Final Exam — Cumulative (Weeks 1–15) · Objectives 1–8

Course: Introduction to Statistics (MATH 11) · Silver Oak University (fictional sample) · Prof. Rivera
Scope: Cumulative — all eight objectives, Weeks 1–15 (foundations & data types · summarizing one variable · relationships between two variables · probability & random variables · normal & sampling distributions · confidence intervals · hypothesis tests · simple linear regression).
Format: 25 items, 100 points (4 each) · application-skewed · mixed item types (multiple-choice, multiple-answer, true/false, matching).
Points: 100 · Assignment group: Final (30% of the course grade) · Window: opens at the start of Module 16 (finals week); due 4 days later.

This is the human-readable exam with its vetted answer key and one-line feedback. The import-ready Classic QTI 1.2 is in L-final-week-16-qti.xml (generated by a validated Python script — parses with 25 items, every single-answer item exactly one correct). The item-bank / coverage note and the Canvas placement block are at the bottom of this file.

This is the live exam. Its paired ungraded rehearsal — O-practice-final-week-16.md — mirrors this blueprint with fresh variants and shares none of these items.

Embedded z-table (cumulative area to the LEFT of z) — supplied on any item that needs it; every numeric item is engineered to land on these:

z	−2.5	−2.0	−1.5	−1.0	−0.5	0	+0.5	+1.0	+1.5	+2.0	+2.5
area to LEFT	.0062	.0228	.0668	.1587	.3085	.5000	.6915	.8413	.9332	.9772	.9938

Embedded critical values (supplied in the stems; no table lookup): confidence z*: 90% → 1.645, 95% → 1.960, 99% → 2.576. Confidence t*: df 9 (95%) → 2.262, df 15 (95%) → 2.131, df 24 (95%) → 2.064.

---

Blueprint (items → objective)

Coverage is proportional to teaching time: Obj 1 ≈ 2 · Obj 2 ≈ 3 · Obj 3 ≈ 2 · Obj 4 ≈ 4 · Obj 5 ≈ 3 · Obj 6 ≈ 4 · Obj 7 ≈ 4 · Obj 8 ≈ 3. No trick questions; all arithmetic is pre-computed; every single-answer item has exactly one correct option.

#	Type	Concept	Objective	Week
1	Multiple choice	Population vs. sample (identify the sample)	1	1
2	Matching	Observational vs. experiment & levels of measurement	1	1
3	Multiple choice	Compute the median	2	3
4	Multiple choice	IQR from a five-number summary	2	3
5	Multiple choice	Compute a sample standard deviation	2	3
6	Multiple choice	Interpret a correlation r	3	4
7	Multiple choice	Conditional proportion (two-way table)	3	4
8	Multiple choice	Conditional probability (two-way table)	4	5
9	Multiple choice	Compute an expected value E(X)	4	6
10	Multiple choice	Binomial probability	4	7
11	Multiple answer	Binomial mean, SD & normal approximation	4	7
12	Multiple choice	Compute a z-score	5	9
13	Multiple choice	Empirical rule (within 2 SD)	5	9
14	Multiple choice	Standard error & a CLT probability	5	10
15	Multiple choice	Construct a 95% confidence interval for a mean (t)	6	11
16	Multiple choice	Margin of error for a proportion (z)	6	12
17	Multiple choice	Required sample size for a proportion (round up)	6	12
18	Multiple answer	Correct interpretation of a confidence interval	6	11–12
19	Multiple choice	Write the alternative hypothesis Hₐ	7	13
20	Multiple choice	p-value vs. α — the decision	7	13
21	Multiple choice	Compute a one-sample t-statistic	7	14
22	Matching	Type I vs. Type II error & the test menu	7	13–14
23	Multiple choice	Interpret a regression slope in context	8	15
24	Multiple choice	Predict ŷ and compute a residual	8	15
25	Multiple choice	r² from r & inference for the slope	8	15

Objective totals: Obj 1 = 2 (8 pts) · Obj 2 = 3 (12 pts) · Obj 3 = 2 (8 pts) · Obj 4 = 4 (16 pts) · Obj 5 = 3 (12 pts) · Obj 6 = 4 (16 pts) · Obj 7 = 4 (16 pts) · Obj 8 = 3 (12 pts) → 25 items, 100 points.

---

Questions, key, and feedback

Objective 1 — Foundations & Types of Data (Week 1)

Q1 (MC). A national gym chain wants to estimate the average number of visits per month among its 48,000 members. Analysts pull the entry records of a randomly chosen 600 members. In this study, what is the sample?
- A. All 48,000 members
- B. The 600 members whose entry records were pulled ✅
- C. The average number of monthly visits
- D. Members who visited at least once
Feedback: The sample is the part actually measured — the 600 randomly chosen members. The 48,000 are the population.

