WilcoxonTests.Rmd

---
title: "Wilcoxon Tests"
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    theme: cerulean
    css: styles.css
---


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```{r, include=FALSE}
library(mosaic)
```


----

Wilcoxon tests allow for the testing of hypotheses about the value of the the *median* without assuming the test statistic follows any parametric distribution. They are often seen as nonparametric alternatives to the various t tests. However, they can also be used on ordinal data (data that is not quite quantitative, but is ordered) unlike t tests which require quantitative data.

----


### Wilcoxon Signed-Rank Test {.tabset .tabset-fade .tabset-pills}

<div style="float:left;width:125px;" align=center>
<img src="./Images/QuantY.png" width=35px;>
</div>

For testing hypotheses about the value of the median of (1) one sample of quantitative data or (2) one set of differences from paired data. 

#### Overview 

<div style="padding-left:125px;">
The nonparametric equivalent of the paired-samples t test as well as the one-sample t test. 

Best for smaller sample sizes where the distribution of the data is not normal. The t test is more appropriate when the data is normal or when the sample size is large. 

While the test will work in most scenarios it suffers slightly when ties (repeated values) are present in the data. If *many* ties are present in the data, the test is not appropriate. If only a few ties are present, the test is still appropriate.

**Hypotheses**

Originally created to test hypotheses about the value of the median, but works as well for the mean when the distribution of the data is symmetrical.

<div style="padding-left:15px;">

<div style="float:right;font-size:.8em;background-color:lightgray;padding:5px;border-radius:4px;"><a style="color:darkgray;" href="javascript:showhide('wilcoxonsignedranklatex')">Math Code</a></div>


<div id="wilcoxonsignedranklatex" style="display:none;">
```{}
$$
  H_0: \text{median of differences} = 0
$$

$$
  H_a: \text{median of differences} \neq 0
$$
```
</div>


$H_0: \text{median of differences} = 0$

$H_a: \text{median of differences} \ \left\{\underset{<}{\stackrel{>}{\neq}}\right\} \ 0$
</div>

**Examples**: [sleep](./Analyses/Wilcoxon Tests/Examples/SleepPairedWilcoxon.html), [CornHeights](./Analyses/Wilcoxon Tests/Examples/CornHeightsPairedWilcoxon.html)

</div>


----

#### R Instructions

<div style="padding-left:125px;">

**Console** Help Command: `?wilcox.test()`

##### Paired Data


`wilcox.test(Y1, Y2, mu = YourNull, alternative = YourAlternative, paired = TRUE, conf.level = 0.95)`

* `Y1` must be a "numeric" vector. One set of measurements from the pair.
* `Y2` also a "numeric" vector. Other set of measurements from the pair.
* `YourNull` is the numeric value from your null hypothesis for the median of differences from the paired data. Usually zero.
* `YourAlternative` is one of the three options: `"two.sided"`, `"greater"`, `"less"` and should correspond to your alternative hypothesis.
* The value for `conf.level = 0.95` can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $1-\alpha$.


**Example Code** 

Hover your mouse over the example codes to learn more.

<a href="javascript:showhide('wilcoxonSignedRank')">
<div class="hoverchunk">
<span class="tooltipr">
wilcox.test(
  <span class="tooltiprtext">'wilcox.test' is a function for non-parametric one and two sample tests.</span>
</span><span class="tooltipr">
sleep\$extra[sleep\$group==1],&nbsp;
  <span class="tooltiprtext">The hours of extra sleep that the group had with drug 2.</span>
  </span><span class="tooltipr">
 sleep\$extra[sleep\$group==2],&nbsp;
  <span class="tooltiprtext">The hours of extra sleep that the same group had with drug 1.</span>
  </span><span class="tooltipr">
mu = 0,&nbsp; 
  <span class="tooltiprtext">The numeric value from the null hypothesis for the median of differences from the paired data is 0 meaning the null hypothesis is $\text{median of differences} = 0$.</span>
</span><span class="tooltipr">
paired=TRUE,&nbsp; 
  <span class="tooltiprtext">This command forces a "paired" samples test to be performed.</span>
</span><span class="tooltipr">
alternative = "two.sided",&nbsp; 
  <span class="tooltiprtext">The alternative hypothesis is "two.sided" meaning the alternative hypothesis is $\text{median of differences} \neq0$.</span>
  </span><span class="tooltipr">
conf.level = 0.95)
  <span class="tooltiprtext">This test has a 0.95 confidence level which corresponds to 1 - $\alpha$.</span>
</span><span class="tooltipr">
&nbsp;&nbsp;&nbsp;&nbsp;  
  <span class="tooltiprtext">Press Enter to run the code if you have typed it in yourself. You can also click here to view the output.</span>
</span><span class="tooltipr" style="float:right;">
&nbsp;...&nbsp; 
  <span class="tooltiprtext">Click to View Output.</span>
</span>
</div>
</a>

<div id="wilcoxonSignedRank" style="display:none;">
<table class="rconsole">
<tr>
<td>
<span class="tooltiprout">
 Wilcoxon signed rank test with continuity correction
    <span class="tooltiprouttext">The phrase "with continuity correction" implies that instead of using the "exact" distribution of the test statistic a "normal approximation" was used instead to compute the p-value. Further, a small correction was made to allow for the change from the "discrete" exact distribution to the "continuous normal distribution" when calculating the p-value.</span>
</span>
</td>
</tr>
</table>
<br/>

<table class="rconsole">
<tr>
<td>
  <span class="tooltiprout">
    data: sleep\$extra[sleep\$group == 1] and sleep\$extra[sleep\$group == 2]
      <span class="tooltiprouttext">This statement of the output just reminds you of the code you used to perform the test. The important thing is recognizing that the first group listed is "Group 1" and the second group listed is "Group 2." This is especially important when using alternative hypotheses of "less" or "greater" as the order is always "Group 1" is "less" than "Group 2" or "Group 1" is "greater" than "Group 2."</span>
</td><td>
  <span class="tooltiprout">
   V = 0,
      <span class="tooltiprouttext">This is the test statistic of the test, i.e., the sum of the ranks from the positive group minus the minimum sum of ranks possible.</span>
</td><td>
  <span class="tooltiprout">
    &nbsp;p-value = 0.009091
      <span class="tooltiprouttext">This is the p-value of the test. If no warning is displayed when the test is run, then this is the "exact" p-value from the non-parametric Wilcoxon Test Statistic distribution. Sometimes a message will appear stating "Cannot compute exact p-value with ties" or other similar messages. In those cases, the p-value is still considered valid even though it is obtained through a normal approximation to the exact distribution.</span>
</td>
</tr><tr>
<td>
  <span class="tooltiprout">
    alternative hypothesis: true location shift is not equal to 0
      <span class="tooltiprouttext">This reports that the alternative hypothesis was "two-sided." If the alternative had been "less" or "greater" the wording would change accordingly.</span>
</td>
</tr>
</table>

</div>

<br>


##### One Sample

`wilcox.test(object, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)`

* `object` must be a "numeric" vector.
