Final Practice Exam (ungraded) · Weeks 1–15 (Objectives 1–8)

Course: Introduction to Statistics (MATH 11) · Silver Oak University (fictional sample) · Prof. Rivera
What this is: a low-stakes rehearsal for the cumulative final. It mirrors the real exam's blueprint — same coverage across all eight objectives, the same item-type mix, length, and application-skewed difficulty — but is built from fresh item-bank variants and shares none of the live final's questions.
Settings: ungraded (0 points) · unlimited attempts · feedback shown after submission · opens before the exam window so you can prepare.

This is the human-readable practice exam with its vetted answer key and feedback (released after submission). The import-ready Classic QTI 1.2 is in O-practice-final-week-16-qti.xml (generated by a validated Python script — parses with 25 items). The Canvas placement block is at the bottom.

Integrity note for students. Every item here is a fresh variant — new numbers and contexts — with a pre-computed, vetted answer. None of these are the live final questions. Working them builds the skill the final tests, honestly. The paired live exam is L-final-week-16.md.

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 9 (90%) → 1.833, df 24 (95%) → 2.064.

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Blueprint (mirrors the final)

Coverage is proportional to teaching time, matching the real exam: Obj 1 ≈ 2 · Obj 2 ≈ 3 · Obj 3 ≈ 2 · Obj 4 ≈ 4 · Obj 5 ≈ 3 · Obj 6 ≈ 4 · Obj 7 ≈ 4 · Obj 8 ≈ 3. (The actual final items are not listed here — only the shared structure.)

#	Type	Concept	Objective	Week
1	Multiple choice	Population vs. sample (identify the sample)	1	1
2	Multiple choice	Parameter vs. statistic	1	1
3	Multiple choice	Compute the mean	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	Marginal proportion (two-way table)	3	4
8	Multiple choice	Compute an expected value E(X)	4	6
9	Multiple choice	Conditional probability (two-way table)	4	5
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 1 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	Standard error of a proportion	6	12
17	Multiple choice	Margin of error for a mean at 90%	6	11
18	Multiple answer	Correct interpretation of a confidence interval	6	11–12
19	Multiple choice	Identify the null hypothesis H₀	7	13
20	Multiple choice	Conclusion in context (fail to reject)	7	14
21	Multiple choice	Compute a one-proportion z-statistic	7	14
22	Multiple choice	Identify a Type II error	7	13
23	Multiple choice	Interpret a regression intercept	8	15
24	Multiple choice	Predict ŷ from the line	8	15
25	Multiple choice	Compute a residual	8	15

Objective totals: Obj 1 = 2 · Obj 2 = 3 · Obj 3 = 2 · Obj 4 = 4 · Obj 5 = 3 · Obj 6 = 4 · Obj 7 = 4 · Obj 8 = 3 → 25 items (ungraded; mirrors the 100-point final's emphasis).

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Questions, key, and feedback

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

Q1 (MC). An orchard owner wants to estimate the average yield of her 9,000 apple trees. She harvests and weighs the fruit from a randomly chosen 300 trees. In this study, what is the sample?
- A. All 9,000 trees
- B. The 300 trees that were harvested and weighed ✅
- C. The average yield per tree
- D. Trees that produced any fruit
Feedback: The sample is the part actually measured — the 300 randomly chosen trees. The 9,000 are the population.

Q2 (MC). A company's complete personnel records show that 38% of all its employees hold a graduate degree. The value 38% is best described as a —
- A. Statistic
- B. Sample
- C. Parameter ✅
- D. Margin of error
Feedback: It describes the entire population (all employees) and comes from the complete records, so it is a parameter. A statistic comes from a sample.

Objective 2 — Summarizing Data (Weeks 2–3)

Q3 (MC). Find the mean of these five values: 10, 20, 30, 40, 50.
- A. 25
- B. 30 ✅
- C. 35
- D. 150
Feedback: Add them (10+20+30+40+50 = 150) and divide by the count (5): 150 ÷ 5 = 30. (Distractor D = the total, before dividing.)

Q4 (MC). A set of delivery distances has the five-number summary Min = 5, Q1 = 20, Median = 28, Q3 = 46, Max = 70. What is the interquartile range (IQR)?
- A. 65
- B. 26 ✅
- C. 18
- D. 28
Feedback: IQR = Q3 − Q1 = 46 − 20 = 26 (the spread of the middle 50%). The range, Max − Min = 65, is the classic distractor (A).

Q5 (MC). Compute the sample standard deviation of these five values: 4, 4, 10, 16, 16. (Their mean is 10; the squared deviations 36, 36, 0, 36, 36 sum to 144.)
- A. √28.8 ≈ 5.37
- B. 6 ✅
- C. 36
- D. 144
Feedback: Sample variance = sum of squared deviations ÷ (n − 1) = 144 ÷ 4 = 36; the SD is √36 = 6. (Dividing by n = 5 gives variance 28.8 — the population formula, distractor A.)

