Last updated: 2026-07-01

Chi-Square Test of Independence

The chi-square test of independence asks whether two categorical variables measured on the same subjects are related or independent. The data live in a contingency table: each cell holds the number of observations that fall into one combination of categories — for a 2×2 table, two categories for each variable. The test compares the counts you actually observed against the counts you would expect if the two variables had no relationship, and returns a statistic, its degrees of freedom, and a p-value. It is the standard tool for questions such as whether a treatment is associated with an outcome, whether a preference depends on group membership, or whether two survey answers move together.

Expected counts under independence

If two variables are independent, the proportion of the sample in any row should not depend on which column you look at. That fixes what each cell should contain. The expected count in a cell is its row total times its column total, divided by the grand total:

$E_{ij} = \frac{(\text{row } i \text{ total})\,(\text{column } j \text{ total})}{N}$

where $N$ is the total number of observations. This is just the fraction of the sample in row $i$ multiplied by the fraction in column $j$, scaled back up to $N$. Crucially, the expected counts come from the table's own margins — you do not supply them, which is what separates this test from a goodness-of-fit test.

The statistic

The test summarizes the gap between observation and expectation with the Pearson statistic

$\chi^2 = \sum \frac{(O - E)^2}{E}$

summed over every cell, where $O$ is an observed count and $E$ its expected count. Each squared deviation is divided by the expected count so that a miss of ten counts weighs heavily when only five were expected but lightly when five thousand were expected. Summing gives a single number that grows as the table drifts away from independence. When the observed counts match the expected counts exactly, every term is zero and $\chi^2 = 0$.

For a 2×2 table the whole sum collapses to a compact algebraic shortcut. Labelling the cells

the statistic equals

$\chi^2 = \frac{N\,(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$

The four factors in the denominator are the two row totals and the two column totals. The cross-product term $ad - bc$ measures how far the table departs from proportionality: it is zero exactly when the rows are proportional, which is the case of perfect independence.

Degrees of freedom and the p-value

The statistic is compared against a chi-square distribution whose shape is fixed by its degrees of freedom. For a contingency table,

$\text{df} = (r - 1)(c - 1)$

where $r$ is the number of rows and $c$ the number of columns. A 2×2 table therefore always has one degree of freedom. Intuitively, once the row and column totals are fixed, filling in a single cell determines all the others, so only one cell is free to vary.

The p-value is the upper-tail probability — the chance of seeing a statistic at least as large as the one observed if the two variables were independent. The test is inherently one-sided in the upper tail: only large values of $\chi^2$ signal an association, while small values mean the table sits close to independence. Compare the p-value against a significance level $\alpha$ (commonly five percent). When $p < \alpha$, the departure from independence is larger than sampling noise comfortably explains, and you conclude the variables are associated. When $p \ge \alpha$, the data are consistent with independence — which is not the same as proving the variables unrelated.

Worked example

Suppose a clinic records whether patients recovered under two treatments:

Here $a = 30$, $b = 10$, $c = 20$, $d = 40$, and $N = 100$. Using the shortcut,

With a 2×2 table the test has one degree of freedom. The upper-tail p-value for $\chi^2 \approx 16.667$ at one degree of freedom is about $0.0000446$. Since $p \lt 0.01$, there is strong evidence that recovery is associated with treatment: patients on Treatment A recovered far more often than independence would predict.

Assumptions

The chi-square approximation rests on a few conditions. Expected counts should be reasonably large — a common rule of thumb requires every expected count to be at least 5. For a 2×2 table with small expected counts, apply Yates's continuity correction or switch to Fisher's exact test. Observations must be independent, each subject counted once and falling into exactly one cell, so the test does not apply to paired or repeated measurements (use McNemar's test for those). Finally, the test works on raw frequencies, not percentages or proportions: convert any rates to counts before entering them.

For comparing one categorical variable against a fixed hypothesized distribution rather than two variables against each other, see the Chi-Square Goodness-of-Fit Test; for turning a z-statistic into a tail probability, see P-Value from Z-Score Calculator.