One sample Wilcoxon signedrank test  overview
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One sample Wilcoxon signedrank test  Binomial test for a single proportion 


Independent variable  Independent variable  
None  None  
Dependent variable  Dependent variable  
One of ordinal level  One categorical with 2 independent groups  
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}: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.  
Alternative 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: $\pi \neq \pi_0$ H_{1} right sided: $\pi > \pi_0$ H_{1} left sided: $\pi < \pi_0$  
Assumptions  Assumptions  

 
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:
 $X$ = number of successes in the sample  
Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true  Sampling distribution of $X$ 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.  Binomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).  
Significant?  Significant?  
For large samples, the table for standard normal probabilities can be used: Two sided:
 Two sided:
 
Example context  Example context  
Is the median mental health score of office workers different from $m_0 = 50$?  Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?  
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...
 Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
 
Jamovi  Jamovi  
TTests > One Sample TTest
 Frequencies > 2 Outcomes  Binomial test
 
Practice questions  Practice questions  