Understanding the Wilcoxon-signed Rank test

[chwala]

Homework Statement

I would like insight on the highlighted part in red.

Relevant Equations

stats

Reference; https://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test#:~:text=The Wilcoxon signed-rank test,-sample Student's t-test.

I have managed to go through the literature (it is pretty straight forward). In general Wilcoxon-rank method applies to data with unequal variances otherwise student t- test would be sufficient.

Now looking at the signed ranks given (check attachment) we have;

$[1,2,3,4,5,6,7,8,9,10,11,-12]$

In my understanding, we shall have $W^{+} =66, W^{-}=12$ and we also know that,

$\dfrac {n(n+1)}{2}=\dfrac {12(13)}{2}=78=[W^{+}+W^{-}]$

therefore it follows that, test statistic $W=12$.

Now to my question, how did they arrive at p-value $\left[\dfrac{55}{2^{12}}\right]$?

Secondly, how did they arrive at the given signed-ranks without the 'ordered absolute value of differences'?

thanks...

 

[chwala]

I think i got it; The p-value is given by;

$P(W=w)=f(w)= \dfrac{c(w)}{2^n}$ where $c(w)$ is the number of possible ways to assign a+ or a- to the first n integers. For our case we have;

$c(w)=\dfrac{n(n+1)}{2}=\dfrac {10×11}{2}=55$.

Therefore,

$P(W=w)=f(w)= \dfrac{55}{2^{12}}$

 

[Ibix]

Note that the t-test can be used (with some modification) for unequal variances. The point about the Wilcoxon test is that it is applicable to data that are not normally distributed, where the t-test assumes normality.

The problem with the uneqal variance test was neatly summed up in (I think) Cochran's book on sample design: a hundred men try a hair restorer product. Fifty go bald and fifty turn into werewolves. Blindly applying the inequal variance test to compare them to a control is just asking "but, were they hairier on average?"