Week 18 — Final Exam (auto-graded) · Introduction to Statistics, Cumulative

Course: Introduction to Statistics (18-week generic edition)
Covers: Weeks 1–17 (Objectives 1–9) — cumulative: 24 items from Weeks 1–8, 36 items from Weeks 10–17
Points: 120 (60 items × 2) · Assignment group: Final (5% of grade) · Due: mid-week of Week 18 (due_offset_days = 4) · One attempt · Closed to AI.

This is the human-readable final with its vetted answer key and feedback — instructor copy; never publish this file to students. The import-ready Classic QTI is in L-final-week-18-qti.xml; the Canvas placement block is at the bottom. Every lookup value a question needs (z, t, or chi-square) is printed inside that question; no table memorization is assessed. No item depends on remembering any course dataset.

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Blueprint

Week	Items	Concepts sampled
W1	3	statistic vs. parameter; sampling designs (matching); observational vs. experiment
W2	3	relative frequency; skew direction; truncated-axis misdirection
W3	3	mean vs. median with an outlier; IQR + 1.5×IQR fences; comparing z-scores
W4	3	reading r; conditional percent from a two-way table; correlation ≠ causation (TF)
W5	3	complement rule; independent AND; conditional probability from counts
W6	3	E(X); linear transformation of mean/SD; continuous exact value (TF)
W7	3	binomial parameters; binomial formula; binomial mean & SD
W8	3	empirical rule; forward normal; inverse normal
W10	4	standard error; CLT statements (select-all); sample-mean probability; SD vs. SE
W11	5	choosing t*; margin of error; building the interval; interpretation; narrower intervals (select-all)
W12	4	SE of p̂; proportion interval; sample size; what the MOE covers (TF)
W13	5	hypotheses; p-value meaning; decision at α; Type I; practical vs. statistical
W14	5	computing t; t-decision; paired recognition; one-sided p; test↔interval duality (TF)
W15	4	one-proportion z; two-proportion output; pooled proportion; two-sample settings (select-all)
W16	4	expected count; goodness-of-fit compute+decide; df for independence; conditions (select-all)
W17	5	slope meaning; residual; r²; slope t-test from output; choose-the-procedure (matching)

Item types: 50 multiple choice, 4 true/false, 4 select-all, 2 matching. Difficulty leans application over recall; distractors are built from the course's named misconceptions. All scenario surfaces are fresh — no graded item from Weeks 1–17 is reused, and the practice final shares zero items with this form.

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Questions, key, and feedback

Weeks 1–8 (24 items)

Q1 (MC · W1). A university transit office wants the proportion of all 18,500 enrolled students who ride the campus shuttle at least once a week. It surveys 500 randomly selected students, and 210 of them (42%) are weekly riders. Which statement is correct?
- A. The 500 surveyed students are the population, and the 18,500 students are the sample
- B. The value 42% is a parameter, because it was computed from genuine survey records
- C. The 42% is a statistic; the proportion among all 18,500 students is the parameter ✅
- D. The value 42% is a census result, so there is no estimation of any kind involved
Feedback: Measured on the sample → statistic; the unseen all-student rate is the parameter. (A is the classic role swap.)

Q2 (Matching · W1). A city ice rink wants feedback from its 6,000 pass holders. Match each plan to the sampling design it uses.
| Plan | Design |
|---|---|
| Draw 300 names at random from the full pass-holder list | Simple random sample |
| Randomly sample pass holders within each age bracket | Stratified sample |
| Randomly pick 3 of the 15 weekly sessions and survey everyone attending those 3 | Cluster sample |
| Take every 20th pass holder from an alphabetized list after a random start | Systematic sample |
Feedback: Stratified samples within every group; cluster takes whole groups — the classic mix-up.

Q3 (MC · W1). A community-garden association looks back at last season's records and notices that plots whose gardeners attended a spring workshop produced more vegetables than plots whose gardeners skipped it. No one was assigned to anything. What kind of study is this, and what does it support?
- A. An observational study — it can show an association, but attending may be tangled with gardener enthusiasm ✅
- B. An experiment — comparing two groups of garden plots is exactly what makes a study an experiment
- C. A census — every plot in the whole garden was measured, so cause and effect has been established
- D. A stratified sample — the garden's plots fall into two natural groups, workshop and non-workshop
Feedback: Nothing was imposed → observational. Enthusiasm is the lurking third variable; comparing groups doesn't make an experiment — imposing a treatment does.