Q2 (Matching). Match each item to the correct description (study design and levels of measurement).
| Item | Correct description |
|---|---|
| Researchers randomly assign patients to a new drug or a placebo, then compare outcomes | Experiment — a treatment is imposed with random assignment |
| Analysts watch enrollment records and note who already chose to join a gym | Observational study — no treatment is assigned |
| A hotel's star rating (1, 2, 3, 4, or 5 stars) | Ordinal — ordered categories with unequal gaps |
| The number of emails a worker received today | Ratio — equal gaps and a true zero |
Feedback: Random assignment of a treatment makes it an experiment; merely recording existing choices is observational. Star ratings are ordered but unequally spaced (ordinal); a count has a true zero (ratio).

Objective 2 — Summarizing Data (Weeks 2–3)

Q3 (MC). Seven daily ride-share fares (in dollars), sorted, are 12, 19, 23, 27, 31, 34, 88. What is the median fare?
- A. 23
- B. 27 ✅
- C. 31
- D. 33.4
Feedback: With seven (odd) sorted values, the median is the middle (4th) value: 27. The mean (≈ 33.4, distractor D) is pulled up by the 88.

Q4 (MC). A set of service times has the five-number summary Min = 18, Q1 = 34, Median = 41, Q3 = 58, Max = 92. What is the interquartile range (IQR)?
- A. 74
- B. 24 ✅
- C. 17
- D. 41
Feedback: IQR = Q3 − Q1 = 58 − 34 = 24 (the spread of the middle 50%). The range, Max − Min = 74, is the classic distractor (A).

Q5 (MC). Compute the sample standard deviation of these five values: 3, 5, 7, 9, 11. (Their mean is 7; the squared deviations 16, 4, 0, 4, 16 sum to 40.)
- A. √8 ≈ 2.83
- B. √10 ≈ 3.16 ✅
- C. 8
- D. 10
Feedback: Sample variance = sum of squared deviations ÷ (n − 1) = 40 ÷ 4 = 10; the SD is √10 ≈ 3.16. (Dividing by n = 5 gives variance 8 — the population formula, distractor A.)

Objective 3 — Exploring Relationships (Week 4)

Q6 (MC). For a sample of used cars, the correlation between a car's age (years) and its resale value (dollars) is r = −0.91. The best description of this relationship is —
- A. A weak negative linear relationship
- B. A strong positive linear relationship
- C. A strong negative linear relationship ✅
- D. No linear relationship
Feedback: The size (0.91, close to 1) means strong; the minus sign means negative. Older cars tend to be worth less.

Q7 (MC). A campus survey of 400 students cross-classifies class standing and tutoring use:

	Used tutoring	Did not	Row total
First-year	176	44	220
Upper-division	90	90	180
Column total	266	134	400

Among the first-year students, what proportion used tutoring?
- A. 176 / 400 = 0.44
- B. 176 / 220 = 0.80 ✅
- C. 90 / 180 = 0.50
- D. 266 / 400 = 0.665
Feedback: "Among first-year students" is a conditional proportion — divide by the first-year row total (220), not the grand total: 176 ÷ 220 = 0.80. (A uses the wrong denominator.)

Objective 4 — Probability & Random Variables (Weeks 5–7)

Q8 (MC). A clinic screens 600 people for a marker. Of the 240 who actually carry the marker, 216 were flagged by the test. What is P(flagged | the person carries the marker)?
- A. 216 / 600 = 0.36
- B. 240 / 600 = 0.40
- C. 216 / 240 = 0.90 ✅
- D. 24 / 240 = 0.10
Feedback: "Given the person carries the marker" restricts you to those 240; 216 were flagged, so 216 ÷ 240 = 0.90 (the test's sensitivity).

Q9 (MC). A prize random variable X has the distribution X = 0 with probability 0.6, X = 5 with probability 0.3, X = 20 with probability 0.1. What is the expected value E(X)?
- A. 3.5 ✅
- B. 8.33
- C. 12.5
- D. 25
Feedback: E(X) = 0(0.6) + 5(0.3) + 20(0.1) = 0 + 1.5 + 2.0 = 3.5. It's the probability-weighted average, not the simple average of the prizes.

Q10 (MC). A student guesses on 4 true/false questions, so the number correct X ~ Binomial(n = 4, p = 0.5). Using C(4, 4) = 1, what is the probability of getting all 4 correct, P(X = 4)?
- A. 1/4 = 0.25
- B. 1/16 = 0.0625 ✅
- C. 4/16 = 0.25
- D. 0.5
Feedback: P(X = 4) = 1 × (0.5)⁴ = 1/16 = 0.0625. Each of the four independent guesses must be right.

