Week 16 — Quiz (auto-graded) · Chi-Square Tests for Categorical Data

Course: Introduction to Statistics (18-week generic edition)
Objective tested: Objective 9 — chi-square goodness-of-fit and independence (expected counts, df, conditions, contributions, conclusions).
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Week 16 · Closed to AI.

This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in F-quiz-week-16-qti.xml; the reusable item-bank entries and the Canvas placement block are at the bottom of this file.

Blueprint

No trick questions; distractors target the Week 16 misconceptions named in the lecture outline.

Questions, key, and feedback

Q1 (MC). A snack company states that its tropical candy mix is 40% pineapple, 30% mango, 20% coconut, and 10% banana. You count the flavors in a large random sample of candies to check the stated mix. Which procedure fits this question?
- A. A chi-square test of independence
- B. A chi-square goodness-of-fit test ✅
- C. A one-sample t-test for a mean
- D. A two-proportion z-test
Feedback: One categorical variable (flavor) compared against a claimed distribution → goodness-of-fit. Independence needs two categorical variables; t and z tools handle means and proportions.

Q2 (MC). An animal shelter surveys a random sample of 500 adopters, recording each adopter's pet type (dog / cat / bird / fish) and housing (house / apartment / dorm). It asks whether pet type is related to housing. Which procedure fits this question?
- A. A chi-square goodness-of-fit test
- B. A paired t-test on matched pairs
- C. A chi-square test of independence ✅
- D. A z-test for one proportion
Feedback: Two categorical variables measured on one sample, asking "are they linked?" → test of independence. No claimed mix exists here — the expected counts come from the margins.

Q3 (MC). A print shop's stock plan assumes 35% of its T-shirt orders are size M. In a random sample of 200 orders, how many size-M orders would be expected if the plan is right?
- A. 70 — the sample size times the claimed proportion ✅
- B. 65 — halfway between the plan and the observed count
- C. 50 — the 200 orders spread evenly across four sizes
- D. 35 — the claimed percentage read directly as a count
Feedback: Expected count = n × claimed proportion = 200 × 0.35 = 70. It comes from the claim and the sample size alone — no observed counts needed, and a percentage is never itself a count.

Q4 (MC). A fruit-chew brand claims its four flavors — cherry, lemon, lime, grape — are equally common. A random sample of 80 chews: 24 cherry, 22 lemon, 18 lime, 16 grape (expected: 20 each). The chi-square statistic is —
- A. 0, because the positive and negative gaps cancel out
- B. 40, the total of the four squared gaps
- C. 12, the total of the four gap sizes ignoring sign
- D. 2.0, the sum of the four (O − E)² ⁄ E terms ✅
Feedback: Contributions: 16⁄20 + 4⁄20 + 4⁄20 + 16⁄20 = 0.8 + 0.2 + 0.2 + 0.8 = 2.0. Squaring stops gaps from canceling, and dividing by E scales each squared gap — skipping the ÷E step is what produces 40.

Q5 (MC). For the fruit-chew test in the previous setting, χ² = 2.0 with df = 3, and the 5% critical value is 7.815. The correct conclusion is —
- A. Reject H₀ — the sample proves the flavor mix differs from the claim
- B. Accept H₀ — the sample proves the claimed mix is exactly correct
- C. Fail to reject H₀ — these counts are consistent with the claimed mix ✅
- D. No verdict is possible, because the statistic is less than 5
Feedback: 2.0 does not exceed 7.815, so the drift is within ordinary luck: fail to reject. "Consistent with" is the careful phrase — surviving the test never proves the claim (fail to reject ≠ accept).

Q6 (Matching). Match each item to its description.
| Item | Correct description |
|---|---|
| Goodness-of-fit test | Compares one categorical variable's counts to a claimed distribution |
| Test of independence | Asks whether two categorical variables are associated |
| Expected count in a two-way table | Row total × column total ÷ grand total |
| Degrees of freedom, goodness-of-fit | Number of categories minus one |
Feedback: One variable vs. a claim → goodness-of-fit (df = k − 1); two variables, linked? → independence, with expected counts built from the margins.

