Sign test  overview
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Sign test  Two sample $z$ test 


Independent variable  Independent variable  
2 paired groups  One categorical with 2 independent groups  
Dependent variable  Dependent variable  
One of ordinal level  One quantitative of interval or ratio level  
Null hypothesis  Null hypothesis  
 $\mu_1 = \mu_2$
$\mu_1$ is the unknown mean in population 1, $\mu_2$ is the unknown mean in population 2  
Alternative hypothesis  Alternative hypothesis  
 Two sided: $\mu_1 \neq \mu_2$ Right sided: $\mu_1 > \mu_2$ Left sided: $\mu_1 < \mu_2$  
Assumptions  Assumptions  
Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another 
 
Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  $z = \dfrac{(\bar{y}_1  \bar{y}_2)  0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1  \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
$\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to H0. The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1  \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1  \bar{y}_2$ is removed from 0. Note: we could just as well compute $\bar{y}_2  \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$  
Sampling distribution of $W$ if H0 were true  Sampling distribution of $z$ if H0 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 a standard normal distribution if the null hypothesis were true.  Standard normal  
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:
 
n.a.  $C\%$ confidence interval for $\mu_1  \mu_2$  
  $(\bar{y}_1  \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
where $z^*$ 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 $\mu_1  \mu_2$ can also be used as significance test.  
n.a.  Visual representation  
  
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 the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1$ = 2 amongst men and $\sigma_2$ = 2.5 amongst women.  
SPSS  n.a.  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
   
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
   
Practice questions  Practice questions  