* `YourNull` is the numeric value from your null hypothesis for the median (even though it says "mu").
* `YourAlternative` is one of the three options: `"two.sided"`, `"greater"`, `"less"` and should correspond to your alternative hypothesis.
* The value for `conf.level = 0.95` can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $1-\alpha$.



**Example Code** 

Hover your mouse over the example codes to learn more.

<a href="javascript:showhide('wilcoxOneSample')">
<div class="hoverchunk">
<span class="tooltipr">
wilcox.test(
  <span class="tooltiprtext">'wilcox.test' is a function for non-parametric one and two sample tests.</span>
</span><span class="tooltipr">
mtcars
  <span class="tooltiprtext">'mtcars' is a dataset. Type 'View(mtcars)' in R to view the dataset.</span>
</span><span class="tooltipr">
$
  <span class="tooltiprtext">The $ allows us to access any variable from the mtcars dataset.</span>
</span><span class="tooltipr">
mpg,&nbsp;
  <span class="tooltiprtext">'mpg' is a quantitative variable (numeric vector) from the mtcars dataset.</span>
</span><span class="tooltipr">
mu = 20,&nbsp; 
  <span class="tooltiprtext"> The numeric value from the null hypothesis is 20 meaning $\mu = 20$. </span>
</span><span class="tooltipr">
alternative = "two.sided",&nbsp; 
  <span class="tooltiprtext"> The alternative is "two.sided" meaning the alternative hypothesis is $\mu\neq20$.</span>
</span><span class="tooltipr">
paired=TRUE,&nbsp; 
  <span class="tooltiprtext">This command forces a "paired" samples test to be performed.</span>
</span><span class="tooltipr">
conf.level = 0.95)
  <span class="tooltiprtext">This test has a 0.95 confidence level which corresponds to 1−α. </span>
</span><span class="tooltipr">
&nbsp;&nbsp;&nbsp;&nbsp;  
  <span class="tooltiprtext">Press Enter to run the code if you have typed it in yourself. You can also click here to view the output.</span>
</span><span class="tooltipr" style="float:right;">
&nbsp;...&nbsp; 
  <span class="tooltiprtext">Click to View Output.</span>
</span>
</div>
</a>

<div id="wilcoxOneSample" style="display:none;">
<table class="rconsole">
<tr>
<td>
<span class="tooltiprout">
 Wilcoxon signed rank test with continuity correction
    <span class="tooltiprouttext">This reports on the type of test performed. The phrase "with continuity correction" implies the normal approximation was used when calculating the p-value of the test.</span>
</span>
</td>
</tr>
</table>
<br/>

<table class="rconsole">
<tr>
<td>
  <span class="tooltiprout">
    data: mtcars\$mpg
      <span class="tooltiprouttext">This print-out reminds us that the mpg column of the mtcars data was used as "Y" in the test.</span>
</td><td>
  <span class="tooltiprout">
   V = 249,
      <span class="tooltiprouttext">The test statistic of the test.</span>
</td><td>
  <span class="tooltiprout">
    &nbsp;p-value = 0.7863
      <span class="tooltiprouttext">The p-value of the test.</span>
</td>
</tr><tr>
<td>
  <span class="tooltiprout">
    alternative hypothesis: true location is not equal to 20
      <span class="tooltiprouttext">The words "not equal" tell us this was a two-sided test. Had it been a one-sided test, either the word "less" or the word "greater" would have appeared instead of "not equal."</span>
</td>
</tr>
</table>

</div>

</div>

----

#### Explanation

<div style="padding-left:125px;">

In many cases it is of interest to perform a hypothesis test about the location of the center of a distribution of data. The Wilcoxon Signed Rank Test allows a nonparametric approach to doing this. 

The Wilcoxon Signed-Rank Test covers two important scenarios.

1. **One sample** of data from a population. (Not very common.)
2. The differences obtained from **paired data**. (Very common.)

The Wilcoxon methods are most easily explained through examples, beginning with the paired data for which the method was originally created. Scroll down for the [One Sample Example](#one) if that is what you are really interested in. However, it is still recommended that you read the paired data example first.



