Sign test  overview
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Sign test  Binomial test for a single proportion  One sample $z$ test for the mean 


Independent variable  Independent variable  Independent variable  
2 paired groups  None  None  
Dependent variable  Dependent variable  Dependent variable  
One of ordinal level  One categorical with 2 independent groups  One quantitative of interval or ratio 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}: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null 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: $\mu \neq \mu_0$ H_{1} right sided: $\mu > \mu_0$ H_{1} left sided: $\mu < \mu_0$  
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  $z = \dfrac{\bar{y}  \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size. The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.  
Sampling distribution of $W$ if H_{0} were true  Sampling distribution of $X$ if H0 were true  Sampling distribution of $z$ 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).  Standard normal distribution  
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:
 
n.a.  n.a.  $C\%$ confidence interval for $\mu$  
    $\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval). The confidence interval for $\mu$ can also be used as significance test.  
n.a.  n.a.  Effect size  
    Cohen's $d$: Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y}  \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$  
n.a.  n.a.  Visual representation  
    
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 the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$  
SPSS  SPSS  n.a.  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
   
Jamovi  Jamovi  n.a.  
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
   
Practice questions  Practice questions  Practice questions  