Can someone explain to me need symmetrical distribution for Wilcoxon signed rank test? So I am confused. I understand that Wilcoxon signed rank test does not need to have normal distribution (hence why it is used when your data is non-normally distributed in place of paired samples t-test). What I don't understand though is that the assumptions say the perform a box plot to see if is symmetrical? Is it looking to see if median's are roughly the same?
I'm not sure what they are referring to.
 A: Let's first consider the test statistic, for two independently drawn measurements, $x_1, \dots, x_N$ and $y_1, \dots, y_N$.
$$W=\sum_i sgn(x_i - y_i)R_i$$
Where $R_i$ is the rank from smallest to largest. (We also need to drop 0's but let's not worry about this here).
Now getting to your question, I assume you are asking why the null hypothesis is that the distribution of differences is symmetric vs. the alternate hypothesis that the distribution of differences is not symmetric.
From a purely heuristic perspective, let's think about what we see when we look at the differences between two measurements of what, under the null, should have the same mean. Simply, we would expect that the differences are symmetric about 0. In particular, it can be shown that the test statistic under the null has a symmetric distribution for small $N$.
Perhaps, easier to see as $N\to\infty$ the test statistic will be approximately normal (a symmetric distribution). This is clear because if the distributions of $x$ and $y$ have the same mean, then we have the sum of mean zero random variables.
There is a little bit of work in actually showing these results and so I will leave you with an introductory reference: https://math.mit.edu/~rmd/650/nonpartests.pdf
A: Here's the kind of nonsense you can get when you try to use the one-sample Wilcoxon signed rank test on data from a highly skewed distribution, such as an exponential population with mean $1.$
This population has median $\eta = -\ln(.5) = 0.69315.$
qexp(.5)
[1] 0.6931472

Look at sample x of size $n = 1000$ from this distribution.
set.seed(2021)
x = rexp(1000)
summary(x)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
0.000806 0.283089 0.691022 1.005858 1.415143 6.687939 

boxplot(x, col="skyblue2", pch=20, horizontal=T)


So the population median is $\eta = 0.69315,$ the
sample median is $H = 0.69102$ and the null hypothesis
$H_0: \eta = 0.69315$ against $H_0: \eta \ne 0.69315.$
is strongly rejected with P-value near $0.$
wilcox.test(x, mu = 0.69315)

        Wilcoxon signed rank test 
        with continuity correction

data:  x
V = 298800, p-value = 1.072e-07
alternative hypothesis: 
 true location is not equal to 0.69315

This is not a fluke and not due to the large sample size. In 100,000 tests on exponential samples of
size $n=100,$ almost 40% led to rejection at the 5% level, Of course, a legitimate test would
have rejected only 5% of the time.
set.seed(1234)
pv = replicate(10^5, wilcox.test(rexp(100), mu = 0.69315)$p.val)
mean(pv <= .05)
[1] 0.38491

