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	One sample Wilcoxon signed-rank test	Binomial test for a single proportion $z$ test for a single proportion $z$ test for the difference between two proportions Goodness of fit test Chi-squared test for the relationship between two categorical variables One sample $z$ test for the mean One sample $t$ test for the mean Paired sample $t$ test Two sample $z$ test Two sample $t$ test - equal variances not assumed Two sample $t$ test - equal variances assumed One way ANOVA Two way ANOVA Pearson correlation Regression (OLS)Logistic regression Mann-Whitney-Wilcoxon test Kruskal-Wallis test Sign test McNemar's test Cochran's Q test Marginal Homogeneity test / Stuart-Maxwell test Friedman test Wilcoxon signed-rank test One sample Wilcoxon signed-rank test Spearman's rho
Independent /column variable	Independent variable
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 of ordinal level
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: $m = m_0$ Here $m$ is the population median, and $m_0$ is the population median according to the null 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: $m \neq m_0$ H₁ right sided: $m > m_0$ H₁ left sided: $m < m_0$
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	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
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.
Sampling distribution of $X^2$ if H₀ were true	Sampling distribution of $W_1$ and of $W_2$ if H₀ were true
Approximately the chi-squared 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.
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 $\alpha$	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 $\alpha$
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 $m_0 = 50$?
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	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 signed-rank test On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank 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
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	T-Tests > One Sample T-Test 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
Practice questions	Practice questions

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