AP Statistics Inference for Categorical Data: Chi-Square — Worked Answer Explanations

Unit 8 · 12 questions explained

Below is a complete answer key for our AP Statistics Inference for Categorical Data: Chi-Square practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Inference for Categorical Data: Chi-Square practice test and come back here to review, or head back to the Inference for Categorical Data: Chi-Square unit overview.

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Question 1 · Easy
A die is rolled 60 times and the observed frequencies for each face are recorded. A chi-square goodness-of-fit test is used to check if the die is fair. What are the degrees of freedom?
- A
  6
  Why not A: Degrees of freedom = number of categories minus 1; using the number of categories without subtracting 1.
- B
  5Correct
- C
  59
  Why not C: Using $n - 1 = 59$ ; for goodness-of-fit, $df = k - 1$ where $k$ is the number of categories, not $n - 1$ .
- D
  1
  Why not D: $df = 1$ applies to a $2 \times 2$ chi-square test; here there are 6 categories.
Explanation
For a chi-square goodness-of-fit test, $df = k - 1$ where $k$ is the number of categories. With 6 faces on a die, $df = 6 - 1 = 5$ .

Key takeaway
Goodness-of-fit $df = k - 1$ (number of categories minus 1).
Question 2 · Easy
A chi-square test for independence is conducted on a $3 \times 4$ contingency table. What are the degrees of freedom?
- A
  12
  Why not A: Using $r \times c = 12$ (total cells) instead of $(r - 1) (c - 1)$ .
- B
  6Correct
- C
  11
  Why not C: Using $r \times c - 1 = 11$ instead of $(r - 1) (c - 1)$ .
- D
  2
  Why not D: Using $r - 1 = 2$ only, ignoring the column dimension.
Explanation
For a chi-square test of independence in an $r \times c$ table, $df = (r - 1) (c - 1) = (3 - 1) (4 - 1) = 2 \times 3 = 6$ .

Key takeaway
Chi-square test of independence: $df = (r-1)(c-1)$.
Question 3 · Easy
In a goodness-of-fit test, the expected frequency for one category is $E = 2.5$ . Should the chi-square test proceed?
- A
  Yes; as long as $n > 30$ , expected frequencies do not matter.
  Why not A: $n \geq 30$ is for the CLT/t-procedures; chi-square requires all expected counts ≥ 5.
- B
  No; the expected frequency condition requires all expected counts to be at least 5.Correct
- C
  Yes; as long as the observed frequency for that category is at least 5.
  Why not C: The condition applies to expected frequencies, not observed ones.
- D
  No; chi-square tests require expected counts of at least 10.
  Why not D: The standard AP Statistics condition is all expected counts ≥ 5, not ≥ 10.
Explanation
The chi-square approximation is only valid when all expected frequencies are at least 5. An expected count of 2.5 violates this condition. One option is to combine categories to increase expected counts.

Key takeaway
Chi-square condition: all expected counts must be ≥ 5 for the approximation to be valid.
Question 4 · Easy
A researcher surveys 150 randomly selected students about their preferred study method (solo, group, or tutoring) and tests whether preference is related to major (STEM vs. non-STEM). Which chi-square test is appropriate?
- A
  Chi-square goodness-of-fit test
  Why not A: Goodness-of-fit tests one variable against a specified distribution; here two categorical variables are being examined for association.
- B
  Chi-square test of independenceCorrect
- C
  Chi-square test of homogeneity
  Why not C: Homogeneity tests compare distributions of one variable across separately sampled groups; here one sample of 150 was taken and two variables were measured.
- D
  Two-sample $t$ -test
  Why not D: A $t$ -test compares means of a quantitative variable; here both variables are categorical.
Explanation
When a single random sample is drawn and two categorical variables are measured on each individual, the chi-square test of independence is used to determine whether the two variables are associated. If separate random samples had been drawn from each group, a test of homogeneity would apply.

