z test for the difference between two proportions  overview
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$z$ test for the difference between two proportions  Sign test 


Independent/grouping variable  Independent variable  
One categorical with 2 independent groups  2 paired groups  
Dependent variable  Dependent variable  
One categorical with 2 independent groups  One of ordinal level  
Null hypothesis  Null hypothesis  
H_{0}: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2. 
 
Alternative hypothesis  Alternative hypothesis  
H_{1} two sided: $\pi_1 \neq \pi_2$ H_{1} right sided: $\pi_1 > \pi_2$ H_{1} left sided: $\pi_1 < \pi_2$ 
 
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$z = \dfrac{p_1  p_2}{\sqrt{p(1  p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. Note: we could just as well compute $p_2  p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$  $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  
Approximately the 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:
 
Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  n.a.  
Regular (large sample):
   
Equivalent to  Equivalent to  
When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels. 
Two sided sign test is equivalent to
 
Example context  Example context  
Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.  Do people tend to score higher on mental health after a mindfulness course?  
SPSS  SPSS  
SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
 Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 
Jamovi  Jamovi  
Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples  $\chi^2$ test of association
 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  