Sign test  overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons 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  Chisquared test for the relationship between two categorical variables 


Independent variable  Independent /column variable  
2 paired groups  One categorical with $I$ independent groups ($I \geqslant 2$)  
Dependent variable  Dependent /row variable  
One of ordinal level  One categorical with $J$ independent groups ($J \geqslant 2$)  
Null hypothesis  Null hypothesis  
 H_{0}: there is no association between the row and column variable More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
 
Alternative hypothesis  Alternative hypothesis  
 H_{1}: there is an association between the row and column variable More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
 
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  $X^2 = \sum{\frac{(\mbox{observed cell count}  \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.  
Sampling distribution of $W$ if H_{0} 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.  Approximately the chisquared distribution with $(I  1) \times (J  1)$ degrees of freedom  
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:

 
Equivalent to  n.a.  
Two sided sign test is equivalent to
   
Example context  Example context  
Do people tend to score higher on mental health after a mindfulness course?  Is there an association between economic class and gender? Is the distribution of economic class different between men and women?  
SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Descriptive Statistics > Crosstabs...
 
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 > Independent Samples  $\chi^2$ test of association
 
Practice questions  Practice questions  