# Wilcoxon signed-rank test - overview

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Wilcoxon signed-rank test | One sample Wilcoxon signed-rank test | Goodness of fit test |
You cannot compare more than 3 methods |
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Independent variable | Independent variable | Independent variable | |

2 paired groups | None | None | |

Dependent variable | Dependent variable | Dependent variable | |

One quantitative of interval or ratio level | One of ordinal level | One categorical with $J$ independent groups ($J \geqslant 2$) | |

Null hypothesis | Null hypothesis | Null hypothesis | |

H_{0}: $m = 0$
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. | H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis. | - H
_{0}: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
- H
_{0}: 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 $\pi_J$
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Alternative hypothesis | Alternative hypothesis | Alternative hypothesis | |

H_{1} two sided: $m \neq 0$H _{1} right sided: $m > 0$H _{1} left sided: $m < 0$
| H_{1} two sided: $m \neq m_0$H _{1} right sided: $m > m_0$H _{1} left sided: $m < m_0$
| - H
_{1}: the population proportions are not all as specified under the null hypothesis
- H
_{1}: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
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Assumptions | Assumptions | Assumptions | |

- 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
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.
| - 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
| - 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
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Test statistic | Test statistic | Test 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:
- 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.
- For each subject, compute the absolute value of the difference score $|\mbox{score}_2 - \mbox{score}_1|$.
- Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
- 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.
- $W_1 = \sum\, R_d^{+}$
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, 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:
- 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.
- For each subject, compute the absolute value of the difference score $|\mbox{score} - m_0|$.
- Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
- 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.
- $W_1 = \sum\, R_d^{+}$
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.
| $X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
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. | |

Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true | Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true | Sampling distribution of $X^2$ if H_{0} 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 $W_1$:
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. | Approximately the chi-squared distribution with $J - 1$ degrees of freedom | |

Significant? | Significant? | Significant? | |

For large samples, the table for standard normal probabilities can be used: Two sided: - Check if $z$ observed in sample is at least as extreme as critical value $z^*$ or
- Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
- Check if $z$ observed in sample is equal to or larger than critical value $z^*$ or
- Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
- Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or
- Find left sided $p$ value corresponding to observed $z$ 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: - Check if $z$ observed in sample is at least as extreme as critical value $z^*$ or
- Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
- Check if $z$ observed in sample is equal to or larger than critical value $z^*$ or
- Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
- Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or
- Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
| - 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 $\alpha$
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Example context | Example context | Example context | |

Is the median of the differences between the mental health scores before and after an intervention different from 0? | Is the median mental health score of office workers different from $m_0 = 50$? | 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 $\pi_{high} = 0.2$? | |

SPSS | SPSS | SPSS | |

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
| 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 > 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)
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Jamovi | Jamovi | Jamovi | |

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
| 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
| 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)
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Practice questions | Practice questions | Practice questions | |