One sample Wilcoxon signed-rank test - overview

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One sample Wilcoxon signed-rank test
$z$ test for the difference between two proportions
Paired sample $t$ test
Chi-squared test for the relationship between two categorical variables
Two way ANOVA
$z$ test for a single proportion
Independent variableIndependent/grouping variableIndependent variableIndependent /column variableIndependent/grouping variablesIndependent variable
NoneOne categorical with 2 independent groups2 paired groupsOne categorical with $I$ independent groups ($I \geqslant 2$)Two categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)None
Dependent variableDependent variableDependent variableDependent /row variableDependent variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull 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.
H0: $\pi_1 = \pi_2$

Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.
H0: $\mu = \mu_0$

Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.
H0: there is no association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H0: the distribution of the dependent variable is the same in each of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H0: the row and column variables are independent
ANOVA $F$ tests:
  • H0 for main and interaction effects together (model): no main effects and interaction effect
  • H0 for independent variable A: no main effect for A
  • H0 for independent variable B: no main effect for B
  • H0 for the interaction term: no interaction effect between A and B
Like in one way ANOVA, we can also perform $t$ tests for specific contrasts and multiple comparisons. This is more advanced stuff.
H0: $\pi = \pi_0$

Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $m \neq m_0$
H1 right sided: $m > m_0$
H1 left sided: $m < m_0$
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1: there is an association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H1: the distribution of the dependent variable is not the same in all of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H1: the row and column variables are dependent
ANOVA $F$ tests:
  • H1 for main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect
  • H1 for independent variable A: there is a main effect for A
  • H1 for independent variable B: there is a main effect for B
  • H1 for the interaction term: there is an interaction effect between A and B
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptions
  • 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 $z$ to be approximately normally distributed. Rule of thumb:
    • Significance test: number of successes and number of failures are each 5 or more in both sample groups
    • Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
    • Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
  • 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
  • Difference scores are normally distributed in the population
  • Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
  • Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb:
    • 2 $\times$ 2 table: all four expected cell counts are 5 or more
    • Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more
  • There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
  • Within each of the $I \times J$ populations, 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 $I \times J$ populations
  • For each of the $I \times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another
  • Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)
  • Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
    • Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
    • Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
    • Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
  • Sample is a simple random sample from the population. That is, observations are independent of one another
If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.
Test statisticTest statisticTest 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.
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2.
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).

The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.
For main and interaction effects together (model):
  • $F = \dfrac{\mbox{mean square model}}{\mbox{mean square error}}$
For independent variable A:
  • $F = \dfrac{\mbox{mean square A}}{\mbox{mean square error}}$
For independent variable B:
  • $F = \dfrac{\mbox{mean square B}}{\mbox{mean square error}}$
For the interaction term:
  • $F = \dfrac{\mbox{mean square interaction}}{\mbox{mean square error}}$
Note: mean square error is also known as mean square residual or mean square within.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.
n.a.n.a.n.a.n.a.Pooled standard deviationn.a.
----$ \begin{aligned} s_p &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - (I \times J)}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $ -
Sampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $F$ if H0 were trueSampling distribution of $z$ 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.
Approximately the standard normal distribution$t$ distribution with $N - 1$ degrees of freedomApproximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedomFor main and interaction effects together (model):
  • $F$ distribution with $(I - 1) + (J - 1) + (I - 1) \times (J - 1)$ (df model, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable A:
  • $F$ distribution with $I - 1$ (df A, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable B:
  • $F$ distribution with $J - 1$ (df B, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For the interaction term:
  • $F$ distribution with $(I - 1) \times (J - 1)$ (df interaction, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
Here $N$ is the total sample size.
Approximately the standard normal distribution
Significant?Significant?Significant?Significant?Significant?Significant?
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
Two sided: Right sided: Left sided: 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 $\alpha$
  • 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 $\alpha$
Two sided: Right sided: Left sided:
n.a.Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$$C\%$ confidence interval for $\mu$n.a.n.a.Approximate $C\%$ confidence interval for $\pi$
-Regular (large sample):
  • $(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$
    where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
With plus four method:
  • $(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$
    where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
$\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20).