Q4 (MC · W2). Volunteers at a thrift store sort one afternoon's 200 donated items and find that 50 of the items are books. What share of the afternoon's donations were books?
- A. 0.50
- B. 0.20
- C. 0.75
- D. 0.25 ✅
Feedback: Relative frequency = count ÷ total = 50 ⁄ 200 = 0.25.

Q5 (MC · W2). At an airport security checkpoint, most travelers get through in 5 to 15 minutes, but a small number are stuck far longer. A histogram of these wait times would be —
- A. Symmetric, because the waits pile up close to a single peak
- B. Skewed to the right, with a thin tail stretching toward the long waits ✅
- C. Skewed to the left, because the tall peak sits at the shorter waits
- D. Uniform, because any wait from five minutes on upward is possible
Feedback: Skew is named for the tail — the long-wait tail points right.

Q6 (MC · W2). A drone-delivery ad compares completion rates with a bar chart: its own 97% next to a rival's 95%. The chart's vertical axis begins at 94%. What is the problem?
- A. Completion rates are parts of one whole, so a pie chart was required here
- B. Starting the axis at 94% makes a 2-point gap look several times larger than it is ✅
- C. Nothing — trimming the axis below 94% simply removes wasted empty space
- D. Bar charts cannot properly display percentages, only raw counts of deliveries
Feedback: Truncated axes exaggerate honest differences — bars start at zero, or the zoom gets labeled loudly.

Q7 (MC · W3). Five climbers logged session lengths of 40, 50, 60, 70, and 130 minutes at a climbing gym. Which pair of summaries is correct?
- A. Mean 70 minutes; median 60 minutes ✅
- B. Mean 60 minutes; median 70 minutes
- C. Mean 70 minutes; median 70 minutes
- D. Mean 85 minutes; median 65 minutes
Feedback: 350 ⁄ 5 = 70; the sorted middle value is 60. The 130 pulls the mean, not the median.

Q8 (MC · W3). An airport posts a checkpoint's weekly wait report: shortest wait 4 minutes, first quartile 12, median 16, third quartile 20, longest wait 45. Under the 1.5 × IQR rule, which statement is correct?
- A. IQR = 41, and nothing can be flagged without first computing the mean
- B. IQR = 8, and no wait is flagged, because 45 sits below the upper fence of 60
- C. IQR = 8, and the 45-minute wait is flagged, because it exceeds the upper fence of 32 ✅
- D. IQR = 16, and every wait longer than 20 minutes is flagged as an outlier
Feedback: IQR = 20 − 12 = 8; fences at 12 − 12 = 0 and 20 + 12 = 32; 45 > 32 → flagged (investigate, don't delete).

Q9 (MC · W3). In one trivia league, Priya scored 87 points on a night when the league average was 75 with SD 8. In a different league, Marcus scored 92 when the average was 72 with SD 10. Relative to their own leagues, who had the stronger night?
- A. Priya — a smaller standard deviation always means a better relative performance
- B. Neither — scores from two different trivia leagues can never be compared at all
- C. Priya — her z-score works out to 1.5, which beats Marcus's z-score of 1.25
- D. Marcus — his z-score of 2.0 beats Priya's z-score of 1.5 on the shared ruler ✅
Feedback: (87 − 75)⁄8 = 1.5; (92 − 72)⁄10 = 2.0. z-scores exist precisely to compare across different scales.

Q10 (MC · W4). Across 40 winter days, a greenhouse logs the outdoor temperature and the hours its heater runs; the correlation is r = −0.88. The best description of the relationship is —
- A. A strong linear association in which colder days go with more heater hours ✅
- B. A weak association, because the value of r is negative rather than positive
- C. Heater hours fall by exactly 0.88 for every one-degree rise in temperature
- D. No usable association, because r cannot be applied to weather-related data
Feedback: Sign = direction (negative: one up, the other down), size = strength (0.88 is strong). r is not a slope.

Q11 (MC · W4). A cooking school's records show 80 weekday sign-ups, of which 20 later canceled, and 40 weekend sign-ups, of which 16 later canceled. What percent of the weekend sign-ups canceled?
- A. 13%, dividing the 16 weekend cancellations by all 120 sign-ups
- B. 30%, because 36 of the 120 total sign-ups ended in cancellation
- C. 40%, dividing the 16 weekend cancellations by the 40 weekend sign-ups ✅
- D. 44%, dividing the 16 weekend cancellations by all 36 cancellations
Feedback: Conditional = inside the named group: 16 ⁄ 40 = 40%. The denominator is the whole game.

