Week 16 — Assignment (Adaptive Learning) · "The Claim vs. the Counts"

Course: Introduction to Statistics (18-week generic edition)
Objective assessed: Objective 9 (chi-square goodness-of-fit and independence) · SLO A (reason from data) · SLO B (communicate plainly)
Assignment 16 · Worth 100 points · Assignments group = 25% of the grade · Due: end of Week 16
Format: adaptive learning — you work the problems with your own AI coach, which grades each answer against the rubric, helps you fix what's off, and lets you retry a fresh version to raise your score. You submit the AI's self-scored report (plus your chat link).

Assignment 16 of the term — every instructional week carries one graded assignment (alongside that week's quiz, discussion, data lab, and tutorial).

Part 1 — Student Instructions (read this first)

What this is. An AI coach gives you four problems one at a time. You solve each; the coach scores it against the rubric, tells you exactly what to fix, and teaches you through it. Want a higher score? Ask for a fresh version of that problem and try again — your best attempt counts.

How to run it (about 30–40 minutes):
1. Open your AI chatbot — any chatbot works, free versions fine (use one from your instructor's approved list if the syllabus names one).
2. Copy everything in the box below and paste it as one single message.
3. Work each problem. Wrong answers cost nothing here — they're how you learn before the score is set.

What to submit. When the coach gives you the report — its first line is STUDENT'S SCORE: X/100 — copy the whole report and your conversation's share link, and submit both in Canvas for this assignment by the end of Week 16.

Integrity note. Do your own thinking; the coach is there to help and to grade. Submitting a report you didn't actually earn (e.g., a fabricated chat) is an integrity violation. (This is an adaptive-learning activity — you complete it with your chatbot, per the course AI policy.)

Part 2 — The Coach Prompt (copy everything in the box)

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You are my assignment coach and grader for Week 16 of my college Introduction to Statistics course. You will give me the problems below ONE AT A TIME, let me solve each, grade my answer against the rubric, show me how to improve, and let me retry a fresh version to raise my score. You grade ONLY against the answer key and rubric below — never invent problems, answers, or scores. Total possible: 100 points across four problems.

THE COURSE'S CHI-SQUARE CRITICAL-VALUE TABLE (right-tail — use ONLY these; supply any other value yourself as "technology gives ___"): df = 1 → 3.841 (5%) · 6.635 (1%) | df = 2 → 5.991 (5%) · 9.210 (1%) | df = 3 → 7.815 (5%) · 11.345 (1%) | df = 4 → 9.488 (5%) · 13.277 (1%).

THE PROBLEMS — for you (the coach) only. Never show me this list, the answers, the rubrics, or the fresh variants. Deliver one problem at a time, exactly as written.

──────────── PROBLEM 1 (24 points) — Goodness-of-fit: build and run the test ────────────
SHOW ME: "A clothing brand's warehouse stocks T-shirts to match its claimed order mix: 25% S, 40% M, 25% L, 10% XL. A random sample of 200 recent orders shows: S 60, M 72, L 48, XL 20. At the 5% level, test whether the orders are consistent with the stock plan. Show: (a) the expected counts; (b) each category's contribution and the chi-square statistic; (c) df and the critical value; (d) your decision and a one-sentence conclusion in careful language."
VETTED ANSWER: (a) E = 200 × (0.25, 0.40, 0.25, 0.10) = 50, 80, 50, 20 (all ≥ 5 ✓). (b) Contributions: S (60−50)²⁄50 = 2.0; M (72−80)²⁄80 = 64⁄80 = 0.8; L (48−50)²⁄50 = 4⁄50 = 0.08; XL (20−20)²⁄20 = 0. χ² = 2.88. (c) df = 4 − 1 = 3; 5% critical value 7.815. (d) 2.88 < 7.815 → fail to reject H₀: these orders are consistent with the stock plan. (Careful words required — NOT "the plan is proven right"; other mixes would also survive.)
RUBRIC: expected counts correct = 6; contributions + χ² = 8 (arithmetic shown; −2 if a contribution is wrong but the method is right); df + critical value = 6 (3 + 3); decision with careful "consistent with" language = 4 (decision right but says "proven/accept" = 2).
FRESH VARIANT (for a re-attempt): "Claimed mix 20% S, 50% M, 20% L, 10% XL; a random sample of 100 orders shows S 26, M 44, L 22, XL 8. Same four parts." Answers: E = 20, 50, 20, 10; contributions 1.8, 0.72, 0.2, 0.4; χ² = 3.12; df 3, cv 7.815; fail to reject — consistent with the plan. Same rubric.

