Chi-squared test for the relationship between two categorical variables overview

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Chi-squared test for the relationship between two categorical variables	$z$ test for a single proportion	Marginal Homogeneity test / Stuart-Maxwell test
Independent /column variable	Independent variable	Independent variable
One categorical with $I$ independent groups ($I \geqslant 2$)	None	2 paired groups
Dependent /row variable	Dependent variable	Dependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)	One categorical with 2 independent groups	One categorical with $J$ independent groups ($J \geqslant 2$)
Null hypothesis	Null hypothesis	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	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: for each category $j$ of the dependent variable, $\pi_j$ for the first paired group = $\pi_j$ for the second paired group. Here $\pi_j$ is the population proportion in category $j.$
Alternative hypothesis	Alternative hypothesis	Alternative hypothesis
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	H₁ two sided: $\pi \neq \pi_0$ H₁ right sided: $\pi > \pi_0$ H₁ left sided: $\pi < \pi_0$	H₁: for some categories of the dependent variable, $\pi_j$ for the first paired group $\neq$ $\pi_j$ for the second paired group.
Assumptions	Assumptions	Assumptions
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	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 of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Test statistic	Test statistic	Test statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$ 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.	$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$ 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.	Computing the test statistic is a bit complicated and involves matrix algebra. Unless you are following a technical course, you probably won't need to calculate it by hand.
Sampling distribution of $X^2$ if H₀ were true	Sampling distribution of $z$ if H₀ were true	Sampling distribution of the test statistic if H₀ were true
Approximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom	Approximately the standard normal distribution	Approximately the chi-squared distribution with $J - 1$ degrees of freedom
Significant?	Significant?	Significant?
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 $\alpha$	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 $\alpha$	If we denote the test statistic as $X^2$: 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 $\alpha$
n.a.	Approximate $C\%$ confidence interval for $\pi$	n.a.
-	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)	-
n.a.	Equivalent to	n.a.
-	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 there an association between economic class and gender? Is the distribution of economic class different between men and women?	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.	Subjects are asked to taste three different types of mayonnaise, and to indicate which of the three types of mayonnaise they like best. They then have to drink a glass of beer, and taste and rate the three types of mayonnaise again. Does drinking a beer change which type of mayonnaise people like best?
SPSS	SPSS	SPSS
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	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 > 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 Marginal Homogeneity test
Jamovi	Jamovi	n.a.
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	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	-
Practice questions	Practice questions	Practice questions

Chi-squared test for the relationship between two categorical variables - overview