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How do you compute expected counts in a two-way table under the assumption of no association?

Topic 8.4 Expected Counts in Two-Way Tables: compute the expected count for each cell of a two-way table under the null hypothesis using the row total times column total divided by the grand total.

A focused answer to AP Statistics Topic 8.4, on computing expected counts in a two-way table under the null of no association, using row total times column total over the grand total, and why this formula encodes independence.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The expected-count formula
Why the formula encodes "no association"
Expected counts and the conditions
Try this

What this topic is asking

The College Board (Topic 8.4) wants you to compute expected counts in a two-way table under the null of no association: for each cell,

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

This is the bridge from a one-variable goodness-of-fit test to the two-variable tests of homogeneity and independence.

The expected-count formula

This single formula generates the expected count for each cell. Multiply the cell's row total by its column total, divide by the grand total. Because the formula reuses the observed marginal totals, the expected table has identical margins to the observed table, a reliable check: if your expected row and column sums do not match the observed margins, you have made an arithmetic error.

Why the formula encodes "no association"

The derivation shows the formula is not arbitrary: it is exactly what "the two variables are independent" predicts. If opinion and age were unrelated, the proportion favoring would be the same in every age group (equal to the overall row proportion), so each cell's expected count is the grand total scaled by the two marginal proportions. The expected table is therefore the "no-association" template against which the observed table is compared. The same formula serves both homogeneity (same distribution across groups) and independence (no association on one sample), because both nulls produce identical expected counts.

Expected counts and the conditions

The expected counts you compute here are exactly what the large-counts condition for a two-way chi-square test checks: every expected count must be at least $5$ . They are also the denominators in the chi-square statistic $\sum (O - E)^2 / E$ of Topic 8.6. So Topic 8.4 is the shared computational core of the homogeneity and independence tests. Computing all expected counts carefully, and verifying each is at least $5$ , is a graded step before any chi-square value is found. If some expected count falls below $5$ , the chi-square approximation is unreliable and the analysis may need categories combined.

Expected counts for a two-way table

A study cross-classifies $400$ students by whether they passed (pass, fail) and study method (A, B, C). Observed margins: pass total $280$ , fail total $120$ ; method A total $150$ , method B total $120$ , method C total $130$ . Compute the full table of expected counts under no association.

step 1 Set up the formula

For each cell, $E = \dfrac{(\text{row total}) \times (\text{column total})}{400}$ . Rows are pass/fail; columns are methods A, B, C.

step 2 Pass row (row total $280$ )

$E_{\text{pass, A}} = \dfrac{280 \times 150}{400} = \dfrac{42000}{400} = 105$ .
$E_{\text{pass, B}} = \dfrac{280 \times 120}{400} = \dfrac{33600}{400} = 84$ .
$E_{\text{pass, C}} = \dfrac{280 \times 130}{400} = \dfrac{36400}{400} = 91$ .

step 3 Fail row (row total $120$ )

$E_{\text{fail, A}} = \dfrac{120 \times 150}{400} = 45$ .
$E_{\text{fail, B}} = \dfrac{120 \times 120}{400} = 36$ .
$E_{\text{fail, C}} = \dfrac{120 \times 130}{400} = 39$ .

step 4 Check the margins and the condition

Pass row of expected counts: $105 + 84 + 91 = 280$ (matches). Method A column: $105 + 45 = 150$ (matches). All expected counts ( $105, 84, 91, 45, 36, 39$ ) are at least $5$ , so the large-counts condition is satisfied.

Try this

Q1. A cell has row total $60$ , column total $90$ , grand total $300$ . Find the expected count. [1 point]

Cue. $E = \dfrac{60 \times 90}{300} = \dfrac{5400}{300} = 18$ .

Q2. What null hypothesis are these expected counts computed under? [1 point]

Cue. No association (independence/homogeneity): the distribution of one variable is the same across categories of the other.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2018 (style)1 marksSection I (multiple choice). In a two-way table, a cell's row total is

80

, its column total is

50

, and the grand total is

200

. The expected count for that cell is (A)

130

(B)

20

(C)

40

(D)

25

Show worked answer →

The correct answer is (B).

The expected count is $\dfrac{\text{row total} \times \text{column total}}{\text{grand total}} = \dfrac{80 \times 50}{200} = \dfrac{4000}{200} = 20$ .

(A) adds the totals. (C) and (D) misapply the formula. The expected count is $20$ .

AP 2021 (style)3 marksSection II (free response). A survey of

300

people cross-classifies opinion (favor, oppose) by age group (young, old). The row totals are favor

180

, oppose

120

; the column totals are young

150

, old

150

. (a) Compute the expected count for the (favor, young) cell. (b) Explain what assumption this expected count is computed under. (c) Explain in words why the formula uses the row and column totals.

Show worked answer →

A 3-point expected-counts question.

(a) (1 point) $E_{\text{favour, young}} = \dfrac{180 \times 150}{300} = \dfrac{27000}{300} = 90$ .
(b) (1 point) It is computed under the null hypothesis of no association (independence) between opinion and age: the proportion favoring is assumed the same across age groups.
(c) (1 point) Under independence, the expected proportion in a cell is (row proportion) times (column proportion); multiplying by the grand total and simplifying gives row total times column total over grand total. The marginal totals carry the overall proportions the null spreads evenly across the table.

Markers reward the correct expected count, naming the independence/no-association assumption, and explaining the role of the marginal totals.

Related dot points

Sources & how we know this

AP Statistics Course and Exam Description — College Board (2020)