Two sample z test  overview
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Two sample $z$ test  Sign test 


Independent/grouping variable  Independent variable  
One categorical with 2 independent groups  2 paired groups  
Dependent variable  Dependent variable  
One quantitative of interval or ratio level  One of ordinal level  
Null hypothesis  Null hypothesis  
H_{0}: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2. 
 
Alternative hypothesis  Alternative hypothesis  
H_{1} two sided: $\mu_1 \neq \mu_2$ H_{1} right sided: $\mu_1 > \mu_2$ H_{1} left sided: $\mu_1 < \mu_2$ 
 
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$z = \dfrac{(\bar{y}_1  \bar{y}_2)  0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1  \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ 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, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis. 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 $\mu_2 > \mu_1$.  $W = $ number of difference scores that is larger than 0  
Sampling distribution of $z$ if H_{0} were true  Sampling distribution of $W$ if H_{0} were true  
Standard normal distribution  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?  
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:
 
$C\%$ confidence interval for $\mu_1  \mu_2$  n.a.  
$(\bar{y}_1  \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
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 $\mu_1  \mu_2$ can also be used as significance test.    
Visual representation  n.a.  
  
n.a.  Equivalent to  
 
Two sided sign test is equivalent to
 
Example context  Example context  
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.  Do people tend to score higher on mental health after a mindfulness course?  
n.a.  SPSS  
  Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 
n.a.  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
 
Practice questions  Practice questions  