Objective 3 — Exploring Relationships (Week 4)

Q6 (MC). For a sample of students, the correlation between hours of sleep and a next-day alertness score is r = +0.88. The best description of this relationship is —
- A. A weak positive linear relationship
- B. A strong positive linear relationship ✅
- C. A strong negative linear relationship
- D. No linear relationship
Feedback: The size (0.88, close to 1) means strong; the plus sign means positive. More sleep tends to go with higher alertness.

Q7 (MC). A survey of 500 residents cross-classifies neighborhood and bike ownership:

	Owns a bike	No bike	Row total
Downtown	180	120	300
Suburb	120	80	200
Column total	300	200	500

What proportion of all residents own a bike (a marginal proportion)?
- A. 180 / 300 = 0.60
- B. 300 / 500 = 0.60 ✅
- C. 120 / 200 = 0.60
- D. 180 / 500 = 0.36
Feedback: "Of all residents" uses the grand total (500) as the denominator — that's a marginal proportion: 300 ÷ 500 = 0.60. (A and C are conditional proportions within a single row, which happen to round elsewhere; D uses the wrong cell.)

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

Q8 (MC). A payout random variable X has the distribution X = 0 with probability 0.5, X = 10 with probability 0.4, X = 100 with probability 0.1. What is the expected value E(X)?
- A. 36.67
- B. 14 ✅
- C. 55
- D. 110
Feedback: E(X) = 0(0.5) + 10(0.4) + 100(0.1) = 0 + 4 + 10 = 14. It's the probability-weighted average, not the simple average of the payouts.

Q9 (MC). A test is given to 800 people. Of the 250 who actually have the condition, 200 tested positive. What is P(tests positive | the person has the condition)?
- A. 200 / 800 = 0.25
- B. 250 / 800 = 0.3125
- C. 200 / 250 = 0.80 ✅
- D. 50 / 250 = 0.20
Feedback: "Given the person has the condition" restricts you to those 250; 200 tested positive, so 200 ÷ 250 = 0.80 (the test's sensitivity).

Q10 (MC). Each of 2 manufactured parts is defective with probability 0.4, independently. Let X = the number defective, X ~ Binomial(n = 2, p = 0.4). Using C(2, 0) = 1, what is P(X = 0) (neither is defective)?
- A. 0.16
- B. 0.48
- C. 0.36 ✅
- D. 0.6
Feedback: P(X = 0) = 1 × (0.4)⁰ × (0.6)² = (0.6)² = 0.36. (Distractor D = 0.6 is just (1 − p), forgetting the square.)

Q11 (Multiple answer — select all that apply). A fair coin is flipped 400 times and X = the number of heads, so X ~ Binomial(n = 400, p = 0.5). Select all statements that are true. (Mean = np; SD = √[np(1 − p)].)
- A. The mean number of heads is 200 ✅
- B. The standard deviation is 10 ✅
- C. The standard deviation is 100
- D. The normal approximation is appropriate here (np ≥ 10 and n(1 − p) ≥ 10) ✅
- E. The mean number of heads is 400
Feedback: Mean = 400 × 0.5 = 200 (A). Variance = 400 × 0.5 × 0.5 = 100, so SD = √100 = 10 (B true; C is the variance, not the SD). np = 200 and n(1 − p) = 200 are both ≥ 10, so the normal approximation applies (D). E doubles the mean.

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

Q12 (MC). A fitness test is normal with mean 60 and standard deviation 12. An athlete scores 84. What is the athlete's z-score?
- A. +1.0
- B. +2.0 ✅
- C. +2.5
- D. −2.0
Feedback: z = (84 − 60) ÷ 12 = 24 ÷ 12 = +2.0 — two standard deviations above the mean.

Q13 (MC). IQ-style scores are normal with mean 100 and standard deviation 15. About what percentage of people score between 85 and 115?
- A. 68% ✅
- B. 95%
- C. 99.7%
- D. 50%
Feedback: 85 and 115 are exactly 1 SD below and above the mean (100 ∓ 1·15), so this is the within-1-SD band → 68%.

Q14 (MC). A population has mean μ = 200 and standard deviation σ = 40. For random samples of size n = 64, the standard error of the sample mean is σ/√n = 40 ÷ 8 = 5. Using z = (x̄ − μ) ÷ (σ/√n) and the embedded table, the probability that the sample mean is greater than 205 is about —
- A. .1587 ✅
- B. .8413
- C. .0228
- D. .50
Feedback: z = (205 − 200) ÷ 5 = +1.0; "greater than" is the area to the RIGHT = 1 − .8413 = .1587. Note you divide by the standard error (5), not by σ = 40. (.8413 is the left/"less than" area.)

