AP Statistics Topic 3.15: Carrying Out a Chi-Square Test for Homogeneity or Independence

01 · From a table to evidence

The concept in 60 seconds

A chi-square test compares observed counts with the counts predicted by the null hypothesis. A small mismatch is ordinary sample variation. A sufficiently large overall mismatch provides evidence against the null model.

Two designs use the same calculation. Homogeneity compares $1$ categorical response distribution across independently sampled populations or randomly assigned treatments. Independence investigates association between $2$ categorical variables in $1$ sampled population. Your design determines which name and conclusion fit.

Visual guide 1: follow the inference chain

1 · Fit the nullExpected counts

Predict each cell from the row total, column total and grand total, after justifying the design.

2 · Combine mismatch $1$ $\chi^{2}$ statistic

Square each $O-E$ difference, divide by $E$ and add every interior-cell contribution.

3 · Assess extremenessRight-tail $p$ -value

Use the chi-square curve with $\operatorname{df}=(r – 1)(c – 1)$ . Find the area from the statistic to the right.

4 · Answer the questionContextual conclusion

Compare $p$ with the preselected $\alpha$ and describe evidence about the populations or association.

A calculation is part of the inference. Keep the populations, hypothesis and design attached to every step.

The key chain: observed counts → expected counts under $H_{0}$ → nonnegative cell contributions → $1$ $\chi^{2}$ statistic → a right-tail $p$ -value → a contextual decision.

The test uses counts of individuals, rather than percentages alone. Each individual contributes to $1$ row-and-column combination. All studies on this page are fictional teaching examples; the numerical results are calculated from the stated data.

Quick check: do homogeneity and independence use different statistic formulas?

No. Both use $\chi^2=\sum_{\text{interior cells}}\frac{(O-E)^2}{E}$ with $\operatorname{df}=(r – 1)(c – 1)$ . Their study designs, hypotheses and population conclusions differ.

02 · Keep the question attached to the counts

Follow the three-school preference study

Researchers select independent SRSs without replacement from Schools A, B and C. The samples contain $100, 120,\text{ and }80$ students, respectively, from populations of $5{,}000, 6{,}000,\text{ and }4{,}000$ students. Each student chooses exactly $1$ preferred notes format: digital only, printed only or mixed.

The investigative question is: Is the distribution of preferred notes format the same across the $3$ school populations? Use a chi-square test for homogeneity. Choose $\alpha =0.05$ before evaluating the result.

Observed preferred notes formats from independent random samples. Each student appears $1\text{ time}$ .
School population	Digital only	Printed only	Mixed	Total
School A	$60$	$25$	$15$	$100$
School B	$54$	$36$	$30$	$120$
School C	$36$	$29$	$15$	$80$
Total	$150$	$90$	$60$	$300$

$H_{0}$ : The distribution of preferred notes format is the same among all students at Schools A, B and C.

$H_{a}$ : Those population distributions are not all the same; at least $1$ differs.

The null does not require $\frac13$ of students to prefer each category. It requires a common distribution across schools, whose category shares may be unequal. Unequal sample sizes also mean that equal population distributions do not predict equal raw counts.

Verify the setup before calculating a $p$ -value

Randomization and independent observations: Independent SRSs are specified, and each student contributes $1$ response. The samples do not consist of the same students measured repeatedly.
$10\%$ condition for each sampled population: $100\le 0.10(5{,}000)=500$ ; $120\le 0.10(6{,}000)=600$ ; $80\le 0.10(4{,}000)=400$ .
Expected counts: The next section calculates every interior expectation. The minimum is $16$ , so all $9$ exceed $5$ .

Use the same $E > 5$ criterion as Topic 3.14, following the Fall 2026 AP course framework. Some other texts use $E\ge 5$ ; an exact $5$ does not pass the strict wording used here. The sampling $10\%$ check is not required merely for random assignment. A separate finite-population sampling stage can still require its own check.

Quick check: may we combine all

3

school populations for

1

10\%

check?

No. These samples were separately selected from $3$ populations. Check each sample against its own population. A large combined population can hide an overly large sampling fraction in $1$ school.

03 · What counts would the null model predict?

Calculate every expected cell count

Use the table’s margins to calculate the expected count in each interior cell:

E=\frac{\text{row total}\times\text{column total}}{\text{grand total}}

For School A’s digital-only cell, the row total is $100$ , the column total is $150$ and the grand total is $300$ :

E=\frac{100\times150}{300}=50

Another way to see this: the pooled digital proportion is $\frac{150}{300}=0.50$ . Under the common-distribution null, $50\%$ of a sample of $100$ is an expected $50$ digital preferences. The combined printed and mixed proportions are $0.30$ and $0.20$ .

