Week 12 — Quiz (auto-graded) · Confidence Intervals for a Proportion

Course: Introduction to Statistics (18-week generic edition)
Objective tested: Objective 6 — one-proportion z-intervals (conditions, construction, interpretation), sample size, and the margin of error in the media.
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Week 12 · 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-12-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 12 misconceptions named in the lecture outline.

Questions, key, and feedback

Q1 (MC). A streaming-video service pulls a random sample of 200 accounts that came up for renewal last quarter; 128 renewed. The value 128 ÷ 200 = 0.64 is best described as —
- A. The population proportion p of all accounts that renew
- B. The sample proportion p̂ — a statistic estimating p ✅
- C. The margin of error of the renewal survey
- D. A parameter describing all of the service's accounts
Feedback: Successes ÷ sample size computed from a sample is p̂ — measured, not true. The all-accounts renewal rate is the parameter p, which 0.64 estimates. (Week 1's hat, still on duty.)

Q2 (MC). A property-management group asks a random sample of 150 apartment units whether they set out recycling; 90 do, so p̂ = 0.60. What is the standard error of p̂, √(0.60 × 0.40 ⁄ 150)?
- A. 0.04 ✅
- B. 0.0016
- C. 0.0033
- D. 0.0196
Feedback: 0.60 × 0.40 = 0.24; 0.24 ⁄ 150 = 0.0016; √0.0016 = 0.04. Distractor B stops before the square root; C and D take the square root of the wrong piece.

Q3 (MC). At a campus, 225 of 300 randomly sampled students have turned on the safety app's alert feature — p̂ = 0.75, and the standard error works out to exactly 0.025. Which is the 95% confidence interval for the proportion of all students with alerts on?
- A. (0.725, 0.775)
- B. (0.709, 0.791)
- C. (0.701, 0.799) ✅
- D. (0.686, 0.814)
Feedback: ME = 1.96 × 0.025 = 0.049, so 0.75 ± 0.049 → (0.701, 0.799). Option A forgot the multiplier entirely; B used the 90% multiplier (1.645); D used the 99% multiplier (2.576).

Q4 (MC). An online retailer computes a 95% confidence interval for the proportion of its orders that get returned: (0.08, 0.12). Which interpretation is correct?
- A. About 95% of the retailer's orders are returned at rates between 8% and 12%
- B. There is a 95% probability that the true return proportion lies inside (0.08, 0.12)
- C. We are 95% confident the interval 0.08 to 0.12 captures the true proportion of all orders returned ✅
- D. If new samples were drawn, 95% of them would give p̂ exactly between 0.08 and 0.12
Feedback: The 95% belongs to the method — about 19 of every 20 random samples yield a capturing interval. A puts people where a proportion belongs; B assigns probability to the fixed p; D describes p̂'s, not intervals.

Q5 (MC). A boutique checks a random sample of 40 online orders and finds 4 were returned. Why can't a one-proportion z-interval for the return rate be built from this sample?
- A. The sample was not selected using random chance
- B. The large-counts condition fails — only 4 successes, fewer than 10 ✅
- C. The population of orders is not at least 10 times the sample size
- D. No interval is possible because n is below 30, the minimum for inference
Feedback: Counting "returned" as a success: 4 < 10, so the z machinery isn't reliable here. The sample was random; the 10% condition is fine; and "n ≥ 30" is a rule of thumb from the means world, not this week's condition.

Q6 (Matching). Match each piece of this week's machinery to its meaning.
| Machinery | Correct meaning |
|---|---|
| The sample proportion p̂ | The interval's center — successes divided by sample size |
| The margin of error | z × SE — the reach the interval extends on each side of the center |
| A 95% confidence level | The method's long-run capture rate — about 19 of every 20 random samples yield an interval containing p |
| The planning value p = 0.5 | The conservative choice that maximizes p(1 − p) when computing a required sample size |
Feedback: Center ± reach, a promise about the method, and the safe planning value — the week's whole anatomy in four rows.

Q7 (MC). A city wants to estimate the proportion of households joining its curbside compost pilot, to within ±0.05 at 95% confidence, with no prior estimate. Using n = p(1 − p)(z ⁄ ME)², how many households must be sampled?
- A. 271
- B. 384
- C. 1,537
- D. 385 ✅
Feedback: n = 0.25 × (1.96 ⁄ 0.05)² = 0.25 × 1536.64 = 384.16 → round up* → 385. Option B rounds down (breaks the promised margin); A uses the 90% multiplier; C forgets the 0.25.

Q8 (True / False). "A poll's reported margin of error (say, ±3 points) accounts for problems like leading question wording and nonresponse, as well as random sampling error."
- True
- False ✅
Feedback: False. The ± prices random sampling luck only. Wording, nonresponse, and undercoverage ride outside the margin — which is why a biased poll can be precisely wrong.

Q9 (Multiple answer — select all that apply). Which of the following are conditions that must be checked before building a one-proportion z-interval?
- A. The data come from a random sample ✅
- B. The sample contains at least 10 successes and at least 10 failures ✅
- C. The population is at least 10 times the sample size ✅
- D. The population's distribution is approximately normal
- E. The sample proportion is exactly 0.5
Feedback: Random, large counts, and the 10% condition are the password. "Population approximately normal" belongs to the means world — a yes/no population has no bell to check — and p̂ = 0.5 is a planning convenience, never a requirement.

Q10 (MC). A regional outlet reports a poll on a school-bond question: 48% support, margin of error ±3 percentage points (95% confidence). Which statement follows from the report?
- A. Support is guaranteed to fall somewhere between 45% and 51% of all voters
- B. The plausible range runs from about 45% to 51%, so a majority can't be ruled out ✅
- C. Exactly 48% of all of the region's voters currently support the school bond
- D. The bond is sure to fail, because the poll's point estimate landed below 50%
Feedback: Read the whole interval: 48 ± 3 → 45% to 51%, which straddles 50% — so majority support is neither established nor ruled out. "Guaranteed" and "exactly" overclaim; "sure to fail" ignores the interval's upper half.

Answer key (quick reference)

Quality gate (self-checked): each single-answer item has exactly one correct option; the multiple-answer item's three correct conditions are the only true options listed; every computation re-verified in the week's math script (0.24⁄150 = 0.0016 → SE 0.04; 1.96 × 0.025 = 0.049 → (0.701, 0.799) with 1.645/2.576 distractor intervals confirmed; 0.25 × (1.96⁄0.05)² = 384.16 → 385; 48 ± 3 → 45 to 51); no positional pattern in the MC key (B A C C B D B — no letter above 3 of 7) and option lengths within each item are comparable; no item asserts a fact outside the Week 12 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=12 · objective=6 · topic=ci-for-a-proportion and deposited in Item Bank: Week 12 — Confidence Intervals for a Proportion with idents w12q1–w12q10. The final (Week 18) and per-term variant updates draw fresh variants from this bank's concepts — never these live stems. (Tags: w12q1 phat-statistic, w12q2 standard-error, w12q3 construct-interval, w12q4 interpret-confidence, w12q5 large-counts, w12q6 interval-anatomy, w12q7 sample-size, w12q8 moe-coverage, w12q9 conditions, w12q10 media-poll-reading.)

Canvas placement block

canvas_object    = Quizzes::Quiz
title            = "Week 12 Quiz — Confidence Intervals for a Proportion"
assignment_group = "Quizzes"
points_possible  = 10
grading_type     = points
due_offset_days  = 6        # end of the module's week
published        = true
shuffle_answers  = true