Q7 (MC). A test of independence is run on a 4 × 3 table (four pet types by three housing types). The degrees of freedom are —
- A. 12, the number of cells in the table
- B. 6, from (4 − 1) × (3 − 1) ✅
- C. 11, the number of cells minus one
- D. 5, from (4 − 1) + (3 − 1)
Feedback: df = (r − 1)(c − 1) = 3 × 2 = 6. With the margins fixed, only six cells are free — "cells minus one" is the classic wrong turn.

Q8 (True / False). "If a goodness-of-fit test fails to reject H₀, the claimed distribution has been proven correct."
- True
- False ✅
Feedback: False. Failing to reject means the counts are compatible with the claim — many nearby distributions would survive the same test. Not guilty ≠ proven innocent.

Q9 (Multiple answer — select all that apply). Which of the following are required conditions for a chi-square test?
- A. The data are counts of individuals in categories, not percentages ✅
- B. Every expected count is at least 5 ✅
- C. The data come from a random or representative sample ✅
- D. Every observed count is at least 5
- E. The variable's histogram is approximately bell-shaped
Feedback: Random data, raw counts, and expected counts ≥ 5. The ≥ 5 rule polices expected counts (observed zeros are legal), and chi-square makes no bell-shape demand on the data.

Q10 (MC). A barbershop tests the claim that its appointments are spread evenly across the five weekdays. From a sample of 150 appointments (expected: 30 per day), the observed counts and contributions are: Mon 12 (contribution 10.8), Tue 30 (0.0), Wed 33 (0.3), Thu 36 (1.2), Fri 39 (2.7); χ² = 15.0. Fri has the most appointments. Which day supplies the strongest evidence against the even-spread claim?
- A. Fri, because it has the largest observed count
- B. Tue, because its contribution is exactly zero
- C. It cannot be determined without the p-value
- D. Mon, because its contribution 10.8 is the largest ✅
Feedback: Contributions rank surprise, not size: Mon sits 18 below its expected 30 and carries 10.8 of the 15.0. A big count near its expected value (Fri, 2.7) contributes far less — rank by (O − E)² ⁄ E, never by O.

Answer key (quick reference)

Quality gate (self-checked): each single-answer item has exactly one correct option; the multiple-answer item's three conditions are the only true requirements listed (D swaps observed for expected; E imports a normality demand chi-square doesn't make); no positional pattern in the key (B C A D C · B · D) and no length giveaway (options within each item are comparable lengths); every computed value re-verified in tools/checks/w16_math.py (Q3: 200 × 0.35 = 70; Q4: contributions 0.8/0.2/0.2/0.8 → χ² = 2.0, squared-gap total 40, gap-size total 12; Q7: (4−1)(3−1) = 6; Q10: contributions 10.8/0.0/0.3/1.2/2.7 → χ² = 15.0 with Mon largest while Fri has the largest count); no item asserts a fact outside the Week 16 course definitions; no scenario reuses the tutorial, practice, chapter, lab, or assignment surfaces.

Item-bank entries (for variants + the final)

All ten items are tagged week=16 · objective=9 · topic=chi-square-tests and deposited in Item Bank: Week 16 — Chi-Square Tests for Categorical Data with idents w16q1–w16q10. The final (Week 18) and per-term variant updates draw fresh variants from this bank's concepts — never these live stems. (Tags: w16q1 gof-recognition, w16q2 independence-recognition, w16q3 expected-count, w16q4 compute-chi-square, w16q5 decision-language, w16q6 test-definitions, w16q7 df-independence, w16q8 fail-to-reject, w16q9 conditions, w16q10 contributions.)

Canvas placement block

canvas_object    = Quizzes::Quiz
title            = "Week 16 Quiz — Chi-Square Tests for Categorical Data"
assignment_group = "Quizzes"
points_possible  = 10
grading_type     = points
due_offset_days  = 6        # end of the module's week
published        = true
shuffle_answers  = true