Sign test  overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons (max. of 3) by clicking on the dropdown button in the righthand column. To practice with a specific method click the button at the bottom row of the table
Sign test  Binomial test for a single proportion  Goodness of fit test 
You cannot compare more than 3 methods 

Independent variable  Independent variable  Independent variable  
2 paired groups  None  None  
Dependent variable  Dependent variable  Dependent variable  
One of ordinal level  One categorical with 2 independent groups  One categorical with $J$ independent groups ($J \geqslant 2$)  
Null hypothesis  Null hypothesis  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  Alternative hypothesis  
 H_{1} two sided: $\pi \neq \pi_0$ H_{1} right sided: $\pi > \pi_0$ H_{1} left sided: $\pi < \pi_0$ 
 
Assumptions  Assumptions  Assumptions  


 
Test statistic  Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  $X$ = number of successes in the sample  $X^2 = \sum{\frac{(\mbox{observed cell count}  \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.  
Sampling distribution of $W$ if H_{0} were true  Sampling distribution of $X$ if H0 were true  Sampling distribution of $X^2$ if H_{0} were true  
The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $P$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $P = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $nP = n \times 0.5$ and standard deviation $\sqrt{nP(1P)} = \sqrt{n \times 0.5(1  0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W  n \times 0.5}{\sqrt{n \times 0.5(1  0.5)}}$$ follows approximately the standard normal distribution if the null hypothesis were true.  Binomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).  Approximately the chisquared distribution with $J  1$ degrees of freedom  
Significant?  Significant?  Significant?  
If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
 Two sided:

 
Equivalent to  n.a.  n.a.  
Two sided sign test is equivalent to
     
Example context  Example context  Example context  
Do people tend to score higher on mental health after a mindfulness course?  Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?  Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?  
SPSS  SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
 Analyze > Nonparametric Tests > Legacy Dialogs > Chisquare...
 
Jamovi  Jamovi  Jamovi  
Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA  Friedman
 Frequencies > 2 Outcomes  Binomial test
 Frequencies > N Outcomes  $\chi^2$ Goodness of fit
 
Practice questions  Practice questions  Practice questions  