##### Paired Data Example

<div style="padding-left:15px;">



<div style="color:#a8a8a8;">
  
Note: the data for this example comes from the original 1945 paper [Individual Comparison by Ranking Methods](http://sci2s.ugr.es/keel/pdf/algorithm/articulo/wilcoxon1945.pdf) by Frank Wilcoxon. 

</div>
  
###### Background
  
Height differences "between cross- and self- fertilized corn plants of the same pair" were collected. The experiment hypothesized that the center of the distribution of the height differences would be zero, with the alternative being that the center was not zero. The result of the data collection was 15 height differences: 
  
<div style="padding-left:15px;">
    
**Differences**: 14, 56, 60, 16, 6, 8, -48, 49, 24, 28, 29, 41, -67, 23, 75 
  
</div>
    
###### Step 1
    
The first step of the Wilcoxon Signed Rank Test is to order the differences from smallest *magnitude* to largest *magnitude*. Negative signs are essentially ignored at this point and only magnitudes of the numbers matter. 
  
<div style="padding-left:15px;">
    
**Sorted Differences**: 6, 8, 14, 16, 23, 24, 28, 29, 41, -48, 49, 56, 60, -67, 75 
</div>
    
###### Step 2
    
The next step is to rank the ordered values. Negative signs are attached to the ranks corresponding to negative numbers.
  
<div style="padding-left:15px;">
    
|&nbsp;           |  |  |  |    |   |   |   |   |   |    |   |   |   |    |   |
|-----------------|--|--|--|----|---|---|---|---|---|----|---|---|---|----|---|
|**Differences**: | 6| 8| 14| 16| 23| 24| 28| 29| 41| -48| 49| 56| 60| -67| 75 | 
|**Ranks**:       | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9| -10| 11 | 12 | 13 | -14 | 15 |
    
</div>
    
Note that the ranks will always be of the form $1, 2, \ldots, n$. In this case, $n=15$.
    
###### Step 3
    
The ranks are then put into two groups.
    
| Negative Ranks | Positive Ranks |
|----------------|----------------|
| -10, -14       | 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15 |
      
###### Step 4
      
One of the groups is summed, usually the group with the fewest observations. Only the absolute values of the ranks are summed.
    
<div style="padding-left:15px;">
      
**Sum of Negative Ranks**: $\left|-10\right| + \left|-14\right| = 24$
      
The sum of the ranks becomes the *test statistic* of the Wilcoxon Test. The test statistic is sometimes called $W$ or $V$ or $U$. 
    
</div>
      
###### Step 5
      
The $p$-value of the test is then obtained by computing the probability of the test statistic being as extreme or more extreme than the one obtained. This is done by first computing the probability of all possible values the test statistic could have obtained using mathematical counting techniques. This is a very tedious process that only a mathematician would enjoy pursuing. However, the end result is fairly easily understood. If you are interested, read the details.
    
<div style="padding-left:30px; padding-right:15px;">
      
      
<a href="javascript:showhide('uniquename')">**Details**</a>
      
<div id="uniquename" style="display:none;">
      
When there are $n=15$ ranks, the possible sums of ranks range from 0 to 120 and hit every integer in between, i.e., $1, 2, 3, \ldots, 120$. (Note, if summing the negative ranks these sums would technically all be negative.) 
    
To verify that $120$ is the largest sum possible for $n=15$ ranks, note that: 
      
* $1+15 = 16$, 
    
* $2 + 14 = 16$, 
    
* $3+13 = 16$, 
    
* $4+12=16$, 
    
* $5+11=16$, 
    
* $6+10=16$,  
    
* $7+9=16$,
    
* and finally that $8 = \frac{16}{2}$.
    
Thus, there are 7 sums of 16 and one sum of $\frac{16}{2}$. This could be said in a mathematically equivalent way by stating there are $\frac{14}{2}$ sums of 16 and one sum of $\frac{16}{2}$. By multiplication this gives 
$$
      \frac{14}{2}\cdot\frac{16}{1} + \frac{1}{1}\cdot\frac{16}{2} = \frac{14\cdot16 + 1\cdot16}{2} = \frac{15\cdot16}{2} = \frac{n(n+1)}{2} = 120
$$
      
The probability of each sum occurring is computed by counting all of the ways a certain sum can occur (combinations) and dividing by the total number of sums possible. (There are 32,768 total different groups of ranks possible when there are $n=15$ ranks.) 
    
For example, a sum of 1 can happen only one way, only the rank of 1 is in the group. A sum of 2 can also only happen 1 way. The sum of 3 however, can happen two ways: we could have the ranks of 1 and 2 in the group, or just the rank of 3 in the group. A similar counting technique is then implemented for each possible sum. After all the calculations are performed, the distribution of possible sums looks like what is shown in the following plot, where the red bars show those sums that are as extreme or more extreme than a sum of $24$ (or its opposite of $120-24=96$).
    
</div>
      

</div>
      
      
      
```{r, echo=FALSE}
tmp <- cbind(rep(c(0,1), each=32768/2),
                 rep(c(0,1), each=32768/4, times=2),
                 rep(c(0,1), each=32768/8, times=4),
                 rep(c(0,1), each=32768/16, times=8),
                 rep(c(0,1), each=32768/32, times=16),
                 rep(c(0,1), each=32768/64, times=32),
                 rep(c(0,1), each=32768/128, times=64),
                 rep(c(0,1), each=32768/256, times=128),
                 rep(c(0,1), each=32768/512, times=256),
                 rep(c(0,1), each=32768/1024, times=512),
                 rep(c(0,1), each=32768/2048, times=1024),
                 rep(c(0,1), each=32768/4096, times=2048),
                 rep(c(0,1), each=32768/8192, times=4096),
                 rep(c(0,1), each=32768/16384, times=8192),
                 rep(c(0,1), each=1, times=16384))
tmp2 <- tmp%*%1:15
tmp3 <- sapply(0:120, function(s) sum(tmp2 == s))
plot(0:120, tmp3/sum(tmp3), col='skyblue2', type='h', ylab="Probability", xlab="Sum of Ranks", main="Probability Distribution of Possible Sums")
points(c(0:24,96:120), tmp3[c(1:25,97:121)]/sum(tmp3), col='firebrick', type='h')
text(11,.01, "Observed Sum")
lines(c(24,24),c(0.009,0.003))
```
    
Computing the probabilities of all possible sums creates a distribution of the test statistic (shown in the plot above). Note that the test statistic is obtained in Step 4 (above) by taking the sum of the ranks. Once the distribution of the test statistic is established, the $p$-value of the test can be calculated as the combined probability of possible sums that are as extreme or more extreme than the one observed. 
