Sign test  overview
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Sign test  Binomial test for a single proportion  One sample Wilcoxon signedrank test  Sign test 


Independent variable  Independent variable  Independent variable  Independent variable  
2 paired groups  None  None  2 paired groups  
Dependent variable  Dependent variable  Dependent variable  Dependent variable  
One of ordinal level  One categorical with 2 independent groups  One of ordinal level  One of ordinal level  
Null hypothesis  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}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis. 
 
Alternative 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: $\pi < \pi_0$  H_{1} two sided: $m \neq m_0$ H_{1} right sided: $m > m_0$ H_{1} left sided: $m < m_0$ 
 
Assumptions  Assumptions  Assumptions  Assumptions  



 
Test statistic  Test statistic  Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  $X$ = number of successes in the sample  Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic.
In order to compute each of the test statistics, follow the steps below:
 $W = $ number of difference scores that is larger than 0  
Sampling distribution of $W$ if H_{0} were true  Sampling distribution of $X$ if H0 were true  Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true  Sampling distribution of $W$ 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).  Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1  \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true. Sampling distribution of $W_2$: If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true. If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used. Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.  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.  
Significant?  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:
 For large samples, the table for standard normal probabilities can be used: Two sided:
 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.  n.a.  Equivalent to  
Two sided sign test is equivalent to
    
Two sided sign test is equivalent to
 
Example context  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 median mental health score of office workers different from $m_0 = 50$?  Do people tend to score higher on mental health after a mindfulness course?  
SPSS  SPSS  SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
 Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
 Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 
Jamovi  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
 TTests > One Sample TTest
 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
 
Practice questions  Practice questions  Practice questions  Practice questions  