Week 18 — Practice Final (ungraded, unlimited attempts) · Rehearse the Real Thing

Course: Introduction to Statistics (18-week generic edition)
Covers: Weeks 1–17, in the same shape as the real final (≈12 of 30 items from Weeks 1–8, ≈18 from Weeks 10–17)
Points: 30 (1 each) · Assignment group: Practice exercises (0% of grade) · Attempts: unlimited · feedback shown after each submission · AI allowed here (it's practice)

This is the human-readable practice final with its vetted answer key and feedback. The import-ready Classic QTI is in O-practice-exam-week-18-qti.xml; the Canvas placement block is at the bottom.

Read this first: the practice exam shares ZERO items with the real exam — it practices the same skills, not the same questions. Memorizing these 30 answers earns nothing on the final; being able to re-derive them earns everything. Take it cold, score yourself, then aim the exam-prep tutorial at whatever you missed, and retake as often as you like.

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Blueprint (mirrors the final)

Weeks	Items	Skills sampled
W1	2	statistic vs. parameter; sampling designs
W2	1	choosing a display
W3	2	resistant center; computing s
W4	1	reading r
W5	2	general addition rule; independence has no memory
W6	1	expected value
W7	1	binomial mean & SD
W8	2	normal-model facts; forward z
W10	2	standard error; probability for a sample mean
W11	3	choosing t*; building an interval; interpretation
W12	2	proportion interval; sample size
W13	2	two-sided hypotheses; Type II
W14	2	t + decision; paired design
W15	2	one-proportion z; choosing a procedure
W16	2	computing chi-square; independence setting
W17	3	prediction; slope CI; choose-the-tool matching

Item types: 25 multiple choice, 2 true/false, 2 select-all, 1 matching — the same mix as the final.

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Questions, key, and feedback

Q1 (MC · W1). An ice rink selects a random sample of 250 of its 4,000 season-pass holders; 175 of them (70%) say they would attend a learn-to-skate clinic. The number 70% is best described as —
- A. A parameter, since it describes the rink's pass holders
- B. A statistic computed from the sample of 250 pass holders ✅
- C. A census result covering every one of the pass holders
- D. The margin of error of the learn-to-skate clinic survey
Feedback: Computed from a sample → statistic. The all-pass-holder rate it estimates is the parameter.

Q2 (MC · W1). A transit office wants opinions from riders of every campus-shuttle line, so it draws a separate random sample of riders within each of the five lines. This design is a —
- A. Cluster sample
- B. Convenience sample
- C. Systematic sample
- D. Stratified sample ✅
Feedback: Sampling within every group = stratified; grabbing whole groups = cluster.

Q3 (MC · W2). A cooking school tallies how many students chose each workshop type — pastry, grill, vegan, or sauces. Which display fits these data best?
- A. A bar chart with separated bars, one bar per workshop type ✅
- B. A histogram with touching bars along a single number line
- C. A stem-and-leaf plot built from the four workshop names
- D. A boxplot of the five-number summary of the four counts
Feedback: Workshop type is categorical → separated bars. Touching bars are for a quantitative number line.

Q4 (MC · W3). Sale prices at a thrift store are strongly right-skewed: most items cost a few dollars, while a handful of collectibles cost far more. The best summary of a typical price is —
- A. The mean, because it uses every dollar of every single item
- B. The range, because it spans the cheapest to the priciest item
- C. The median, because the rare collectible prices barely budge it ✅
- D. The largest mode, because the expensive items matter the most
Feedback: Skewed data → resistant center. The mean chases the collectible tail.

Q5 (MC · W3). Five community-garden plots produced 5, 5, 20, 35, and 35 kg of tomatoes; the mean is 20 kg. The sample standard deviation is —
- A. 225 kg
- B. 15 kg ✅
- C. 30 kg
- D. 9 kg
Feedback: Squared deviations 225 + 225 + 0 + 225 + 225 = 900; 900 ⁄ 4 = 225; √225 = 15. (225 is the variance, in squared units.)

Q6 (Select all · W4). Over 40 days, a greenhouse team finds r = −0.85 between vent opening (cm) and afternoon humidity (%). Which conclusions are sound? Select all that apply.
- A. The linear association between the two variables is strong ✅
- B. Wider vent openings tend to go along with lower humidity ✅
- C. r is measured in percentage points of afternoon humidity
- D. Swapping the two axes would change r from −0.85 to +0.85
- E. This linear association is stronger than one with r = +0.55 ✅
Feedback: Sign = direction, size = strength; r is unitless and symmetric in the two variables.

