How do I decide between a test for independence and a test for homogeneity on the FRQ?

Look at the sampling design. One single sample classified by two categorical variables calls for a test of independence. Two or more separate samples or groups compared on one categorical variable calls for a test of homogeneity. The calculations are identical; only the hypotheses and conclusion wording change, so choosing correctly mainly protects your hypothesis and conclusion points.

Do I have to show the expected-count calculation by hand?

You should show at least one expected count explicitly, such as row total×column totalgrand total, to demonstrate the large counts condition. You may then reference calculator output for the full statistic and P-value, but always label the test name, ^2 value, degrees of freedom, and P-value clearly.

What happens to my score if I choose the wrong chi-square procedure?

You typically lose the procedure-identification and possibly the hypothesis points, but if your mechanics — conditions, statistic, and conclusion — are internally consistent with your chosen method, you can still earn partial credit. That is why naming the correct test first is worth careful reading of the prompt.

Can a chi-square test ever prove two variables are independent?

No. Failing to reject H_0 means there is not convincing evidence of an association, which is not the same as proving independence. Always phrase a fail-to-reject conclusion as insufficient evidence of a difference or association, never as confirmation that the variables are unrelated.

AP-STATS-8-FRQ

U8 FRQ Practice

Master AP Statistics Unit 8 FRQs: attack chi-square goodness-of-fit and two-way table problems with full inference structure and complete scoring language.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U8 FRQ Practice, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Unit 8 free-response questions reward more than a correct test statistic — they reward a complete, communicated chi-square analysis. On the AP exam a chi-square FRQ almost always asks you to identify the right procedure, check conditions, compute a statistic and

P

-value, and state a conclusion in context. This lesson is pure attack strategy: how to read the prompt, decide which of the three chi-square tests fits, and write each step so a reader can follow your reasoning.

You already know the mechanics of goodness-of-fit and two-way table tests. Here you'll assemble them into the four-part inference template graders expect and learn the phrases that earn — and lose — points.

Recognizing which chi-square test the FRQ wants

The first decision on any Unit 8 FRQ is procedure choice, and graders often award a point just for naming the correct test. Read for how many variables and samples appear.

Clue in prompt	Procedure	Degrees of freedom
One sample, one categorical variable compared to claimed proportions/distribution	Goodness-of-fit	$k-1$
One sample classified by two variables	Test for independence	$(r-1)(c-1)$
Several independent samples (or groups), one variable	Test for homogeneity	$(r-1)(c-1)$

The practical tell: if the problem gives you a set of hypothesized percentages that should sum to

100\%

, it is goodness-of-fit. If it gives a two-way table, decide between independence and homogeneity by asking whether the data came from one sample cross-classified (independence) or from separate groups being compared (homogeneity).

A common misconception is that the two-way table tests are computed differently — they are not. The expected counts, statistic, and

P

-value calculations are identical; only the hypotheses and conclusion wording differ. So spend your energy getting the hypothesis statement and conclusion context right, because that is where the procedure-specific points live.

Writing hypotheses that earn the point

Hypotheses must be stated in words and in context — symbols alone rarely earn full credit on chi-square FRQs.

For goodness-of-fit, write

H_0

as the specific distribution: for example,

H_0

: the distribution of blood types in the population is

45\%

40\%

11\%

4\%

AB. The alternative is that at least one of these proportions differs from the stated value — not that all of them differ. Writing 'all proportions are different' is a frequent error that costs the point.

For independence,

H_0

states the two variables are independent (not associated) in the population, and

H_a

states they are associated. For homogeneity,

H_0

states the distribution of the categorical variable is the same across all populations/groups, and

H_a

states at least one distribution differs.

Always define the population. Saying 'the proportions are the same' without naming the groups or the variable leaves the response ungrounded in context. Name the variable, name the groups, and match the wording to the test you selected in the previous step.

Conditions and computations graders look for

Every chi-square FRQ requires condition checks, and vague statements lose points. State three things: random sampling/assignment (cite how the data were collected), independence (the

10\%

condition if sampling without replacement, or independent groups), and the large-counts condition — all expected counts at least

5

. You must show or reference the expected counts, not just assert the condition holds.