Objective 6 — Confidence Intervals (Weeks 11–12)

Q15 (MC). A random sample of n = 25 orders has mean weight x̄ = 80 ounces and sample SD s = 20, so the standard error is s/√n = 20 ÷ 5 = 4. Using the supplied t* = 2.064 (95%, df 24), the 95% confidence interval for the mean weight is —
- A. (76, 84)
- B. (71.7, 88.3) ✅
- C. (60, 100)
- D. (40, 120)
Feedback: ME = t*·SE = 2.064 × 4 = 8.26; interval = 80 ± 8.26 = (71.7, 88.3). (A uses just the SE without multiplying by t*; C and D are too wide.)

Q16 (MC). A poll has p̂ = 0.30 and n = 2100. The standard error (SE = √(p̂(1−p̂)/n)) is — (note: 0.30 × 0.70 = 0.21, and 0.21 ÷ 2100 = 0.0001)
- A. 0.21
- B. 0.0001
- C. 0.01 ✅
- D. 0.30
Feedback: √(0.0001) = 0.01. 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.)

Q17 (MC). For a 90% confidence interval for a mean, a sample has standard error SE = 5 and the supplied t* = 1.833 (90%, df 9). The margin of error is —
- A. 5
- B. 2.73
- C. 9.17 ✅
- D. 1.833
Feedback: ME = t*·SE = 1.833 × 5 = 9.17. (A is just the SE — you must multiply by t*; D is the critical value alone.)

Q18 (Multiple answer — select all that apply). A 95% confidence interval for the proportion of customers who would recommend a product is (0.62, 0.68). Which statements are correct interpretations?
- A. We are 95% confident that the true proportion of all customers who would recommend the product is between 62% and 68%. ✅
- B. If we repeated this polling many times, about 95% of the intervals we built would contain the true proportion. ✅
- C. 95% of customers would recommend the product at a rate between 62% and 68%.
- D. There is a 95% probability that the true proportion falls in this particular interval (0.62, 0.68).
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).

Objective 7 — Hypothesis Testing (Weeks 13–14)

Q19 (MC). A cereal box has long averaged 350 grams. A quality engineer suspects the filling machine has drifted and the average is now different. For a hypothesis test, what is the null hypothesis (H₀)?
- A. μ = 350 grams ✅
- B. μ ≠ 350 grams
- C. μ > 350 grams
- D. The engineer's suspicion is correct
Feedback: H₀ is the "no change / status quo" claim and always carries the equals idea — the mean is still 350. The engineer's suspicion is the alternative. (B–D drop the "=", which H₀ must keep.)

Q20 (MC). A two-sample test compares the mean delivery times of two couriers. H₀: the two means are equal; Hₐ: they differ. With α = 0.05, technology reports p = 0.22. What is the best conclusion?
- A. There is significant evidence the two couriers differ
- B. There is not enough evidence that the two couriers' mean delivery times differ ✅
- C. The two couriers are proven to be identical
- D. The first courier is definitely faster
Feedback: 0.22 > 0.05 → fail to reject H₀, so there isn't enough evidence of a difference. (That is not proof the couriers are identical — "absence of evidence is not evidence of absence.")

Q21 (MC). A one-proportion z-test has p̂ = 0.55, p₀ = 0.50, n = 100. The standard error is √(p₀(1−p₀)/n) = √(0.50·0.50/100) = √0.0025 = 0.05. What is z = (p̂ − p₀)/SE?
- A. z = 0.05
- B. z = 1.0 ✅
- C. z = 10
- D. z = 0.5
Feedback: z = (0.55 − 0.50)/0.05 = 0.05/0.05 = 1.0. (The standard error uses p₀ = 0.50, and the values go in as decimals.)

Q22 (MC). A medical screen uses H₀: "the patient is healthy." The test fails to reject H₀ for a patient who is actually sick — telling a sick patient they are fine. This is a —
- A. Type I error
- B. Type II error ✅
- C. Correct decision
- D. Practical-significance error
Feedback: Failing to reject a false H₀ is a Type II error (false negative) — a miss, like letting the guilty go free. A Type I error would be alarming a healthy patient (rejecting a true H₀).

Objective 8 — Linear Regression & Inference (Week 15)

Q23 (MC). A regression line predicting a delivery's cost (y, dollars) from its distance (x, miles) is ŷ = 8 + 3x. The intercept of 8 represents —
- A. The predicted cost for a delivery of 0 miles (a base/fixed cost) ✅
- B. The increase in cost per extra mile
- C. The number of deliveries in the study
- D. The longest distance in the data
Feedback: The intercept is the predicted y when x = 0 — here, the predicted cost at 0 miles (a base fee). (B is the slope.)

