A chi-square test of independence asks one question of a contingency table: are the two categorical variables related, or independent? The procedure is short — build a table of expected counts, sum up a single ratio, look up a p-value — but it relies on getting the expected counts right. This walkthrough builds the entire test from a 2-by-3 table.
What the Test Actually Tests
The chi-square test of independence works with two categorical variables measured on the same subjects. Examples: major and political affiliation; brand preference and age group; treatment and recovery outcome. The data sits in a contingency table where each cell counts how many subjects fall into that row-and-column combination.
The hypotheses are about the relationship between the two variables:
- H₀: the two variables are independent in the population.
- H₁: the two variables are associated (not independent).
"Independent" here has its usual probability meaning: knowing the row category gives you no information about the column category. If that is true in the population, the proportions inside each row should be the same as the overall column proportions, up to random sampling variation.
Expected Counts: The Heart of the Test
Every cell in the contingency table gets an expected count — what you would expect to see in that cell if the null hypothesis (independence) were true. The formula is:
E = (row total × column total) / grand total
That formula comes straight from the independence definition. If row category and column category are independent, P(row r AND column c) = P(row r) × P(column c). Multiply by the grand total n to convert the joint probability into an expected count: E = n × (row r total / n) × (column c total / n) = (row total × column total) / n.
Once you have observed counts O and expected counts E for every cell, the test statistic is
χ² = Σ (O − E)² / E
with degrees of freedom df = (rows − 1) × (columns − 1). Large χ² means observed counts diverge from what independence predicts; small χ² means they match closely.
The Worked Example
A survey of 300 students asks for their preferred study spot (Library, Café, Home) and their year (Freshman, Junior). The observed table:
- Freshmen — Library: 60, Café: 30, Home: 30. Row total = 120.
- Juniors — Library: 50, Café: 90, Home: 40. Row total = 180.
- Column totals — Library: 110, Café: 120, Home: 70. Grand total = 300.
Is there an association between year and preferred study spot? Test at α = 0.05.
Step 1 — Expected counts. For each cell, E = (row total × column total) / 300.
- Freshmen × Library: (120 × 110) / 300 = 44
- Freshmen × Café: (120 × 120) / 300 = 48
- Freshmen × Home: (120 × 70) / 300 = 28
- Juniors × Library: (180 × 110) / 300 = 66
- Juniors × Café: (180 × 120) / 300 = 72
- Juniors × Home: (180 × 70) / 300 = 42
A quick sanity check: each row of expected counts should sum to the original row total (44 + 48 + 28 = 120; 66 + 72 + 42 = 180), and each column to the original column total. They do.
Step 2 — Check the condition. A standard requirement is that every expected count is at least 5. The smallest here is 28, well above 5, so chi-square is valid.
Step 3 — Compute (O − E)² / E for each cell.
- (60 − 44)² / 44 = 256 / 44 ≈ 5.82
- (30 − 48)² / 48 = 324 / 48 = 6.75
- (30 − 28)² / 28 = 4 / 28 ≈ 0.14
- (50 − 66)² / 66 = 256 / 66 ≈ 3.88
- (90 − 72)² / 72 = 324 / 72 = 4.50
- (40 − 42)² / 42 = 4 / 42 ≈ 0.10
Step 4 — Sum the contributions.
χ² ≈ 5.82 + 6.75 + 0.14 + 3.88 + 4.50 + 0.10 = 21.19
Step 5 — Degrees of freedom and p-value. df = (2 − 1)(3 − 1) = 2. For χ² = 21.19 with 2 df, the p-value is far below 0.001 (the 0.05 critical value for df = 2 is 5.99).
Step 6 — Decision and conclusion. Reject H₀. There is statistically significant evidence at the 0.05 level that year and preferred study spot are associated. Looking at the largest contributions, Freshmen are over-represented at the Library and under-represented at the Café; Juniors are the mirror image. That is where the association comes from.
Conditions and Common Mistakes
The chi-square test of independence is valid when:
- The data are counts of subjects, not percentages or averages.
- The sample is random and the subjects are independent of each other.
- Every expected count is at least 5. This is the rule of thumb that keeps the chi-square distribution a good approximation. If any expected count is small, combine categories or use Fisher's exact test for a 2-by-2 table.
The most common mistakes are computational. Students subtract on the wrong cells, divide by O instead of E, or forget to compute the expected counts at all. Write expected counts in a separate table next to the observed table — keeping them visually separate cuts down on errors.
The other frequent slip is interpretation. A significant chi-square says the two variables are associated, not which categories drive it. To say "Freshmen prefer the Library," you have to inspect the cells with the largest (O − E)² / E contributions, as we did above.
Goodness-of-Fit Is a Different Test
The chi-square goodness-of-fit test compares one categorical variable to a hypothesized distribution (e.g., are M&M colors equally common?). It uses the same χ² formula but has df = k − 1, where k is the number of categories, and there is only one row of data. The test of independence always has at least two rows and two columns, and its df is (rows − 1)(columns − 1). Same statistic, different setups — make sure you know which one the problem is asking for.
Getting Help
Chi-square tests sit inside the same hypothesis-testing framework as everything else, so setting up a hypothesis test is the right place to refresh the structure of H₀, H₁, decision rule, and conclusion. For an example of a test on means rather than counts, the one-way ANOVA walkthrough covers a comparable multi-group setup.
Conclusion
A chi-square test of independence checks whether two categorical variables are related. Build the expected-counts table from row totals × column totals / grand total, sum (O − E)² / E across all cells, compare to a chi-square distribution with (rows − 1)(columns − 1) df. In the study-spot example χ² = 21.19 on 2 df, far past any reasonable cutoff, so year of study and preferred study spot are not independent in this sample. The whole test is mechanical once you can write the expected-counts table cleanly.
