z test for the difference between two proportions  overview
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$z$ test for the difference between two proportions  One sample $z$ test for the mean 


Independent/grouping variable  Independent variable  
One categorical with 2 independent groups  None  
Dependent variable  Dependent variable  
One categorical with 2 independent groups  One quantitative of interval or ratio 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.  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  
H_{1} two sided: $\pi_1 \neq \pi_2$ H_{1} right sided: $\pi_1 > \pi_2$ H_{1} left sided: $\pi_1 < \pi_2$  H_{1} two sided: $\mu \neq \mu_0$ H_{1} right sided: $\mu > \mu_0$ H_{1} left sided: $\mu < \mu_0$  
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 $\pi_2 > \pi_1.$  $z = \dfrac{\bar{y}  \mu_0}{\sigma / \sqrt{N}}$
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 $z$ if H_{0} were true  Sampling distribution of $z$ if H_{0} were true  
Approximately the standard normal distribution  Standard normal distribution  
Significant?  Significant?  
Two sided:
 Two sided:
 
Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  $C\%$ confidence interval for $\mu$  
Regular (large sample):
 $\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
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.  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.  Visual representation  
  
Equivalent to  n.a.  
When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels.    
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.  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  n.a.  
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...
   
Jamovi  n.a.  
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
   
Practice questions  Practice questions  