Compute an expected count in a two-way table as

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

. In goodness-of-fit, expected count

= n p_i

where

p_i

is the hypothesized proportion. Show at least one expected-count calculation.

Then report the statistic

\chi^2 = \sum \frac{(O-E)^2}{E}

along with the degrees of freedom and the

P

-value. You may use calculator output, but you must state the test name,

\chi^2

value,

df

, and

P

-value. Listing only a number with no label is risky. Rounding to two or three decimals is fine as long as your conclusion follows from the value you reported.

The four-part conclusion and communication points

AP inference conclusions follow a reliable structure: compare

P

-value to

\alpha

, make a reject/fail-to-reject decision, and interpret in context. A complete conclusion has four moving parts.

Part	What to write
Compare	'Since $P = 0.02 < \alpha = 0.05$ ...'
Decide	'...we reject $H_0$ .'
Conclude	'There is convincing evidence that...'
Context	'...the distribution of X differs across the groups.'

Two phrasing pitfalls: never say you 'accept

H_0

' — say 'fail to reject.' And never claim you 'proved' the alternative; chi-square gives evidence, not proof. If you fail to reject, state there is 'not convincing evidence' of the association or difference, not that the variables are 'definitely independent.'

Finally, many Unit 8 FRQs add a follow-up part: identify which cell contributed most to

\chi^2

, or describe the direction of a difference using observed versus expected counts. Answer these by comparing

O

and

E

directly — a cell where observed far exceeds expected is over-represented. Always tie the interpretation back to the context described in the prompt to secure the communication point.

Managing multi-part FRQs and time

Unit 8 FRQs are frequently multi-part: part (a) might ask for a graph or expected counts, part (b) the full test, part (c) an interpretation or a caution about generalizing. Treat each part independently — a wrong number in part (a) usually does not cost you again in part (b) if your method is correct (follow-through credit).

Budget roughly the standard timing per full FRQ and do not over-write. Graders reward clear, labeled work, not paragraphs of hedging. Show the expected-count formula once, report the statistic and

P

-value clearly, and keep the conclusion to the four parts above.

Watch for parts that ask about scope of inference: if data came from a random sample you can generalize to that population; if from a randomized experiment you can discuss cause; if neither, you must say inference is limited. A chi-square test never by itself establishes causation from observational data — flagging that in a final part often earns an easy point that rushed students skip.

Key terms

Chi-square goodness-of-fit test.: A test comparing one sample's observed category counts to a hypothesized distribution, using $df = k-1$ .
Test for independence.: A chi-square test on one sample classified by two categorical variables, assessing whether the variables are associated in the population.
Test for homogeneity.: A chi-square test comparing the distribution of one categorical variable across two or more independent samples or groups.
Expected count.: The count predicted under $H_0$ ; equals $np_i$ for goodness-of-fit or $\frac{\text{row total}\times\text{column total}}{\text{grand total}}$ in a two-way table.
Large counts condition.: The requirement that every expected count is at least $5$ for the chi-square approximation to be valid.
Chi-square statistic.: $\chi^2 = \sum \frac{(O-E)^2}{E}$ , measuring total discrepancy between observed and expected counts.
Component (cell contribution).: The value $\frac{(O-E)^2}{E}$ for a single cell; large components identify categories driving a significant result.

Worked example

A researcher randomly samples 300 adults and records their primary news source (TV, Online, Print) and age group (18–39, 40+). The two-way table is: TV — 40 (18–39), 80 (40+); Online — 90 (18–39), 30 (40+); Print — 10 (18–39), 50 (40+). At

\alpha = 0.05

, is there convincing evidence of an association between age group and primary news source?

Because one random sample was classified by two variables, this is a chi-square test for independence.

State hypotheses.

H_0

: Age group and primary news source are independent in the population of adults.

H_a

: Age group and primary news source are associated.