    
For this example, it turns out that the probability of getting a sum of (the absolute value of) negative ranks as extreme or more extreme than $24$ is $p=0.04126$ (the sum of the probabilities of the red bars in the plot above). Thus, at the $\alpha=0.05$ level we would reject the null hypothesis that the center of the distribution of differences is zero. We conclude that the center of the distribution is greater than zero because the sum of negative ranks is much smaller than we expected under the zero center hypothesis (the null). Thus, there is sufficient evidence to conclude that the centers of the distributions of "cross- and self-fertilized corn plants" heights are not equal. One is greater than the other. Notice how the following dot plot shows that the differences are in favor of the cross-fertilized plants (the first group in the subtraction) being taller. This is true even though two self-fertilized plants were much taller than their cross-fertilized counterpart (the two negative differences).
    
```{r, echo=FALSE, fig.height = 4}
stripchart(c(14,56, 60, 16, 6, 8, -48, 49, 24, 28, 29, 41, -67, 23, 75), method="stack", pch=16, xlab="Differences", main="Cross- vs. Self-Fertilized Corn Plant Height Differences")
```
    
###### Comment 
    
If the distribution of differences is symmetric, then the hypotheses can be written as 
$$
  H_0: \mu = 0
$$
$$
  H_a: \mu \neq 0
$$
      
If the distribution is skewed, then the hypotheses technically refer to the median instead of the mean and should be written as
    
$$
      H_0: \text{median} = 0
$$
$$
      H_a: \text{median} \neq 0
$$
      

</div>



##### One Sample Example {#one}

<div style="padding-left:15px;">






The idea behind the one sample Wilcoxon Signed Rank test is nearly identical to the paired data. The only change is that the median must be subtracted from all observed values to obtain the *differences*. Note that the mean is equal to the median when data is symmetric.

###### Background

Suppose we are interested in testing to see if the median hourly wage of BYU-Idaho students during their off-track employment is equal to the minimum wage in Idaho, $7.25 an hour as of January 1st, 2015. Five randomly sampled hourly wages from BYU-Idaho Math 221B students provides the following data.

<div style="padding-left:15px;">

**Wages**: $6.00, $9.00, $8.10, $18.00, $10.45 

</div>

```{r, include=FALSE}
wages <- c(6.00, 9.00, 8.10, 18.00, 10.45)
```

The differences are then obtained by subtracting the hypothesized value for the median (or mean if the data is symmetric) from all observations.

<div style="padding-left:15px;">

```{r, include=FALSE}
diff <- wages - 7.25
```

**Differences**: -1.25, 1.75, 0.85, 10.75, 3.20  

<div style="color:#a8a8a8;">
Note: from this point down, the wording of this example is identical to the paired data example (above) with the numbers changed to match $n=5$. It is useful to continue reading to reinforce the idea of the Wilcoxon Signed Rank Test, but no new knowledge will be presented.
</div>

</div>



###### Step 1

The first step of the Wilcoxon Signed Rank Test is to order the differences from smallest *magnitude* to largest *magnitude*. Negative signs are essentially ignored at this point and only magnitudes of the numbers matter. 

<div style="padding-left:15px;">

**Sorted Differences**: 0.85, -1.25, 1.75, 3.20, 10.75

</div>

###### Step 2

The next step is to rank the ordered values. Negative signs are attached to the ranks corresponding to negative numbers.

<div style="padding-left:15px;">

**Ranks**: 1, -2, 3, 4, 5

</div>

Note that the ranks will always be of the form $1, 2, \ldots, n$. In this case, $n=5$.

###### Step 3

The ranks are then put into two groups.

| Negative Ranks | Positive Ranks |
|----------------|----------------|
| -2      | 1, 3, 4, 5 |

###### Step 4

One of the groups is summed, usually the group with the fewest observations.

<div style="padding-left:15px;">

**Sum of Negative Ranks**: $\left|-2\right| = 2$

</div>

###### Step 5

The $p$-value of the test is then obtained by computing the probabilities of all possible sums using mathematical counting techniques. This is a very tedious process that only a mathematician would enjoy pursuing. However, the end result is fairly easily understood. If you are interested, read the details.

<div style="padding-left:30px; padding-right:15px;">

**Details**

When there are $n=5$ ranks, the possible sums of ranks range from 0 to 15 and hit every integer in between, i.e., $1, 2, 3, \ldots, 15$. (Note, if summing the negative ranks these sums would technically all be negative.) 

To verify that $15$ is the largest sum possible for $n=5$ ranks, note that: 

* $1+5 = 6$, 

* $2 + 4 = 6$, 

* and finally $3 = \frac{6}{2}$.

