mu\\
-\sigma^2/2
\end{array}\right),\left(\begin{array}{cc}
\sigma^2_W & \tau \\
\tau & \sigma^2/2
\end{array}\right)\right]\\
$$
under the null, then $$W_n\to_d N\left(\mu+\tau,\sigma^2_W\right)\quad\text{under the alternative}$$
For quadratic mean differentiable densities, local asymptotic normality and contiguity are automatically satisfied, which in turn implies Le Cam lemma.
Using this lemma, we only need to compute $\mathrm{cov}(W_n,\Delta_n)$ under the null. $\Delta_n$ obeys LAN $$\Delta_n\approx \frac{h}{\sqrt{n}}\sum_{i=1}^{n}l(X_i)-\frac{1}{2}h^2I_0$$ where $l$ is score function, $I_0$ is information matrix.
Then, for instance, for signed test $S_n$
$$\mathrm{cov}(\sqrt{n}(S_n-1/2),\Delta_n)=-h\mathrm{cov}\left(1\{X_i>0\},\frac{f'}{f}(X_i)\right)=h\int_0^\infty f'=hf(0)$$
Best Answer
Wilcoxon signed rank test and all such univariate analyses (ANOVA, t-Test, etc.) can be analyzed using regression. You will probably need to do a log transformation in the percentages if they are your outcome (DV) and/or use logistic regression if the arm is a dichotomous variable.