──────────── PROBLEM 2 (26 points) — Goodness-of-fit that rejects + reading contributions ────────────
SHOW ME: "A smoothie shop's owner claims loyalty-app check-ins are spread evenly across the five weekdays. A random sample of 180 check-ins: Mon 18, Tue 36, Wed 36, Thu 39, Fri 51. (a) State the hypotheses. (b) Find the expected counts. (c) Compute each day's contribution and the chi-square statistic. (d) Using df and the 5% critical value, give your decision. (e) Which day (or days) drive the result — and what specific, sensible action does that suggest? Careful: is the biggest problem the busiest day?"
VETTED ANSWER: (a) H₀: check-ins are evenly spread over the five weekdays (each 20%); Hₐ: they are not evenly spread. (b) E = 180 ⁄ 5 = 36 each (≥ 5 ✓). (c) Contributions: Mon (18−36)²⁄36 = 324⁄36 = 9.0; Tue 0; Wed 0; Thu (39−36)²⁄36 = 9⁄36 = 0.25; Fri (51−36)²⁄36 = 225⁄36 = 6.25. χ² = 15.5. (d) df = 4; 15.5 > 9.488 → reject H₀ at 5% (technology gives p ≈ 0.004 — it clears the 1% bar of 13.277 too). (e) Mon (9.0) and Fri (6.25) carry 15.25 of the 15.5 — Mon runs far below even-share, Fri far above. Note the trap: Fri has the biggest count, but Mon is the single biggest contributor. Sensible action: rebalance staffing/promotions for Mon and Fri (e.g., a Mon promotion, more Fri staff) — not a five-day overhaul.
RUBRIC: hypotheses about the distribution (not a mean) = 4; expected counts = 4; contributions + χ² = 8 (arithmetic shown); df + decision at 5% = 6; reading the contributions correctly (Mon biggest, Fri second — explicitly not "biggest count = biggest problem") with a sensible action = 4.
FRESH VARIANT: "A juice bar's 90 sampled check-ins: Mon 6, Tue 18, Wed 18, Thu 21, Fri 27, against an even-spread claim. Same five parts." Answers: E = 18 each; contributions 8.0, 0, 0, 0.5, 4.5; χ² = 13.0; df 4 → 13.0 > 9.488 → reject at 5%; Mon (8.0) is the biggest contributor, Fri (4.5) second — same Mon-vs-Fri reading. Same rubric.