Thus, there are 2 sums of 6 and one sum of $\frac{6}{2}$. This could be said in a mathematically equivalent way by stating there are $\frac{4}{2}$ sums of 6 and one sum of $\frac{6}{2}$. By multiplication this gives 
$$
  \frac{4}{2}\cdot\frac{6}{1} + \frac{1}{1}\cdot\frac{6}{2} = \frac{4\cdot6 + 1\cdot6}{2} = \frac{5\cdot6}{2} = \frac{n(n+1)}{2} = 15
$$

The probability of each sum occurring is computed by counting all of the ways a certain sum can occur (combinations) and dividing by the total number of sums possible. (There are 32 total different groups of ranks possible when there are $n=5$ ranks.) 

For example, a sum of 1 can happen only one way, only the rank of 1 is in the group. A sum of 2 can also only happen 1 way. The sum of 3 however, can happen two ways: we could have the ranks of 1 and 2 in the group, or just the rank of 3 in the group. A similar counting technique is then implemented for each possible sum. After all the calculations are performed, the distribution of possible sums looks like what is shown in the following plot, where the red bars show those sums that are as extreme or more extreme than a sum of $2$ (or its opposite of $15-2=13$).

```{r, echo=FALSE}
tmp <- cbind(rep(c(0,1), each=32/2),
             rep(c(0,1), each=32/4, times=2),
             rep(c(0,1), each=32/8, times=4),
             rep(c(0,1), each=32/16, times=8),
             rep(c(0,1), each=32/32, times=16))
tmp2 <- tmp%*%1:5
tmp3 <- sapply(0:15, function(s) sum(tmp2 == s))
plot(0:15, tmp3/sum(tmp3), col='skyblue2', type='h', ylab="Probability", xlab="Sum of Ranks", main="Probability Distribution of Possible Sums", ylim=c(0,.1), xlim=c(-1,16))
points(c(0:2,13:15), tmp3[c(1:3,14:16)]/sum(tmp3), col='firebrick', type='h')
text(1,.08, "Observed Sum = 2")
lines(c(2,2),c(0.035,0.075))
arrows(2,0.075,2,0.035, angle=30, length=.05)
```

Computing the probabilities of all possible sums creates a distribution of the test statistic (shown in the plot above). Note that the test statistic is obtained in Step 4 (above) by taking the sum of the ranks. Once the distribution of the test statistic is established, the $p$-value of the test can be calculated as the combined probability of possible sums that are as extreme or more extreme than the one observed. 

</div>

For this example, it turns out that the probability of getting a sum of negative ranks as extreme or more extreme than $-2$ is $p=0.1875$ (the sum of the probabilities of the red bars in the plot above). Thus, at the $\alpha=0.05$ level we would fail to reject the null hypothesis that the center of the distribution of differences is zero. We will continue to assume the null hypothesis was true, that the median off-track hourly wage of BYU-Idaho students is the same as the Idaho minimum wage.


###### Final Comment

Notice that when the sample size is large the distribution of the test statistic can be approximated by a normal distribution. Most software applications use this approximation when the sample size is over $50$, i.e., for $n\geq50$ because computing all the possible sums of ranks becomes incredibly time consuming. Thus, for large sample sizes the results will be almost identical whether the Wilcoxon Tests or a t Test is used.







</div>

</div>

----








### Wilcoxon Rank Sum (Mann-Whitney) Test {.tabset .tabset-fade .tabset-pills} 

<div style="float:left;width:125px;" align=center>
<img src="./Images/QuantYQualXg2.png" width=58px;>
</div>

For testing the equality of the medians of two (possibly different) distributions of a quantitative variable. 


#### Overview 

<div style="padding-left:125px;">

The nonparametric equivalent of the Independent Samples t Test. Can also be used when data is ordered (ordinal) but does not have an exact measurement. For example, first place, second place, and so on. 

The Independent Samples t Test is more appropriate when the distributions are normal, or when the sample size for each sample is large. 

The test is negatively affected when there are ties (repeated values) present in the data, but the results are still useful if there are relatively few ties. 


**Hypotheses**

Originally designed to test for the equality of medians from two identically shaped distributions. 

<div style="padding-left:15px;">


<div style="float:right;font-size:.8em;background-color:lightgray;padding:5px;border-radius:4px;"><a style="color:darkgray;" href="javascript:showhide('wilcoxonranksumlatex')">Math Code</a></div>


<div id="wilcoxonranksumlatex" style="display:none;">
```{}
$$
  H_0: \text{difference in medians} = 0
$$

$$
  H_a: \text{difference in medians} \neq 0
$$
```
</div>

$H_0: \text{difference in medians} = 0$

$H_a: \text{difference in medians} \neq 0$

</div>

However, the test also allows for the more general hypotheses that one distribution is *stochastically greater* than the other. 

<div style="padding-left:15px;">

$H_0: \text{the distributions are stochastically equal}$

$H_a: \text{one distribution is stochastically greater than the other}$

</div>

If these hypotheses are used, then the distributions do not have to be identically distributed.

(Note: Men's heights are *stochastically greater* than women's heights because men are generally taller than women, but not all men are taller than all women.) 


**Examples**: [BugSpray](./Analyses/Wilcoxon Tests/Examples/BugSprayWilcoxonRankSum.html), [MoralIntegration](./Analyses/Wilcoxon Tests/Examples/MoralIntegration.html)

</div>

----



#### R Intstructions

<div style="padding-left:125px;">
**Console** Help Command: `?wilcox.test()`

There are two ways to perform the test.

