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	Chi-squared test for the relationship between two categorical variables
Independent/grouping variable	Independent variable	Independent /column variable
One categorical with 2 independent groups	None	One categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variable	Dependent variable	Dependent /row variable
One categorical with 2 independent groups	One categorical with 2 independent groups	One categorical with $J$ independent groups ($J \geqslant 2$)
Null hypothesis	Null hypothesis	Null 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.	H₀: $\pi = \pi_0$ Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.	H₀: there is no association between the row and column variable More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable: H₀: the distribution of the dependent variable is the same in each of the $I$ populations If there is one random sample of size $N$ from the total population: H₀: the row and column variables are independent
Alternative hypothesis	Alternative hypothesis	Alternative hypothesis
H₁ two sided: $\pi_1 \neq \pi_2$ H₁ right sided: $\pi_1 > \pi_2$ H₁ left sided: $\pi_1 < \pi_2$	H₁ two sided: $\pi \neq \pi_0$ H₁ right sided: $\pi > \pi_0$ H₁ left sided: $\pi < \pi_0$	H₁: there is an association between the row and column variable More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable: H₁: the distribution of the dependent variable is not the same in all of the $I$ populations If there is one random sample of size $N$ from the total population: H₁: the row and column variables are dependent
Assumptions	Assumptions	Assumptions
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 size is large enough for $z$ to be approximately normally distributed. Rule of thumb: Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10 Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more Sample is a simple random sample from the population. That is, observations are independent of one another If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.	Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb: 2 $\times$ 2 table: all four expected cell counts are 5 or more Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
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.	$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$ Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.
Sampling distribution of $z$ if H₀ were true	Sampling distribution of $z$ if H₀ were true	Sampling distribution of $X^2$ if H₀ were true
Approximately the standard normal distribution	Approximately the standard normal distribution	Approximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom
Significant?	Significant?	Significant?
Two sided: Check if $z$ observed in sample is at least as extreme as critical value $z^$ or Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $z$ observed in sample is equal to or larger than critical value $z^$ or Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$	Two sided: Check if $z$ observed in sample is at least as extreme as critical value $z^$ or Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $z$ observed in sample is equal to or larger than critical value $z^$ or Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$	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 $\alpha$
Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$	Approximate $C\%$ confidence interval for $\pi$	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)	Regular (large sample): $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$ 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) With plus four method: $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$ where $p_{plus} = \dfrac{X + 2}{N + 4}$ 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 to	Equivalent to	n.a.
When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels.	When testing two sided: goodness of fit test, with a categorical variable with 2 levels. When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.	-
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 there an association between economic class and gender? Is the distribution of economic class different between 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 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 > Binomial... Put your dichotomous variable in the box below Test Variable List Fill in the value for $\pi_0$ in the box next to Test Proportion If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution	Analyze > Descriptive Statistics > Crosstabs... Put one of your two categorical variables in the box below Row(s), and the other categorical 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
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 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 > 2 Outcomes - Binomial test Put your dichotomous variable in the white box at the right Fill in the value for $\pi_0$ in the box next to Test value Under Hypothesis, select your alternative hypothesis Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution	Frequencies > Independent Samples - $\chi^2$ test of association Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
Practice questions	Practice questions	Practice questions

z test for the difference between two proportions - overview