Chi-square test calculator for goodness of fit and independence in a contingency table: χ², p-value, expected counts and Cramér's V.
آخر تحديث أمثلة تم التحقق منها: 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.
مصدر التحقق: 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)
مصدر التحقق: Classic textbook example (χ² ≈ 0.47, p ≈ 0.93); exact χ² with Python fractions, p from A&S 26.4.4
مصدر التحقق: All E = 25, χ² = 4·25/25 = 4; p = erfc(√2) = 0.0455003 (Python math.erfc); V = √(4/100)
مصدر التحقق: 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.
تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 5. مثلًا، يجري التحقق من «Is the die fair? (defaults)» بالرجوع إلى 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).
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).
تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 5. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.
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