**Option 1:**

`wilcox.test(Y ~ X, data = YourData, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)`

* `Y` must be a "numeric" vector from `YourData` that represents the data for both samples.
* `X` must be a "factor" or "character" vector from `YourData` that represents the group assignment for each observation. There can only be two groups specified in this column of data.
* `YourNull` is the numeric value from your null hypothesis for the difference in medians from the two groups.
* `YourAlternative` is one of the three options: `"two.sided"`, `"greater"`, `"less"` and should correspond to your alternative hypothesis.
* The value for `conf.level = 0.95` can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $1-\alpha$.



**Example Code** 

Hover your mouse over the example codes to learn more.

<a href="javascript:showhide('wilcoxonRankSum')">
<div class="hoverchunk">
<span class="tooltipr">
wilcox.test(
  <span class="tooltiprtext">'wilcox.test' is a function for non-parametric one and two sample tests.</span>
</span><span class="tooltipr">
length&nbsp;
  <span class="tooltiprtext">'length' is a quantitative variable (numeric vector).</span>
</span><span class="tooltipr">
~&nbsp;
  <span class="tooltiprtext">'~' is the tilde symbol.</span>
</span><span class="tooltipr">
sex,&nbsp;
  <span class="tooltiprtext">'sex' is a 'factor' or 'character' vector that represents the group assignment for each observation. There are two groups.</span>
</span><span class="tooltipr">
data=KidsFeet,&nbsp; 
  <span class="tooltiprtext">'KidsFeet' is a dataset in library(mosaic). Type View(KidsFeet) to view it.</span>
</span><span class="tooltipr">
mu = 0,&nbsp; 
  <span class="tooltiprtext">The numeric value from the null hypothesis for the difference in medians from the two groups is 0 meaning the null hypothesis is $\text{difference in medians} = 0$</span>
</span><span class="tooltipr">
alternative = "two.sided",&nbsp; 
  <span class="tooltiprtext"> The alternative is "two-sided" meaning the alternative hypothesis is $\text{difference in medians} \neq 0$.</span>
  </span><span class="tooltipr">
conf.level = 0.95)
  <span class="tooltiprtext">This test has a 0.95 confidence level which corresponds to $1-\alpha$</span>
</span><span class="tooltipr">
&nbsp;&nbsp;&nbsp;&nbsp;  
  <span class="tooltiprtext">Press Enter to run the code if you have typed it in yourself. You can also click here to view the output.</span>
</span><span class="tooltipr" style="float:right;">
&nbsp;...&nbsp; 
  <span class="tooltiprtext">Click to View Output.</span>
</span>
</div>
</a>

<div id="wilcoxonRankSum" style="display:none;">
<table class="rconsole">
<tr>
<td>
<span class="tooltiprout">
  Wilcoxon rank sum test with continuity correction
    <span class="tooltiprouttext">This states the test that was performed and that a normal approximation to the test statistic was used instead of the exact distribution.</span>
</span>
</td>
</tr>
</table>
<br/>

<table class="rconsole">
<tr>
<td>
  <span class="tooltiprout">
    data: length by sex
      <span class="tooltiprouttext">This states that the length column was split into the two groups found in the "sex" column. Unfortunately, it forgets to remind us that the test used the KidsFeet data set.</span>
</td><td>
  <span class="tooltiprout">
   V = 252,
      <span class="tooltiprouttext">The test statistic of the test. In this case, the sum of the ranks from the alphabetically first group minus the minimum sum of ranks possible.</span>
</td><td>
  <span class="tooltiprout">
    &nbsp;p-value = 0.0836
      <span class="tooltiprouttext">The p-value of the test.</span>
</td>
</tr><tr>
<td>
  <span class="tooltiprout">
    alternative hypothesis: true location shift is not equal to 0
      <span class="tooltiprouttext">This reports that the test used an alternative hypothesis of "not equal" (a two-sided test). Further, the phrase "true location shift" emphasizes that the Wilcoxon Rank Sum Test is testing to see if one distribution is shifted higher or lower than the other.</span>
</td>
</tr>
</table>

</div>

<br>

**Option 2:**

`wilcox.test(object1, object2, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)`

* `object1` must be a "numeric" vector that represents the first sample of data.
* `obejct2` must be a "numeric" vector that represents the second sample of data.
* `YourNull` is the numeric value from your null hypothesis for the difference in medians from the two groups.
* `YourAlternative` is one of the three options: `"two.sided"`, `"greater"`, `"less"` and should correspond to your alternative hypothesis.
* The value for `conf.level = 0.95` can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $1-\alpha$.




