Sign test  overview
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Sign test  Friedman test 


Independent variable  Independent/grouping variable  
2 paired groups  One within subject factor ($\geq 2$ related groups)  
Dependent variable  Dependent variable  
One of ordinal level  One of ordinal level  
Null hypothesis  Null hypothesis  
 H_{0}: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups
Usually the related groups are the different measurement points. Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.  
Alternative hypothesis  Alternative hypothesis  
 H_{1}: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups  
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  $Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i  3 \times N(k + 1)$
Here $N$ is the number of 'blocks' (usually the subjects  so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$. Note: if ties are present in the data, the formula for $Q$ is more complicated.  
Sampling distribution of $W$ if H_{0} were true  Sampling distribution of $Q$ 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.  If the number of blocks $N$ is large, approximately the chisquared distribution with $k  1$ degrees of freedom.
For small samples, the exact distribution of $Q$ should be used.  
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:
 If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
 
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 a difference in depression level between measurement point 1 (preintervention), measurement point 2 (1 week postintervention), and measurement point 3 (6 weeks postintervention)?  
SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
 
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
 ANOVA > Repeated Measures ANOVA  Friedman
 
Practice questions  Practice questions  