Q11 (Multiple answer — select all that apply). A fair coin is flipped 144 times and X = the number of heads, so X ~ Binomial(n = 144, p = 0.5). Select all statements that are true. (Mean = np; SD = √[np(1 − p)].)
- A. The mean number of heads is 72 ✅
- B. The standard deviation is 6 ✅
- C. The standard deviation is 36
- D. The normal approximation is appropriate here (np ≥ 10 and n(1 − p) ≥ 10) ✅
- E. The mean number of heads is 144
Feedback: Mean = 144 × 0.5 = 72 (A). Variance = 144 × 0.5 × 0.5 = 36, so SD = √36 = 6 (B true; C is the variance, not the SD). np = 72 and n(1 − p) = 72 are both ≥ 10, so the normal approximation applies (D). E doubles the mean.

Objective 5 — Normal & Sampling Distributions (Weeks 9–10)

Q12 (MC). A standardized placement test is normal with mean 500 and standard deviation 100. A student scores 750. What is the student's z-score?
- A. +1.5
- B. +2.0
- C. +2.5 ✅
- D. −2.5
Feedback: z = (750 − 500) ÷ 100 = 250 ÷ 100 = +2.5 — two and a half standard deviations above the mean.

Q13 (MC). Battery lifetimes are normal with mean 200 hours and standard deviation 20 hours. About what percentage of batteries last between 160 and 240 hours?
- A. 68%
- B. 95% ✅
- C. 99.7%
- D. 50%
Feedback: 160 and 240 are exactly 2 SDs below and above the mean (200 ∓ 2·20), so this is the within-2-SD band → 95%.

Q14 (MC). A population has mean μ = 500 and standard deviation σ = 30. For random samples of size n = 36, the standard error of the sample mean is σ/√n = 30 ÷ 6 = 5. Using z = (x̄ − μ) ÷ (σ/√n) and the embedded table, the probability that the sample mean is less than 490 is about —
- A. .0228 ✅
- B. .9772
- C. .1587
- D. .50
Feedback: z = (490 − 500) ÷ 5 = −2.0; the area to the LEFT of −2.0 is .0228. Note you divide by the standard error (5), not by σ = 30. (.9772 is the right-tail/"above" value.)

Objective 6 — Confidence Intervals (Weeks 11–12)

Q15 (MC). A random sample of n = 16 customers has mean spend x̄ = 42 dollars and sample SD s = 8, so the standard error is s/√n = 8 ÷ 4 = 2. Using the supplied t* = 2.131 (95%, df 15), the 95% confidence interval for the mean spend is —
- A. (40, 44)
- B. (37.7, 46.3) ✅
- C. (34, 50)
- D. (26, 58)
Feedback: ME = t*·SE = 2.131 × 2 = 4.26; interval = 42 ± 4.26 = (37.7, 46.3). (A uses just the SE without multiplying by t*; C and D are too wide.)

Q16 (MC). A poll finds p̂ = 0.50 with n = 2500, so the standard error is √(p̂(1−p̂)/n) = √(0.25/2500) = 0.01. For a 95% interval the supplied z* = 1.960. The margin of error is —
- A. 0.01
- B. 0.0196 ✅
- C. 0.005
- D. 0.196
Feedback: ME = z*·SE = 1.960 × 0.01 = 0.0196. (A is just the SE — you must multiply by z*; D is off by a factor of 10.)

Q17 (MC). A pollster wants a 95% interval (z* = 1.960) with a target margin of error 0.05 and, having no prior estimate, uses the worst-case p̂ = 0.5: n = (z*/ME)²·p̂(1−p̂) = (1.960/0.05)² × 0.25 = (39.2)² × 0.25 = 1536.64 × 0.25 = 384.16. The required sample size is —
- A. 384
- B. 385 ✅
- C. 1537
- D. 768
Feedback: A sample size is always rounded UP to the next whole person, so 384.16 → 385. (A rounds down — which would miss the target margin; C forgets to multiply by p̂(1−p̂) = 0.25.)

Q18 (Multiple answer — select all that apply). A 95% confidence interval for the mean nightly sleep of students is (6.8, 7.4) hours. Which statements are correct interpretations?
- A. We are 95% confident that the true mean nightly sleep is between 6.8 and 7.4 hours. ✅
- B. If we repeated this sampling many times, about 95% of the intervals we built would contain the true mean. ✅
- C. 95% of students sleep between 6.8 and 7.4 hours per night.
- D. There is a 95% probability that the true mean falls in this particular interval (6.8, 7.4).
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).