**Example Code**

<a href="javascript:showhide('wilcoxonRankSum2')">
<div class="hoverchunk">
<span class="tooltipr">
wilcox.test(
  <span class="tooltiprtext">'wilcox.test' is a function for non-parametric one and two sample tests.</span>
</span><span class="tooltipr">
KidsFeet\$length[KidsFeet\$sex&nbsp;==&nbsp;"B"],&nbsp;
  <span class="tooltiprtext">A numeric vector of foot length for the first sample of data or for the group of boys.</span>
</span><span class="tooltipr">
KidsFeet\$length[KidsFeet$sex&nbsp;==&nbsp;"G"],&nbsp;
  <span class="tooltiprtext">A numeric vector of foot length for the second sample of data or for the group of girls.</span>
</span><span class="tooltipr">
mu = 0,&nbsp; 
  <span class="tooltiprtext">The numeric value from the null hypothesis for the difference in medians from the two groups is 0 meaning the null hypothesis is $\text{difference in medians} = 0$</span>
</span><span class="tooltipr">
alternative = "two.sided",&nbsp; 
  <span class="tooltiprtext"> The alternative is "two-sided" meaning the alternative hypothesis is $\text{difference in medians} \neq 0$.</span>
  </span><span class="tooltipr">
conf.level = 0.95)
  <span class="tooltiprtext">This test has a 0.95 confidence level which corresponds to $1-\alpha$</span>
</span><span class="tooltipr">
&nbsp;&nbsp;&nbsp;&nbsp;  
  <span class="tooltiprtext">Press Enter to run the code if you have typed it in yourself. You can also click here to view the output.</span>
</span><span class="tooltipr" style="float:right;">
&nbsp;...&nbsp; 
  <span class="tooltiprtext">Click to View Output.</span>
</span>
</div>
</a>

<div id="wilcoxonRankSum2" style="display:none;">
<table class="rconsole">
<tr>
<td>
<span class="tooltiprout">
  Wilcoxon rank sum test with continuity correction
    <span class="tooltiprouttext">This states the type of test performed.</span>
</span>
</td>
</tr>
</table>
<br/>

<table class="rconsole">
<tr>
<td>
  <span class="tooltiprout">
    data: KidsFeet\$length[KidsFeet\$sex == "B"] and KidsFeet\$length[KidsFeet\$sex == "G"]
      <span class="tooltiprouttext">Reminds you what data was used for the test and points out that the first set of data listed is "Group 1" and the second set of data listed is "Group 2."</span>
</td><td>
  <span class="tooltiprout">
   V = 252,
      <span class="tooltiprouttext">The test statistic of the test.</span>
</td><td>
  <span class="tooltiprout">
    &nbsp;p-value = 0.0836
      <span class="tooltiprouttext">The p-value of the test.</span>
</td>
</tr><tr>
<td>
  <span class="tooltiprout">
    alternative hypothesis: true location shift is not equal to 0
      <span class="tooltiprouttext">The alternative hypothesis of the test. The phrase "not equal" implies a "two-sided" alternative was used.</span>
</td>
</tr>
</table>

</div>




</div>

----


#### Explanation

<div style="padding-left:125px;">


In many cases it is of interest to perform a hypothesis test concerning the equality of the centers of two (possibly different) distributions. In other words, an independent samples test. The Wilcoxon Rank Sum Test allows a nonparametric approach to doing this. It is often considered the nonparametric equivalent of the independent samples t test.

The method is most easily explained through an example. The theory behind it is very similar to the theory behind the Wilcoxon Signed-Rank Test.

<br />

##### Independent Samples Data Example

<div style="padding-left:15px;">

<div style="color:#a8a8a8;">

Note: the data for this example comes from the original 1945 paper [Individual Comparison by Ranking Methods](http://sci2s.ugr.es/keel/pdf/algorithm/articulo/wilcoxon1945.pdf) by Frank Wilcoxon. 

</div>

###### Background

The percent of flies (bugs) killed from two different concentrations of a certain spray were recorded from 16 different trials, 8 trials per treatment concentration. The experiment hypothesized that the center of the distributions of the percent killed by either concentration were the same. In other words, that both treatments were equally effective. The alternative hypothesis was that the treatments differed in their effectiveness.

<div style="padding-left:15px;">

| Spray Concentration | Percent Killed |
|---------------------|----------------|
| A | 68, 68, 59, 72, 64, 67, 70, 74 |
| B | 60, 67, 61, 62, 67, 63, 56, 58 |


```{r, echo=FALSE, render='asis'}
kill <- cbind(PercentKillA = c(68,68,59,72,64,67,70,74),
              PercentKillB = c(60,67,61,62,67,63,56,58))
```


</div>

###### Step 1

The first step of the Wilcoxon Rank Sum Test is to order all the data from smallest *magnitude* to largest *magnitude*, while keeping track of the group. 

<div style="padding-left:15px;">

| Sorted Data | | | | | | | | | | | | | | | | | 
|-------------|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|
| Percent Killed | 56| 58| 59| 60| 61| 62| 63| 64| 67| 67| 67| 68| 68| 70| 72| 74|
| Concentration | B| B| A| B| B| B| B| A| A| B| B| A| A| A| A| A| 


</div>

###### Step 2

The next step is to rank the ordered values. 

<div style="padding-left:15px;">

| Sorted Data | | | | | | | | | | | | | | | | | 
|-------------|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|
| Percent Killed | 56| 58| 59| 60| 61| 62| 63| 64| 67| 67| 67| 68| 68| 70| 72| 74|
| Concentration | B| B| A| B| B| B| B| A| A| B| B| A| A| A| A| A| 
| Rank          | 1| 2| 3| 4| 5| 6| 7| 8| 9|10|11|12|13|14|15|16|

</div>

Note that the ranks will always be of the form $1, 2, \ldots, n$. In this case, $n=16$.

Any ranks that are tied need to have the average rank assigned to each of those that are tied.

| Sorted Data | | | | | | | | | | | | | | | | | 
|-------------|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|
| Percent Killed | 56| 58| 59| 60| 61| 62| 63| 64| <span style="color:#B22222;">67</span>| <span style="color:#B22222;">67</span>| <span style="color:#B22222;">67</span>| <span style="color:#7EC0EE;">68</span>| <span style="color:#7EC0EE;">68</span>| 70| 72| 74|
| Concentration | B| B| A| B| B| B| B| A| A| B| B| A| A| A| A| A| 
| Rank          | 1| 2| 3| 4| 5| 6| 7| 8| <span style="color:#B22222;">10</span>|<span style="color:#B22222;">10</span>|<span style="color:#B22222;">10</span>|<span style="color:#7EC0EE;">12.5</span>|<span style="color:#7EC0EE;">12.5</span>|14|15|16|

###### Step 3

The ranks are then returned to their original groups.

