Week 17 — Quiz (auto-graded) · Linear Regression with Inference + Course Synthesis

Course: Introduction to Statistics (18-week generic edition)
Objective tested: Objective 8 — the least-squares line, r², residuals, extrapolation, and inference for the slope (plus one synthesis item spanning Objectives 6–9).
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Week 17 · 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-17-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 17 misconceptions named in the lecture outline (slope without "predicted/on average," r² as accuracy, causal leaps, extrapolation, "accept H₀," df and t* misuse).

Questions, key, and feedback

Q1 (MC). A mechanic's shop fits a least-squares line to data on its customers' cars: ŷ = 3.8 + 1.6x, where x = engine size (liters) and y = fuel use (liters per 100 km). What does the slope 1.6 mean?
- A. Every car with a one-liter-larger engine uses exactly 1.6 more liters per 100 km
- B. Each additional liter of engine size predicts, on average, 1.6 more liters of fuel per 100 km ✅
- C. Larger engines cause fuel use to rise by 1.6 liters per 100 km, all else being equal
- D. About 1.6% of the variation in fuel use is explained by engine size
Feedback: The slope is a predicted, on-average change in y per one-unit step in x. "Exactly" (A) erases the scatter; "cause" (C) claims more than an observational fit can; (D) confuses the slope with r².

Q2 (MC). Using the same line (ŷ = 3.8 + 1.6x), what is the predicted fuel use for a car with a 2.5-liter engine?
- A. 7.8 liters per 100 km ✅
- B. 5.4 liters per 100 km
- C. 9.5 liters per 100 km
- D. 4.0 liters per 100 km
Feedback: ŷ = 3.8 + 1.6(2.5) = 3.8 + 4.0 = 7.8. Multiply the slope by x first, then add the intercept. (B adds without multiplying; C multiplies the intercept; D forgets the intercept.)

Q3 (MC). One car in the data has a 2.5-liter engine and actually used 8.4 liters per 100 km. Its residual is —
- A. −0.6, and the car sits below the fitted line
- B. 0.6% of the variation left unexplained by the line
- C. +8.4, because the actual value is what the line missed
- D. +0.6, and the car used more fuel than the line predicted ✅
Feedback: Residual = actual − predicted = 8.4 − 7.8 = +0.6 — a positive miss, so the point sits above the line: the car used more than predicted.

Q4 (MC). For these cars the correlation between engine size and fuel use is r = 0.8. Which statement correctly uses r²?
- A. 80% of the cars fall exactly on the fitted regression line
- B. About 64% of the variation in fuel use is explained by engine size ✅
- C. The line's fuel-use predictions are correct 64% of the time
- D. Fuel use rises by 64% for each additional liter of engine size
Feedback: r² = (0.8)² = 0.64 — the share of variation in y explained by the linear relationship. It is never an accuracy rate (C) or a percent-change (D). r² is a share, not a grade.

Q5 (MC). A music school regresses audition score on weekly practice hours for n = 17 violin students. The output shows, for the slope: Coef = 4.0, SE Coef = 1.6, P = 0.025. Using the course t-table (df 15, 95%: t* = 2.131), what is the correct conclusion at α = 0.05?
- A. t = 2.5 > 2.131, so reject H₀ — the data give evidence of a real linear relationship ✅
- B. t = 2.5 > 2.131, so accept H₀ — practice hours have no effect on audition scores
- C. t = 0.4 < 2.131, so fail to reject H₀ — the slope is too small to matter here
- D. P = 0.025 means there is only a 2.5% chance that H₀ is true, so reject it
Feedback: t = Coef ÷ SE = 4.0 ÷ 1.6 = 2.5; df = 17 − 2 = 15; 2.5 > 2.131 → reject (and P = 0.025 < 0.05 agrees). "Accept H₀" is never a legal conclusion (B), C divides the wrong way, and D misreads what a p-value is a probability of.

