Chi-square test calculator for goodness of fit and independence in a contingency table: χ², p-value, expected counts and Cramér's V.
更新日 検証済みの例：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 |
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.
照合元：χ² by hand: Σ(O − 20)²/20 = 76/20; p from the A&S 26.4.4 closed form for odd ν (pyref.chi2_sf_int); χ²₀.₉₅,₅ = 11.070 (χ² table)
照合元：Classic textbook example (χ² ≈ 0.47, p ≈ 0.93); exact χ² with Python fractions, p from A&S 26.4.4
照合元：All E = 25, χ² = 4·25/25 = 4; p = erfc(√2) = 0.0455003 (Python math.erfc); V = √(4/100)
照合元：Expected counts R·C/N and χ² with Python 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.
精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。 独立した出典の解答と照合した計算例：5。 例えば、「Is the die fair? (defaults)」はχ² by hand: Σ(O − 20)²/20 = 76/20; p from the A&S 26.4.4 closed form for odd ν (pyref.chi2_sf_int); χ²₀.₉₅,₅ = 11.070 (χ² table)と照合しています。
NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.5.15 Chi-square goodness-of-fit test; NIST/SEMATECH e-Handbook, §7.4.5 contingency tables / test of independence; Abramowitz & Stegun, Handbook of Mathematical Functions, §26.4 (χ² probability function).
この計算機には、独立した出典の解答を使った計算例が 5 件あります。テストに組み込まれており、ここでも実行できます。
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