Chi-Square Goodness of Fit Test

What is the chi-square goodness of fit test?

The $\chi^2$ test checks whether observed frequencies match expected frequencies. Is the die fair? Are the survey responses distributed as expected?

The formula

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

where $O_i$ = observed count and $E_i$ = expected count for each category.

Step-by-step: Is this die fair?

Data: 120 rolls. Observed: $\{18, 22, 16, 20, 24, 20\}$. If fair, expected: $120/6 = 20$ for each face.

Step 1 — State hypotheses.

$H_0$: Die is fair (all faces equally likely). $H_a$: Die is not fair.

Step 2 — Compute $\chi^2$:

$\chi^2 = \frac{(18-20)^2}{20} + \frac{(22-20)^2}{20} + \frac{(16-20)^2}{20} + \frac{(20-20)^2}{20} + \frac{(24-20)^2}{20} + \frac{(20-20)^2}{20}$

$= \frac{4}{20} + \frac{4}{20} + \frac{16}{20} + \frac{0}{20} + \frac{16}{20} + \frac{0}{20} = 0.2 + 0.2 + 0.8 + 0 + 0.8 + 0 = 2.0$

Step 3 — Degrees of freedom: $df = \text{categories} - 1 = 6 - 1 = 5$.

Step 4 — Find the p-value. From a $\chi^2$ table with $df = 5$: $\chi^2 = 2.0$ gives $p \approx 0.85$.

Step 5 — Decision. $p = 0.85 > 0.05$. Fail to reject $H_0$. No evidence the die is unfair — the deviations are within normal random variation.

Interpreting the $\chi^2$ value

• Large $\chi^2$ (small p-value): observed is far from expected → reject $H_0$.
• Small $\chi^2$ (large p-value): observed is close to expected → fail to reject.

Try it in the visualization

Adjust the observed counts for each face. The bar chart compares observed vs expected. $\chi^2$ and the p-value update live. Adjust counts to make the die obviously unfair and see $\chi^2$ spike.