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Lesson 108 — Chi-squared test: goodness of fit and independence

Chi-squared statistics: asymptotic distribution, degrees of freedom, goodness-of-fit test and independence test in contingency tables. Yates correction, Cramér's V.

Used in: 3rd year High School · Stochastik LK German · H2 Statistics Singapore · Math B Japanese — inferential statistics

\chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}

The chi-squared statistic measures the total discrepancy between observed counts $O_i$ and expected counts $E_i$ under $H_0$ . The larger $\chi^2$ , the more evidence against the null hypothesis. For goodness of fit, it tests whether data follow a prescribed distribution; for independence, it tests whether two categorical variables are independent in an $r \times c$ table.

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Rigorous notation, full derivation, hypotheses

Rigorous definition

Chi-squared distribution

"Chi-squared distributions have an additivity property: if $X_1 \sim \chi^2_{n_1}$ and $X_2 \sim \chi^2_{n_2}$ are independent, then $X_1 + X_2 \sim \chi^2_{n_1 + n_2}$ ." — OpenStax Statistics, §11.1

Goodness-of-fit test

Definition· Hypotheses and goodness-of-fit statistic

Given a sample of $n$ observations categorized into $k$ classes:

Hypotheses: $H_0: p_i = p_i^0 \quad (i = 1, \ldots, k) \qquad \text{vs.} \qquad H_1: \text{at least one } p_i \neq p_i^0$

Expected frequencies: $E_i = n \, p_i^0$ .

Statistic: $\chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}$

Degrees of freedom: $df = k - 1 - r$ , where $r$ is the number of parameters estimated from the data (typically $r = 0$ for fully specified distributions).

Under $H_0$ , $\chi^2 \xrightarrow{d} \chi^2_{df}$ as $n \to \infty$ (Pearson's asymptotic result, 1900).

Independence test in $r \times c$ table

"Expected frequencies for an independence test are calculated assuming that the row proportions are equal across all columns. If the null hypothesis is true (variables independent), this assumption is satisfied." — OpenIntro Statistics, §6.4

Assumptions for validity (Cochran's rule)

Yates correction ( $2 \times 2$ table)

Effect size: Cramér's V

Chi-squared curve with df = 5. The yellow region to the right of the critical value is the rejection region for H0 at level alpha = 5%.

Worked examples

Example— 1· Goodness of fit: fair die?

Problem. A die is rolled 120 times. Counts per face: 15, 22, 17, 25, 19, 22. At the 5% level, is the die fair?

Strategy. Goodness-of-fit test: $H_0$ is that each face has probability 1/6. Expected: $E_i = 120/6 = 20$ for each face. $df = 6 - 1 = 5$ .

Solution.

$\chi^2 = \frac{(15-20)^2}{20} + \frac{(22-20)^2}{20} + \frac{(17-20)^2}{20} + \frac{(25-20)^2}{20} + \frac{(19-20)^2}{20} + \frac{(22-20)^2}{20}$

$= \frac{25 + 4 + 9 + 25 + 1 + 4}{20} = \frac{68}{20} = 3,40$

Critical value: $\chi^2_{0,05;\,5} = 11,07$ . Since $3,40 < 11,07$ , we do not reject $H_0$ .

Verification. The p-value associated with $\chi^2 = 3,40$ with $df = 5$ is approximately 0.64 — well above 0.05. The largest discrepancy was on face 4 (+5), but within what is expected from sampling variation.

Source. OpenStax Statistics, §11.3, Example 11.3 — CC-BY.

Example— 2· Independence: gender and opinion on affirmative action

Problem. Survey of 400 Brazilian voters records gender (Male/Female) and opinion on racial quotas (Favorable/Neutral/Against). Observed table:

	Favorable	Neutral	Against	Total
Male	80	40	80	200
Female	120	50	30	200
Total	200	90	110	400

Test independence at $\alpha = 5\%$ .

Strategy. Table $2 \times 3$ : $df = (2-1)(3-1) = 2$ .

Solution.

$E_{M,Fav} = 200 \times 200 / 400 = 100$ ; $E_{M,Neu} = 200 \times 90/400 = 45$ ; $E_{M,Con} = 200 \times 110/400 = 55$ .

$E_{F,Fav} = 100$ ; $E_{F,Neu} = 45$ ; $E_{F,Con} = 55$ .

$\chi^2 = \frac{(80-100)^2}{100} + \frac{(40-45)^2}{45} + \frac{(80-55)^2}{55} + \frac{(120-100)^2}{100} + \frac{(50-45)^2}{45} + \frac{(30-55)^2}{55}$

$= 4,00 + 0,56 + 11,36 + 4,00 + 0,56 + 11,36 = 31,84$

Critical value: $\chi^2_{0,05;\,2} = 5,99$ . Since $31,84 \gg 5,99$ , we reject $H_0$ — there is significant association.

Verification. $V = \sqrt{31,84/(400 \times 1)} \approx 0,282$ — medium effect. The largest contribution comes from the "Against" category (25 points out of 31.84).

Source. OpenIntro Statistics, §6.4, Exercise 6.29 — CC-BY-SA.

Example— 3· Goodness of fit to Poisson distribution

Problem. A call center records the number of calls per hour over 200 hours: 0 calls (30 hours), 1 call (60 hours), 2 calls (54 hours), 3 calls (36 hours), 4 or more (20 hours). Test whether the data follow Poisson at 5%.

Strategy. Estimate $\hat{\lambda} = (0 \cdot 30 + 1 \cdot 60 + 2 \cdot 54 + 3 \cdot 36 + 4 \cdot 20)/200 = 328/200 = 1,64$ . Since we estimated 1 parameter, $df = 5 - 1 - 1 = 3$ .