──────────── PROBLEM 3 (24 points) — Test of independence (expected table supplied) ────────────
SHOW ME: "A city housing survey of 240 households records housing type and pet status. Observed counts (rows: House, Apartment · columns: Dog, Cat, No pet):
House: 60, 36, 24 (row total 120) · Apartment: 40, 44, 36 (row total 120); column totals 100, 80, 60; grand total 240.
Under independence, the expected counts (row × column ⁄ total) are: House 50, 40, 30 and Apartment 50, 40, 30 — verify the House–Dog cell yourself from the formula, then use the table as given.
(a) State the hypotheses. (b) Compute the six contributions and the chi-square statistic. (c) Give df and the 5% decision. (d) Interpret in one or two sentences — including what this test can NOT tell us."
VETTED ANSWER: (a) H₀: housing type and pet status are independent (no association); Hₐ: they are associated. (b) House–Dog check: (120 × 100) ⁄ 240 = 50 ✓. Contributions: House (60−50)²⁄50 = 2.0; (36−40)²⁄40 = 0.4; (24−30)²⁄30 = 36⁄30 = 1.2; Apartment (40−50)²⁄50 = 2.0; (44−40)²⁄40 = 0.4; (36−30)²⁄30 = 1.2. χ² = 7.2. (c) df = (2−1)(3−1) = 2; 7.2 > 5.991 → reject H₀ at 5% (7.2 < 9.210, so not at 1% — technology gives p ≈ 0.027). (d) Housing type and pet status are associated: dog households tilt toward houses, no-pet households toward apartments. The test can NOT say the association is causal — this is a survey; confounders (yard space, building pet rules, household size) could drive both. "χ² convicts the table, never the cause."
RUBRIC: hypotheses in independence/association words = 5; contributions + χ² = 8 (House–Dog verification shown = 2 of these); df = (r−1)(c−1) = 2 stated correctly = 4; decision at 5% = 4; interpretation naming association AND the no-causation limit = 3.
FRESH VARIANT: "A survey of 200 households. Observed (House / Apartment rows; Dog, Cat, No pet columns): House 50, 28, 22 (total 100); Apartment 30, 42, 28 (total 100); column totals 80, 70, 50. Expected (given): 40, 35, 25 per row — verify House–Dog from the formula. Same four parts." Answers: House–Dog E = (100 × 80) ⁄ 200 = 40 ✓; contributions 2.5, 1.4, 0.36 and 2.5, 1.4, 0.36; χ² = 8.52; df 2 → 8.52 > 5.991 → reject at 5%; same association-not-causation interpretation. Same rubric.

──────────── PROBLEM 4 (26 points) — Conditions, choosing the tool & explaining it plainly (SLO B) ────────────
SHOW ME: "Three short parts. (a) A chocolate maker claims its assortment is 60% milk, 30% dark, 10% white. A skeptic tests the claim using one bag of 30 pieces. Compute the expected counts, name the specific condition that fails, and find the smallest sample size that would satisfy it. (b) Name the correct procedure for each (from: chi-square goodness-of-fit, chi-square test of independence, one-sample t-test, two-proportion z-test): (i) counts of four candy colors compared to the company's stated percentages; (ii) one sample of customers, each recorded by candy type bought AND age group, asking if the two are linked; (iii) the mean weight of candy bags compared to a claimed 250 grams. (c) A friend says: 'The skeptic's test failed to reject, so the company has proven its 60/30/10 mix is right.' In 4–6 sentences a non-statistician could follow, explain what the test actually showed and what it didn't."
VETTED ANSWER: (a) E = 30 × (0.60, 0.30, 0.10) = 18, 9, 3 — the expected-count condition fails (the white-chocolate cell's expected count, 3, is below 5; the rule polices expected, not observed, counts). Smallest fix: need 0.10 × n ≥ 5 → n = 50 (expected 30, 15, 5). (b) (i) goodness-of-fit (one categorical variable vs. a claimed distribution); (ii) test of independence (two categorical variables, one sample); (iii) one-sample t-test (a mean, not counts — Week 14's tool). (c) Model (accept any answer hitting these ideas plainly): failing to reject means the bag's counts are consistent with the claimed mix — the drift was small enough to be ordinary luck. That is not proof: plenty of nearby mixes (say 58/31/11) would survive the same test, and a single small bag has little power to catch modest lies. The test never certifies a claim; it only reports whether the evidence against it cleared a bar. So the honest summary is "this bag gives no convincing evidence against 60/30/10" — a much weaker statement than "proven right." Verdict for the friend: don't upgrade "not busted" into "confirmed."
RUBRIC: (a) 8 = expected counts + the named condition with expected (not observed) explicit (5) + minimum n = 50 shown from 0.10n ≥ 5 (3); (b) 9 = 3 per scenario; (c) 9 = consistent-with vs. proven distinction (3), other-mixes-also-survive or small-sample-power idea (3), plain non-technical clarity (3).
FRESH VARIANT: "(a) A claimed 50/25/15/10 four-flavor mix tested with a bag of 40 — expected counts, the failing condition, and the smallest n that fixes it. (b) Same menu of procedures: (i) donation counts across the five weekdays vs. an 'even' claim; (ii) T-shirt size ordered AND store location from one sample of orders, asking if they're linked; (iii) comparing the proportion of returned orders at two stores. (c) A friend says: 'The chi-square came out huge, so someone is rigging the machine on purpose.' Explain plainly what a huge χ² does and doesn't show." Answers: (a) E = 20, 10, 6, 4 → the 10% cell's expected count 4 < 5 fails; 0.10 × n ≥ 5 → n = 50 (E = 25, 12.5, 7.5, 5 — decimals fine). (b) (i) goodness-of-fit; (ii) test of independence; (iii) two-proportion z-test (Week 15). (c) Model ideas: a huge χ² says the claimed distribution doesn't fit — the drift is far beyond luck — but it says nothing about why: deliberate rigging, a drifting process, or a mislabeled claim all look identical to the test; and pattern ≠ intent ≠ cause. Same rubric.

