z test for the difference between two proportions - overview

This page offers structured overviews of one or more selected methods. Add additional methods for comparisons (max. of 3) by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table

$z$ test for the difference between two proportions
McNemar's test
Two sample $t$ test - equal variances assumed
You cannot compare more than 3 methods
Independent/grouping variableIndependent variableIndependent/grouping variable
One categorical with 2 independent groups2 paired groupsOne categorical with 2 independent groups
Dependent variableDependent variableDependent variable
One categorical with 2 independent groupsOne categorical with 2 independent groupsOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesis
H0: $\pi_1 = \pi_2$

Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.

Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:

  1. First score of pair is 0, second score of pair is 0
  2. First score of pair is 0, second score of pair is 1 (switched)
  3. First score of pair is 1, second score of pair is 0 (switched)
  4. First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the null hypothesis are:

  • H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
  • H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)

H0: $\mu_1 = \mu_2$

Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$

The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the alternative hypothesis are:

  • H1: $\pi_1 \neq \pi_2$
  • H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)

H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
AssumptionsAssumptionsAssumptions
  • Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
    • Significance test: number of successes and number of failures are each 5 or more in both sample groups
    • Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
    • Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Within each population, the scores on the dependent variable are normally distributed
  • The standard deviation of the scores on the dependent variable is the same in both populations: $\sigma_1 = \sigma_2$
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
Test statisticTest statisticTest 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 $\pi_2 > \pi_1.$
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
$t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$
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 H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $t$ if H0 were true
Approximately the standard normal distribution

If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.

If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.

$t$ distribution with $n_1 + n_2 - 2$ degrees of freedom
Significant?Significant?Significant?
Two sided: Right sided: Left sided: For test statistic $X^2$:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
  • Check if $b$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$n.a.$C\%$ confidence interval for $\mu_1 - \mu_2$
Regular (large sample):
  • $(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$
    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)
With plus four method:
  • $(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$
    where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, and 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)
-$(\bar{y}_1 - \bar{y}_2) \pm t^* \times s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$
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
--
Two sample t test - equal variances assumed
Equivalent toEquivalent toEquivalent to
When testing two sided: chi-squared 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):
  • two sided two sample $t$ test is equivalent to ANOVA $F$ test when $I$ = 2
  • two sample $t$ test is equivalent to $t$ test for contrast when $I$ = 2
  • two sample $t$ test is equivalent to $t$ test multiple comparisons when $I$ = 2
OLS regression with one categorical independent variable with 2 levels:
  • two sided two sample $t$ test is equivalent to $F$ test regression model
  • two sample $t$ test is equivalent to $t$ test for regression coefficient $\beta_1$
Example contextExample contextExample context
Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?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.
SPSSSPSSSPSS
SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:

Analyze > Descriptive Statistics > Crosstabs...
  • Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s)
  • Click the Statistics... button, and click on the square in front of Chi-square
  • Continue and click OK
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the McNemar test
Analyze > Compare Means > Independent-Samples T Test...
  • Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable
  • Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow
  • Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2
  • Continue and click OK
JamoviJamoviJamovi
Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared 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
  • Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
Frequencies > Paired Samples - McNemar test
  • Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
T-Tests > Independent Samples T-Test
  • Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
  • Under Tests, select Student's (selected by default)
  • Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questions