One sample Wilcoxon signedrank test  overview
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One sample Wilcoxon signedrank test  $z$ test for the difference between two proportions  One sample $t$ test for the mean 


Independent variable  Independent/grouping variable  Independent variable  
None  One categorical with 2 independent 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  
H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.  H_{0}: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.  H_{0}: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.  
Alternative hypothesis  Alternative hypothesis  Alternative hypothesis  
H_{1} two sided: $m \neq m_0$ H_{1} right sided: $m > m_0$ H_{1} left sided: $m < m_0$  H_{1} two sided: $\pi_1 \neq \pi_2$ H_{1} right sided: $\pi_1 > \pi_2$ H_{1} left sided: $\pi_1 < \pi_2$  H_{1} two sided: $\mu \neq \mu_0$ H_{1} right sided: $\mu > \mu_0$ H_{1} left sided: $\mu < \mu_0$  
Assumptions  Assumptions  Assumptions  


 
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:
 $z = \dfrac{p_1  p_2}{\sqrt{p(1  p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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, $\mu_0$ 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 H_{0} were true  Sampling distribution of $z$ if H_{0} were true  Sampling distribution of $t$ if H_{0} 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 freedom  
Significant?  Significant?  Significant?  
For large samples, the table for standard normal probabilities can be used: Two sided:
 Two sided:
 Two sided:
 
n.a.  Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  $C\%$ confidence interval for $\mu$  
  Regular (large sample):
 $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N1}$ 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.  
  When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels.    
Example context  Example context  Example 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 $\mu_0 = 50$?  
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...
 SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
 Analyze > Compare Means > OneSample T Test...
 
Jamovi  Jamovi  Jamovi  
TTests > One Sample TTest
 Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared 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
 TTests > One Sample TTest
 
Practice questions  Practice questions  Practice questions  