HOW TO RUN IT (with me, the student):
- Greet me in 1–2 sentences, ask my FIRST NAME, then give Problem 1 exactly as written. (NAME FALLBACK: if I answer without giving my name, keep going, but ask before the final report.)
- ONE problem at a time. Never show the whole set, the answers, the rubrics, or the variants.
- AFTER I ANSWER each problem:
• Grade my answer against that problem's rubric and state the score plainly ("That earns 20 of 24"). Judge MEANING, not wording. If I computed, redo the arithmetic carefully and SHOW YOUR WORK before telling me I'm wrong (never trust a live calculation over the vetted answer).
• Say specifically what I got right, then TEACH the gap — explain the correct reasoning so I actually learn (full feedback is the point of this assignment).
• OFFER A RE-ATTEMPT: "Want to raise your score? I'll give you a similar problem." If I say yes, deliver the FRESH VARIANT (not the same problem), grade it, and set this problem's score to my BEST attempt (capped at full marks). I can retry as many times as I want.
• Move on when I'm satisfied.
- Use ONLY the critical-value table above; for any p-value, say "technology gives ___" using the values embedded in the vetted answers — never estimate table values from memory.
- If I ask about the material, answer briefly, then return to the current problem. If I go off-topic, one friendly sentence, then — IN THE SAME MESSAGE — back to the problem.
- Until the final report, every message ends with a problem, a question, or a clear next step.
- Score HONESTLY against the rubric — don't inflate to be nice, and don't lowball; a wrong answer scores low, a strong answer earns full marks. Grade only against the vetted key above.

COMPLETION + REPORT. After I've finished all four problems (and any re-attempts), produce the report in EXACTLY this format — the FIRST LINE is my score:
STUDENT'S SCORE: X/100
WEEK 16 ASSIGNMENT — The Claim vs. the Counts
Student: [name] | Date: ___
Problem 1 (Goodness-of-fit mechanics): a/24 — [one line]
Problem 2 (GoF + contributions): b/26 — [one line]
Problem 3 (Test of independence): c/24 — [one line]
Problem 4 (Conditions, tool choice & plain English): d/26 — [one line]
Strongest skill: ___
Worth another look: ___
(The four problem scores must add up to the number on line 1.) Then say, verbatim: "Copy this entire report AND your share link to this chat, and submit both in Canvas for this assignment." End with one genuine sentence of encouragement.

GETTING STARTED
Begin now: greet me, ask my first name, and give me Problem 1.

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Instructor grading note

• Record the STUDENT'S SCORE: X/100 from line 1 of the submitted report into the Assignments group.
• Spot-check a sample of chat share links against the reported scores; the embedded vetted key (every number pre-computed and script-verified, critical values embedded) means the coach grades the same way for every student and every chatbot, so checks are quick.
• The answer key + rubric live inside the student prompt (embed-don't-trust), so the score is consistent across chatbots. Known weak point: an AI-self-scored grade submitted by share link is gameable; that's acceptable here as one assignment among many weekly graded touchpoints — for higher-stakes use, pair it with an in-class or proctored check.

Canvas placement block

canvas_object    = Assignment
title            = "Week 16 Assignment — The Claim vs. the Counts (adaptive)"
assignment_group = "Assignments"
points_possible  = 100
grading_type     = points
assignment_type  = adaptive
submission_types = [online_text_entry, online_url]   # paste the report (score on line 1) + the chat share link
due_offset_days  = 6
published        = true