AP Statistics · Lesson 11 of 11 · Unit 3

Carrying out a chi-square test for homogeneity or independence

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Carrying out a chi-square test turns raw two-way table counts into a single test statistic and p-value to decide whether categorical variables are associated or whether distributions differ across populations. On the AP Statistics exam, this can appear in multiple-choice questions testing cell contributions or expected counts, and in Section II (for example Question 3 or Question 4), where you must compute the test results, interpret the p-value, and justify a conclusion in context.

By the end you’ll be able to

Calculate the expected counts for a two-way table, assuming the null hypothesis is true.
Calculate the chi-square statistic, its degrees of freedom and its p-value using Table C or technology.
Interpret the p-value of a chi-square test in context, as a probability computed assuming the null hypothesis is true.
Justify a conclusion about the population (independence) or populations (homogeneity) by comparing the p-value with $\alpha$, without accepting the null hypothesis.

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What the exam asks

Calculate expected counts for cells in a two-way table assuming the null hypothesis ( $H_0$ ) is true.
Calculate the chi-square test statistic ( $\chi^2$ ), degrees of freedom ( $df$ ), and p-value using technology or Table C.
Identify which cell contributes the most to the $\chi^2$ statistic and explain the direction of the discrepancy.
Interpret the p-value of a chi-square test in context.
Justify a conclusion in context by comparing the p-value to the significance level $\alpha$ .

Core ideas

Expected counts

Although a chi-square test for homogeneity and a chi-square test for independence answer different investigative questions, their mechanics are identical. Assuming $H_0$ is true, the expected count for any interior cell in a two-way table is:

\text{Expected count} = \frac{\text{row total} \times \text{column total}}{\text{table total}}

Do not round expected counts to whole numbers; keep at least two decimal places so you can verify the large counts condition (all expected counts $\ge 5$ , or $> 5$ ) and compute the test statistic accurately.

The chi-square statistic and degrees of freedom

The chi-square test statistic measures how far the observed counts fall from the counts expected under $H_0$ , scaled by the expected counts:

\chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}

Each term $\frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}$ is a cell’s component (or contribution) to the $\chi^2$ statistic. Larger discrepancies between observed and expected counts produce larger components and a larger overall value.

When conditions are met, $\chi^2$ follows a chi-square distribution with degrees of freedom based on the number of rows ( $r$ ) and columns ( $c$ ) of categorical categories (excluding total rows and columns):

df = (\text{rows} - 1)(\text{columns} - 1)

Finding and interpreting the p-value

Because any departure from expected counts—whether observed counts are higher or lower than expected—makes $(\text{Observed} - \text{Expected})^2$ positive, chi-square tests are always upper-tail tests.

P-value calculation: $P(\chi^2 \ge \text{observed } \chi^2)$ with $df = (r-1)(c-1)$ , found via graphing calculator ( or a matrix test) or approximated as an interval using Table C.