Expected counts under the common-distribution null. Use the $9$ interior cells in the statistic.
School population	Digital only	Printed only	Mixed	Total
School A	$50$	$30$	$20$	$100$
School B	$60$	$36$	$24$	$120$
School C	$40$	$24$	$16$	$80$
Total	$150$	$90$	$60$	$300$

The expected table preserves the observed row totals, column totals and grand total. For example, School A’s expectations sum to $50+30+20=100$ . The digital column sums to $50+60+40=150$ . These checks can reveal an arithmetic or data-entry error.

What to remember about $E$

An observed count $O$ is what happened; an expected count $E$ is predicted by the null model using the margins.
Expected counts can be fractional. They describe model-based counts, so they need not be whole numbers.
Keep full precision for the statistic. Rounding $E$ too early can change the final $p$ -value.
Check every interior expected cell. The margins are totals, rather than additional response categories.

For an independence test: the same formula predicts a cell count when the $2$ categorical variables are independent. The table arithmetic stays the same, while the null claim refers to the $2$ variables in $1$ population.

Quick check: why is School B’s expected digital count

60

rather than

50

B has a sample of $120$ , compared with A’s $100$ . Applying the common digital share $0.50$ gives $120 \times 0.50=60$ . A homogeneity null predicts the same proportions, rather than the same counts.

04 · Combine all the cell mismatches

Build the chi-square statistic

The Pearson chi-square statistic adds $1$ contribution from every interior cell:

\chi^2=\sum_{\text{interior cells}}\frac{(O-E)^2}{E}

Read the formula as: subtract expected from observed, square that difference, divide by the expected count, and add the results. The sum uses all $9$ school-study cells.

Visual guide 2: unpack

1

cell contribution

Observed

60

; expected

50

$\text{Difference}=10$

$O-E=60-50$ . A positive difference is an observed excess in this sample cell.

Remove the sign $\text{Squared difference}=100$

$(60-50)^2=100$ . An equally large shortage would have the same square.

Adjust for

E

$\text{Contribution}=2$

$\frac{100}{50}=2$ . Add this to the other $8$ contributions for the school statistic.

A contribution measures this cell’s squared, scaled mismatch with its fitted null expectation. It is not a $p$ -value or an independent test decision.

For A’s digital cell, $O=60$ and $E=50$ , so the contribution is $\frac{(60-50)^2}{50}=2$ . For A’s printed cell, $O=25$ and $E=30$ , giving $\frac{(25-30)^2}{30}=\frac{25}{30}\approx0.833333$ .

$9$ school-study contributions. Displayed contributions are rounded; $\chi^{2}$ uses full precision.
Cell	Observed $O$	Expected $E$	Contribution $\frac{(O-E)^2}{E}$
A: Digital only	$60$	$50$	$2.000000$
A: Printed only	$25$	$30$	$0.833333$
A: Mixed	$15$	$20$	$1.250000$
B: Digital only	$54$	$60$	$0.600000$
B: Printed only	$36$	$36$	$0.000000$
B: Mixed	$30$	$24$	$1.500000$
C: Digital only	$36$	$40$	$0.400000$
C: Printed only	$29$	$24$	$1.041667$
C: Mixed	$15$	$16$	$0.062500$

$\chi^{2}=7.6875$ , using the unrounded cell contributions.

Why square and divide?

Squaring prevents positive and negative differences from cancelling. Under $H_{0}$ the margins are preserved, so signed $O-E$ differences can sum to $0$ even when many cells differ. Both an excess and a shortage can add evidence against the null.

Dividing by $E$ adjusts for the count scale. A difference of $10$ with $E=50$ contributes $2$ ; the same difference with $E=100$ contributes $1$ . Comparing only the raw differences misses this scaling.

Visual guide 3: see how the

9

contributions add up

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Carrying Out a Chi-Square Test for Homogeneity or Independence

By the end of this lesson, you should be able to:

The concept in 60 seconds

Follow the three-school preference study

Verify the setup before calculating a $p$ -value

Calculate every expected cell count

What to remember about $E$

Build the chi-square statistic

Why square and divide?

Main Menu

Quick Links

The concept in 60 seconds

Follow the three-school preference study

Verify the setup before calculating a pp-value

Calculate every expected cell count

What to remember about EE

Build the chi-square statistic

Why square and divide?

Main Menu

Quick Links​

Verify the setup before calculating a $p$ -value

What to remember about $E$

Quick Links