Goodness of Fit Calculators
0 calculators tagged with “Goodness of Fit”
All Calculators
No calculators found for this topic.
Chi-Square Goodness of Fit Formula
χ² = Σ [(O − E)² / E]
O = observed count; E = expected count in each category. Degrees of freedom = number of categories − 1.
Example: 100 pea plants expected in 3:1 ratio (75 purple : 25 white), but 68 purple and 32 white observed. χ² = (68−75)²/75 + (32−25)²/25 = 0.653 + 1.96 = 2.613. df = 1. Critical value at α = 0.05 is 3.841. Since 2.613 < 3.841, fail to reject H₀ — data are consistent with 3:1.
Interpreting P-values
- p > 0.05: Data fit the model — fail to reject H₀
- p 0.01–0.05: Marginal evidence against the model
- p < 0.01: Significant departure — model may not apply
- p < 0.001: Strong evidence — model is rejected
Key Assumptions
All expected counts must be ≥ 5. Observations must be independent. Data must be counts, not proportions or continuous values. For small expected counts, merge categories or use Fisher's exact test.
Other Goodness of Fit Tests
Kolmogorov-Smirnov (KS): for continuous distributions. Shapiro-Wilk: specifically for normality. Anderson-Darling: more sensitive to tail deviations than KS.
Glossary
Frequently Asked Questions
Step 1: State null hypothesis (data fit the expected model). Step 2: Calculate expected counts E = N × expected proportion per category. Step 3: Compute χ² = Σ[(O − E)² / E]. Step 4: df = number of categories − 1. Step 5: Compare χ² to critical value from table (or compute p-value). If χ² > critical value (or p < 0.05), reject H₀. Verify all expected counts ≥ 5 before running the test.
In genetics, chi-square goodness of fit tests whether observed offspring phenotype ratios match Mendelian predictions. Example: cross Aa × Aa, observe 310 dominant and 90 recessive out of 400. Expected: 300 and 100 (3:1 ratio). χ² = (310−300)²/300 + (90−100)²/100 = 0.333 + 1.0 = 1.333. df = 1, p > 0.05 — data are consistent with Mendelian 3:1 ratio. This is routinely used in undergraduate genetics labs.
All expected counts (E) must be ≥ 5 for the chi-square test to be valid. When expected counts are smaller, the chi-square distribution does not adequately approximate the true sampling distribution, inflating Type I error. Solutions: merge low-frequency categories until each has E ≥ 5; use Fisher's exact test for 2×2 tables; or use a simulation (permutation) test for larger tables with small counts.
Goodness of fit tests whether one categorical variable matches a specified theoretical distribution (e.g., 3:1 phenotype ratio). Test of independence uses a two-way contingency table to test whether two categorical variables are associated. Both use χ² = Σ[(O−E)²/E], but expected values differ: from the theoretical model for GoF; from row × column marginals for independence. Degrees of freedom also differ: (categories − 1) for GoF; (rows − 1)(columns − 1) for independence.