Chisquared test for the relationship between two categorical variables  overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the righthand column. To practice with a specific method click the button at the bottom row of the table
Chisquared test for the relationship between two categorical variables
One categorical with $I$ independent groups ($I \geqslant 2$)
None
Dependent /row variable
Dependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)
One categorical with 2 independent groups
Null hypothesis
Null hypothesis
There is no association between the row and column variable
More precise statement:
If there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable: 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: The row and column variables are independent
$\pi = \pi_0$
$\pi$ is the population proportion of "successes"; $\pi_0$ is the population proportion of successes according to the null hypothesis
Alternative hypothesis
Alternative hypothesis
There is an association between the row and column variable More precise statement:
If there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable: 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: The row and column variables are dependent
Two sided: $\pi \neq \pi_0$
Right sided: $\pi > \pi_0$
Left sided: $\pi < \pi_0$
Assumptions
Assumptions
Sample size is large enough for $X^2$ to be approximately chisquared 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
$X^2 = \sum{\frac{(\mbox{observed cell count}  \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
where 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
$z = \dfrac{p  \pi_0}{\sqrt{\dfrac{\pi_0(1  \pi_0)}{N}}}$
$p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size
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 $\alpha$
n.a.
Approximate $C\%$ confidence interval for $\pi$

Regular (large sample):
$p \pm z^* \times \sqrt{\dfrac{p(1  p)}{N}}$
where $z^*$ 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 $z^*$ 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

When testing two sided: goodness of fit test, with 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
Is there an association between economic class and gender? Is the distribution of economic class different between men and women?
Is the proportion smokers amongst office workers different from $\pi_0 = .2$? Use the normal approximation for the sampling distribution of the test statistic.
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 Chisquare
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
Jamovi
Jamovi
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