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

Course: Introduction to Statistics (MATH 11) · Silver Oak University (fictional sample) · Prof. Rivera
Objective tested: Objective 6 — construct and interpret confidence intervals for proportions (one-proportion z-interval; margin of error; choosing a sample size; correct interpretation).
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Module 12.

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.

Embed-don't-trust: every z* value is supplied in the stem and every margin/endpoint/sample-size is pre-computed below. No item requires a z-table lookup. The supplied values used are 90% → 1.645, 95% → 1.960, 99% → 2.576.

Blueprint

No trick questions; distractors target the Week 12 misconceptions named in the lecture outline (t/df-vs-z for a proportion, forgetting the large-counts check, "more confidence is always better," getting stuck on the sample size when p̂ is unknown / rounding down, and the two CI misreadings). All arithmetic is pre-computed and vetted.

Questions, key, and feedback

Q1 (MC). You want a confidence interval for a population proportion from one random sample. Which critical value should you use?
- A. The t-distribution, with degrees of freedom df = n − 1
- B. z*, with no degrees of freedom ✅
- C. The binomial coefficient
- D. Either t or z — they always give the same answer
Feedback: A proportion uses z* (from our supplied table) and has no degrees of freedom. The t-with-df method was last week's interval for a mean, where we estimated the spread with s. (A is the classic carry-over trap.)

Q2 (MC). Which formula gives a confidence interval for a population proportion?
- A. p̂ ± z*·(p̂(1−p̂)/n)
- B. p̂ ± z*·√(p̂(1−p̂)/n) ✅
- C. p̂ ± t*·(s/√n)
- D. p̂ ± z*·√(p̂/n)
Feedback: The interval is p̂ ± z*·√(p̂(1−p̂)/n); the √(p̂(1−p̂)/n) is the standard error. (A forgets the square root; C is the mean formula; D drops the (1−p̂).)

Q3 (True / False). Before trusting a one-proportion z-interval, the large-counts (success–failure) condition requires at least 10 expected successes and 10 expected failures — that is, n·p̂ ≥ 10 and n(1−p̂) ≥ 10.
- True ✅
- False
Feedback: True. The normal-model interval is trustworthy only when both n·p̂ and n(1−p̂) are at least 10 (along with random sampling and independence). It mainly bites for small n or a very rare/very common trait.

Q4 (MC). A poll has p̂ = 0.20 and n = 100. The standard error (SE = √(p̂(1−p̂)/n)) is — (note: 0.20 × 0.80 = 0.16, and 0.16 ÷ 100 = 0.0016)
- A. 0.16
- B. 0.0016
- C. 0.04 ✅
- D. 0.20
Feedback: √(0.0016) = 0.04. Don't stop at the value inside the root — take its square root. (A is p̂(1−p̂); B is the variance before the square root.)

Q5 (MC). A poll finds p̂ = 0.40 with n = 600, so the standard error is 0.02. Using the supplied z* = 1.960 (95%), the 95% confidence interval for the true proportion is —
- A. (0.38, 0.42)
- B. (0.361, 0.439) ✅
- C. (0.34, 0.46)
- D. (0.20, 0.60)
Feedback: ME = z*·SE = 1.960 × 0.02 = 0.039; interval = 0.40 ± 0.039 = (0.361, 0.439) — about 36% to 44%. (A uses just the SE without multiplying by z*; C and D are too wide.)

Q6 (MC). A survey finds p̂ = 0.25 with n = 300, so the standard error is 0.025. For a 95% interval the supplied z* = 1.960. The margin of error is —
- A. 0.025
- B. 0.049 ✅
- C. 0.0125
- D. 0.49
Feedback: ME = z*·SE = 1.960 × 0.025 = 0.049. (A is just the SE — you must multiply by z*; D is off by a factor of 10.)

