Hypothesis tests

Chi-Square Test Calculator

Test whether two categorical variables are associated, or whether observed counts match the distribution you expected.

Also called chi square calculator, chi-squared test calculator, Pearson's chi-square calculator.

Updated August 2026Runs in your browser — nothing is uploadedVerified against R

Rows are one variable, columns are the other. The test asks whether the row distribution changes from column to column.

Paste the table of countsOne row per line, cells separated by tabs, commas or spaces. A word at the start of a line is used as that row's label. Any size, 2×2 upward.

Significance level αConventionally 0.05. Decide before you look at the data.

Apply Yates's continuity correction (2×2 only)

Makes a 2×2 test more conservative. Modern practice is to leave it off and use Fisher's exact test when counts are small.

The calculation runs in your browser, so this box needs JavaScript. The formula, the worked example and the interpretation below do not.

How to read this result

χ² adds up how far each cell sits from what independence predicts, scaled by how large that cell was expected to be. It is always positive and always upper-tailed: any departure in any direction makes it larger, which is why there is no two-tailed version to choose.

A significant result says the variables are associated. It does not say how, or how strongly. That is what the standardised residuals table is for: cells beyond about ±2 are the ones driving the result, and their sign tells you whether that combination occurred more or less often than independence predicts.

Cramér's V is the effect size. χ² grows with the sample size — double every count and χ² doubles while the pattern is unchanged. V divides that out, giving a number between 0 and 1 that is comparable across studies. Around 0.1 is a weak association, 0.3 moderate, 0.5 strong.

Watch the expected counts, not the observed ones. The approximation behind the p-value needs expected counts of about 5 or more. Small observed counts are fine; small expected counts are not, and for a 2×2 table the fix is Fisher's exact test, which needs no approximation at all.

The formula

χ^{2} = \sum \frac{(O - E)^{2}}{E} E_{i j} = \frac{(row i total) (column j total)}{n}

$O$: the observed count in a cell
$E$: the count expected if the two variables were independent
$n$: the total number of observations

Degrees of freedom: (rows − 1)(columns − 1) for a test of independence, categories − 1 for goodness of fit.

Effect size

V = \sqrt{\frac{χ^{2}}{n \cdot \min (r - 1, c - 1)}}

Cramér's V. For a 2×2 table it is the phi coefficient, and it equals the absolute correlation between the two binary variables.

Worked example

Does a training course change pass rates?

One hundred candidates, half of whom took a preparation course. Of the 50 who took it, 35 passed; of the 50 who did not, 20 passed. That looks like a large difference — but with 50 people per group, how large a difference does chance produce?

	passed	failed
course	35	15
no course	20	30

Work out the margins.
rows 50 and 50; columns 55 passed and 45 failed; n = 100
Expected count for each cell: row total × column total ÷ n.
course/passed = 50 × 55 / 100 = 27.5; course/failed = 22.5; and the same for the second row
For each cell, square the gap and divide by the expected count.
(35 − 27.5)² / 27.5 = 56.25 / 27.5 = 2.045 (15 − 22.5)² / 22.5 = 2.500
Add all four terms. By symmetry the second row contributes the same as the first.
χ² = 2 × (2.045 + 2.500) = 9.091
Degrees of freedom: (2 − 1)(2 − 1) = 1. Look χ² up in the upper tail.
p = 0.00257
Scale it into an effect size.
V = √(9.091 / (100 × 1)) = 0.302

χ² 9.09df 1p 0.0026Cramér's V 0.30

The pass rate is 70% with the course and 40% without. The test says that gap is unlikely to be chance; it says nothing about whether the course caused it, since the people who chose to take it may differ in other ways.

Checked against R's chisq.test(matrix, correct = FALSE).

The calculator above is loaded with these numbers by the Load the worked example button.

Assumptions, and when to use something else

Counts, not percentages or means. Feeding percentages into a chi-square test produces a number that depends on whether you typed 70 or 0.70, which is a clear sign it is meaningless.
Each observation appears in exactly one cell. One hundred people in a 2×2 table means one hundred counts in total. Repeated measurements of the same people belong in McNemar's test.
Expected counts of about 5 or more. The classic guidance is that no expected count should be below 5; the modern, more relaxed version allows up to 20% of cells below 5 provided none is below 1.
Independent observations. Clustered sampling — several members of the same household, several measurements from the same machine — breaks the test in a way that makes the p-value too small.

A 2×2 table with small countsFisher's exact testComputes the exact probability by enumerating every table with the same margins. No minimum expected count, no approximation.
The same subjects measured twiceMcNemar's testPaired yes/no data — before and after, two raters on the same items. Chi-square would treat the pairs as independent and get it wrong.
Comparing two proportions and you want a direction and an intervalz-test for proportionsGives the same p-value for a 2×2 table plus the difference in proportions and a confidence interval for it.
You want the strength of the associationOdds ratioFor a 2×2 table, the odds ratio with its confidence interval says how much more likely the outcome is in one group.

Questions people ask

What is the difference between a chi-square test of independence and goodness of fit?

Independence tests one sample cross-classified by two variables — are they related? Goodness of fit tests one variable against a distribution you specify — do these counts match what I expected? The arithmetic is nearly identical; the degrees of freedom differ, and so does the question.

What if my expected counts are below 5?

For a 2×2 table, use Fisher's exact test, which is exact and has no such requirement. For a larger table, combine sparse categories into meaningful groups — but combine them for a reason, not to make the number work.

Should I use Yates's continuity correction?

Usually not. It was designed to make the continuous chi-square distribution better approximate discrete counts in a 2×2 table, and it overcorrects: the resulting test is noticeably conservative. If the counts are small enough for the correction to matter, they are small enough to warrant Fisher's exact test instead.

Can chi-square tell me the direction of the association?

Not on its own — χ² is always positive and symmetric in that sense. The standardised residuals table shows which cells are over- and under-represented, and for a 2×2 table the odds ratio gives the direction and the magnitude together.

How many rows and columns can I use?

Any number. Paste a 2×2 or a 7×5; the calculator works out the degrees of freedom and the expected counts from the shape of what you paste.