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  Cochran's Q 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 categorical with 2 independent groups  
Null hypothesis  Null hypothesis  
 H_{0}: $\pi_1 = \pi_2 = \ldots = \pi_I$
Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$  
Alternative hypothesis  Alternative hypothesis  
 H_{1}: not all population proportions are equal  
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  If a failure is scored as 0 and a success is scored as 1:
$Q = k(k  1) \dfrac{\sum_{groups} \Big (\mbox{group total}  \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k  \mbox{block total})}$ Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores. Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.  
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 (usually the number of subjects) is large, approximately the chisquared distribution with $k  1$ 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:
 If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
 
Equivalent to  Equivalent to  
Two sided sign test is equivalent to
 Friedman test, with a categorical dependent variable consisting of two independent groups.  
Example context  Example context  
Do people tend to score higher on mental health after a mindfulness course?  Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?  
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
 Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to:
ANOVA > Repeated Measures ANOVA  Friedman
 
Practice questions  Practice questions  