Chi-square test of independence (and homogeneity)

Tests whether two categorical variables are associated in an r × c contingency table by comparing observed counts with those expected under independence.

E_{ij} = \frac{R_i\,C_j}{N},\qquad X^2 = \sum_{i,j}\frac{(O_{ij} - E_{ij})^2}{E_{ij}},\qquad df = (r-1)(c-1)

$O_{ij}, E_{ij}$: Observed and expected counts in row i, column j.
$R_i, C_j$: Row i total and column j total.
$N$: Grand total.
$r, c$: Numbers of rows and columns.

\frac{O_{ij} - E_{ij}}{\sqrt{E_{ij}(1 - R_i/N)(1 - C_j/N)}}

Adjusted standardised residual: roughly N(0, 1) under H₀; values beyond about ±2 show which cells drive the association.

$R_i/N, C_j/N$: Row and column proportions.

When to use

Two categorical variables measured on the same units (independence), or one categorical outcome compared across several groups (homogeneity) — the arithmetic is identical. For 2×2 tables with small expected counts use Fisher's exact test.

Assumptions

Independent units, each in exactly one cell; counts, not percentages.
Expected counts mostly ≥ 5 (the engine reports how many are below 5).
Yates' continuity correction is optional (2×2 only) and off by default.

How to interpret

A small p-value is evidence of association; it says nothing about direction or cause.
Describe the association with conditional percentages (row or column percentages) and adjusted residuals.
Effect size: Cramér's $V = \sqrt{X^2/(N(\min(r,c)-1))}$ (0 = no association, 1 = perfect).

Common mistakes

Using row percentages or means instead of counts.
Counting the same person in several cells (e.g. multiple-response questions).
Interpreting a significant result in a huge sample as a strong association — check V.

Example

Preferred plan (A, B, C) by region: North 20, 30, 25; South 30, 20, 25 (N = 150). All $E_{ij} = 25$ . $X^2 = 4.0$ on 2 df, $p = 0.135$ , $V = 0.163$ . Adjusted residuals are ±1.73 for plans A and B: no convincing evidence that preferences differ by region.