# Understanding the Wilcoxon-signed Rank test

• chwala
In summary, the Wilcoxon signed-rank test is a statistical test used for data with unequal variances. It is applicable to non-normally distributed data and provides a p-value based on the number of possible ways to assign positive or negative signs to the first n integers. This test is useful for comparing data sets with unequal variances, unlike the t-test which assumes normality and may not be appropriate for all types of data.

#### chwala

Gold Member
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...

#### Attachments

• wilcoxon signed rank test1.pdf
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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}}##

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?"

chwala

## 1. What is the Wilcoxon-signed Rank test?

The Wilcoxon-signed Rank test is a non-parametric statistical test used to determine whether there is a significant difference between two related samples. It is often used when the data does not meet the assumptions of a parametric test, such as when the data is not normally distributed.

## 2. When should the Wilcoxon-signed Rank test be used?

The Wilcoxon-signed Rank test should be used when working with paired or matched data, where each observation in one sample is related to a specific observation in the other sample. It is also appropriate when the data is not normally distributed or when the sample size is small.

## 3. How does the Wilcoxon-signed Rank test work?

The Wilcoxon-signed Rank test works by ranking the absolute differences between the paired observations and then calculating the sum of the ranks for each group. The test then compares the sums of the ranks to determine if there is a significant difference between the two samples.

## 4. What are the assumptions of the Wilcoxon-signed Rank test?

The Wilcoxon-signed Rank test does not assume that the data is normally distributed, but it does assume that the paired differences are independent and that the data is at least ordinal. It also assumes that the paired differences are continuous or at least approximately continuous.

## 5. How do you interpret the results of the Wilcoxon-signed Rank test?

The results of the Wilcoxon-signed Rank test will provide a p-value, which indicates the probability of obtaining the observed results if the null hypothesis (no difference between the two samples) is true. A p-value less than 0.05 is typically considered statistically significant, indicating that there is a significant difference between the two samples.