Q6 (Multiple answer — select all that apply). Which of the following are conditions to check before running inference for a regression slope?
- A. The scatterplot shows a roughly linear pattern ✅
- B. The residual plot shows random scatter with roughly equal spread ✅
- C. The explanatory variable x follows a normal distribution
- D. The residuals are roughly normal, with no extreme outliers ✅
- E. The sample slope b is larger than 1
Feedback: The LINE check: Linear, Independent, Normal residuals, Equal spread. It's the residuals that should look normal — x never needs to be (C) — and the size of b is a finding, not a condition (E).

Q7 (MC). The engine-size line (ŷ = 3.8 + 1.6x) was fitted on cars with engines from 1.0 to 3.0 liters. A truck has a 6.2-liter engine. What should the analyst do?
- A. Predict 13.72 liters per 100 km — with r = 0.8 the line is strong enough to use anywhere
- B. Predict 13.72 liters per 100 km, but report it as a cautious underestimate for the truck
- C. Refuse to use this line — 6.2 liters is far outside the fitted range, so this is extrapolation ✅
- D. Double the prediction for a 3.1-liter engine, because 6.2 liters is exactly twice 3.1
Feedback: Outside the fitted range the line is fiction: the pattern itself may change, and no r makes extrapolation safe. (The arithmetic 3.8 + 1.6 × 6.2 = 13.72 is correct — and beside the point.)

Q8 (True / False). A slope t-test gives p = 0.32, so the data prove there is no relationship between the two variables.
- True
- False ✅
Feedback: False twice over. Fail to reject ≠ accept: p = 0.32 means the data couldn't rule out the flat line, not that the flat line is true. And the t-test only examines a linear relationship — a curved one could hide behind a flat slope.

Q9 (MC). For the violin study (slope b = 4.0, SE = 1.6, n = 17), which is the 95% confidence interval for the slope, using t* = 2.131?
- A. (2.4, 5.6)
- B. (0.86, 7.14)
- C. (0.59, 7.41) ✅
- D. (1.9, 6.1)
Feedback: b ± t*·SE = 4.0 ± 2.131 × 1.6 = 4.0 ± 3.41 → (0.59, 7.41). (A is b ± SE; B used z* = 1.96; D added ±2.131 without multiplying by SE.) Note the interval excludes 0 — matching Q5's rejection.

Q10 (Matching). Match each research question to the procedure it calls for.

The table below pairs each scenario with its correct procedure.

Feedback: Name the answer's shape first: a mean → t; two percents → two-proportion z; counts in a two-way table → chi-square independence; predicting one number from another → regression and its slope test.

Answer key (quick reference)

Quality gate (self-checked): each single-answer item has exactly one correct option; the multiple-answer item's three LINE conditions are the only true options listed; no positional pattern in the key (B A D B A · C · C) and no letter carries more than two of the seven MC items; option lengths within each item are comparable (no length giveaway); every numeric claim (7.8; +0.6; 0.64; t = 2.5 vs. 2.131 with P = 0.025; 13.72; the interval (0.59, 7.41)) is re-verified in tools/checks/w17_math.py; every table value an item needs is stated in the stem; no item asserts a fact outside the Week 17 course definitions; no scenario reuses the tutorial, practice, chapter, lab, or assignment surfaces (different contexts and numbers throughout).

Item-bank entries (for variants + the final)

All ten items are tagged week=17 · objective=8 · topic=linear-regression-inference-synthesis and deposited in Item Bank: Week 17 — Linear Regression with Inference + Course Synthesis with idents w17q1–w17q10. The final (Week 18) and per-term variant updates draw fresh variants from this bank's concepts — never these live stems. (Tags: w17q1 slope-interpretation, w17q2 prediction, w17q3 residual, w17q4 r-squared, w17q5 slope-t-test, w17q6 conditions-LINE, w17q7 extrapolation, w17q8 fail-to-reject-vs-accept, w17q9 slope-CI, w17q10 choosing-procedure.)

Canvas placement block

canvas_object    = Quizzes::Quiz
title            = "Week 17 Quiz — Linear Regression with Inference + Course Synthesis"
assignment_group = "Quizzes"
points_possible  = 10
grading_type     = points
due_offset_days  = 6        # end of the module's week
published        = true
shuffle_answers  = true