Q24 (MC). A regression line is ŷ = 10 + 2x. Predict ŷ when x = 15.
- A. 25
- B. 32
- C. 40 ✅
- D. 300
Feedback: Plug in x = 15: ŷ = 10 + 2(15) = 10 + 30 = 40. Prediction is just arithmetic. (A adds only one 15-step's worth; D multiplies incorrectly.)

Q25 (MC). For the line ŷ = 5 + 4x, a data point has x = 6 and an observed y of 25. The residual (observed − predicted) is —
- A. −4 ✅
- B. +4
- C. 29
- D. 25
Feedback: Predicted ŷ = 5 + 4(6) = 5 + 24 = 29. Residual = observed − predicted = 25 − 29 = −4 (the point is 4 below the line). (B reverses the subtraction; C is the predicted value, not the residual.)

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Answer key (quick reference)

Q	Answer	Q	Answer
1	B	14	A (.1587)
2	C	15	B (71.7, 88.3)
3	B (30)	16	C (0.01)
4	B (26)	17	C (9.17)
5	B (6)	18	A, B
6	B	19	A (μ = 350)
7	B (300/500 = 0.60)	20	B (fail to reject)
8	B (14)	21	B (z = 1.0)
9	C (200/250 = 0.80)	22	B (Type II)
10	C (0.36)	23	A
11	A, B, D	24	C (40)
12	B (+2.0)	25	A (−4)
13	A (68%)

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

• Structure: 25 items mirroring the final's emphasis — 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 live exam's blueprint exactly (ungraded).
• Single-answer integrity: every multiple-choice item (Q1–Q10, Q12–Q17, Q19–Q25) has exactly one correct option; the multiple-answer items key Q11 → A, B, D and Q18 → A, B (the remaining options must be left unselected).
• Arithmetic pre-computed and independently re-verified (Python verify_practice.py): Q3 mean 150/5 = 30; Q4 IQR 46−20 = 26; Q5 sample SD √(144/4) = √36 = 6; Q7 marginal 300/500 = 0.60; Q8 0+4+10 = 14; Q9 200/250 = 0.80; Q10 (0.6)² = 0.36; Q11 mean 200, var 100, SD 10, np = 200 ≥ 10; Q12 (84−60)/12 = 2.0; Q13 85/115 = μ ∓ 1σ → 68%; Q14 SE 40/√64 = 5, z = (205−200)/5 = +1.0, right area 1−.8413 = .1587; Q15 ME 2.064×4 = 8.26 → (71.7, 88.3); Q16 SE √(0.21/2100) = √0.0001 = 0.01; Q17 ME 1.833×5 = 9.17; Q21 SE √(0.25/100) = 0.05, z = (0.55−0.50)/0.05 = 1.0; Q24 ŷ = 10+2·15 = 40; Q25 residual 25−(5+24) = −4. All checks PASS.
• Table/critical-value alignment: every standardized item lands exactly on the embedded z-table (z = +2.0 → .9772/.0228, z = +1.0 → .8413/.1587) and every CI/test item uses a supplied t* or z* (no lookup): t* = 2.064 (df 24), t* = 1.833 (df 9, 90%).
• QTI parse confirmation: O-practice-final-week-16-qti.xml parses as imsqti_xmlv1p2 with 25 items; every single-answer respcondition sets SCORE = 100 on exactly one option; the two multiple-answer items award 100 only for the exact correct-set selection.
• Integrity vs. the live final: 0 items are shared with L-final-week-16.md (verified by full stem-plus-options comparison; every shared concept slot uses different numbers and contexts).
• No content outside the Weeks 1–15 course definitions; no hallucinated facts.

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Item-bank & coverage note

All 25 items are fresh variants assembled from the Week 1–15 item banks per Prompt O — drawn so that they mirror the final's blueprint while sharing none of its live items — tagged course=MATH11 · practice=final · weeks=1–15 · objectives=1–8 and deposited back into the banks for future per-term ($39) regenerations. Each term's update regenerates fresh practice variants alongside the live final, and the practice form continues to share none of the live exam's items.

Canvas placement block

canvas_object             = Quizzes::Quiz
title                     = "Final Practice Exam (ungraded)"
assignment_group          = "Practice exercises"
points_possible           = 0
grading_type              = not_graded
allowed_attempts          = unlimited
show_feedback             = true        # released after submission
available_from_offset_days = -5        # opens 5 days before the exam window
due_offset_days           = 4         # on or before the final's due date
published                 = true
shuffle_answers           = true
provenance                = "~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com"