What Is a Chi-Square Test and How Do You Use It?

If you work with categorical data—like pass/fail counts, defect categories, or survey responses—you often need to know whether what you observe matches what you expect. The chi-square test (χ² test) is the standard statistical tool for exactly that. It tells you whether the differences between observed and expected frequencies are due to chance or something real.

What It Is

The chi-square test is a nonparametric hypothesis test used with count data. It comes in two common forms:

• Goodness-of-fit test: Checks whether a sample distribution matches a hypothesized distribution (e.g., "Are defects evenly spread across shifts")
• Test of independence: Checks whether two categorical variables are related (e.g., "Is product failure independent of the production line")

The test compares observed counts (O) with expected counts (E) under the null hypothesis. If the gap is too large, you reject the null hypothesis.

How It Works

The core formula is:

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

Where:

• \(O\) = observed frequency in a category
• \(E\) = expected frequency in that category (calculated from the hypothesis or from row/column totals)
• Sum over all categories or cells

After calculating χ², you compare it to a critical value from the chi-square distribution with the appropriate degrees of freedom (df):

• Goodness-of-fit: \(df = k - 1\) (k = number of categories)
• Independence: \(df = (r - 1)(c - 1)\) (r = rows, c = columns)

You also need the p-value. For a 95% confidence level, the typical significance threshold is α = 0.05. If the p-value is less than 0.05, you reject the null hypothesis.

Key assumption: Each expected frequency should be at least 5 (or at least 80% of cells should meet this for larger tables).

A Worked Illustrative Example

Example data (illustrative only): A factory inspects 200 parts from three shifts. The hypothesis is that defects are evenly distributed (33.3% per shift).

Shift | Observed defects (O) | Expected defects (E)

• Day | 30 | 20
• Evening | 15 | 20
• Night | 15 | 20

Calculate:

\[
\chi^2 = \frac{(30-20)^2}{20} + \frac{(15-20)^2}{20} + \frac{(15-20)^2}{20} = 5 + 1.25 + 1.25 = 7.5
\]

Degrees of freedom: \(df = 3 - 1 = 2\). Using a chi-square table, the critical value for df = 2 at α = 0.05 is 5.991. Since 7.5 > 5.991, we reject the null hypothesis—the defect distribution is not even across shifts.

Common Pitfalls

• Using counts, not percentages or averages. The test requires raw frequencies.
• Ignoring the expected frequency rule. Small expected values inflate χ² and invalidate results.
• Confusing correlation with causation. A significant independence test shows association, not cause.
• Forgetting to state hypotheses clearly. Always define H₀ and H₁ before running the test.

Get the Answer Fast

Manual calculation is fine for small tables, but for real-world data you want speed and accuracy. Use the free chi-square test calculator at https://www.6sq.com/tools/chi_square/ to compute χ², degrees of freedom, and the p-value in seconds—so you can focus on the decision, not the arithmetic.