Chi-Square Distribution Calculator

The chi-square distribution is a family of right-skewed probability distributions, one for each number of degrees of freedom. It shows up constantly in statistics: goodness-of-fit tests, tests of independence in contingency tables, and variance inference all rely on it.

The distribution is defined only for non-negative values. With small degrees of freedom it is heavily right-skewed; as degrees of freedom grow it approaches a normal distribution, centered around df and with variance 2·df.

This calculator takes a chi-square statistic and degrees of freedom, then returns the right-tail p-value — the probability of observing a chi-square value at least as large as yours, assuming the null hypothesis is true. A p-value below 0.05 is conventionally “statistically significant.”

The p-value is computed using the Wilson-Hilferty approximation, which converts a chi-square quantile to a standard normal quantile:

z = [(χ²/df)^(1/3) - (1 - 2/(9·df))] / sqrt(2/(9·df))

p-value = 1 - Φ(z), where Φ is the standard normal CDF.

This approximation is accurate to three decimal places for df ≥ 2 and reasonable chi-square values, which covers virtually all practical hypothesis tests.

Common critical values for reference:

• df=1: χ² = 3.84 gives p = 0.05; χ² = 6.63 gives p = 0.01
• df=2: χ² = 5.99 gives p = 0.05; χ² = 9.21 gives p = 0.01
• df=5: χ² = 11.07 gives p = 0.05; χ² = 15.09 gives p = 0.01

The chart plots the chi-square PDF for your chosen degrees of freedom, shading the right-tail region that corresponds to the p-value.

Computing the χ² statistic from data

If you have observed counts and need to compute the chi-square statistic itself before entering it above, the formula is:

χ² = Σ (O − E)² / E

where O is the observed count in each category, E is the expected count under the null hypothesis, and the sum runs over all categories. Degrees of freedom for a goodness-of-fit test = (number of categories − 1).

Worked example — is a die fair? A die is rolled 60 times. Under a fair-die null hypothesis, each face is expected 10 times. Observed counts: 8, 12, 11, 7, 13, 9.

• Each contribution: (8−10)²/10 = 0.4; (12−10)²/10 = 0.4; (11−10)²/10 = 0.1; (7−10)²/10 = 0.9; (13−10)²/10 = 0.9; (9−10)²/10 = 0.1
• Sum: χ² = 2.8
• df = 6 − 1 = 5
• The critical value at α = 0.05 for df = 5 is 11.07; 2.8 is well below, so we fail to reject the null. The die looks fair.

A second example — survey preferences. 200 people are asked their favourite season; the null hypothesis is equal preference (50 each). Observed: Spring 65, Summer 55, Autumn 45, Winter 35.

• χ² = 15²/50 + 5²/50 + (−5)²/50 + (−15)²/50 = 4.5 + 0.5 + 0.5 + 4.5 = 10.0
• df = 4 − 1 = 3; critical value at α = 0.05 is 7.815
• χ² = 10.0 > 7.815, so we reject the null. There is a real seasonal preference.

Rule of thumb that catches a lot of bad chi-square work: each expected count E should be at least 5. With smaller expected counts the chi-square approximation breaks down and a Fisher exact test is the better tool.