Week 11 — Quiz (auto-graded) · Confidence Intervals for Means
Course: Introduction to Statistics (MATH 11) · Silver Oak University (fictional sample) · Prof. Rivera
Objective tested: Objective 6 — construct and interpret confidence intervals for means (t-distribution; constructing a CI; margin of error; correct interpretation).
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Module 11.
This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in
F-quiz-week-11-qti.xml; the reusable item-bank entries and the Canvas placement block are at the bottom of this file.Embed-don't-trust: every t* value is supplied in the stem and every margin/endpoint is pre-computed below. No item requires a t-table lookup. The supplied values used are 95%: df 9 → 2.262, df 15 → 2.131, df 24 → 2.064, df ∞ → 1.960; 90%: df 9 → 1.833, df ∞ → 1.645.
Blueprint
| # | Type | Concept | Objective |
|---|---|---|---|
| 1 | Multiple choice | When to use t vs z (σ unknown) | 6 |
| 2 | Multiple choice | Degrees of freedom (df = n − 1) | 6 |
| 3 | True / False | t-distribution is wider than z (fatter tails) | 6 |
| 4 | Multiple choice | Identify the CI-for-a-mean formula | 6 |
| 5 | Multiple choice | Compute the standard error | 6 |
| 6 | Multiple choice | Construct a full 95% confidence interval | 6 |
| 7 | Multiple choice | Compute the margin of error | 6 |
| 8 | Multiple choice | What shrinks the margin of error | 6 |
| 9 | Multiple choice | Compute the margin of error at 90% | 6 |
| 10 | Multiple answer | Correct interpretation vs. the misinterpretations | 6 |
No trick questions; distractors target the Week 11 misconceptions named in the lecture outline (z-vs-t, forgetting √n, "more confidence is always better," 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 mean. You know the sample mean and the sample standard deviation s, but you do not know the population standard deviation σ. Which model should you use?
- A. The z (standard normal) model
- B. The t-distribution ✅
- C. The binomial model
- D. Either one — they always give the same answer
Feedback: With σ unknown (the usual case), use the t-distribution with df = n − 1. z is only for the rare case σ is known. (A is the classic z-vs-t trap.)
Q2 (MC). You collect a sample of n = 25 to build a confidence interval for the mean. The degrees of freedom are —
- A. 25
- B. 24 ✅
- C. 26
- D. 5
Feedback: For a one-sample mean, df = n − 1 = 25 − 1 = 24. That's the row you'd read in the t* table.
Q3 (True / False). Compared with the standard normal (z) distribution, the t-distribution has fatter tails, so for the same confidence level its critical value is larger than z.
- True ✅
- False
Feedback: True. The t-distribution is wider (fatter tails) to account for s being an estimate, so t* > z for the same confidence — and t* shrinks toward z as df grows.
Q4 (MC). Which formula gives a confidence interval for a population mean (σ unknown)?
- A. x̄ ± t*·s
- B. x̄ ± t*·(s/√n) ✅
- C. x̄ ± z·(s·√n)
- D. x̄ ± t*·(s/n)
Feedback: The interval is x̄ ± t*·(s/√n); the s/√n is the standard error. (D forgets the square root; A forgets to divide by √n entirely.)
Q5 (MC). A sample has standard deviation s = 15 and size n = 25. The standard error (SE = s/√n) is —
- A. 15
- B. 0.6
- C. 3 ✅
- D. 75
Feedback: √25 = 5, and 15 ÷ 5 = 3. Divide by √n, not n. (B divides by n; D multiplies.)
Q6 (MC). A random sample of n = 25 has mean x̄ = 68 and sample SD s = 15, so the standard error is 3. Using the supplied t* = 2.064 (95%, df 24), the 95% confidence interval for the mean is —
- A. (65, 71)
- B. (61.8, 74.2) ✅
- C. (38, 98)
- D. (66.0, 70.0)
Feedback: ME = t*·SE = 2.064 × 3 = 6.19; interval = 68 ± 6.19 = (61.8, 74.2). (A uses just the SE; D uses 2/3 of it — both forget to multiply by t*.)
Q7 (MC). A random sample of n = 16 has sample SD s = 40, so the standard error is 10. For a 95% interval the supplied t* = 2.131 (df 15). The margin of error is —
- A. 10
- B. 21.31 ✅
- C. 85.24
- D. 2.5
Feedback: ME = t*·SE = 2.131 × 10 = 21.31. (A is just the SE — you must multiply by t*.)
Q8 (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. Increasing the sample standard deviation s
- D. Switching from z to t
Feedback: A larger n shrinks the margin (√n is in the denominator). Higher confidence and larger s both widen it. (B is the "more confidence is always better" trap — it costs width.)
Q9 (MC). For a 90% confidence interval, a sample has standard error SE = 4 and the supplied t* = 1.833 (90%, df 9). The margin of error is —
- A. 4
- B. 2.18
- C. 7.33 ✅
- D. 1.833
Feedback: ME = t*·SE = 1.833 × 4 = 7.33. (Multiply the supplied critical value by the standard error.)
Q10 (Multiple answer — select all that apply). A 95% confidence interval for the mean weekly study hours is (12, 16). Which statements are correct interpretations?
- A. We are 95% confident that the true mean weekly study hours is between 12 and 16. ✅
- B. If we repeated this sampling many times, about 95% of the intervals we built would contain the true mean. ✅
- C. 95% of students study between 12 and 16 hours per week.
- D. There is a 95% probability that the true mean falls in this particular interval (12, 16).
Feedback: A and B are correct — both describe confidence about the mean and the method's long-run capture rate. C is the "95% of the data" misread (a CI is about the mean, 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)
| Q | Answer |
|---|---|
| 1 | B |
| 2 | B |
| 3 | True |
| 4 | B |
| 5 | C |
| 6 | B |
| 7 | B |
| 8 | A |
| 9 | C |
| 10 | A, B |
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: Q5 SE = 15/√25 = 3; Q6 ME = 2.064×3 = 6.19 → (61.8, 74.2); Q7 ME = 2.131×10 = 21.31; Q9 ME = 1.833×4 = 7.33. Every t* value is supplied in its stem (no lookup), and no item asserts a fact outside the Week 11 course definitions.
Item-bank entries (for variants + the midterm/final)
All ten items are tagged course=MATH11 · week=11 · objective=6 · topic=confidence-intervals-means and deposited in Item Bank: Week 11 — Confidence Intervals for Means. The final (Week 16) and the per-term variant updates draw fresh items from this bank. (Tags: q1 t-vs-z, q2 degrees-of-freedom, q3 t-distribution-shape, q4 ci-formula, q5 standard-error, q6 construct-ci, q7 margin-of-error, q8 me-drivers, q9 margin-of-error-90, q10 ci-interpretation.)
Canvas placement block
canvas_object = Quizzes::Quiz
title = "Week 11 Quiz — Confidence Intervals for Means"
assignment_group = "Quizzes"
points_possible = 10
grading_type = points
due_offset_days = 6 # 6 days after module start (Sun Nov 15)
published = true
shuffle_answers = true
provenance = "~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com"
F-quiz-week-11-qti.xml) ships inside the course's .imscc package — it lands in the Canvas gradebook on import.~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com