Objective 7 — Hypothesis Testing (Weeks 13–14)

Q19 (MC). A manufacturer claims its new battery lasts, on average, longer than the industry standard of 500 hours. You want to test that claim with one sample. The correct alternative hypothesis (Hₐ) is —
- A. μ = 500
- B. μ ≠ 500
- C. μ > 500 ✅
- D. μ < 500
Feedback: The claim is one-directional ("longer than 500"), so Hₐ: μ > 500. H₀ would be μ = 500. (Hₐ never carries an "="; "longer" points only up.)

Q20 (MC). A test is conducted at significance level α = 0.05, and the data produce p = 0.03. What is the correct decision?
- A. Reject H₀ ✅
- B. Fail to reject H₀
- C. Accept H₀ as proven true
- D. Increase α until the result changes
Feedback: The rule is p ≤ α → reject H₀, and 0.03 ≤ 0.05. A test never "accepts H₀ as proven," and α is fixed before seeing the data. (C and D are both classic errors.)

Q21 (MC). A one-sample t-test has x̄ = 45, μ₀ = 42, s = 10, n = 25. The standard error is s/√n = 10 ÷ 5 = 2. What is the test statistic t = (x̄ − μ₀)/(s/√n)?
- A. t = 0.3
- B. t = 1.5 ✅
- C. t = 7.5
- D. t = 0.6
Feedback: SE = 10/√25 = 2, so t = (45 − 42)/2 = 1.5. (A drops the √n and divides by s = 10; the denominator is the standard error s/√n, not s.)

Q22 (Matching). Match each term to its correct description.
| Term | Correct description |
|---|---|
| Type I error | Rejecting H₀ when it is actually true (false positive) |
| Type II error | Failing to reject H₀ when it is actually false (false negative) |
| One-sample t-test | Compares a sample mean to one fixed number (σ unknown) |
| One-proportion z-test | Compares a sample proportion / rate to one fixed value |
Feedback: Type I = false alarm (convict the innocent); Type II = a miss (free the guilty). A mean against a fixed number is a one-sample t-test; a rate against a fixed value is a one-proportion z-test.

Objective 8 — Linear Regression & Inference (Week 15)

Q23 (MC). A regression line predicting a home's monthly heating cost (y, dollars) from its size (x, hundreds of square feet) is ŷ = 30 + 6x. The slope of 6 means —
- A. A 0-square-foot home is predicted to cost $6 to heat
- B. For each additional 100 square feet, the predicted monthly heating cost rises by about $6 ✅
- C. The correlation between size and cost is 6
- D. The home has 6 rooms
Feedback: The slope is the change in the predicted y for each one-unit increase in x — here, +$6 of heating cost per extra 100 square feet (units matter). (A confuses the slope with the intercept; C confuses it with r, which can't exceed 1.)

Q24 (MC). A regression line is ŷ = 12 + 5x. A data point has x = 8 and an observed y of 60. The residual (observed − predicted) is —
- A. −8
- B. +8 ✅
- C. 52
- D. 60
Feedback: Predicted ŷ = 12 + 5(8) = 12 + 40 = 52. Residual = observed − predicted = 60 − 52 = +8 (the point is 8 above the line). (A reverses the subtraction; C is the predicted value, not the residual.)

Q25 (MC). A regression of sales on advertising reports a correlation of r = 0.7 and a p-value of 0.01 for the slope (tested at α = 0.05). Which statement is correct?
- A. r² = 0.49, and because 0.01 < 0.05 we fail to reject H₀: slope = 0
- B. r² = 0.49 (about 49% of the variation in sales is explained by the line), and because 0.01 < 0.05 we reject H₀: slope = 0 ✅
- C. r² = 0.70, and the slope is not significant
- D. r² = 0.84, and the slope is significant
Feedback: r² = r² = 0.7 × 0.7 = 0.49 — about 49% of the variation in y is explained. For the slope, p = 0.01 < α = 0.05, so we reject H₀: slope = 0 (the slope is significant). (A flips the decision; C repeats r as r²; D miscomputes r².)