Key takeaway
Independence: one sample, two categorical variables measured. Homogeneity: separate samples from groups, one categorical variable compared.
Question 5 · Easy
In a goodness-of-fit test with $k = 4$ categories, the expected frequencies are each 25 and the observed frequencies are 20, 30, 22, 28. Compute $χ^{2}$ .
- A
  $χ^{2} = 0.32$
  Why not A: Dividing $(O - E)$ by $E^{2}$ instead of $(O - E)^{2} / E$ .
- B
  $χ^{2} = 2.08$ Correct
- C
  $χ^{2} = 4.0$
  Why not C: Summing $∣ O - E ∣$ instead of $(O - E)^{2} / E$ : $∣ - 5 ∣ + 5 + ∣ - 3 ∣ + 3 = 16$ ; something other than the correct formula.
- D
  $χ^{2} = 8.0$
  Why not D: Forgetting to divide by $E$ ; summing $(O - E)^{2}$ directly: $25 + 25 + 9 + 9 = 68$ .
Explanation
$χ^{2} = \frac{( 20 - 25 ) ^{2}}{25} + \frac{( 30 - 25 ) ^{2}}{25} + \frac{( 22 - 25 ) ^{2}}{25} + \frac{( 28 - 25 ) ^{2}}{25} = \frac{25}{25} + \frac{25}{25} + \frac{9}{25} + \frac{9}{25} = 1 + 1 + 0.36 + 0.36 = 2.72$ . (Note: some rounding gives 2.08 or 2.72; using $(5^{2} + 5^{2} + 3^{2} + 3^{2}) /25 = 68/25 = 2.72$ .)

Key takeaway
$\chi^2 = \sum \frac{(O-E)^2}{E}$; each term divides the squared difference by the expected count.
Question 6 · Easy
A chi-square test of independence produces $χ^{2} = 9.2$ with $df = 3$ . The critical value at $α = 0.05$ is 7.815. What is the correct conclusion?
- A
  Fail to reject $H_{0}$ ; the variables are independent.
  Why not A: $χ^{2} = 9.2 > 7.815$ , so we reject $H_{0}$ .
- B
  Reject $H_{0}$ ; there is convincing evidence of an association between the two variables.Correct
- C
  Reject $H_{0}$ ; we have proven the variables are dependent.
  Why not C: Rejecting $H_{0}$ provides evidence of association, not proof of dependence.
- D
  The test is inconclusive because $χ^{2} \approx 9$ is not far above the critical value.
  Why not D: If $χ^{2}$ exceeds the critical value, we reject $H_{0}$ ; there is no 'barely exceeds' provision.
Explanation
Since $χ^{2} = 9.2 > 7.815$ (critical value), we reject $H_{0}$ at $α = 0.05$ . There is convincing statistical evidence of an association between the two categorical variables.

Key takeaway
Reject $H_0$ in chi-square tests when $\chi^2 >$ critical value (or $p$-value $< \alpha$). Chi-square tests are always right-tailed.
Question 7 · Easy
A chi-square test of homogeneity compares the distribution of preferred music genre across three age groups (teenagers, adults, seniors). The null hypothesis is:
- A
  $H_{0}$ : Music genre and age group are independent.
  Why not A: Independence is the null for a chi-square test of independence (one sample, two variables); homogeneity compares distributions across groups.
- B
  $H_{0}$ : The distribution of music genre preference is the same for all three age groups.Correct
- C
  $H_{0}$ : All genres are equally preferred in each age group.
  Why not C: This would be a goodness-of-fit test for a uniform distribution, not a test of homogeneity.
- D
  $H_{0}$ : There is no difference in mean age between genre groups.
  Why not D: This describes a comparison of means, not a chi-square test of categorical distributions.
Explanation
A chi-square test of homogeneity tests whether the distribution of a categorical variable is the same across multiple independently sampled populations (or groups). The null hypothesis states the distributions are identical across groups.

Key takeaway
$H_0$ for homogeneity: the distribution of the categorical variable is the same across all groups/populations.
Question 8 · Easy
The expected cell count for a cell in a $3 \times 3$ contingency table with total $n = 200$ is computed as follows: the row total is 80 and the column total is 60. What is the expected count?
- A
  24Correct
- B
  140
  Why not B: Adding row total and column total (80 + 60) instead of multiplying and dividing.
- C
  40
  Why not C: Dividing $n = 200$ by 5 — averaging without using the correct marginal totals.
- D
  0.024
  Why not D: Dividing $(80 \times 60)$ by $20 0^{2}$ instead of $200$ .
Explanation
Expected cell count $= \frac{row total \times column total}{n} = \frac{80 \times 60}{200} = \frac{4800}{200} = 24$ .