Q7 (MC · W5). On any drone route, P(wind delay) = 0.20, P(battery swap) = 0.15, and P(both happen) = 0.05. The probability that a route has a wind delay or a battery swap is —
- A. 0.35
- B. 0.05
- C. 0.30 ✅
- D. 0.03
Feedback: General addition rule: 0.20 + 0.15 − 0.05 = 0.30 — subtract the overlap so it isn't counted twice.

Q8 (True/False · W5). A trivia host's buzzer randomly picks one of four teams each round, every round independent. Team Curie hasn't been picked for six straight rounds, so the buzzer is now more likely than one-in-four to pick them next round.
- True
- False ✅
Feedback: Independent draws have no memory — streaks create no debts (the gambler's fallacy).

Q9 (MC · W6). On a pickup run, a thrift-store truck collects X pieces of furniture, where P(0) = 0.5, P(1) = 0.3, and P(2) = 0.2. The expected value of X is —
- A. 0.7 pieces ✅
- B. 1.0 piece
- C. 0.2 pieces
- D. 1.5 pieces
Feedback: 0(0.5) + 1(0.3) + 2(0.2) = 0.7 — a long-run average, not a guarantee (and not a possible value).

Q10 (MC · W7). A drone fleet completes 90% of deliveries on the first attempt, independently. Across 100 deliveries, the mean and standard deviation of the number completed on the first attempt are —
- A. Mean 90; SD 9
- B. Mean 90; SD 3 ✅
- C. Mean 10; SD 3
- D. Mean 90; SD 0.9
Feedback: Mean = np = 90; SD = √(100 × 0.9 × 0.1) = √9 = 3.

Q11 (Select all · W8). A greenhouse's overnight low temperatures follow an approximately normal distribution. Which statements must hold for this model? Select all that apply.
- A. The curve is symmetric about its mean ✅
- B. About 95% of nights fall within 2 SDs of the mean ✅
- C. The total area under the curve equals 1 ✅
- D. Every single night's low falls within 3 SDs of the mean
- E. The curve leans toward the warmer nights
Feedback: Symmetry, total area 1, and 68–95–99.7 are the model's guarantees; 99.7% within 3 SDs is not 100%.

Q12 (MC · W8). Tomatoes from one community garden weigh approximately N(120, 16) grams. Using the course z-table (the area left of z = 1.25 is 0.8944), what fraction of tomatoes weigh more than 140 grams?
- A. 0.8944
- B. 0.25
- C. 0.8413
- D. 0.1056 ✅
Feedback: z = (140 − 120) ⁄ 16 = 1.25; "more than" = 1 − 0.8944 = 0.1056.

Q13 (MC · W10). Minute-by-minute CO2 readings in a greenhouse have standard deviation 6 ppm. For random samples of 36 readings, the standard error of the sample mean is —
- A. 1 ppm ✅
- B. 6 ppm
- C. 0.17 ppm
- D. 36 ppm
Feedback: SE = σ ⁄ √n = 6 ⁄ 6 = 1 — the ruler for averages, not individuals.

Q14 (MC · W10). One campus-shuttle loop takes 30 minutes on average, give or take a standard deviation of 5. An auditor will average a random sample of 25 loop times. How likely is that average to come out above 31 minutes? (The course z-table puts the area left of z = 1 at 0.8413.)
- A. 0.8413
- B. 0.5793
- C. 0.1587 ✅
- D. 0.0228
Feedback: SE = 5 ⁄ √25 = 1; z = (31 − 30) ⁄ 1 = 1 → 1 − 0.8413 = 0.1587.

Q15 (MC · W11). A researcher wants a 99% confidence interval for a mean from a random sample of n = 25 ice-rink rental invoices. From the course t-table (df 24 row: 1.711 / 2.064 / 2.797 for 90% / 95% / 99%), t is —
- A. 2.064
- B. 2.797 ✅
- C. 1.711
- D. 2.576
Feedback: n = 25 → df 24; the 99% column reads 2.797. (2.576 is the large-sample z, not this row.)

Q16 (MC · W11). Sixteen randomly sampled meal-prep sessions at a cooking studio average 30 minutes with s = 8. With SE = 2 and t = 2.131 (df 15, 95%), the 95% confidence interval for the mean session length is —
- A. (28.0, 32.0) minutes
- B. (26.1, 33.9) minutes
- C. (22.0, 38.0) minutes
- D. (25.74, 34.26) minutes ✅
Feedback: ME = 2.131 × 2 = 4.262 → 30 ± 4.262. (Option B is the z = 1.96 shortcut — too narrow.)

