One sample Wilcoxon signed-rank test overview

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One sample Wilcoxon signed-rank test	Cochran's Q test	One sample $t$ test for the mean
Independent variable	Independent/grouping variable	Independent variable
None	One within subject factor ($\geq 2$ related groups)	None
Dependent variable	Dependent variable	Dependent variable
One of ordinal level	One categorical with 2 independent groups	One quantitative of interval or ratio level
Null hypothesis	Null hypothesis	Null hypothesis
H₀: $m = m_0$ Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.	H₀: $\pi_1 = \pi_2 = \ldots = \pi_I$ Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$	H₀: $\mu = \mu_0$ Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
Alternative hypothesis	Alternative hypothesis	Alternative hypothesis
H₁ two sided: $m \neq m_0$ H₁ right sided: $m > m_0$ H₁ left sided: $m < m_0$	H₁: not all population proportions are equal	H₁ two sided: $\mu \neq \mu_0$ H₁ right sided: $\mu > \mu_0$ H₁ left sided: $\mu < \mu_0$
Assumptions	Assumptions	Assumptions
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 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another	Scores are normally distributed in the population Sample is a simple random sample from the population. That is, observations are independent of one another
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 $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. 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. 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.	If a failure is scored as 0 and a success is scored as 1: $Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$ Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores. Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.	$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$ Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $s$ is the sample standard deviation, and $N$ is the sample size. 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 $\mu_0$.
Sampling distribution of $W_1$ and of $W_2$ if H₀ were true	Sampling distribution of $Q$ if H₀ were true	Sampling distribution 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 the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom	$t$ distribution with $N - 1$ 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 $\alpha$ Right sided: 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 $\alpha$ Left sided: 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 $\alpha$	If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$: 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 $\alpha$	Two sided: Check if $t$ observed in sample is at least as extreme as critical value $t^$ or Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $t$ observed in sample is equal to or larger than critical value $t^$ or Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $t$ observed in sample is equal to or smaller than critical value $t^*$ or Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$
n.a.	n.a.	$C\%$ confidence interval for $\mu$
-	-	$\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$ 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.
n.a.	n.a.	Effect size
-	-	Cohen's $d$: Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean $\bar{y}$ is removed from $\mu_0.$
n.a.	n.a.	Visual representation
-	-
n.a.	Equivalent to	n.a.
-	Friedman test, with a categorical dependent variable consisting of two independent groups.	-
Example context	Example context	Example context
Is the median mental health score of office workers different from $m_0 = 50$?	Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?	Is the average mental health score of office workers different from $\mu_0 = 50$?
SPSS	SPSS	SPSS
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 > K Related Samples... Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables Under Test Type, select Cochran's Q test	Analyze > Compare Means > One-Sample T Test... Put your variable in the box below Test Variable(s) Fill in the value for $\mu_0$ in the box next to Test Value
Jamovi	Jamovi	Jamovi
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 Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to: ANOVA > Repeated Measures ANOVA - Friedman Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures	T-Tests > One Sample T-Test Put your variable in the box below Dependent Variables Under Hypothesis, fill in the value for $\mu_0$ in the box next to Test Value, and select your alternative hypothesis
Practice questions	Practice questions	Practice questions

One sample Wilcoxon signed-rank test - overview