Chi-Square Independence and Homogeneity

Definition A contingency table with $r$ rows and $c$ columns counts units classified by two categorical variables. Let $O_{ij}$ be the observed count in row $i$ and column $j$, $O_{i+}$ the total of row $i$, $O_{+j}$ the total of column $j$ and $n$ the grand total. When every row and column total is positive, the expected count of the cell is

It is the count the cell would have if every row had the column proportions $O_{+j}/n$ of the whole table, and it estimates the mean of $O_{ij}$ under the null hypothesis of independence or of homogeneity. The table of expected counts has the same row totals and the same column totals as the observed table, and its entries need not be whole numbers. The chi-square approximation in the tests below needs expected counts that are not too small. A common guideline asks for all of them to be at least 5, but such cutoffs are rules of thumb, not theorems. For a sparse table, an exact method such as Fisher’s exact test or a simulation is the better choice.

Definition Suppose $n$ units are sampled independently from one population, for example by simple random sampling from a large population, and each unit is classified by a row variable $G$ with $r$ categories and a column variable $Y$ with $c$ categories, so that it lands in exactly one cell. The test of independence has null hypothesis

for every $i$ and $j$, and alternative $H_1$ that the variables are associated. When every category of $G$ has positive probability, the null hypothesis says that the conditional distribution of $Y$ given $G = i$ is the same for every $i$. The statistic is

If $H_0$ holds and the expected counts are large, $X^2$ has approximately the chi-square distribution with

degrees of freedom, and the p-value is the upper tail $P(\chi^2_{\nu} \ge x)$ at the observed value $x$. The count $(r - 1)(c - 1)$ is the number of interior cells that can be chosen freely once the margins are fixed. Clustered observations, weighted survey samples and paired outcomes fall outside these conditions and need other methods.

Definition Suppose independent random samples of fixed sizes $n_1, \dots, n_r$ are drawn from $r$ populations, and each sampled unit falls into exactly one of $c$ categories. Let $p_{ij}$ be the probability of category $j$ in population $i$. The test of homogeneity has null hypothesis

for every $j$, so that all the populations share one category distribution, and alternative $H_1$ that at least two populations differ in at least one category. It uses the same expected counts $E_{ij}$, the same statistic $X^2$ and the same approximate chi-square reference with $(r - 1)(c - 1)$ degrees of freedom as the test of independence. With two populations and two categories, $X^2$ equals the square of the pooled two-proportion $z$ statistic.

Definition The Pearson residual of the cell in row $i$ and column $j$ is

Its square is the contribution of the cell to $X^2$, so the squared residuals add to $X^2$. A positive residual marks a cell with more units than the null hypothesis predicts, and a negative residual a cell with fewer. Under $H_0$ with large counts, Pearson residuals have variances below 1 and are correlated with each other, so they are not independent standard normal values. An adjusted residual divides $O_{ij} - E_{ij}$ by an estimate of its own standard error instead, and it is approximately standard normal under $H_0$ when the counts are large. Scanning many cells for the largest residual is a multiple comparison problem, so a formal claim about a single cell needs a follow-up procedure planned for that purpose.

Definition For a table with $r \ge 2$ rows and $c \ge 2$ columns and every row and column total positive, Cramér’s V is

It satisfies $0 \le V \le 1$. It equals $0$ exactly when $X^2 = 0$, that is, when every observed count equals its expected count, and it equals $1$ exactly when the category of one variable determines the category of the other, as in a 2 by 2 table with all its counts on one diagonal. Multiplying every count by the same positive factor multiplies $X^2$ by that factor and leaves $V$ unchanged.