The confidence interval for $\mu$ can also be used as significance test.
--Regular (large sample):
  • $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
    where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
With plus four method:
  • $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
    where $p_{plus} = \dfrac{X + 2}{N + 4}$ and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
n.a.n.a.Effect sizen.a.Effect sizen.a.
--Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0.$
-
  • Proportion variance explained $R^2$:
    Proportion variance of the dependent variable $y$ explained by the independent variables and the interaction effect together:
    $$ \begin{align} R^2 &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}} \end{align} $$ $R^2$ 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 $\eta^2$:
    Proportion variance of the dependent variable $y$ explained by an independent variable or interaction effect:
    $$ \begin{align} \eta^2_A &= \dfrac{\mbox{sum of squares A}}{\mbox{sum of squares total}}\\ \\ \eta^2_B &= \dfrac{\mbox{sum of squares B}}{\mbox{sum of squares total}}\\ \\ \eta^2_{int} &= \dfrac{\mbox{sum of squares int}}{\mbox{sum of squares total}} \end{align} $$ $\eta^2$ 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:
    $$ \begin{align} \omega^2_A &= \dfrac{\mbox{sum of squares A} - \mbox{degrees of freedom A} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_B &= \dfrac{\mbox{sum of squares B} - \mbox{degrees of freedom B} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_{int} &= \dfrac{\mbox{sum of squares int} - \mbox{degrees of freedom int} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \end{align} $$ $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2$. Only for balanced designs (equal sample sizes).

  • Proportion variance explained $\eta^2_{partial}$: $$ \begin{align} \eta^2_{partial\,A} &= \frac{\mbox{sum of squares A}}{\mbox{sum of squares A} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,B} &= \frac{\mbox{sum of squares B}}{\mbox{sum of squares B} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,int} &= \frac{\mbox{sum of squares int}}{\mbox{sum of squares int} + \mbox{sum of squares error}} \end{align} $$
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n.a.n.a.Visual representationn.a.n.a.n.a.
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Paired sample t test
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n.a.n.a.n.a.n.a.ANOVA tablen.a.
----
two way ANOVA table
-
n.a.Equivalent toEquivalent ton.a.Equivalent toEquivalent to
-When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels.
  • One sample $t$ test on the difference scores.
  • Repeated measures ANOVA with one dichotomous within subjects factor.
-OLS regression with two categorical independent variables and the interaction term, transformed into $(I - 1)$ + $(J - 1)$ + $(I - 1) \times (J - 1)$ code variables.
  • When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
  • When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
Example contextExample contextExample contextExample contextExample contextExample context
Is the median mental health score of office workers different from $m_0 = 50$?Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.Is the average difference between the mental health scores before and after an intervention different from $\mu_0 = 0$?Is there an association between economic class and gender? Is the distribution of economic class different between men and women?Is the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender?Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
SPSSSPSSSPSSSPSSSPSSSPSS
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
SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:

Analyze > Descriptive Statistics > Crosstabs...
  • Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s)
  • Click the Statistics... button, and click on the square in front of Chi-square
  • Continue and click OK
Analyze > Compare Means > Paired-Samples T Test...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
Analyze > Descriptive Statistics > Crosstabs...
  • Put one of your two categorical variables in the box below Row(s), and the other categorical variable in the box below Column(s)
  • Click the Statistics... button, and click on the square in front of Chi-square
  • Continue and click OK
Analyze > General Linear Model > Univariate...
  • Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s)
Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
  • Put your dichotomous variable in the box below Test Variable List
  • Fill in the value for $\pi_0$ in the box next to Test Proportion
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
JamoviJamoviJamoviJamoviJamoviJamovi
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
Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:

Frequencies > Independent Samples - $\chi^2$ test of association
  • Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
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 Hypothesis, select your alternative hypothesis
Frequencies > Independent Samples - $\chi^2$ test of association
  • Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
ANOVA > ANOVA
  • Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors
Frequencies > 2 Outcomes - Binomial test
  • Put your dichotomous variable in the white box at the right
  • Fill in the value for $\pi_0$ in the box next to Test value
  • Under Hypothesis, select your alternative hypothesis
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
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