One sample Wilcoxon signed-rank test - overview

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One sample Wilcoxon signed-rank test
Mann-Whitney-Wilcoxon test
Goodness of fit test
Wilcoxon signed-rank test
Independent variableIndependent/grouping variableIndependent variableIndependent variable
NoneOne categorical with 2 independent groupsNone2 paired groups
Dependent variableDependent variableDependent variableDependent variable
One of ordinal levelOne of ordinal levelOne categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
H0: $m = m_0$

Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in both populations:
  • H0: the population median for group 1 is equal to the population median for group 2
Else:
Formulation 1:
  • H0: the population scores in group 1 are not systematically higher or lower than the population scores in group 2
Formulation 2:
  • H0: P(an observation from population 1 exceeds an observation from population 2) = P(an observation from population 2 exceeds observation from population 1)
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
  • H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
  • H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$, the probability of drawing an observation from condition $J$ is $\pi_J$
H0: $m = 0$

Here $m$ is the population median of the difference scores. A difference score is the difference between the first score of a pair and the second score of a pair.

Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $m \neq m_0$
H1 right sided: $m > m_0$
H1 left sided: $m < m_0$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in both populations:
  • H1 two sided: the population median for group 1 is not equal to the population median for group 2
  • H1 right sided: the population median for group 1 is larger than the population median for group 2
  • H1 left sided: the population median for group 1 is smaller than the population median for group 2
Else:
Formulation 1:
  • H1 two sided: the population scores in group 1 are systematically higher or lower than the population scores in group 2
  • H1 right sided: the population scores in group 1 are systematically higher than the population scores in group 2
  • H1 left sided: the population scores in group 1 are systematically lower than the population scores in group 2
Formulation 2:
  • H1 two sided: P(an observation from population 1 exceeds an observation from population 2) $\neq$ P(an observation from population 2 exceeds an observation from population 1)
  • H1 right sided: P(an observation from population 1 exceeds an observation from population 2) > P(an observation from population 2 exceeds an observation from population 1)
  • H1 left sided: P(an observation from population 1 exceeds an observation from population 2) < P(an observation from population 2 exceeds an observation from population 1)
  • H1: the population proportions are not all as specified under the null hypothesis
or equivalently
  • H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
H1 two sided: $m \neq 0$
H1 right sided: $m > 0$
H1 left sided: $m < 0$
AssumptionsAssumptionsAssumptionsAssumptions
  • The population distribution of the scores is symmetric
  • Sample is a simple random sample from the population. That is, observations are independent of one another
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
  • Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
  • Sample is a simple random sample from the population. That is, observations are independent of one another
  • The population distribution of the difference scores is symmetric
  • Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Note: sometimes it considered sufficient for the data to be measured on an ordinal scale, rather than an interval or ratio scale. However, since the test statistic is based on ranked difference scores, we need to know whether a change in scores from, say, 6 to 7 is larger than/smaller than/equal to a change from 5 to 6. This is impossible to know for ordinal scales, since for these scales the size of the difference between values is meaningless.
Test statisticTest statisticTest statisticTest statistic
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
  1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
  2. For each subject, compute the absolute value of the difference score $|\mbox{score} - m_0|$.
  3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
  4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:

  • $W_1 = \sum\, R_d^{+}$
    or
    $W_1 = \sum\, R_d^{-}$
    That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
    • Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
    • If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
  • $W_2 = \sum\, \mbox{sign}_d \times R_d$
    That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
Two different types of test statistics can be used; both will result in the same test outcome. The first is the Wilcoxon rank sum statistic $W$: The second type of test statistic is the Mann-Whitney $U$ statistic:
  • $U = W - \dfrac{n_1(n_1 + 1)}{2}$
where $n_1$ is the sample size of group 1.

Note: we could just as well base W and U on group 2. This would only 'flip' the right and left sided alternative hypotheses. Also, tables with critical values for $U$ are often based on the smaller of $U$ for group 1 and for group 2.
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
  1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}_2 - \mbox{score}_1)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
  2. For each subject, compute the absolute value of the difference score $|\mbox{score}_2 - \mbox{score}_1|$.
  3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
  4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:

  • $W_1 = \sum\, R_d^{+}$
    or
    $W_1 = \sum\, R_d^{-}$
    That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
    • tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
    • If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
  • $W_2 = \sum\, \mbox{sign}_d \times R_d$
    That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
Sampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $W$ and of $U$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were true
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.

Sampling distribution of $W$:
For large samples, $W$ is approximately normally distributed with mean $\mu_W$ and standard deviation $\sigma_W$ if the null hypothesis were true. Here $$ \begin{aligned} \mu_W &= \dfrac{n_1(n_1 + n_2 + 1)}{2}\\ \sigma_W &= \sqrt{\dfrac{n_1 n_2(n_1 + n_2 + 1)}{12}} \end{aligned} $$ Hence, for large samples, the standardized test statistic $$ z_W = \dfrac{W - \mu_W}{\sigma_W}\\ $$ follows approximately the standard normal distribution if the null hypothesis were true. Note that if your $W$ value is based on group 2, $\mu_W$ becomes $\frac{n_2(n_1 + n_2 + 1)}{2}$.

Sampling distribution of $U$:
For large samples, $U$ is approximately normally distributed with mean $\mu_U$ and standard deviation $\sigma_U$ if the null hypothesis were true. Here $$ \begin{aligned} \mu_U &= \dfrac{n_1 n_2}{2}\\ \sigma_U &= \sqrt{\dfrac{n_1 n_2(n_1 + n_2 + 1)}{12}} \end{aligned} $$ Hence, for large samples, the standardized test statistic $$ z_U = \dfrac{U - \mu_U}{\sigma_U}\\ $$ follows approximately the standard normal distribution if the null hypothesis were true.

For small samples, the exact distribution of $W$ or $U$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_W$ and $\sigma_U$ is more complicated.
Approximately the chi-squared distribution with $J - 1$ degrees of freedomSampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
Significant?Significant?Significant?Significant?
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
n.a.Equivalent ton.a.n.a.
-If there are no ties in the data, the two sided Mann-Whitney-Wilcoxon test is equivalent to the Kruskal-Wallis test with an independent variable with 2 levels ($I$ = 2).--
Example contextExample contextExample contextExample context
Is the median mental health score of office workers different from $m_0 = 50$?Do men tend to score higher on social economic status than women? Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?Is the median of the differences between the mental health scores before and after an intervention different from 0?
SPSSSPSSSPSSSPSS
Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:

Analyze > Nonparametric Tests > One Sample...
  • On the Objective tab, choose Customize Analysis
  • On the Fields tab, specify the variable for which you want to compute the Wilcoxon signed-rank test
  • On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank test)'. Fill in your $m_0$ in the box next to Hypothesized median
  • Click Run
  • Double click on the output table to see the full results
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Independent Samples...
  • Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
  • Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow
  • Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2
  • Continue and click OK
Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square...
  • Put your categorical variable in the box below Test Variable List
  • Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Wilcoxon test
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T-Tests > One Sample T-Test
  • Put your variable in the box below Dependent Variables
  • Under Tests, select Wilcoxon rank
  • Under Hypothesis, fill in the value for $m_0$ in the box next to Test Value, and select your alternative hypothesis
T-Tests > Independent Samples T-Test
  • Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
  • Under Tests, select Mann-Whitney U
  • Under Hypothesis, select your alternative hypothesis
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
  • Put your categorical variable in the box below Variable
  • Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
T-Tests > Paired Samples T-Test
  • Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
  • Under Tests, select Wilcoxon rank
  • Under Hypothesis, select your alternative hypothesis
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