Chi-Square Goodness-of-Fit Test
Do observed counts match a claimed distribution? (χ²GOF-Test)
Results
Fail to reject H₀: χ² = 4.32, p-value = 0.2289
The p-value (0.2289) is greater than α = 0.05, so the data do not give enough evidence to conclude that the population does not follow the claimed distribution.
Results
| χ² test statistic | 4.32 |
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| df = k − 1 | 3 |
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| p-value | 0.2289 |
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| Critical value | 7.815 |
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| Decision at α = 0.05 | Fail to reject H₀ |
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On the TI-84
χ²GOF-Test L1, L2, 3
Show steps
- Hypotheses. H₀: the population follows the claimed distribution Ha: it does not
- Expected counts. E = n × (claimed proportion) for each category: 25, 25, 25, 25.
- Test statistic. χ² = Σ (O − E)² ÷ E = 4.32, with df = 4 − 1 = 3.
- p-value. Right-tailed test: area to the right of the test statistic: p-value = 0.2289. Since 0.2289 > 0.05, fail to reject H₀.
- Critical-value method. At α = 0.05, the rejection region is χ² ≥ 7.815. The test statistic χ² = 4.32 is not in the rejection region, so we fail to reject H₀. (Same decision as the p-value method.)
- Conclusion. The p-value (0.2289) is greater than α = 0.05, so the data do not give enough evidence to conclude that the population does not follow the claimed distribution.
About this calculator
When to use this
A goodness-of-fit test checks whether the counts in one categorical variable match a claimed distribution, such as "a die is fair" or "customers are spread evenly across the week."
How it works
For each category, compare the observed count O with the count you'd expect, E, if the claim were true. The statistic χ² = Σ(O − E)²/E is small when they match and large when they don't, so the test is always right-tailed. Every expected count should be at least 5.
More chi-square & anova calculators: χ² Independence · One-Way ANOVA