Q17 (True/False · W11). A 95% confidence interval for the MEAN weight of bags dropped at a donation center runs from 5.2 to 6.4 kg, which means about 95% of the individual bags weigh between 5.2 and 6.4 kg.
- True
- False ✅
Feedback: The interval speaks about the mean, never about individual bags — individuals spread far wider.

Q18 (MC · W12). Twenty of 100 randomly sampled drone deliveries needed a second attempt, so p̂ = 0.20 with SE = 0.04. The 95% confidence interval for the true redelivery proportion is —
- A. (0.16, 0.24)
- B. (0.134, 0.266)
- C. (0.15, 0.25)
- D. (0.122, 0.278) ✅
Feedback: 0.20 ± 1.96 × 0.04 = 0.20 ± 0.0784. (Option B used the 90% multiplier.)

Q19 (MC · W12). A city wants to estimate the share of residents who would buy an ice-rink season pass, to within plus-or-minus 0.02 at 95% confidence, using the conservative p = 0.5. The required sample size n = p(1 − p)(z/ME)² is —
- A. 1,201
- B. 97
- C. 2,401 ✅
- D. 601
Feedback: 0.25 × (1.96 ⁄ 0.02)² = 0.25 × 9,604 = 2,401. Halving a margin costs four times the sample.

Q20 (MC · W13). A climbing gym's website says intro classes average 12 students. A coach thinks the true average is different — maybe higher, maybe lower — and will test using a random sample of class rosters. The hypotheses are —
- A. H₀: μ = 12 vs. Hₐ: μ ≠ 12 ✅
- B. H₀: x̄ = 12 vs. Hₐ: x̄ ≠ 12
- C. H₀: μ ≠ 12 vs. Hₐ: μ = 12
- D. H₀: μ = 12 vs. Hₐ: μ > 12
Feedback: Two-sided suspicion → ≠, and hypotheses describe the parameter μ, never the sample's x̄.

Q21 (MC · W13). A quiz platform claims 40% of teams renew their season registration. Suppose the true renewal rate is actually well below 40%, but a test of H₀: p = 0.40 returns a p-value of 0.18, so the analyst fails to reject. This outcome is —
- A. A Type I error — a false alarm about the platform's renewal claim
- B. A Type II error — the test missed a claim that really is false ✅
- C. Correct by definition, because a p-value above 0.05 ends the analysis
- D. A calculation error, because a p-value cannot exceed the true rate
Feedback: Failing to catch a false H₀ is the miss — Type II. (Type I would be a false alarm on a true H₀.)

Q22 (MC · W14). An airline claims its bag-drop kiosks take 4.0 minutes on average. From n = 25 random uses, x̄ = 4.5 minutes with s = 1.25. The t-statistic, and the two-sided α = 0.05 decision against t = 2.064 (df 24), are —
- A. t = 2.0; fail to reject H₀ — not convincing evidence against the 4-minute claim ✅
- B. t = 2.0; reject H₀ — the kiosks are proven to be slower than advertised
- C. t = 0.4; fail to reject H₀ — the sample sits within one SE of the claim
- D. t = 10.0; reject H₀ — the sample mean is ten standard errors above 4.0
Feedback:* SE = 1.25 ⁄ 5 = 0.25; t = 0.5 ⁄ 0.25 = 2.0 < 2.064 → fail to reject (which is not "the claim is proven").

Q23 (MC · W14). Ten community-garden plots are measured for yield the season before adding compost and again the season after, so each plot supplies a before number and an after number. The right way to test whether compost changed mean yield is —
- A. A two-sample t-test comparing the before group with the after group
- B. A paired analysis — a one-sample t-test on each plot's after-minus-before difference ✅
- C. A one-proportion z-test on the share of the plots that improved
- D. A chi-square goodness-of-fit test comparing the two seasons' totals
Feedback: Same plot, two linked measurements → subtract first. Treating the columns as independent buries the signal.

Q24 (MC · W15). A drone operator advertises that 90% of packages arrive on time. Regulators sample 100 deliveries and find 84 on time. With SE = √(0.90 × 0.10 ⁄ 100) = 0.03, the test statistic is —
- A. z = +2.0
- B. z = −0.06
- C. z = −6.0
- D. z = −2.0 ✅
Feedback: z = (0.84 − 0.90) ⁄ 0.03 = −2.0 — the SE comes from the claimed p₀ = 0.90.

