Chi Square Test of Independence Calculator

Hypothesis Test

Chi-Square Test of Independence

Enter your observed counts in the contingency table below, then calculate.

Row Variable Name

Column Variable Name

Observed Frequencies

Significance Level (α)

Categorical Data Analysis

What Is a Chi-Square Test of Independence?

The chi-square test of independence determines whether two categorical variables are statistically associated or are independent of one another. It works by comparing the frequencies you actually observed in a contingency table to the frequencies you would expect to see if the two variables had no relationship at all. A large gap between observed and expected counts points to a real association; a small gap is consistent with independence.

When to Use It

You have two categorical variables measured on the same set of subjects (e.g., gender and product preference).
Your data is organized as counts in a contingency table, not percentages or continuous measurements.
Each subject or observation falls into exactly one cell of the table.
You want to know if the variables are related, not how strongly one predicts a numeric outcome.
Expected cell counts are large enough for the approximation to hold — generally at least 5 in most cells.

The Formula

\[ \chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \] \[ E_{ij} = \frac{(\text{Row}_i\text{ Total}) \times (\text{Column}_j\text{ Total})}{\text{Grand Total}} \] \( df = (r - 1)(c - 1) \)

\(O_{ij}\) is the observed count in row \(i\), column \(j\); \(E_{ij}\) is the count expected under independence; \(r\) and \(c\) are the number of rows and columns.

Worked Example

A researcher surveys 100 students on gender and their preferred study resource:

	Textbook	Video	Study Group
Male	15	25	10
Female	20	10	20

Row totals: 50, 50. Column totals: 35, 35, 30. Grand total: 100.
Expected counts, e.g. \(E_{11} = 50 \times 35 / 100 = 17.5\); computed the same way for all 6 cells (17.5, 17.5, 15, 17.5, 17.5, 15).
Chi-square contributions: 0.3571 + 3.2143 + 1.6667 + 0.3571 + 3.2143 + 1.6667 = 10.4762.
Degrees of freedom: \(df = (2-1)(3-1) = 2\).
Critical value at \(\alpha = 0.05\), df = 2, is 5.9915. Since 10.4762 > 5.9915 (p = 0.0053), we reject the null hypothesis and conclude gender and preferred study resource are associated.

Need only a critical value for a given df and α, without running a full test? Use our Chi-Square Critical Value Calculator. For very small samples where expected counts are too low for this test to be reliable, statisticians often turn to Fisher's Exact Test instead.

Your Results

Chi-Square (χ²)

p-value

Critical Value

Total N

Cramer's V

Decision

Step 1 — Row, Column, and Grand Totals

Step 2 — Expected Frequencies

Step 3 — Chi-Square Contribution per Cell

Step 4 — Sum the Contributions

Step 5 — Degrees of Freedom

Step 6 — Compare to the Critical Value

Chi-square distribution Rejection region Test statistic Critical value

Chi Square Test of Independence Calculator