Q7 (MC). Holding everything else fixed, which change makes the margin of error smaller (a narrower interval)?
- A. Increasing the sample size n ✅
- B. Increasing the confidence level from 95% to 99%
- C. Moving p̂ from 0.20 toward 0.50
- D. Switching the reported value from a decimal to a percentage
Feedback: A larger n shrinks the margin (n is under the square root in the denominator). Higher confidence (bigger z*) and a p̂ nearer 0.5 (bigger p̂(1−p̂)) both widen it. (B is the "more confidence is always better" trap — it costs width.)

Q8 (MC). A pollster wants a 95% interval with a target margin of error but has no prior estimate of the true proportion. To get the safe (largest) required sample size, which value of p̂ should they plug into n = (z*/ME)²·p̂(1−p̂)?
- A. 0.0
- B. 0.5 ✅
- C. 1.0
- D. The exact true proportion, which must be known first
Feedback: Use the worst case p̂ = 0.5 — it maximizes p̂(1−p̂) at 0.25, giving the largest n you could need, so you're safe whatever the true proportion turns out to be. (D is the "I'm stuck without p̂" trap — you're never stuck.)

Q9 (MC). For 95% confidence (z* = 1.960) with a target margin of error 0.04 and worst-case p̂ = 0.5: (z*/ME)² = (1.960/0.04)² = (49)² = 2401, and 2401 × 0.25 = 600.25. The required sample size is —
- A. 600
- B. 601 ✅
- C. 2401
- D. 1200
Feedback: A sample size is always rounded UP to the next whole person, so 600.25 → 601. (A rounds down — which would miss the target margin; C forgets to multiply by p̂(1−p̂) = 0.25.)

Q10 (Multiple answer — select all that apply). A 95% confidence interval for the proportion of voters who approve a measure is (0.46, 0.54). Which statements are correct interpretations?
- A. We are 95% confident that the true proportion of all voters who approve is between 46% and 54%. ✅
- B. If we repeated this polling many times, about 95% of the intervals we built would contain the true proportion. ✅
- C. 95% of voters approve at a rate between 46% and 54%.
- D. There is a 95% probability that the true proportion falls in this particular interval (0.46, 0.54).
Feedback: A and B are correct — both describe confidence about the proportion and the method's long-run capture rate. C is the "95% of people" misread (a CI is about the overall rate, not individuals); D is the "95% chance for this fixed interval" misread (the interval is already decided; the 95% belongs to the procedure).

Answer key (quick reference)

Quality gate (self-checked): each single-answer item has exactly one correct option; the multiple-answer item (Q10) lists both correct interpretations and both named misreadings as distractors. Arithmetic re-verified: Q4 SE = √(0.20×0.80/100) = √0.0016 = 0.04; Q5 ME = 1.960×0.02 = 0.0392 → (0.361, 0.439); Q6 ME = 1.960×0.025 = 0.049; Q9 (1.960/0.04)² × 0.25 = 2401 × 0.25 = 600.25 → 601. Every z* value is supplied in its stem (no lookup), and no item asserts a fact outside the Week 12 course definitions.

Item-bank entries (for variants + the midterm/final)

All ten items are tagged course=MATH11 · week=12 · objective=6 · topic=confidence-intervals-proportions and deposited in Item Bank: Week 12 — Confidence Intervals for Proportions. The final (Week 16) and the per-term variant updates draw fresh items from this bank. (Tags: q1 z-not-t-proportion, q2 ci-formula-proportion, q3 large-counts-condition, q4 standard-error-proportion, q5 construct-ci, q6 margin-of-error, q7 me-drivers, q8 worst-case-phat, q9 sample-size, q10 ci-interpretation.)

Canvas placement block

canvas_object   = Quizzes::Quiz
title           = "Week 12 Quiz — Confidence Intervals for Proportions"
assignment_group = "Quizzes"
points_possible = 10
grading_type    = points
due_offset_days = 6        # 6 days after module start (Sun Nov 22)
published       = true
shuffle_answers = true
provenance      = "~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com"