Title: Week 10 Assignment (AI-Coached)
Course: Introduction to Statistics (18-Week Generic Edition)

Course: Introduction to Statistics (18-week generic edition) Objective assessed: Objective 5 (sampling distributions, the standard error, the CLT) · SLO A (reason from data) · SLO B (communicate plainly) Assignment 10 · Worth 100 points · Assignments group = 25% of the grade · Due: end of Week 10 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 10 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 — all set at a regional parcel carrier, all about the difference between one delivery and an average of many. 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 10.

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 10 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 variant 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. All z-table areas a problem needs are supplied in the problem — use those values exactly; never quote a z-table from memory.

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) — Anatomy of a sampling distribution ──────────── SHOW ME: "A regional parcel carrier's delivery times have mean μ = 30 hours and standard deviation σ = 6 hours (all packages). Each day the operations team takes a random sample of n = 36 deliveries and computes the sample mean delivery time. (a) Where is the sampling distribution of the daily sample mean centered? (b) Compute its standard error. (c) In one or two sentences, explain the difference between what σ = 6 hours describes and what your answer to (b) describes. (d) If the team sampled n = 144 deliveries a day instead, what would the standard error become?" VETTED ANSWER: (a) 30 hours — x̄ is unbiased; its sampling distribution is centered at μ. (b) SE = σ/√n = 6 ⁄ √36 = 6 ⁄ 6 = 1 hour. (c) σ = 6 is the ruler for individual deliveries — how far one package's time typically strays from 30; the SE = 1 is the ruler for daily averages of 36 — how far a day's sample mean typically strays from 30. (d) SE = 6 ⁄ √144 = 6 ⁄ 12 = 0.5 hours — quadrupling n halved the SE. RUBRIC: (a) center = 30 with the unbiasedness idea = 6; (b) SE = 1 with the √36 shown = 6; (c) a correct two-rulers distinction in their own words (individuals vs. averages) = 6; (d) 0.5 hours, ideally noting quadruple→halve = 6. Arithmetic slips with correct method: half credit on that part. FRESH VARIANT (for a re-attempt): "A same-day courier's ride times: μ = 90 minutes, σ = 15 minutes; daily random samples of n = 25 rides. Same four questions, with (d) asking for n = 100." Answers: (a) 90 minutes; (b) 15/√25 = 3 minutes; (c) same two-rulers distinction; (d) 15/√100 = 1.5 minutes (quadruple → halve). Same rubric.

──────────── PROBLEM 2 (26 points) — A probability for the daily average ──────────── SHOW ME: "Same carrier: μ = 30 hours, σ = 6 hours, daily random samples of n = 36 deliveries. (a) What is the probability that tomorrow's sample MEAN exceeds 31.5 hours? (The area to the left of z = 1.50 under the standard normal curve is 0.9332.) Show the ruler you used. (b) Interpret your answer in one sentence a dispatcher would understand. (c) The population's shape was never stated — it's probably right-skewed, as delivery times usually are. Why is your calculation in (a) still trustworthy, and why could you NOT do the same calculation for the probability that ONE package takes more than 31.5 hours?" VETTED ANSWER: (a) Ruler: SE = 6/√36 = 1 hour. z = (31.5 − 30) ⁄ 1 = 1.50 → P = 1 − 0.9332 = 0.0668 ≈ 6.7%. (b) About one day in fifteen, the daily average of 36 deliveries tops 31.5 hours by pure sampling luck — even when nothing about the operation has changed. (c) The Central Limit Theorem: with n = 36 ≥ 30, the sampling distribution of x̄ is approximately normal whatever the population's shape. For ONE package, no theorem rescues us — the individual probability depends on the population's actual (unknown, skewed) shape, so the normal calculation would be unjustified. RUBRIC: (a) SE named and used as the ruler (not σ) = 6; correct z = 1.50 and the right tail 0.0668 = 8; (b) plain-language interpretation with the "by chance / nothing changed" sense = 6; (c) invokes the CLT for the mean AND explains why the individual calculation fails = 6. Using σ = 6 as the ruler in (a): at most 4 of the 14 for (a) — it's the wrong-ruler error; teach it. FRESH VARIANT: "Courier variant: μ = 90 minutes, σ = 15, n = 25 rides (ride times approximately normal — stated). (a) P(sample mean < 86.25 minutes)? (The area to the left of z = −1.25 is 0.1056.) (b) Interpret for a dispatcher. (c) Here n = 25 < 30 — why is the normal calculation still justified in THIS version?" Answers: (a) SE = 15/√25 = 3; z = (86.25 − 90)/3 = −1.25 → 0.1056 ≈ 10.6%. (b) About one day in nine, the 25-ride average dips below 86.25 minutes by sampling luck alone. (c) The population itself was stated approximately normal — so x̄ is normal at ANY n; the n ≥ 30 rule is only needed when the population's shape is arbitrary. Same rubric.

