Chisquared test for the relationship between two categorical variables  overview
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Chisquared test for the relationship between two categorical variables
One categorical with $I$ independent groups ($I \geqslant 2$)
None
One categorical with 2 independent groups
Dependent /row variable
Dependent variable
Dependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)
One of ordinal level
One of ordinal level
Null hypothesis
Null hypothesis
Null hypothesis
H_{0}: 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:
H_{0}: 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:
H_{0}: the row and column variables are independent
H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
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:
H_{0}: the population median for group 1 is equal to the population median for group 2
Else:
Formulation 1:
H_{0}: the population scores in group 1 are not systematically higher or lower than the population scores in group 2
Formulation 2:
H_{0}:
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 hypothesis
Alternative hypothesis
Alternative hypothesis
H_{1}: 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:
H_{1}: 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:
H_{1}: the row and column variables are dependent
H_{1} two sided: $m \neq m_0$
H_{1} right sided: $m > m_0$
H_{1} left sided: $m < m_0$
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:
H_{1} two sided: the population median for group 1 is not equal to the population median for group 2
H_{1} right sided: the population median for group 1 is larger than the population median for group 2
H_{1} left sided: the population median for group 1 is smaller than the population median for group 2
Else:
Formulation 1:
H_{1} two sided: the population scores in group 1 are systematically higher or lower than the population scores in group 2
H_{1} right sided: the population scores in group 1 are systematically higher than the population scores in group 2
H_{1} left sided: the population scores in group 1 are systematically lower than the population scores in group 2
Formulation 2:
H_{1} 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)
H_{1} 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)
H_{1} 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)
Assumptions
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
The population distribution of the scores is symmetric
Sample is a simple random sample from the population. That is, observations 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 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.
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic.
In order to compute each of the test statistics, follow the steps below:
For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}  m_0)$. The sign is 1 if the difference is larger than zero, 1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
For each subject, compute the absolute value of the difference score $\mbox{score}  m_0$.
Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:
$W_1 = \sum\, R_d^{+}$
or
$W_1 = \sum\, R_d^{}$
That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{}$. So if you are using such a table, pick the smaller one.
If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{}$, the right and left sided alternative hypotheses 'flip').
$W_2 = \sum\, \mbox{sign}_d \times R_d$
That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
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 MannWhitney $U$ 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 $W_1$ and of $W_2$ if H_{0} were true
Sampling distribution of $W$ and of $U$ if H_{0} were true
Approximately the chisquared distribution with $(I  1) \times (J  1)$ degrees of freedom
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here
$$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$
$$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$
Hence, if $N_r$ is large, the standardized test statistic
$$z = \frac{W_1  \mu_{W_1}}{\sigma_{W_1}}$$
follows approximately the standard normal distribution if the null hypothesis were true.
Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here
$$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$
Hence, if $N_r$ is large, the standardized test statistic
$$z = \frac{W_2}{\sigma_{W_2}}$$
follows approximately the standard normal distribution if the null hypothesis were true.
If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.
Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
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.
Find $p$ value corresponding to observed $X^2$ 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:
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 $\alpha$
For large samples, the table for standard normal probabilities can be used:
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.
n.a.
Equivalent to


If there are no ties in the data, the two sided MannWhitneyWilcoxon test is equivalent to the KruskalWallis test with an independent variable with 2 levels ($I$ = 2).
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 median mental health score of office workers different from $m_0 = 50$?
Do men tend to score higher on social economic status than women?
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 Chisquare
Continue and click OK
Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
On the Objective tab, choose Customize Analysis
On the Fields tab, specify the variable for which you want to compute the Wilcoxon signedrank test
On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signedrank test)'. Fill in your $m_0$ in the box next to Hypothesized median
Click Run
Double click on the output table to see the full results
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
Jamovi
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
TTests > One Sample TTest
Put your variable in the box below Dependent Variables
Under Tests, select Wilcoxon rank
Under Hypothesis, fill in the value for $m_0$ in the box next to Test Value, and select your alternative hypothesis
TTests > Independent Samples TTest
Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
Under Tests, select MannWhitney U
Under Hypothesis, select your alternative hypothesis