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How do you compute the chi-square statistic and P-value and conclude a test of homogeneity or independence?

Topic 8.6 Carrying Out a Chi-Square Test for Homogeneity or Independence: compute the chi-square statistic from a two-way table, find the P-value using (rows minus 1)(columns minus 1) degrees of freedom, and state a conclusion in context.

A focused answer to AP Statistics Topic 8.6, on computing the chi-square statistic from a two-way table, finding the P-value with (r minus 1)(c minus 1) degrees of freedom, and stating a conclusion in context, with a full worked test.

Generated by Claude Opus 4.811 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 statistic and degrees of freedom
From statistic to conclusion
What a significant result does and does not say
Try this

What this topic is asking

The College Board (Topic 8.6) wants you to carry out and conclude a chi-square test of homogeneity or independence: compute the chi-square statistic from the two-way table, find the P-value with $(r - 1)(c - 1)$ degrees of freedom, compare to $\alpha$ , and state a conclusion in context.

The statistic and degrees of freedom

The statistic is the same $\sum (O - E)^2 / E$ as goodness of fit, now summed over every cell of the table. The only computational difference from the one-variable test is the degrees of freedom: $(r - 1)(c - 1)$ instead of $k - 1$ . For a $2 \times 2$ table, $df = 1$ ; for a $3 \times 4$ table, $df = 6$ . As before, the chi-square distribution is right-skewed, so the P-value is the upper tail only, never doubled.

From statistic to conclusion

The decision rule is standard; the conclusion wording follows the design. For homogeneity, conclude about the distribution differing across groups ("there is convincing evidence the distribution of preference differs among the regions"). For independence, conclude about an association between two variables ("there is convincing evidence that exercise habit and sleep quality are associated"). Matching the conclusion to the test type is graded, and a homogeneity conclusion phrased as "associated" (or vice versa) loses the contextual mark.

What a significant result does and does not say

A rejected test says the distributions differ (or the variables are associated) somewhere in the table, but does not say which cells or how. To discuss where the effect lies, inspect the individual cell contributions $(O - E)^2 / E$ : the largest terms identify the cells driving the result. And, as always, a chi-square test of independence shows association, not causation; only a randomised experiment supports a causal claim. Likewise, a non-significant result does not prove the distributions are identical or the variables independent, it only fails to find evidence against the null. Reporting these limits is part of a complete answer.

Complete test of homogeneity

Separate random samples of $150$ students from each of two campuses are classified by transport mode (walk, bus, car). Campus A: walk $60$ , bus $50$ , car $40$ . Campus B: walk $45$ , bus $45$ , car $60$ . Test at $\alpha = 0.05$ whether the transport-mode distribution is the same on both campuses.

step 1 Hypotheses and conditions

$H_0$ : the distribution of transport mode is the same on both campuses. $H_a$ : it is not the same for both. Each campus is a random sample; expected counts (below) are all $\ge 5$ .

step 2 Expected counts

Row totals: A $= 150$ , B $= 150$ ; column totals: walk $= 105$ , bus $= 95$ , car $= 100$ ; grand total $= 300$ .
$E_{A,\text{walk}} = \dfrac{150 \times 105}{300} = 52.5$ , $E_{A,\text{bus}} = \dfrac{150 \times 95}{300} = 47.5$ , $E_{A,\text{car}} = \dfrac{150 \times 100}{300} = 50$ . By symmetry, campus B has the same expected counts $52.5, 47.5, 50$ .

step 3 Chi-square statistic

\chi^2 = \frac{(60-52.5)^2}{52.5} + \frac{(50-47.5)^2}{47.5} + \frac{(40-50)^2}{50} + \frac{(45-52.5)^2}{52.5} + \frac{(45-47.5)^2}{47.5} + \frac{(60-50)^2}{50}.

= \frac{56.25}{52.5} + \frac{6.25}{47.5} + \frac{100}{50} + \frac{56.25}{52.5} + \frac{6.25}{47.5} + \frac{100}{50} \approx 1.071 + 0.132 + 2.0 + 1.071 + 0.132 + 2.0 = 6.41.

df = (2-1)(3-1) = 2

step 4 P-value

Upper tail: $\text{P-value} = P(\chi^2_2 > 6.41) \approx 0.041$ .

step 5 Conclusion in context

Since P-value $0.041 < 0.05 = \alpha$ , reject $H_0$ . There is convincing evidence that the distribution of transport mode differs between the two campuses (homogeneity does not hold).

Try this

Q1. A $4 \times 3$ table is tested. Find the degrees of freedom. [1 point]

Cue. $df = (4-1)(3-1) = 3 \times 2 = 6$ .

Q2. A test of independence is significant. State two things it does NOT establish. [2 points]

Cue. It does not establish causation (association is not causation), and it does not identify which specific cells differ (inspect the largest $(O-E)^2/E$ contributions for that).

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). A two-way table has

3

rows and

4

columns. The chi-square degrees of freedom are (A)

12

(B)

7

(C)

6

(D)

11

Show worked answer →

The correct answer is (C).

For a two-way table, $df = (\text{rows} - 1)(\text{columns} - 1) = (3-1)(4-1) = 2 \times 3 = 6$ .

(A) is rows times columns. (B) is rows plus columns. (D) is the total cells minus one. The correct value is $6$ .

AP 2022 (style)4 marksSection II (free response). A single random sample of

200

adults is classified by exercise habit (regular/irregular) and sleep quality (good/poor). Observed: regular and good

70

, regular and poor

30

, irregular and good

40

, irregular and poor

60

. Test at

\alpha = 0.05

whether exercise habit and sleep quality are associated. Compute the chi-square statistic and P-value (use

df = 1

), and conclude in context (justify in context).

Show worked answer →

A 4-point test of independence.

(1) (1 point) $H_0$ : exercise habit and sleep quality are independent; $H_a$ : they are associated. One random sample; expected counts checked $\ge 5$ .
(2) (1 point) Row totals: regular $100$ , irregular $100$ ; column totals: good $110$ , poor $90$ ; grand total $200$ . Expected: regular-good $= \dfrac{100 \times 110}{200} = 55$ , regular-poor $= 45$ , irregular-good $= 55$ , irregular-poor $= 45$ . All $\ge 5$ .
(3) (1 point) $\chi^2 = \dfrac{(70-55)^2}{55} + \dfrac{(30-45)^2}{45} + \dfrac{(40-55)^2}{55} + \dfrac{(60-45)^2}{45} = \dfrac{225}{55} + \dfrac{225}{45} + \dfrac{225}{55} + \dfrac{225}{45} \approx 4.09 + 5.00 + 4.09 + 5.00 = 18.18$ , $df = 1$ . P-value $= P(\chi^2_1 > 18.18) < 0.001$ .
(4) (1 point) Since P-value $< 0.001 < 0.05$ , reject $H_0$ . There is convincing evidence that exercise habit and sleep quality are associated.

Markers reward expected counts, the chi-square sum, $df = (2-1)(2-1) = 1$ , the P-value, and a contextual conclusion.

Related dot points

Sources & how we know this

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