Chi-Squared Statistic Calculator

Test statistic

For k categories with observed counts O_i and expected counts E_i (under the null hypothesis), Pearson's chi-squared statistic is:

\chi^2 = Σ_i=1^k ((O_i - E_i)^2)/(E_i)

Degrees of freedom

For a goodness-of-fit test of k categories with r parameters estimated from the data:

\mathrmdf = k - r - 1

Uniform null hypothesis

If you have no specific expectation, the simplest null hypothesis is uniform: every category is expected to have the same count — i.e. the mean of the observed counts. Then each E_i = (Σ O_i)/k.

Decision rule

Compare the statistic to a chi-squared distribution with df degrees of freedom. At a 5% significance level, reject the null if χ² exceeds the upper-95-pp quantile. Smaller p-values mean stronger evidence against the null.

Caveats

Each expected count should typically be ≥ 5 for the chi-squared approximation to be valid. For smaller expectations, use Fisher's exact test or combine categories.

Reading the p-Value

The p-value is the probability of seeing a statistic at least as large as the one observed *if the null hypothesis were true*. It is not the probability that the null hypothesis is correct, and it says nothing about the size of any effect — a very large sample can make a trivial deviation statistically significant.

The conventional threshold is α = 0.05, so p below 0.05 leads to rejecting the null. With three degrees of freedom the corresponding critical value is 7.815: any χ² above that is significant at the 5% level. For reference, the 5% critical values are 3.841 for df = 1, 5.991 for df = 2, 9.488 for df = 4 and 11.070 for df = 5. Failing to reject is not evidence that the null is true — only that this data does not contradict it.

p = P(\chi^2_df > \chi^2_\textobserved)

Degrees of Freedom

For a goodness-of-fit test across k categories, df = k - 1. One degree is lost because the expected counts are constrained to sum to the same total as the observed counts: once any k-1 of the deviations are known, the last is determined.

If the expected distribution has parameters estimated from the same data, subtract one further degree for each. Fitting a Poisson distribution and estimating its mean from the data, for example, gives df = k - 2. For a contingency table testing independence between a variable with r rows and one with c columns, df = (r-1)(c-1). Getting df wrong changes the p-value substantially — with χ² = 7.6, the p-value is 0.006 at df = 1 but 0.107 at df = 4.

\textdf = k - 1 - (\textparameters estimated)

Assumptions and When Not to Use It

Four conditions must hold. The observations must be counts of independent events, not measurements or percentages — running the test on proportions instead of raw frequencies is the single most common misuse. The categories must be mutually exclusive and cover every observation. Each expected count should be at least 5. And the sample must be reasonably large, since the chi-squared distribution is only an asymptotic approximation to the true discrete sampling distribution.

When expected counts fall below 5, the options are to merge adjacent categories, apply Yates's continuity correction for a 2×2 table, or switch to Fisher's exact test, which computes the exact probability rather than approximating it. At the opposite extreme, with very large samples almost any departure becomes significant, so report an effect size — Cramér's V for contingency tables — alongside the p-value rather than relying on significance alone.

Worked Example

Inputs

• o1: 10
• o2: 12
• o3: 8
• o4: 15
• uniformExpected: 1

Result: chiSquare: 2.3778 total: 45 expected: 11.25 degreesOfFreedom: 3 pValue: 0.4978

Explanation

Total = 45; expected per category (uniform) = 45/4 = 11.25. χ² = ((10−11.25)²/11.25) + ((12−11.25)²/11.25) + ((8−11.25)²/11.25) + ((15−11.25)²/11.25) = (1.5625 + 0.5625 + 10.5625 + 14.0625)/11.25 = 26.75/11.25 ≈ 2.3778. With df = k−1 = 3, the 5% critical value is 7.815 and p ≈ 0.498 — far from significant, so there is no evidence against the uniform expectation. Category 4 (observed 15 against 11.25 expected) contributes the most to the statistic, but not nearly enough to matter.

Second Scenario

Inputs

• o1: 13.5
• o2: 12
• o3: 8
• o4: 15
• uniformExpected: 1

Result: chiSquare: 2.3778 total: 45 expected: 11.25 degreesOfFreedom: 3 pValue: 0.4978

Explanation

This scenario uses different inputs (o1 = 13.5, o2 = 12, o3 = 8, o4 = 15, uniformExpected = 1) to show how changing one variable affects the chi-squared statistic result. Run the calculator above with these values to get the exact updated output with step-by-step work.