Q12 (True/False · W4). Across 60 community-garden plots, mulch depth and tomato yield show a strong positive correlation — so a gardener who adds mulch has proof that her yield will rise.
- True
- False ✅
Feedback: Observational association ≠ causation — diligent gardeners may both mulch more and tend more. Only random assignment earns the causal verb.

Q13 (MC · W5). A trivia host's question generator draws categories with P(history) = 0.30, P(science) = 0.25, P(movies) = 0.25, and P(sports) = 0.20. The probability that the next question is not a movies question is —
- A. 0.25
- B. 0.75 ✅
- C. 0.55
- D. 0.80
Feedback: Complement rule: 1 − 0.25 = 0.75.

Q14 (MC · W5). Each of an ice rink's evening public-skate sessions sells out with probability 0.3, independently of the others. The probability that both of the next two evening sessions sell out is —
- A. 0.60
- B. 0.30
- C. 0.15
- D. 0.09 ✅
Feedback: Independent AND → multiply: 0.3 × 0.3 = 0.09. (Adding gives the 0.60 trap.)

Q15 (MC · W5). One morning, 160 bags pass through a standard security lane and 8 are pulled for extra screening, while 40 bags pass through a priority lane and 6 are pulled. Given that a bag went through the priority lane, what is the probability it was pulled?
- A. 0.05
- B. 0.07
- C. 0.43
- D. 0.15 ✅
Feedback: Shrink the world to the 40 priority-lane bags: 6 ⁄ 40 = 0.15. (0.05 answers a different question — the standard lane's.)

Q16 (MC · W6). Let X be the number of guests a cooking-class registrant brings along. Studio records give P(0) = 0.3, P(1) = 0.4, P(2) = 0.2, and P(3) = 0.1. The expected value E(X) is —
- A. 1.1 guests ✅
- B. 1.5 guests
- C. 0.4 guests
- D. 1.0 guest
Feedback: 0(.3) + 1(.4) + 2(.2) + 3(.1) = 1.1 — a long-run average that needn't be a possible value.

Q17 (MC · W6). A delivery drone consumes X battery units per trip, with mean 4 and standard deviation 1.5. The operator's energy cost per trip is Y = 2X + 3 dollars. The mean and standard deviation of Y are —
- A. Mean $11; SD $6.00
- B. Mean $11; SD $3.00 ✅
- C. Mean $8; SD $3.00
- D. Mean $11; SD $1.50
Feedback: Mean: 2(4) + 3 = 11. SD: 2(1.5) = 3 — adding a constant shifts the center but never the spread.

Q18 (True/False · W6). A greenhouse sensor records air temperature as a continuous random variable, so the probability that tomorrow's mid-afternoon reading equals one exact value — say 21.4638 degrees — is 0.
- True ✅
- False
Feedback: True — for continuous variables, probability is area over an interval; one exact point has zero width.

Q19 (MC · W7). About 10% of campus-shuttle riders request the bike-rack stop, independently of one another. Let X count how many of the next 20 riders request it. Then X is binomial with —
- A. n = 10, p = 0.20
- B. n = 20, p = 0.90
- C. n = 20, p = 0.10 ✅
- D. n = 2, p = 0.10
Feedback: n = the fixed number of trials (20 riders); p = each trial's success chance (0.10).

Q20 (MC · W7). A climber sends a particular bouldering problem on 50% of her attempts, independently. In 4 attempts, what is the probability of exactly 2 sends?
- A. 0.375 ✅
- B. 0.25
- C. 0.5
- D. 0.0625
Feedback: C(4, 2)(0.5)²(0.5)² = 6 ⁄ 16 = 0.375. Forgetting the "ways" factor C(4, 2) gives the 0.0625 trap.

Q21 (MC · W7). Experience shows 25% of the small appliances donated to a thrift store still work, independently. For a batch of 48 donated appliances, the mean and standard deviation of the number that work are —
- A. Mean 12; SD 3 ✅
- B. Mean 12; SD 9
- C. Mean 24; SD 3
- D. Mean 12; SD 6
Feedback: np = 48 × 0.25 = 12; σ = √(48 × 0.25 × 0.75) = √9 = 3. (9 is the variance.)

