z test for the difference between two proportions  overview
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$z$ test for the difference between two proportions  $z$ test for a single proportion  Two sample $t$ test  equal variances assumed 
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Independent/grouping variable  Independent variable  Independent/grouping variable  
One categorical with 2 independent groups  None  One categorical with 2 independent groups  
Dependent variable  Dependent variable  Dependent variable  
One categorical with 2 independent groups  One categorical with 2 independent groups  One quantitative of interval or ratio level  
Null hypothesis  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}: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.  H_{0}: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.  
Alternative hypothesis  Alternative hypothesis  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: $\pi \neq \pi_0$ H_{1} right sided: $\pi > \pi_0$ H_{1} left sided: $\pi < \pi_0$  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  Assumptions  


 
Test statistic  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 $\pi_2 > \pi_1.$  $z = \dfrac{p  \pi_0}{\sqrt{\dfrac{\pi_0(1  \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.  $t = \dfrac{(\bar{y}_1  \bar{y}_2)  0}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} = \dfrac{\bar{y}_1  \bar{y}_2}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s_p$ is the pooled standard deviation, $n_1$ 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 $s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1  \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1  \bar{y}_2$ 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 $\mu_2 > \mu_1$.  
n.a.  n.a.  Pooled standard deviation  
    $s_p = \sqrt{\dfrac{(n_1  1) \times s^2_1 + (n_2  1) \times s^2_2}{n_1 + n_2  2}}$  
Sampling distribution of $z$ if H_{0} were true  Sampling distribution of $z$ if H_{0} were true  Sampling distribution of $t$ if H_{0} were true  
Approximately the standard normal distribution  Approximately the standard normal distribution  $t$ distribution with $n_1 + n_2  2$ degrees of freedom  
Significant?  Significant?  Significant?  
Two sided:
 Two sided:
 Two sided:
 
Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  Approximate $C\%$ confidence interval for $\pi$  $C\%$ confidence interval for $\mu_1  \mu_2$  
Regular (large sample):
 Regular (large sample):
 $(\bar{y}_1  \bar{y}_2) \pm t^* \times s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$
where the critical value $t^*$ is the value under the $t_{n_1 + n_2  2}$ distribution with the area $C / 100$ between $t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). The confidence interval for $\mu_1  \mu_2$ can also be used as significance test.  
n.a.  n.a.  Effect size  
    Cohen's $d$: Standardized difference between the mean in group $1$ and in group $2$: $$d = \frac{\bar{y}_1  \bar{y}_2}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ the two sample means are removed from each other.  
n.a.  n.a.  Visual representation  
    
Equivalent to  Equivalent to  Equivalent to  
When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels. 
 One way ANOVA with an independent variable with 2 levels ($I$ = 2):
 
Example context  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 proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.  Is the average mental health score different between men and women? Assume that in the population, the standard deviation of mental health scores is equal amongst men and women.  
SPSS  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 > Binomial...
 Analyze > Compare Means > IndependentSamples T Test...
 
Jamovi  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
 Frequencies > 2 Outcomes  Binomial test
 TTests > Independent Samples TTest
 
Practice questions  Practice questions  Practice questions  