ChiTest

Chi-Square Calculator

χ² = Σ((observed − expected)² / expected).

Observed values

Expected values

χ² statistic

2.9

degrees of freedom: 3

What you entered

Categories: 4

1
Category 1: (observed − expected)² ÷ expected
(18 − 20)² ÷ 20= 0.2
2
Category 2: (observed − expected)² ÷ expected
(22 − 20)² ÷ 20= 0.2
3
Category 3: (observed − expected)² ÷ expected
(15 − 20)² ÷ 20= 1.25
4
Category 4: (observed − expected)² ÷ expected
(25 − 20)² ÷ 20= 1.25
5
Sum across all categories
0.2 + 0.2 + 1.25 + 1.25= 2.9

Result

χ² statistic: 2.9

χ² = 2.9 with 3 degrees of freedom — the larger this value, the more your observed counts diverge from what was expected.

Chi-square statistic for goodness of fit

Chi-square sums the squared difference between observed and expected counts, divided by the expected count, across every category.

Observed	Expected	Chi-square	Degrees of freedom
45, 55	50, 50	1.000	1
20, 30, 50	25, 25, 50	2.000	2
10, 20, 30, 40	25, 25, 25, 25	20.000	3

Degrees of freedom is the number of categories minus 1. Compare the statistic against the critical value for that df: at 0.05 significance the thresholds are 3.84 for 1 df, 5.99 for 2 and 7.81 for 3. On that basis the first two rows are consistent with the expected distribution and the third, at 20.000 against a threshold of 7.81, is clearly not. Chi-square needs reasonably large expected counts in every category - the usual guidance is at least 5 - and it works on raw counts, never on percentages or proportions.

Measuring how 'surprised' the data is

Chi-square sums up, category by category, how far your actual observed counts deviate from what you'd expect if there were no real effect — the bigger the number, the less likely the difference is due to random chance alone.

Where it's used

A/B testing, survey analysis, and checking whether a die or coin is fair all use chi-square to test whether an observed pattern could plausibly be random noise.

Frequently asked questions

I rolled a die 60 times and got: 8, 12, 9, 11, 10, 10 for faces 1–6. Is the die fair?

Expected: 10 each. χ² = (8−10)²/10 + (12−10)²/10 + … = 0.4+0.4+0.1+0.1+0+0 = 1.0. With 5 degrees of freedom, the critical value at p=0.05 is 11.07. Since 1.0 < 11.07, there's no evidence the die is unfair — the variation is normal randomness.

How is chi-square different from a t-test?

Chi-square tests categorical/count data (did more people choose A vs B?). T-tests compare numerical means (is group A's average score higher than group B's?). Use chi-square for frequencies and proportions; use t-tests for measurements and averages.

What are degrees of freedom in a chi-square test?

df = (number of categories − 1) for a goodness-of-fit test. For a contingency table: df = (rows − 1) × (columns − 1). More degrees of freedom shift the critical value higher, meaning you need a larger χ² to reach significance.

My website A/B test shows 45/500 vs 55/500 conversions. Is this significant?

Expected: 50 each. χ² = (45−50)²/50 + (55−50)²/50 + (455−450)²/450 + (445−450)²/450 = 0.5+0.5+0.056+0.056 = 1.11. With df=1, critical value at p=0.05 is 3.84. Since 1.11 < 3.84, the difference is NOT statistically significant — you need more data.