Q25 (MC · W15). Researchers want to know whether the completion rate of a six-week cooking course differs between two unrelated schools, using one random sample of enrollees from each school. The procedure that fits is —
- A. A two-proportion z-test ✅
- B. A paired t-test on enrollees
- C. A one-sample t-test for a mean
- D. A goodness-of-fit test
Feedback: Two independent groups, each producing a completion percentage → two-proportion z.

Q26 (MC · W16). A seed packet promises a zinnia mix of 50% red, 30% yellow, and 20% orange. In a greenhouse trial, 100 sprouted plants show 40 red, 36 yellow, and 24 orange (expected: 50 / 30 / 20). The chi-square statistic is —
- A. 20 — the three gap sizes simply added up, ignoring their signs
- B. 1.33 — the average of the three categories' contributions
- C. 0 — the gaps above and below the claim cancel each other out
- D. 4.0 — the sum of the three (O − E) squared / E contributions ✅
Feedback: 100⁄50 + 36⁄30 + 16⁄20 = 2.0 + 1.2 + 0.8 = 4.0. Squaring stops the cancel; dividing by E scales the surprise.

Q27 (MC · W16). An ice rink records each visitor's ticket type (public skate / lesson / hockey) and age group (youth / adult / senior), then asks whether ticket type is related to age group. The procedure that fits is —
- A. A chi-square goodness-of-fit test against an even three-way split
- B. A two-sample t-test comparing the ticket types across age groups
- C. A chi-square test of independence on the 3 × 3 table of counts ✅
- D. A one-proportion z-test on the share of visitors who are youths
Feedback: Two categorical variables from one sample, asking "related?" → independence, df = (3−1)(3−1) = 4.

Q28 (MC · W17). For 30 community-garden plots, the fitted line ŷ = 1.5 + 0.4x predicts tomato yield (kg) from compost added (kg). The predicted yield for a plot given 10 kg of compost is —
- A. 4.0 kg
- B. 5.5 kg ✅
- C. 15.4 kg
- D. 1.9 kg
Feedback: ŷ = 1.5 + 0.4(10) = 5.5 — and 10 kg sits inside the fitted range, so the prediction is legitimate.

Q29 (MC · W17). A shuttle office regresses ride time on the number of stops for n = 26 runs; the output gives slope b = 2.5 with SE = 1.0. Using t = 2.064 (df 24), the 95% confidence interval for the true slope is —
- A. (1.5, 3.5)
- B. (2.4, 2.6)
- C. (0.54, 4.46)
- D. (0.44, 4.56) ✅
Feedback:* 2.5 ± 2.064 × 1.0 = (0.436, 4.564); df = n − 2 = 24. Zero is outside → the slope is significant at 5%.

Q30 (Matching · W17). Match each study plan to the tool it needs.
| Study plan | Tool |
|---|---|
| Estimate the proportion of climbing-gym members who boulder weekly | One-proportion z-interval |
| Test whether mean airport screening time differs from a posted 11 minutes | One-sample t-test |
| Check a claimed 50/30/20 mix of trivia team sizes against sampled counts | Chi-square goodness-of-fit test |
| Compare mean thrift-store purchase amounts between two independent branches | Two-sample t-test |
Feedback: Name the answer's shape — proportion, mean, counts — then count the samples. The shape names the tool.

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Answer key (quick reference)

Q	Answer		Q	Answer		Q	Answer
1	B		11	A, B, C		21	B
2	D		12	D		22	A
3	A		13	A		23	B
4	C		14	C		24	D
5	B		15	B		25	A
6	A, B, E		16	D		26	D
7	C		17	False		27	C
8	False		18	D		28	B
9	A		19	C		29	D
10	B		20	A		30	see pairs above

Quality gate (self-checked, checker-enforced): every single-answer item has exactly one correct option; every computation was pre-verified by script (tools/checks/w18_math.py); the answer-key sequence carries no positional pattern, no letter runs over three, and no letter over 40%; the keyed option is not systematically the longest or shortest; and the automated leakage screen (tools/check_exam.py) confirms zero stem overlap with the live final (token similarity < 0.45 on all 60 × 30 pairs) and no echo of any weekly quiz item. The zero-shared-items promise in the header is machine-verified, not just intended.

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Canvas placement block

canvas_object    = Quizzes::Quiz
title            = "Week 18 Practice Final — Rehearse the Real Thing"
assignment_group = "Practice exercises"
points_possible  = 30
grading_type     = points          # 0-weight group; shows a score, counts nothing
allowed_attempts = unlimited
due_offset_days  = 4               # aligned with the final; stays open for late review
shuffle_answers  = true
show_correct_answers = true        # feedback after each submission — that's the point
published        = true