Chi-Square Test for Homogeneity

Unit 8 · Inference for Categorical Data: Chi-Square

What AP Stats asks here

Homogeneity asks whether the distribution of a categorical variable is the same across several populations sampled separately. Arithmetically the test is identical to the chi-square independence test: $E_{ij}$ comes from the same formula, df = $(r-1)(c-1)$ , the p-value is right-tailed. The only thing that changes is the sampling design and the conclusion sentence.

Homogeneity test

Expected count

E_{ij} = \dfrac{(\text{row total}_i)(\text{col total}_j)}{\text{grand total}}

Test statistic

\chi^2 = \sum \dfrac{(O_{ij} - E_{ij})^2}{E_{ij}}

Degrees of freedom

(r - 1)(c - 1)

Design diagnostic

Homogeneity

\text{separate samples from each population; row totals fixed in advance}

AP Tip: Look at how the data was collected. If the researcher pre-set sample sizes for each subpopulation, it's homogeneity. If they took one sample and classified two ways, it's independence.

Caution: Rejecting H₀ says the distributions differ; it does not say one population caused another's distribution. No causal language belongs in a chi-square conclusion.

Type 1

Recognize homogeneity design

Homogeneity samples each population separately, fixing their sample sizes in advance. Independence draws one sample and lets both marginals fall out.

Example 1

Which scenario calls for a chi-square test for homogeneity? (A) A market researcher samples 800 adults and records age group and brand preference. (B) A researcher samples 100 students from each of three schools and asks about lunch preference. (C) A hospital records every ER patient and classifies by time of day and outcome. (D) A pollster samples 2000 voters and records both state and candidate.

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Type 2

Expected counts and chi-square

Use the same formula as the independence test. df depends only on the table shape: $(r - 1)(c - 1)$ .

Example 2

Three schools, each surveyed with

n = 100

, classified by 4 lunch types. The grand total is 300. For the Pizza column total of 108, what is the expected count for each school in the Pizza cell, and what is df for the test? (A) Expected ≈ 36, df = 6 (B) Expected ≈ 36, df = 12 (C) Expected ≈ 27, df = 6 (D) Expected ≈ 108, df = 11

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Type 3

Conclusion language

Conclude about whether the populations share the same distribution. Never claim causation.

Example 3

A chi-square test for homogeneity across three schools rejects

H_0

\alpha = 0.05

. Which conclusion is appropriate? (A) School A causes students to prefer pizza. (B) There is convincing evidence at the 5% level that the distribution of lunch preferences differs across at least one pair of the three schools. (C) The distributions are identical across the three schools. (D) The data prove that all three schools have different lunch preferences.

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