# 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
One sample $z$ test for the mean
Independent variableIndependent/grouping variableIndependent variable
NoneOne categorical with 2 independent groupsNone
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.
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, and $\mu_0$ is the population mean according to the null hypothesis.
Alternative 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: $\mu < \mu_0$
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 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
• 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 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 $\pi_2 > \pi_1.$
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
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 $z$ 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 distributionStandard normal distribution
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:
n.a.Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$$C\% confidence interval for \mu -Regular (large sample): • (p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval) With plus four method: • (p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}} where p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}, p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}, and the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval) \bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval). The confidence interval for \mu can also be used as significance test. n.a.n.a.Effect size --Cohen's d: Standardized difference between the sample mean and \mu_0:$$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 representation -- 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.- Example 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 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$\sigma = 3.$SPSSSPSSn.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 - JamoviJamovin.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
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