| Ranks of Spray A | Ranks of Spray B |
|----------------|----------------|
| 3, 8, 10, 12.5, 12.5, 14, 15, 16 | 1, 2, 4, 5, 6, 7, 10, 10 |

###### Step 4

The ranks are summed for one of the groups. (It does not matter which group.)

<div style="padding-left:15px;">

**Sum of Ranks for Spray A**: 
$$
 3+8+10+12.5+12.5+14+15+16 = 91
$$

</div>

###### Step 5

The $p$-value of the test is then obtained by computing the probabilities of all possible sums that one group can achieve using mathematical counting techniques. This is a very tedious process that only a mathematician would enjoy pursuing. However, the end result is fairly easily understood. If you are interested, read the details.

<div style="padding-left:30px; padding-right:15px;">

<hr />

**Details**

When there are $n=16$ ranks, with just $8$ of the ranks assigned to one group, the possible sums of ranks range from $36$ to $100$ and include every integer in between, i.e., $36, 37, 38, \ldots, 100$.

Note that the smallest sum would be obtained if the ranks 1-8 were in the group. The largest sum would be obtained if the ranks 9-16 were in the group. 

The probability of each possible sum occurring is computed by counting all of the ways a certain sum can occur (combinations) and dividing by the total number of sums possible. (There are 12,870 total different sets of ranks possible when there are $n=16$ ranks and $8$ are assigned to one group.) 

The distribution of possible sums looks like what is shown in the following plot, where the red bars show those sums that are as extreme or more extreme than a sum of $91$ (or its opposite of $136-91=45$).

```{r, echo=FALSE}
tmp <- cbind(rep(c(0,1), each=65536/2),
             rep(c(0,1), each=65536/4, times=2),
             rep(c(0,1), each=65536/8, times=4),
             rep(c(0,1), each=65536/16, times=8),
             rep(c(0,1), each=65536/32, times=16),
             rep(c(0,1), each=65536/64, times=32),
             rep(c(0,1), each=65536/128, times=64),
             rep(c(0,1), each=65536/256, times=128),
             rep(c(0,1), each=65536/512, times=256),
             rep(c(0,1), each=65536/1024, times=512),
             rep(c(0,1), each=65536/2048, times=1024),
             rep(c(0,1), each=65536/4096, times=2048),
             rep(c(0,1), each=65536/8192, times=4096),
             rep(c(0,1), each=65536/16384, times=8192),
             rep(c(0,1), each=65536/32768, times=16384),
             rep(c(0,1), each=1, times=32768))
tmp1 <- apply(tmp,1,sum) == 8
tmp1 <- tmp[tmp1,]
tmp2 <- tmp1%*%1:16
tmp3 <- sapply(36:100, function(s) sum(tmp2 == s))
plot(36:100, tmp3/sum(tmp3), col='skyblue2', type='h', ylab="Probability", xlab="Sum of Ranks", main="Probability Distribution of Possible Sums")
points(c(36:45,91:100), tmp3[c(1:10,56:65)]/sum(tmp3), col='firebrick', type='h')
text(93,.017, "Observed Sum")
arrows(91,0.016,91,0.003, length=0.1, angle=20)
```

Computing the probabilities of all possible sums creates a distribution of the test statistic (shown in the plot above). Note that the test statistic is obtained in Step 4 (above) by taking the sum of the ranks. Once the distribution of the test statistic is established, the $p$-value of the test can be calculated as the combined probability of possible sums that are as extreme or more extreme than the one observed. 

<hr />

</div>

For this example, it turns out that the probability of getting a sum of negative ranks as extreme or more extreme than $91$ is $p=0.01476$ (the sum of the probabilities of the red bars in the plot above). Thus, at the $\alpha=0.05$ level we would reject the null hypothesis that the difference in the center of the distributions is zero. We conclude that the Spray Concentration A is more effective at killing flies (bugs).

###### Comment 

The hypotheses for the Wilcoxon Rank Sum Test are difficult to write out in simple mathematical statements. The test is often referred to as a test of medians, but goes deeper than this. Technically, it allows us to determine if one distribution is stochastically larger than another. In other words, if one distribution typically gives larger values than does another distribution. If the distributions are identically shaped and have the same spread, then this implies the medians (and means) are different. 

Thus, the hypotheses for the test can be written mathematically as
$$
  H_0: \text{difference in medians} = 0
$$
$$
  H_a: \text{difference in medians} \neq 0
$$
or even as
$$
  H_0: \mu_1-\mu_2 = 0
$$
$$
  H_a: \mu_1-\mu_2 \neq 0
$$

However, it is important to remember that the estimated difference in location parameters that results from the test does not provide a measurement on either of these things. However, the $p$-value can lead us to determine whether to reject, or fail to reject the null, whichever of the above hypotheses is used.

As stated in the R help file for this test `?wilcox.test()` "the estimator for the difference in location parameters does not estimate the difference in medians (a common misconception) but rather the median of the difference between a sample from [the first population] and a sample from [the second population]." These are technical details that most people ignore without encountering too much difficulty. However, it does remind us that the t test is more easily interpreted whenever it is appropriate to use that test.

</div>

<br />

##### Final Comment

Notice that when the sample size is large the distribution of the test statistic can be approximated by a normal distribution. Most software applications use this approximation when the sample size is over $50$, i.e., for $n\geq50$ because computing all the possible sums of ranks becomes incredibly time consuming. Thus, for large sample sizes the results will be almost identical whether the Wilcoxon Tests or a t Test is used.

</div>

----


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