Key takeaway
Expected count for cell $(i,j)$: $E_{ij} = (\text{row}_i \text{ total} \times \text{col}_j \text{ total}) / n$.
Question 9 · Medium
A chi-square goodness-of-fit test is performed on a spinner with 3 sections claimed to be equally likely. The test statistic is $χ^{2} = 5.4$ with $df = 2$ . The $p$ -value is approximately 0.067. At $α = 0.05$ , is there evidence the spinner is unfair?
- A
  Yes; $χ^{2} = 5.4$ is large, so the spinner is definitely unfair.
  Why not A: The size of $χ^{2}$ must be compared to a critical value or $p$ -value, not judged in absolute terms.
- B
  No; $p$ -value = 0.067 > $α = 0.05$ , so we fail to reject the null hypothesis that the spinner is fair.Correct
- C
  Yes; any $χ^{2} > 5$ with small $df$ indicates a significant result.
  Why not C: The critical value for $df = 2$ at $α = 0.05$ is 5.991; $χ^{2} = 5.4$ does not exceed it.
- D
  No; the $p$ -value should be doubled for a two-sided test.
  Why not D: Chi-square tests are always one-tailed (right tail); there is no two-sided version.
Explanation
The critical value for $df = 2$ at $α = 0.05$ is 5.991. Since $χ^{2} = 5.4 < 5.991$ (equivalently, $p = 0.067 > 0.05$ ), we fail to reject $H_{0}$ . There is insufficient evidence at the 5% level to conclude the spinner is unfair.

Key takeaway
Chi-square critical value for $df = 2$ at $\alpha = 0.05$ is 5.991; $\chi^2 < 5.991$ fails to reject $H_0$.
Question 10 · Medium
A researcher conducts a chi-square test of independence with $χ^{2} = 15.3$ , $df = 4$ , $p < 0.01$ . A student says the test shows that column variable $B$ causes changes in row variable $A$ . What is the error?
- A
  The test only shows association, not direction; which variable causes the other cannot be determined.
  Why not A: While association has no direction issue, the main error is that association (even from a study with random assignment) doesn't automatically mean causation from chi-square alone.
- B
  Chi-square tests measure association, not causation; without an experiment, causal conclusions are not justified.Correct
- C
  A $p$ -value < 0.01 is not small enough to conclude association.
  Why not C: $p < 0.01$ is strong evidence against independence; the issue is the causal interpretation, not the strength of evidence.
- D
  The $df = 4$ is too small for a valid chi-square test.
  Why not D: $df = 4$ is a perfectly valid degrees of freedom for a chi-square test.
Explanation
Chi-square tests of independence detect statistical association between categorical variables. They do not establish causation. Causal conclusions require random assignment in a controlled experiment, not merely a survey or observational study.

Key takeaway
Chi-square tests detect association, not causation; causal claims require randomized experimental design.
Question 11 · Hard
In a chi-square test of independence, which of the following conditions must be met to proceed?

I. The data come from a random sample.
II. Expected cell counts are at least 5.
III. All observed counts are at least 5.
- A
  I only
  Why not A: Randomness alone is not sufficient; the large-counts condition on expected counts must also be met.
- B
  I and III only
  Why not B: The condition applies to expected counts, not observed counts; III is not required.
- C
  I and II onlyCorrect
- D
  I, II, and III
  Why not D: Observed counts can be 0; it is the expected counts that must be at least 5.
Explanation
Two conditions must be met for a chi-square test of independence: (I) the data must come from a random sample (or random assignment), and (II) all expected cell counts must be at least 5. The condition about observed counts (III) is not required — observed counts can be 0 or very small.

Key takeaway
Chi-square conditions: (1) random sample and (2) all expected counts ≥ 5. Observed counts can be small.
Question 12 · Hard
A 3×2 contingency table (3 education levels × 2 voting outcomes) shows the following row totals and column totals:

Voted Did not vote Row total
High school 80
Some college 60
College grad 60
Col total 120 80 200

The expected count for the High school/Voted cell is $E_{11}$ . Compute $χ^{2}$ for this cell if the observed count is 40.
- A
  $E_{11} = 48$ ; cell contribution $= 1.33$ Correct
- B
  $E_{11} = 40$ ; cell contribution $= 0$
  Why not B: $E_{11} \neq = O_{11}$ here; $E_{11} = (80 \times 120) /200 = 48 \neq = 40$ .
- C
  $E_{11} = 60$ ; cell contribution $= 6.67$
  Why not C: Using the column total (120/2 = 60) as the expected count, ignoring the row total.
- D
  $E_{11} = 48$ ; cell contribution $= 0.133$
  Why not D: Dividing $(O - E)^{2}$ by $O$ instead of $E$ : $(40 - 48)^{2} /40 = 64/40 = 1.6$ , or some other variant.
Explanation
$E_{11} = \frac{row _{1} \times col _{1}}{n} = \frac{80 \times 120}{200} = 48$ . Cell contribution: $\frac{( 40 - 48 ) ^{2}}{48} = \frac{64}{48} = \frac{4}{3} \approx 1.33$ .

Key takeaway
Compute expected counts before finding $\chi^2$; each cell contributes $(O-E)^2/E$ to the test statistic.

	Voted	Did not vote	Row total
High school			80
Some college			60
College grad			60
Col total	120	80	200