Chi-square test calculator for goodness of fit and independence in a contingency table: χ², p-value, expected counts and Cramér's V.
By AdminDrafted with AI writing toolsUpdated Checked examples: 5
Fail to reject H₀ at α = 0.05. If every category were equally likely, a result at least this extreme would turn up with probability 0.5786. The p-value is not the probability that H₀ is true, and not rejecting H₀ does not show it is true: the sample may be too small to detect a real effect. The largest contribution comes from category 1 (observed 15, expected 20).
| Category | Observed | Expected | (O − E)²/E |
|---|---|---|---|
| 1 | 15 | 20 | 1.25 |
| 2 | 22 | 20 | 0.2 |
| 3 | 18 | 20 | 0.2 |
| 4 | 25 | 20 | 1.25 |
| 5 | 17 | 20 | 0.45 |
| 6 | 23 | 20 | 0.45 |
H₀: all category probabilities are equal. H₁: at least one differs.
Fail to reject H₀
Equivalently, compare χ² with the critical value 11.0705.
Pearson's chi-square test compares observed counts with the counts a hypothesis predicts: χ² = Σ(O − E)²/E. The goodness-of-fit test takes the expected counts from equal or given proportions and has k − 1 degrees of freedom, one fewer for each parameter estimated from the data. The test of independence sets E = row total × column total ÷ grand total for each cell of a contingency table and has (r − 1)(c − 1) degrees of freedom.
It answers questions such as whether a die is fair or whether a preference depends on region. The default 120 rolls (15, 22, 18, 25, 17, 23) give χ² = 3.8 on 5 df and p = 0.579, far below the 11.07 needed at α = 0.05, so the counts are consistent with a fair die.
The χ² approximation needs expected counts of about 5 or more. Cramér's V, from 0 to 1, measures how strong an association in a table is.
Pearson's chi-square compares observed counts O with expected counts E: χ² = Σ (O − E)² ÷ E. A fair die rolled 120 times expects 20 of each face, so counts of 15, 22, 18, 25, 17 and 23 give χ² = 3.8 on 5 degrees of freedom and p = 0.579: no evidence of bias.
Worked by hand:
Only the upper tail counts, because a large χ² means a poor fit. A fair die would give counts at least this uneven in about 58% of 120-roll experiments.
For goodness of fit, each expected count is the total times the category's hypothesized share: Eᵢ = N × pᵢ. Mendel's 556 pea plants, tested against a 9:3:3:1 ratio, expect 312.75, 104.25, 104.25 and 34.75.
The observed 315, 108, 101 and 32 give χ² = 0.470 on 3 df and p = 0.925, a close fit to the ratio.
For a test of independence, each cell expects E = row total × column total ÷ grand total (NIST).
Take a table with rows 42, 33, 25 and 28, 37, 35: the row totals are 100 and 100, the column totals 70, 70 and 60, and N = 200.
The first two columns then expect 100 × 70 ÷ 200 = 35 in each row, and the third 100 × 60 ÷ 200 = 30. The six contributions sum to χ² = 4.695 on 2 df, so p = 0.0956 and the rows and columns are consistent with independence at α = 0.05.
The 5% critical values for 1, 2, 3 and 5 df are 3.841, 5.991, 7.815 and 11.070 (NIST).
The p-value comes from a large-sample approximation that needs large enough expected counts. Cochran's (1954) rule asks that no expected count fall below 1 and no more than 20% fall below 5.
The rule concerns expected counts, not observed ones, so an observed 0 is fine when its expected count is 5 or more. When the rule fails, merge sparse categories or, for a 2×2 table, use Fisher's exact test.
The statistic is Pearson's without Yates's continuity correction. Each observation must fall in exactly one cell, and observations must be independent.
χ² grows in proportion to the sample size. A 60/40 split tested against equal shares gives:
Entering percentages treats every sample as exactly 100 observations. Expected ratios are different: they can be on any scale, so 3, 2, 5 and 0.3, 0.2, 0.5 give the same expected counts.
A small p-value says an association exists; Cramér's V says how strong it is, from 0 to 1: V = √(χ² ÷ (N × min(r − 1, c − 1))). The 2×2 table 20, 30 and 30, 20 has χ² = 4 with N = 100, so V = 0.2.
Cohen (1988) treats 0.1, 0.3 and 0.5 as small, medium and large for a 2×2 table. The 2×3 table above has V = 0.153. V shows no direction and says nothing about cause.
| Scenario | χ² statistic | p-value |
|---|---|---|
| Default | 3.800000 | 0.578555 |
| Two categories | 4.000000 | 0.045500 |
| Mendel | 0.470024 | 0.925426 |
| Independence | 4.695238 | 0.095597 |
| Two by two | 4.000000 | 0.045500 |
"Default" is the 120 die rolls and "Two categories" is 60 and 40 against equal shares. "Mendel" is the pea counts against 9:3:3:1, "Independence" the 2×3 table and "Two by two" the 20, 30 and 30, 20 table.
For chi-square tails and critical values at other degrees of freedom, use the t, chi-square and F distribution calculator. Means rather than counts belong in the t-test calculator or the one-way ANOVA calculator.
Checked against: χ² by hand: Σ(O − 20)²/20 = 76/20; p from the A&S 26.4.4 closed form for odd ν; χ²₀.₉₅,₅ = 11.070 (χ² table)
Checked against: Classic textbook example (χ² ≈ 0.47, p ≈ 0.93); exact χ² with exact fractions, p from A&S 26.4.4
Checked against: All E = 25, χ² = 4·25/25 = 4; p = erfc(√2) = 0.0455003; V = √(4/100)
Checked against: Expected counts R·C/N and χ² with exact fractions; p = e^(−χ²/2) for ν = 2 (A&S 26.4.5)
It is the probability of a χ² statistic at least as large as the one observed if the null hypothesis holds, such as a fair die or independent rows and columns. For the default rolls p = 0.579: a fair die would give counts at least this uneven in about 58% of 120-roll experiments. A large p-value does not prove the die fair, since small samples can miss a real bias.
Cochran's (1954) rule asks that no expected count be below 1 and no more than 20% of cells be below 5. When it fails, merge sparse categories or, for a 2×2 table, use Fisher's exact test. The rule is about expected counts, not observed ones: an observed 0 is fine when its expected count is 5 or more. The warning here appears whenever any expected count is below 5.
For goodness of fit, df = k − 1 − m, where k is the number of categories and m the number of parameters estimated from the data, so a six-sided die gives 5. For independence, df = (rows − 1) × (columns − 1): 2 for a 2×3 table and 1 for a 2×2 table. The 5% critical values for 1, 2 and 5 df are 3.841, 5.991 and 11.070.
Cramér's V = √(χ² / (N × (min(r, c) − 1))) rescales χ² to a 0-to-1 strength of association that does not grow with the sample size. The 2×2 worked example, χ² = 4 with N = 100, gives V = 0.2. Cohen (1988) treats 0.1, 0.3 and 0.5 as small, medium and large for a 2×2 table; tables with more rows and columns have lower thresholds.
No. χ² grows in proportion to the sample size, so the same percentages from 1,000 observations give 10 times the χ² of 100 observations. Entering percentages treats the sample as exactly 100 and gives the wrong p-value for any other sample size. Expected ratios, by contrast, can be on any scale: 9 3 3 1 is rescaled to the observed total.
This calculator includes 5 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.