z test for the difference between two proportions - overview

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$z$ test for the difference between two proportions
McNemar's test
Mann-Whitney-Wilcoxon test
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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 of ordinal 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)

If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in both populations:
  • H0: the population median for group 1 is equal to the population median for group 2
Else:
Formulation 1:
  • H0: the population scores in group 1 are not systematically higher or lower than the population scores in group 2
Formulation 2:
  • H0: P(an observation from population 1 exceeds an observation from population 2) = P(an observation from population 2 exceeds observation from population 1)
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
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)

If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in both populations:
  • H1 two sided: the population median for group 1 is not equal to the population median for group 2
  • H1 right sided: the population median for group 1 is larger than the population median for group 2
  • H1 left sided: the population median for group 1 is smaller than the population median for group 2
Else:
Formulation 1:
  • H1 two sided: the population scores in group 1 are systematically higher or lower than the population scores in group 2
  • H1 right sided: the population scores in group 1 are systematically higher than the population scores in group 2
  • H1 left sided: the population scores in group 1 are systematically lower than the population scores in group 2
Formulation 2:
  • H1 two sided: P(an observation from population 1 exceeds an observation from population 2) $\neq$ P(an observation from population 2 exceeds an observation from population 1)
  • H1 right sided: P(an observation from population 1 exceeds an observation from population 2) > P(an observation from population 2 exceeds an observation from population 1)
  • H1 left sided: P(an observation from population 1 exceeds an observation from population 2) < P(an observation from population 2 exceeds an observation from population 1)
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
  • 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.
Two different types of test statistics can be used; both will result in the same test outcome. The first is the Wilcoxon rank sum statistic $W$: The second type of test statistic is the Mann-Whitney $U$ statistic:
  • $U = W - \dfrac{n_1(n_1 + 1)}{2}$
where $n_1$ is the sample size of group 1.

Note: we could just as well base W and U on group 2. This would only 'flip' the right and left sided alternative hypotheses. Also, tables with critical values for $U$ are often based on the smaller of $U$ for group 1 and for group 2.
Sampling distribution of $z$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $W$ and of $U$ 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 $b$.

Sampling distribution of $W$:
For large samples, $W$ is approximately normally distributed with mean $\mu_W$ and standard deviation $\sigma_W$ if the null hypothesis were true. Here $$ \begin{aligned} \mu_W &= \dfrac{n_1(n_1 + n_2 + 1)}{2}\\ \sigma_W &= \sqrt{\dfrac{n_1 n_2(n_1 + n_2 + 1)}{12}} \end{aligned} $$ Hence, for large samples, the standardized test statistic $$ z_W = \dfrac{W - \mu_W}{\sigma_W}\\ $$ follows approximately the standard normal distribution if the null hypothesis were true. Note that if your $W$ value is based on group 2, $\mu_W$ becomes $\frac{n_2(n_1 + n_2 + 1)}{2}$.

Sampling distribution of $U$:
For large samples, $U$ is approximately normally distributed with mean $\mu_U$ and standard deviation $\sigma_U$ if the null hypothesis were true. Here $$ \begin{aligned} \mu_U &= \dfrac{n_1 n_2}{2}\\ \sigma_U &= \sqrt{\dfrac{n_1 n_2(n_1 + n_2 + 1)}{12}} \end{aligned} $$ Hence, for large samples, the standardized test statistic $$ z_U = \dfrac{U - \mu_U}{\sigma_U}\\ $$ follows approximately the standard normal distribution if the null hypothesis were true.

For small samples, the exact distribution of $W$ or $U$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_W$ and $\sigma_U$ is more complicated.
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 $\alpha$
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$n.a.n.a.
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)
--
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.If there are no ties in the data, the two sided Mann-Whitney-Wilcoxon test is equivalent to the Kruskal-Wallis test with an independent variable with 2 levels ($I$ = 2).
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?Do men tend to score higher on social economic status than 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 > Nonparametric Tests > Legacy Dialogs > 2 Independent Samples...
  • Put your dependent variable in the box below Test Variable List 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 variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
  • Under Tests, select Mann-Whitney U
  • Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questions