Check conditions. The sample was randomly selected, so the random condition is met. The sample of 300 is clearly less than

10\%

of all adults, so independence holds. Row totals are TV

=120

, Online

=120

, Print

=60

; column totals are 18–39

=140

, 40+

=160

; grand total

=300

. An example expected count for TV/18–39 is

\frac{120\times140}{300}=56

. Computing all expected counts gives values

56, 64, 56, 64, 28, 32

— all at least

5

, so the large counts condition is met.

Compute the statistic.

\chi^2 = \frac{(40-56)^2}{56}+\frac{(80-64)^2}{64}+\frac{(90-56)^2}{56}+\frac{(30-64)^2}{64}+\frac{(10-28)^2}{28}+\frac{(50-32)^2}{32} \approx 4.57+4.00+20.64+18.06+11.57+10.13 = 68.97.

With

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

, the

P

-value is essentially

0

(far less than

0.001

).

Conclude. Since

P < 0.05

, we reject

H_0

. There is convincing evidence that age group and primary news source are associated among adults. The largest components come from the Online and Print cells, showing younger adults favor online sources while older adults favor print more than independence would predict.

Practice questions

A prompt gives observed counts of party affiliation for one random sample of voters and separately for a second random sample from a different state, then asks whether the distribution of affiliation is the same in both states. Which chi-square procedure is appropriate?

Goodness-of-fit test
Test for independence
Test for homogeneity
A two-sample z-test for proportions

Answer: Test for homogeneity

Two separate samples (from two states) are compared on a single categorical variable (party affiliation). Comparing the distribution of one variable across independent groups is the definition of a test for homogeneity. Independence would require one sample cross-classified by two variables, and goodness-of-fit compares one sample to a fixed distribution.

On a goodness-of-fit FRQ a student writes

H_0

: all four proportions are equal to each other, when the prompt gave specific hypothesized percentages of 50, 30, 15, and 5. Explain why this loses the hypothesis point and write a correct

H_0

and

H_a

Answer: The null must state the specific hypothesized distribution, not that the proportions are equal.

Goodness-of-fit tests a claimed distribution, so

H_0

should be: the population proportions are

0.50, 0.30, 0.15,

and

0.05

for the four categories. Writing that all proportions are equal describes a different (and here incorrect) claim. The correct

H_a

is: at least one of the proportions differs from its hypothesized value — not that all differ. Stating 'all proportions differ' would also be wrong.

In a test for independence a student reports

\chi^2 = 12.4

with

df=3

and

P=0.006

, then writes: 'We accept the alternative and conclude the variables cause each other.' Identify two errors and give a correct conclusion at

\alpha=0.05

Answer: You never 'accept' a hypothesis and chi-square does not establish causation.

First, you reject

H_0

rather than 'accept' the alternative — the decision language must be reject or fail to reject. Second, an observational chi-square test shows association, not causation. A correct conclusion: since

P=0.006 < 0.05

, we reject

H_0

; there is convincing evidence that the two variables are associated in the population. No causal claim is justified without a randomized experiment.

FAQ

How do I decide between a test for independence and a test for homogeneity on the FRQ?: Look at the sampling design. One single sample classified by two categorical variables calls for a test of independence. Two or more separate samples or groups compared on one categorical variable calls for a test of homogeneity. The calculations are identical; only the hypotheses and conclusion wording change, so choosing correctly mainly protects your hypothesis and conclusion points.
Do I have to show the expected-count calculation by hand?: You should show at least one expected count explicitly, such as $\frac{\text{row total}\times\text{column total}}{\text{grand total}}$ , to demonstrate the large counts condition. You may then reference calculator output for the full statistic and $P$ -value, but always label the test name, $\chi^2$ value, degrees of freedom, and $P$ -value clearly.
What happens to my score if I choose the wrong chi-square procedure?: You typically lose the procedure-identification and possibly the hypothesis points, but if your mechanics — conditions, statistic, and conclusion — are internally consistent with your chosen method, you can still earn partial credit. That is why naming the correct test first is worth careful reading of the prompt.
Can a chi-square test ever prove two variables are independent?: No. Failing to reject $H_0$ means there is not convincing evidence of an association, which is not the same as proving independence. Always phrase a fail-to-reject conclusion as insufficient evidence of a difference or association, never as confirmation that the variables are unrelated.

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