Q22 (MC · W8). A greenhouse's overnight low temperatures are approximately normal with mean 16 degrees C and SD 2 degrees C. About 95% of nights have lows between —
- A. 14 and 18 degrees C
- B. 10 and 22 degrees C
- C. 15 and 17 degrees C
- D. 12 and 20 degrees C ✅
Feedback: 95% ↔ within 2 SDs: 16 ± 4. (14–18 is the 68% band; 10–22 is the 99.7% band.)

Q23 (MC · W8). Skate-sharpening times at an ice rink are approximately N(10, 2) minutes. Using the course z-table (the area to the left of z = 1.25 is 0.8944), what proportion of sharpenings finish in under 12.5 minutes?
- A. 0.1056
- B. 0.8944 ✅
- C. 0.9332
- D. 0.6915
Feedback: z = (12.5 − 10) ⁄ 2 = 1.25; "under" = the left area itself, 0.8944.

Q24 (MC · W8). Package weights on a drone route are approximately N(2.0, 0.4) kg. The service adds a surcharge to the heaviest 6.68% of packages. (The course z-table gives the area left of z = 1.5 as 0.9332.) The surcharge starts at —
- A. 2.4 kg
- B. 3.0 kg
- C. 2.6 kg ✅
- D. 1.4 kg
Feedback: Top 6.68% ↔ left area 0.9332 ↔ z = 1.5 → x = 2.0 + 1.5(0.4) = 2.6 kg — start at the mean, walk z SDs.

Weeks 10–17 (36 items)

Q25 (MC · W10). Security screening times for individual travelers vary with a standard deviation of 8 minutes. When auditors average n = 16 randomly chosen travelers, the typical wobble of that average — its standard error — is
- A. 8 minutes
- B. 2 minutes ✅
- C. 0.5 minutes
- D. 4 minutes
Feedback: SE = σ ⁄ √n = 8 ⁄ 4 = 2 — averages wobble less than individuals.

Q26 (Select all · W10). Campus-shuttle ride times are strongly right-skewed. A transit analyst plans to average random samples of n = 50 rides. Which statements are true? Select all that apply.
- A. The sampling distribution of the sample mean is centered at the population mean ride time ✅
- B. Averaging 50 rides reshapes the population of individual ride times into a bell curve
- C. The mean of 50 rides varies less from sample to sample than single ride times do ✅
- D. With n = 50, the sampling distribution of the sample mean is approximately normal despite the skew ✅
- E. If the samples came only from early-morning rides, the CLT would remove that bias
Feedback: The CLT fixes the shape of x̄'s distribution — never the population's histogram, and never bias.

Q27 (MC · W10). Bags dropped at a donation center weigh 6.0 kg on average with standard deviation 1.5 kg. For a random sample of 36 bags, what is the probability the sample mean weight exceeds 6.5 kg? (Course z-table: the area left of z = 2 is 0.9772.)
- A. 0.9772
- B. 0.3707
- C. 0.1587
- D. 0.0228 ✅
Feedback: SE = 1.5 ⁄ 6 = 0.25; z = 0.5 ⁄ 0.25 = 2 → 1 − 0.9772 = 0.0228. Extreme averages are rare.

Q28 (MC · W10). Single visit lengths at a climbing gym have standard deviation 30 minutes. An analyst repeatedly draws random samples of 25 visits and records each sample's mean. The typical spread of those sample means is —
- A. 6 minutes — sample means wobble far less than individual visits do ✅
- B. 30 minutes — averaging leaves the spread of the data unchanged
- C. 150 minutes — averaging multiplies the spread by the sample size's square root
- D. 1.2 minutes — the spread shrinks by the full factor of the sample size, 25
Feedback: SE = 30 ⁄ √25 = 6. Two rulers: σ for individuals, σ⁄√n for means (÷√n, not ÷n).

Q29 (MC · W11). A culinary school times a random sample of 10 students' knife-skills drills and wants a 95% confidence interval for the mean time. From the course t-table (df 9 row: 1.833 / 2.262 / 3.250 for 90% / 95% / 99%), the multiplier t is —
- A. 1.833
- B. 3.250
- C. 2.262 ✅
- D. 1.960
Feedback:* n = 10 → df 9; the 95% column reads 2.262. (1.960 is the large-sample z* — too little humility for n = 10.)