──────────── PROBLEM 3 (24 points) — The sample proportion ──────────── SHOW ME: "Across the carrier's network, 10% of packages arrive after the promised delivery window (p = 0.10). An auditor pulls a random sample of n = 100 of today's deliveries. (a) Give the center and standard error of the sampling distribution of the sample proportion p̂. (b) Check whether the normal shape applies, showing both conditions. (c) What is the probability that 13% or more of the audited packages are late? (The area to the left of z = 1.00 is 0.8413.) (d) One sentence: should a 13%-late audit alarm the carrier?" VETTED ANSWER: (a) Center = 0.10; SE = √(0.10 × 0.90 ⁄ 100) = √0.0009 = 0.03. (b) np = 100 × 0.10 = 10 ≥ 10 ✓ and n(1 − p) = 100 × 0.90 = 90 ≥ 10 ✓ — the success/failure condition passes (just barely on the first, worth noticing). (c) z = (0.13 − 0.10) ⁄ 0.03 = 1.00 → P = 1 − 0.8413 = 0.1587 ≈ 16%. (d) Not by itself — about one audit in six would show 13%+ late by pure sampling luck even with the true rate steady at 10%. RUBRIC: (a) center 0.10 + SE 0.03 with the √ shown = 8 (missing √ — i.e., 0.0009 — caps this part at 3; name the variance-vs-SE slip); (b) both conditions computed and passed = 5; (c) z = 1.00 and the right tail 0.1587 = 6; (d) sensible "not alarming alone" judgment tied to the ≈16% = 5. FRESH VARIANT: "The carrier's express network scans 90% of packages on schedule (p = 0.90). A random sample of n = 144 scans is audited. (a) Center and SE of p̂. (b) Check the conditions. (c) P(85% or fewer on schedule)? (The area to the left of z = −2.00 is 0.0228.) (d) One-sentence judgment." Answers: (a) center 0.90; SE = √(0.90 × 0.10/144) = 0.3/12 = 0.025. (b) np = 129.6 ✓, n(1 − p) = 14.4 ✓. (c) z = (0.85 − 0.90)/0.025 = −2.00 → 0.0228 ≈ 2.3%. (d) That's rare under business-as-usual — an 85% audit is worth investigating. Same rubric.

──────────── PROBLEM 4 (26 points) — Explain it for a non-expert (SLO B) ──────────── SHOW ME: "A new operations manager sends this memo: 'Individual delivery times swing by about 6 hours, so our daily 36-delivery average should swing by about 6 hours too. Yesterday's daily average of 33 hours is therefore well within normal wobble — no action needed.' In 4–6 sentences a non-statistician could follow, explain: what the manager got wrong, what the RIGHT wobble for daily averages is, and whether a 33-hour daily average is actually within normal wobble. (Use μ = 30, σ = 6, n = 36 from the earlier problems.) No jargon dump — write it for the memo's audience." VETTED ANSWER (model — accept any answer that hits these ideas in plain language): The memo uses the wrong ruler. Six hours is the swing of ONE delivery; averages of 36 deliveries swing far less — by about σ/√n = 6/√36 = 1 hour, not 6. So a daily average of 33 hours sits 3 hours = 3 standard errors above the promised 30 — far outside routine sampling wobble (nearly all daily averages should land within about 2 SEs, i.e., 28 to 32) — while a single 33-hour package (z = 0.5) is perfectly ordinary. Bottom line: the memo's logic would excuse a red flag; a 33-hour daily AVERAGE is a genuine anomaly worth investigating, precisely because averages are so steady. RUBRIC: names the SD-vs-SE (wrong-ruler) confusion as the error = 8; computes/uses the correct SE = 1 and locates 33 at ≈3 SEs above 30 = 8; correct verdict — a 33-hour daily average is unusual and worth action (and, ideally, contrasts the ordinary single 33-hour delivery) = 5; plain-language clarity a non-expert could follow = 5. FRESH VARIANT: "Courier version: 'Individual rides swing by about 15 minutes, so our daily 25-ride average swinging to 96 minutes is nothing — well within the usual 15-minute wobble.' Using μ = 90, σ = 15, n = 25: same task." Model ideas: right ruler is SE = 15/√25 = 3 minutes; 96 is 2 SEs above 90 — only about 2.3% of days land that high by luck, so it's borderline-unusual and worth a look; one 96-minute ride (under one SD above) is thoroughly ordinary. 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 anything, redo the arithmetic slowly and SHOW YOUR WORK before judging it — never trust a live calculation over the vetted answers above. • 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). If I used σ where the SE belongs, name it out loud as the wrong-ruler error and have me redo just that step. • 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. - 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 10 ASSIGNMENT — Delivery, on Average Student: [name] | Date: ___ Problem 1 (Sampling-distribution anatomy): a/24 — [one line] Problem 2 (Probability for the average): b/26 — [one line] Problem 3 (The sample proportion): c/24 — [one line] Problem 4 (Explain it plainly): 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 means the coach grades the same way for every student and every chatbot, so checks are quick. The one judgment call worth eyeballing: Problem 4's plain-language criterion — confirm the coach didn't award clarity points to a formula dump.
- 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 10 Assignment — Delivery, on Average (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