# One sample Wilcoxon signed-rank test - overview

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One sample Wilcoxon signed-rank test
McNemar's test
One way ANOVA
Independent variableIndependent variableIndependent/grouping variable
None2 paired groupsOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio level
Null 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.

Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:

1. First score of pair is 0, second score of pair is 0
2. First score of pair is 0, second score of pair is 1 (switched)
3. First score of pair is 1, second score of pair is 0 (switched)
4. First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the null hypothesis are:

• H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
• H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)

ANOVA $F$ test:
• H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
$\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
• H0: $\Psi = 0$
$\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
• H0: $\mu_g = \mu_h$
$\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $m \neq m_0$
H1 right sided: $m > m_0$
H1 left sided: $m < m_0$

The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the alternative hypothesis are:

• H1: $\pi_1 \neq \pi_2$
• H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)

ANOVA $F$ test:
• H1: not all population means are equal
$t$ Test for contrast:
• H1 two sided: $\Psi \neq 0$
• H1 right sided: $\Psi > 0$
• H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
• H1 - usually two sided: $\mu_g \neq \mu_h$
AssumptionsAssumptionsAssumptions
• 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 of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
Test 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.
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
ANOVA $F$ test:
• \begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned}
where $N$ is the total sample size, and $I$ is the number of groups.
Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
• $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
• $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
$\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
n.a.n.a.Pooled standard deviation
--\begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}

Here $s^2_i$ is the variance in group $i.$
Sampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $F$ and of $t$ 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.

If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.

If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.

Sampling distribution of $F$:
• $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - I$ degrees of freedom
Significant?Significant?Significant?
For large samples, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
For test statistic $X^2$:
• 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 $\alpha$
If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
• Check if $b$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)

$t$ Test for contrast two sided:
$t$ Test for contrast right sided:
$t$ Test for contrast left sided:

$t$ Test multiple comparisons two sided:
• Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
• Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
• Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
n.a.n.a.$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for $\mu_i$
--Confidence interval for $\Psi$ (contrast):
• $c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}$
where the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for $\mu_g - \mu_h$ (multiple comparisons):
• $(\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}$
where $t^{**}$ depends upon $C$, degrees of freedom ($N - I$), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, $t^{**} = t^* =$ the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$. Note that $n_g$ is the sample size of group $g$, $n_h$ is the sample size of group $h$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for single population mean $\mu_i$:
• $\bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}}$
where $\bar{y}_i$ is the sample mean in group $i$, $n_i$ is the sample size of group $i$, and the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
n.a.n.a.Effect size
--
• Proportion variance explained $\eta^2$ and $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variable: \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} Only in one way ANOVA $\eta^2 = R^2.$ $\eta^2$ (and $R^2$) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.

• Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to: $$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$ $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2.$

• Cohen's $d$:
Standardized difference between the mean in group $g$ and in group $h$: $$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ two sample means are removed from each other.
n.a.n.a.ANOVA table
--
Click the link for a step by step explanation of how to compute the sum of squares.
n.a.Equivalent toEquivalent to
-
OLS regression with one categorical independent variable transformed into $I - 1$ code variables:
• $F$ test ANOVA is equivalent to $F$ test regression model
• $t$ test for contrast $i$ is equivalent to $t$ test for regression coefficient $\beta_i$ (specific contrast tested depends on how the code variables are defined)
Example contextExample contextExample context
Is the median mental health score of office workers different from $m_0 = 50$?Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?Is the average mental health score different between people from a low, moderate, and high economic class?
SPSSSPSSSPSS
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 Related Samples...
• Put the two paired variables in the boxes below Variable 1 and Variable 2
• Under Test Type, select the McNemar test
Analyze > Compare Means > One-Way ANOVA...
• Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor
or
Analyze > General Linear Model > Univariate...
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s)
JamoviJamoviJamovi
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 > Paired Samples - McNemar test
• Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
Practice questionsPractice questionsPractice questions