---

Answer key (quick reference)

Q	Answer	Q	Answer
1	B	14	A (.0228)
2	Random-assign→Experiment / Records→Observational / Star rating→Ordinal / Email count→Ratio	15	B (37.7, 46.3)
3	B (27)	16	B (0.0196)
4	B (24)	17	B (385)
5	B (√10 ≈ 3.16)	18	A, B
6	C	19	C (μ > 500)
7	B (176/220 = 0.80)	20	A (reject H₀)
8	C (216/240 = 0.90)	21	B (t = 1.5)
9	A (3.5)	22	Type I→reject true H₀ / Type II→fail to reject false H₀ / one-sample t→mean vs number / one-proportion z→rate vs value
10	B (1/16 = 0.0625)	23	B
11	A, B, D	24	B (+8)
12	C (+2.5)	25	B
13	B (95%)

---

Quality gate (H5 — self-checked, computer-verified)

• Structure: 25 items, 4 points each, 100 points total; coverage Obj 1 = 2 · Obj 2 = 3 · Obj 3 = 2 · Obj 4 = 4 · Obj 5 = 3 · Obj 6 = 4 · Obj 7 = 4 · Obj 8 = 3 matches the blueprint exactly.
• Single-answer integrity: every multiple-choice item (Q1, Q3–Q10, Q12–Q17, Q19–Q21, Q23–Q25) has exactly one correct option; the matching items (Q2, Q22) pair all four rows one-to-one; the multiple-answer items key Q11 → A, B, D (C and E left unselected) and Q18 → A, B (C and D left unselected).
• Arithmetic pre-computed and independently re-verified (Python verify_final.py): Q3 median (4th of 7) = 27; Q4 IQR 58−34 = 24; Q5 sample SD √(40/4) = √10 ≈ 3.16; Q7 176/220 = 0.80; Q8 216/240 = 0.90; Q9 0+1.5+2.0 = 3.5; Q10 (0.5)⁴ = 1/16 = 0.0625; Q11 mean 72, var 36, SD 6, np = 72 ≥ 10; Q12 (750−500)/100 = 2.5; Q13 160/240 = μ ∓ 2σ → 95%; Q14 SE 30/√36 = 5, z = (490−500)/5 = −2.0, left area .0228; Q15 ME 2.131×2 = 4.26 → (37.7, 46.3); Q16 ME 1.960×0.01 = 0.0196; Q17 (1.960/0.05)²×0.25 = 384.16 → 385; Q21 SE 10/√25 = 2, t = (45−42)/2 = 1.5; Q24 residual 60−(12+40) = +8; Q25 r² = 0.7² = 0.49 and 0.01 < 0.05 → reject. All checks PASS.
• Table/critical-value alignment: every standardized item lands exactly on the embedded z-table (z = +2.5 → .9938 area-left context; z = −2.0 → .0228) and every CI/test item uses a supplied t* or z* (no lookup): t* = 2.131 (df 15), z* = 1.960.
• QTI parse confirmation: L-final-week-16-qti.xml parses as imsqti_xmlv1p2 with 25 items; every single-answer respcondition sets SCORE = 100 on exactly one option; each matching item's four partial-credit blocks add to 100; the two multiple-answer items award 100 only for the exact correct-set selection.
• Integrity vs. the practice final: 0 items are shared with O-practice-final-week-16.md (verified by full stem-plus-options comparison; the maximum overlap is a same-concept slot with different numbers and contexts).
• No content outside the Weeks 1–15 course definitions; no hallucinated facts.

---

Item-bank & coverage note

All 25 items are fresh variants assembled from the Week 1–15 item banks per Prompt L (changed numbers and contexts to reduce answer-sharing with the weekly quizzes and the midterm), tagged course=MATH11 · exam=final · weeks=1–15 · objectives=1–8 and deposited back into the banks for future per-term ($39) regenerations:

Objective	Drawn from banks	Items
1	Week 1 (Foundations & Types of Data)	Q1–Q2
2	Weeks 2–3 (Summarizing Data; Center & Spread)	Q3–Q5
3	Week 4 (Exploring Relationships)	Q6–Q7
4	Weeks 5–7 (Probability; Random Variables; Binomial & Normal)	Q8–Q11
5	Weeks 9–10 (Normal Distribution; Sampling Distributions)	Q12–Q14
6	Weeks 11–12 (CIs for Means; CIs for Proportions)	Q15–Q18
7	Weeks 13–14 (Hypothesis-Testing Foundations; Tests for Means & Proportions)	Q19–Q22
8	Week 15 (Linear Regression & Inference)	Q23–Q25

Each term's update regenerates fresh final variants from these same banks; the paired practice final is regenerated alongside and continues to share none of the live items.

Canvas placement block

canvas_object             = Quizzes::Quiz
title                     = "Final Exam — Cumulative (Weeks 1–15)"
assignment_group          = "Final"
points_possible           = 100
grading_type              = points
available_from_offset_days = 0        # opens at the start of Module 16 (finals week)
due_offset_days           = 4        # 4 days after module start
published                 = true
allowed_attempts          = 1
shuffle_answers           = true
provenance                = "~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com"