Solution.

Poisson probabilities ( $\hat{\lambda} = 1,64$ ): $p_0 = e^{-1,64} \approx 0,1942$ ; $p_1 \approx 0,3185$ ; $p_2 \approx 0,2612$ ; $p_3 \approx 0,1427$ ; $p_{4+} = 1 - \sum_{0}^{3} p_i \approx 0,0834$ .

Expected: $E_0 = 38,8$ ; $E_1 = 63,7$ ; $E_2 = 52,2$ ; $E_3 = 28,5$ ; $E_{4+} = 16,7$ . All $\geq 5$ : OK.

$\chi^2 = \frac{(30-38,8)^2}{38,8} + \frac{(60-63,7)^2}{63,7} + \frac{(54-52,2)^2}{52,2} + \frac{(36-28,5)^2}{28,5} + \frac{(20-16,7)^2}{16,7}$

$\approx 2,00 + 0,21 + 0,06 + 1,97 + 0,65 = 4,89$

Critical value: $\chi^2_{0,05;\,3} = 7,81$ . Since $4,89 < 7,81$ , we do not reject $H_0$ — the data are compatible with Poisson.

Verification. The $df = 3$ (not 4) because we estimated $\lambda$ from the data itself; using the wrong $df$ is the most common pitfall in this type of problem.

Source. OpenStax Statistics, §11.3, Example 11.5 (adapted) — CC-BY.

Example— 4· $2 \\times 2$ table with Yates correction

Problem. A clinical trial compares vaccine vs. placebo in 60 volunteers (30 in each group). Results:

	Protected	Unprotected	Total
Vaccine	22	8	30
Placebo	15	15	30
Total	37	23	60

With Yates correction, test independence at 5%.

Strategy. Table $2 \times 2$ , $df = 1$ . Expected: $E_{V,P} = 37 \times 30/60 = 18,5$ ; $E_{V,NP} = 23 \times 30/60 = 11,5$ ; others equal by symmetry.

Solution.

$\chi^2_Y = \frac{(\lvert 22 - 18,5\rvert - 0,5)^2}{18,5} + \frac{(\lvert 8 - 11,5\rvert - 0,5)^2}{11,5} + \frac{(\lvert 15 - 18,5\rvert - 0,5)^2}{18,5} + \frac{(\lvert 15 - 11,5\rvert - 0,5)^2}{11,5}$

$= \frac{(3,0)^2}{18,5} + \frac{(3,0)^2}{11,5} + \frac{(3,0)^2}{18,5} + \frac{(3,0)^2}{11,5} = \frac{9}{18,5} + \frac{9}{11,5} + \frac{9}{18,5} + \frac{9}{11,5}$

$= 0,486 + 0,783 + 0,486 + 0,783 = 2,538$

Critical value: $\chi^2_{0,05;\,1} = 3,84$ . Since $2,538 < 3,84$ , we do not reject $H_0$ at the 5% level.

Verification. Without Yates correction: $\chi^2 = (3,5)^2 \times (2/18,5 + 2/11,5) = 12,25 \times (0,108 + 0,174) = 3,45$ . Still would not reject, but the correction was conservative as expected.

Source. OpenIntro Statistics, §6.4, Exercise 6.35 (adapted) — CC-BY-SA.

Example— 5· Mendelian genetics: 9:3:3:1 ratio

Problem. A dihybrid cross in peas (Mendel) predicts phenotypes in the ratio 9:3:3:1 (smooth-yellow : smooth-green : wrinkled-yellow : wrinkled-green). Observed in 160 offspring: 90, 30, 27, 13. Test at 5% whether the data follow Mendel.

Strategy. Expected proportions: $p_1 = 9/16$ , $p_2 = 3/16$ , $p_3 = 3/16$ , $p_4 = 1/16$ . $df = 4 - 1 = 3$ (fully specified distribution).

Solution.

$E_1 = 160 \times 9/16 = 90$ ; $E_2 = 160 \times 3/16 = 30$ ; $E_3 = 30$ ; $E_4 = 160 \times 1/16 = 10$ .

$\chi^2 = \frac{(90-90)^2}{90} + \frac{(30-30)^2}{30} + \frac{(27-30)^2}{30} + \frac{(13-10)^2}{10}$

$= 0 + 0 + \frac{9}{30} + \frac{9}{10} = 0,30 + 0,90 = 1,20$

Critical value: $\chi^2_{0,05;\,3} = 7,81$ . Since $1,20 \ll 7,81$ , we do not reject $H_0$ — the data are consistent with Mendel's laws.

Verification. The p-value is approximately 0.75 — the data fit the 9:3:3:1 ratio almost perfectly. The wrinkled-green category ( $E_4 = 10$ ) is exactly at the limit of Cochran's rule; for smaller samples, category merging or Fisher–Freeman–Halton would be recommended.

Source. OpenStax Statistics, §11.3, Example 11.7 (adapted) — CC-BY.

Exercise list

42 exercises · 10 with worked solution (25%)

Application 23Understanding 8Modeling 7Challenge 2Proof 2

Sources

OpenStax Statistics — Illowsky, Dean · CC-BY · Chapter 11 (§11.1–11.5). Primary source for exercises and examples.
OpenIntro Statistics (4th ed) — Diez, Çetinkaya-Rundel, Barr · CC-BY-SA · §6.3–6.4. Conceptual approach and context-based exercises.
Introduction to Modern Statistics — Çetinkaya-Rundel, Hardin · CC-BY-SA · §18–19. Perspective via simulation and modern inference.