# 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
Friedman test
One sample Wilcoxon signed-rank test
Cochran's Q test
One sample $z$ test for the mean
Independent variableIndependent/grouping variableIndependent variableIndependent/grouping variableIndependent variableIndependent/grouping variableIndependent variable
NoneOne categorical with 2 independent groups2 paired groupsOne within subject factor ($\geq 2$ related groups)NoneOne within subject factor ($\geq 2$ related groups)None
Dependent variableDependent variableDependent variableDependent variableDependent variableDependent variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne of ordinal levelOne of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio level
Null hypothesisNull 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: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups

Usually the related groups are the different measurement points. 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: $m = m_0$

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 hypothesisAlternative 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: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups 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 equalH1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptions
• 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 of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
• 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
• Population standard deviation $\sigma$ is known
• Sample is a simple random sample from the population. That is, observations are independent of one another
Test statisticTest 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 $\mu_0$.
$Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$

Here $N$ is the number of 'blocks' (usually the subjects - so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$.

Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$.

Note: if ties are present in the data, the formula for $Q$ is more complicated.
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.
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.
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.

The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
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 $Q$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $Q$ 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 freedomIf the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom.

For small samples, the exact distribution of $Q$ should be used.
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 freedomStandard normal distribution
Significant?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:
If the number of blocks $N$ 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 $\alpha$
For large samples, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
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:
Right sided:
Left sided:
n.a.Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$$C\% confidence interval for \mun.a.n.a.n.a.C\% confidence interval for \mu -Regular (large sample): • (p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval) With plus four method: • (p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}} where p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}, p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}, and the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval) \bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}} where the critical value t^* is the value under the t_{N-1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu can also be used as significance test. ---\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval). The confidence interval for \mu can also be used as significance test. n.a.n.a.Effect sizen.a.n.a.n.a.Effect size --Cohen's d: Standardized difference between the sample mean of the difference scores and \mu_0:$$d = \frac{\bar{y} - \mu_0}{s}$$Cohen's d indicates how many standard deviations s the sample mean of the difference scores \bar{y} is removed from \mu_0. ---Cohen's d: Standardized difference between the sample mean and \mu_0:$$d = \frac{\bar{y} - \mu_0}{\sigma}$$Cohen's$d$indicates how many standard deviations$\sigma$the sample mean$\bar{y}$is removed from$\mu_0.$n.a.n.a.Visual representationn.a.n.a.n.a.Visual representation ----- n.a.Equivalent toEquivalent ton.a.n.a.Equivalent ton.a. -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. --Friedman test, with a categorical dependent variable consisting of two independent groups.- Example contextExample 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 a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)?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$\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is$\sigma = 3.$SPSSSPSSSPSSSPSSSPSSSPSSn.a. 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 > 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 the Friedman 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 > 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 - JamoviJamoviJamoviJamoviJamoviJamovin.a. 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 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 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
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