The chi-square test: a complete guide

Chi-square compares the counts you actually observed with the counts you would expect if nothing were going on. The bigger the gap between observed and expected, the larger the statistic, and the less plausible it becomes that the pattern arose by chance.

The arithmetic is one formula applied to every cell of your table and then summed:

χ² = Σ (O − E)² / E

Each cell contributes an amount proportional to how far it strays from expectation, scaled by how big the expectation was. A discrepancy of 10 matters enormously when you expected 5 and barely at all when you expected 500 — dividing by E is what encodes that.

A third variant, the test of homogeneity, is arithmetically identical to the test of independence and differs only in how the sample was collected — whether you sampled one population and cross-classified, or sampled fixed numbers from several populations. The calculation and the reporting are the same.

When chi-square is not the right test

Chi-square has few assumptions, but they are strict and the third one is violated constantly.

The independence assumption is the one with the least room for negotiation, and it is broken more often than people realise. It requires that each participant contributes exactly one observation to exactly one cell. If you asked 60 people three questions each and cross-tabulated all 180 responses, your table looks like it has 180 independent observations when it really has 60 people answering repeatedly. The statistic is then inflated, sometimes dramatically, and no correction will rescue it. The fix is structural rather than statistical: analyse one response per person, or move to a model that accounts for the repeated measurements, such as a generalised estimating equation or a mixed-effects logistic regression.

The expected-count rule in practice

Note that the rule applies to expected counts, not observed ones. A cell can contain zero observations and still be fine, provided the expected count for that cell is 5 or more. Every statistics package reports the minimum expected count alongside the test — check it every time, and report it.

167 students are asked whether they agree that statistics training should be compulsory. We want to know whether responses differ between undergraduates and postgraduates.

Step 1 — expected counts

E = (row total × column total) / grand total. For undergraduates who agree: (80 × 80) / 167 = 38.3. Repeat for all six cells. The smallest expected count is 17.7, comfortably above 5, so the test is valid.

Step 2 — the cell contributions

Step 3 — degrees of freedom and the decision

df = (r − 1)(c − 1) = (2−1)(3−1) = 2. The critical value for df = 2 at α = .05 is 5.99. Our statistic of 4.96 is smaller, so p = .084 and we do not reject the null.

A department claims its intake is evenly split across four subject backgrounds. Among 120 students, the observed counts are 38, 24, 31 and 27.

If the claim were true, each category would contain 120 / 4 = 30 students.

df = k − 1 = 3, and the critical value is 7.81. Since 3.66 < 7.81, p = .300: the data are consistent with an even split.

Chi-square is heavily influenced by sample size. Multiply every count in a table by ten and the statistic multiplies by ten too, while the pattern of association is unchanged. With a large enough sample, trivial departures from independence become significant. An effect size is what separates the two.

For our worked example: V = √(4.96 / (167 × 1)) = √0.0297 = 0.172 — a small association.

A significant chi-square on a table larger than 2×2 tells you the variables are associated somewhere, but not where. Reporting only the overall test leaves the interesting part unsaid.

Adjusted standardised residuals

For each cell, the adjusted standardised residual measures how far the observed count sits from expectation in standard-error units. Because they are approximately standard normal, they read like z-scores:

Because you are inspecting several cells at once, apply a Bonferroni correction: divide .05 by the number of cells. For a 2×3 table that gives .05/6 = .0083, corresponding to a residual beyond about ±2.64.

The 2×2 table is the most common chi-square in applied research — treated versus untreated, improved versus not — and it has its own set of conventions worth knowing.

Yates' continuity correction

Chi-square uses a continuous distribution to approximate discrete counts, and with only one degree of freedom that approximation is at its roughest. Yates' correction subtracts 0.5 from each absolute difference before squaring, which shrinks the statistic and raises the p-value:

χ²ₖₐₜₑₛ = Σ (|O − E| − 0.5)² / E

Whether to apply it is genuinely contested. The correction is conservative — it can be too conservative, costing real power in moderate samples. The common position is to use Fisher's exact test when expected counts are small, and the uncorrected chi-square when they are comfortably above 5, rather than relying on Yates as a middle path. SPSS reports it by default for 2×2 tables, so state which row of the output you are quoting.

A worked 2×2

Expected counts are 28, 22, 28 and 22 — all well above 5, so no exact test is needed. The uncorrected statistic is χ²(1, N = 100) = 5.844, p = .016.

The odds ratio makes the size of the effect concrete: OR = (34 × 28) / (16 × 22) = 952 / 352 = 2.70. The odds of improvement were about 2.7 times higher in the intervention group.

1. Running chi-square on percentages

The statistic scales with the total, so percentages produce an answer for a sample of 100 regardless of your real n. Always use raw counts.

2. Ignoring the expected-count rule

Report the minimum expected count. If any cell falls below 5, switch to Fisher's exact test or justify merging categories.

3. Using chi-square on paired data

If the same people are measured before and after, the observations are not independent. Use McNemar's test.

4. Reporting the test without an effect size

With a large sample, significance is close to guaranteed. Cramer's V is what tells the reader whether it matters.

5. Concluding causation

Chi-square establishes association between categorical variables. It says nothing about direction or cause.

6. Stopping at the overall test

For tables larger than 2×2, examine adjusted residuals so you can say where the association lies.

7. Binning a continuous variable to force a chi-square

Cutting a continuous outcome into categories throws away information and power. Use the test the measurement level actually supports.

APA 7 wants the symbol italicised, degrees of freedom and sample size in parentheses together, and an exact p-value.