Q30 (MC · W11). A random sample of 25 community-garden plots has harvest weights with s = 5 kg, so the standard error of the mean is 1 kg. For a 95% confidence interval (df 24, t = 2.064), the margin of error is —
- A. 1.000 kg
- B. 5.000 kg
- C. 2.064 kg ✅
- D. 0.413 kg
Feedback:* ME = t* × SE = 2.064 × 1 = 2.064 kg. (Using s = 5 instead of SE = 1 is the classic swap.)

Q31 (MC · W11). Those same 25 garden plots average 40 kg. Combining the pieces (SE = 1 kg, t = 2.064), the 95% confidence interval for the mean harvest of all plots is —
- A. (39.0, 41.0) kg
- B. (37.94, 42.06) kg ✅
- C. (35.0, 45.0) kg
- D. (38.97, 41.03) kg
Feedback:* 40 ± 2.064 → (37.936, 42.064), rounded at the end. (A forgot t*; C used s; D halved the margin.)

Q32 (MC · W11). From a random sample of private-event rentals, an ice rink computes a 95% confidence interval of (82, 94) minutes for the MEAN rental length. Which reading is correct?
- A. About 95% of the rink's individual rentals last between 82 and 94 minutes
- B. There is a 95% chance that the next single rental will run between 82 and 94 minutes
- C. 95% of all future samples would produce a sample mean inside this exact interval
- D. We are 95% confident that the interval 82 to 94 captures the mean length of all rentals ✅
Feedback: Confidence describes the method's capture rate for the mean — never individuals, never one future rental.

Q33 (Select all · W11). A grower wants a tighter estimate of a greenhouse's mean daytime humidity. Which of these moves would shrink the width of her 95% interval? Select all that apply.
- A. Quadrupling the number of sensor readings in the sample ✅
- B. Dropping the confidence level from 99% to 90% ✅
- C. Raising the confidence level from 95% to 99.9%
- D. Using a steadier sensor so the sample standard deviation shrinks ✅
- E. Reporting the same interval in a different unit of measurement
Feedback: Width dials: n up → narrower (by √n); confidence down → narrower; s down → narrower. Certainty costs width.

Q34 (MC · W12). In a random sample of 150 thrift-store donations, 60 are clothing, so p̂ = 0.40. The standard error of p̂, the square root of (0.40 × 0.60 ⁄ 150), equals —
- A. 0.04 ✅
- B. 0.0016
- C. 0.24
- D. 0.008
Feedback: 0.24 ⁄ 150 = 0.0016; √0.0016 = 0.04. (0.0016 is the un-rooted variance — the named trap.)

Q35 (MC · W12). Fifty of 100 randomly sampled trivia-night attendees are first-timers (p̂ = 0.50, SE = 0.05). The 95% confidence interval for the proportion of all attendees who are first-timers is —
- A. (0.45, 0.55)
- B. (0.402, 0.598) ✅
- C. (0.30, 0.70)
- D. (0.418, 0.582)
Feedback: 0.50 ± 1.96(0.05) = 0.50 ± 0.098. (A used 1 SE; D used the 90% multiplier 1.645.)

Q36 (MC · W12). A parks office will survey residents about joining a community-garden waitlist and wants a margin of error of plus-or-minus 0.04 at 95% confidence, with no prior estimate (p = 0.5). Using n = p(1 − p)(z/ME)², the required sample size is —
- A. 600
- B. 385
- C. 1,201
- D. 601 ✅
Feedback: 0.25 × (1.96 ⁄ 0.04)² = 600.25 → sample sizes always round up: 601. (385 answers ±0.05.)

Q37 (True/False · W12). An airport publishes traveler-satisfaction results with a plus-or-minus 4 point margin. That cushion also absorbs distortion caused by slanted question wording and by travelers who decline to respond.
- True
- False ✅
Feedback: False — the margin of error prices random sampling error only. Wording and nonresponse bias ride outside the ±.

Q38 (MC · W13). A courier service advertises that its drone drops average 12 minutes from launch to doorstep. A neighborhood group suspects deliveries actually take longer and gathers a random sample. The group's hypotheses should be —
- 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: H₀ holds the equals sign about the parameter μ; the pre-stated suspicion (longer) is Hₐ. Never hypothesize about x̄.

