Chi-Square Calculator

What Is a Chi-Square Test?

A chi-square test compares observed categorical frequencies with the frequencies expected under a specified null hypothesis. The bigger the gaps between observed and expected counts, relative to the expected counts, the larger the χ² statistic and the smaller the p-value.

The two main uses are the goodness-of-fit test, which asks whether one categorical variable follows a specified distribution (is this die fair?), and the test of independence, which asks whether two categorical variables are associated (does product preference differ by education level?). Both use χ² = Σ (O − E)² / E, but they get the expected counts and the degrees of freedom differently, and they answer different questions.

How to Calculate a Chi-Square Goodness-of-Fit Test

1. List the categories with their observed counts O.

2. Work out the expected count E for each category from the hypothesized distribution: equal counts, given percentages, or a model.

3. For each category compute (O − E)² / E.

4. Add them up to get χ².

5. Degrees of freedom: df = k − 1 for k categories, minus one more for every parameter estimated from the data.

6. Find the p-value, the right-tail area of the chi-square distribution beyond χ², and compare it with α.

Example: a die rolled 120 times gives 15, 22, 18, 25, 16 and 24 for faces 1 to 6. If the die is fair, each face is expected 20 times. χ² = (25 + 4 + 4 + 25 + 16 + 16) / 20 = 4.5 with df = 5, so p ≈ 0.480. At α = 0.05 we fail to reject H₀: the counts are consistent with a fair die (which is not the same as proving it is fair).

How to Calculate a Chi-Square Test of Independence

1. Put the counts in an r × c contingency table and add the row totals, column totals and grand total n.

2. For each cell, the expected count is E = (row total × column total) / n.

3. Compute (O − E)² / E for every cell and add them to get χ².

4. Degrees of freedom: df = (r − 1)(c − 1).

5. Find the p-value and compare it with α.

Example: 50 people in Group A and 50 in Group B; 30 and 15 of them purchased. The column totals are 45 purchased and 55 did not, so each group's expected counts are 50 × 45 / 100 = 22.5 and 27.5. χ² = 2 × 7.5² / 22.5 + 2 × 7.5² / 27.5 ≈ 9.09 with df = 1, p ≈ 0.0026. At α = 0.05 we reject H₀: the data provide evidence that purchasing is associated with the group.

How to Interpret the Chi-Square p-Value

The p-value is the probability of getting a χ² at least as large as yours if the null hypothesis were true. It is always a right-tail probability, because any difference between observed and expected counts, in either direction, makes χ² bigger.

If p < α, reject H₀: the result is statistically significant at that level. If p ≥ α, fail to reject H₀: the data do not provide enough evidence against it. Failing to reject is not proof that H₀ is true; a small sample may simply be unable to detect a difference.

Statistical significance is not practical importance. With thousands of observations a trivial difference can be significant. For a test of independence, Cramér's V (from 0 to 1) describes the strength of the association separately from the p-value.

When to Use a Chi-Square Test

Goodness-of-fit: comparing observed category counts with an expected distribution, such as whether customer visits are spread evenly across weekdays, whether a die or spinner is fair, or whether offspring follow a 9:3:3:1 genetic ratio.

Test of independence: comparing how category membership is distributed across groups, such as whether preferred product tier differs by education level, or whether the purchase rate differs between two website designs.

Chi-square tests are not for continuous measurements, means or correlations between numbers. To compare two numerical variables, use correlation or regression instead. Each observation must be independent and counted in exactly one cell; the same person counted twice, or before-and-after measurements on the same people, need other methods.

Small Expected Counts

Chi-square tests rely on an approximation that may perform poorly with sparse expected counts. A widely taught rule of thumb asks for expected counts of at least 5 in every cell; some texts accept up to 20% of cells below 5 as long as none is below 1. These are guidelines, not laws.

When expected counts are small, combining categories can help if the combined categories still make sense for your question; do not merge categories just to force a test to run. For 2 × 2 tables, Fisher's exact test calculates the p-value exactly instead. In the small-counts example (7 and 1 improved vs not improved in treatment, 2 and 5 in control), every expected count is below 5; the chi-square p-value is about 0.020 while Fisher's exact test gives about 0.041.

Yates' Correction and Fisher's Exact Test

Yates' continuity correction, for 2 × 2 tables only, subtracts 0.5 from each |O − E| before squaring. It makes the test more conservative (a smaller χ², a larger p-value) to compensate for approximating discrete counts with a continuous distribution; in the purchase example it changes χ² from 9.09 to 7.92 and p from 0.0026 to 0.0049. Many statisticians consider it overly conservative, so it is off by default here.

Fisher's exact test computes the exact probability of all 2 × 2 tables with the same row and column totals that are as extreme as, or more extreme than, the observed one. It does not depend on large expected counts. The calculator shows it alongside the chi-square result when you ask for it, never as a silent replacement.