Sign test  overview
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Sign test  Binomial test for a single proportion  Spearman's rho 


Independent variable  Independent variable  Variable 1  
2 paired groups  None  One of ordinal level  
Dependent variable  Dependent variable  Variable 2  
One of ordinal level  One categorical with 2 independent groups  One of ordinal level  
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.  H_{0}: $\rho_s = 0$
Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level. In words, the null hypothesis would be: H_{0}: there is no monotonic relationship between the two variables in the population.  
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: $\pi < \pi_0$  H_{1} two sided: $\rho_s \neq 0$ H_{1} right sided: $\rho_s > 0$ H_{1} left sided: $\rho_s < 0$  
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  $t = \dfrac{r_s \times \sqrt{N  2}}{\sqrt{1  r_s^2}} $ Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores.  
Sampling distribution of $W$ if H_{0} were true  Sampling distribution of $X$ if H0 were true  Sampling distribution of $t$ 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 $t$ distribution with $N  2$ 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:
 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 there a monotonic relationship between physical health and mental health?  
SPSS  SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
 Analyze > Correlate > Bivariate...
 
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
 Regression > Correlation Matrix
 
Practice questions  Practice questions  Practice questions  