Q39 (MC · W13). A climbing gym claims the mean wait for an auto-belay station is 5 minutes. A member runs a test on a random sample of waits and gets p = 0.04. What does 0.04 mean here?
- A. If the true mean wait really were 5 minutes, samples as extreme as this one would occur about 4% of the time ✅
- B. There is only a 4% probability that the climbing gym's five-minute claim is actually true
- C. About 4% of the gym's members end up waiting longer than five minutes at the stations
- D. The sample's average wait must have landed exactly 4% above the claimed five minutes
Feedback: A p-value is computed assuming H₀ — it is never the probability that H₀ is true.

Q40 (MC · W13). A cooking school tests whether its no-show rate differs from the 20% it budgets for, and the software reports p = 0.32. At α = 0.05, the correct move is —
- A. Reject H₀ and conclude that the no-show rate has changed from 20%
- B. Fail to reject H₀ — the data don't give convincing evidence of a rate different from 20% ✅
- C. Accept H₀ and conclude the no-show rate is proven to be exactly 20%
- D. Lower α to 0.32 so that the observed result becomes statistically significant
Feedback: p > α → fail to reject — which is not "accept." And α is chosen before the data, never after.

Q41 (MC · W13). An ice rink claims 85% of its public sessions start on time. A skaters' group tests H₀: p = 0.85 against Hₐ: p < 0.85. In this setting, a Type I error would be —
- A. Failing to find a problem when the true on-time rate really is below 85%
- B. Concluding the on-time rate is below 85% when the true rate really is 85% ✅
- C. Choosing a significance level that is larger than the p-value after testing
- D. Recording sessions as late because the rink's lobby clock ran a few minutes fast
Feedback: Type I = rejecting a true H₀ — the false alarm (probability α). Option A is Type II, the miss.

Q42 (MC · W13). Using 250,000 logged rides, a campus-shuttle team finds its new boarding process cut the mean wait by 6 seconds, with p = 0.0004. The best reading is —
- A. The tiny p-value proves the improvement is large enough to matter to riders
- B. Such a small p-value means there is a 0.04% chance the process was unchanged
- C. The result must be an arithmetic error, because 6 seconds is too small to detect
- D. Statistically significant, but 6 seconds may be too small to matter — huge samples flag tiny effects ✅
Feedback: Significant ≠ important. Ask "how big?" right after "how surprising?"

Q43 (MC · W14). A greenhouse supplier claims its misting system holds mean daytime humidity at 60%. Across n = 16 randomly chosen days, a grower measures x̄ = 57% with s = 8. The t-statistic is —
- A. t = −0.375
- B. t = −3.0
- C. t = −1.5 ✅
- D. t = +1.5
Feedback: SE = 8 ⁄ √16 = 2; t = (57 − 60) ⁄ 2 = −1.5. (−0.375 divided by s instead of SE; the sign follows the data.)

Q44 (MC · W14). The grower finishes her humidity analysis: df = 15, a two-sided alternative, and the course t-table's 95% cutoff of 2.131. At the 5% level, she should —
- A. Fail to reject H₀ — the size of t, 1.5, is under 2.131, so these data don't refute the 60% claim ✅
- B. Reject H₀ — the negative sign of the t-statistic by itself contradicts the supplier's claim
- C. Accept H₀ — the sample has proven the misting system holds humidity at exactly 60%
- D. Reject H₀ — the size of t, 1.5, is greater than the significance level of 0.05
Feedback: |t| below the cutoff → fail to reject (never "accept"). t compares to t*; p compares to α — never mix the pairs.

Q45 (MC · W14). An ice rink asks 14 members of its speed club to skate one timed lap on blade profile R and one on blade profile S, order randomized, recording both laps for each skater. To compare the two profiles, the right analysis is —
- A. A two-sample t-test treating the R laps and the S laps as independent groups
- B. A two-proportion z-test on the share of skaters who preferred each profile
- C. No valid analysis exists, because every skater tried both of the blade profiles
- D. A paired t-test — a one-sample t on each skater's R-minus-S lap difference ✅
Feedback: Same skater, two linked laps → subtract first. Pairing cancels skater-to-skater spread; ignoring it buries the signal.

Q46 (MC · W14). A thrift-store manager had a pre-stated hunch that a remodel raised the store's mean purchase amount. Her random sample yields t = +1.9, and software prints the two-sided p-value, 0.06. For her one-sided alternative, the p-value and the α = 0.05 decision are —
- A. p = 0.12; fail to reject H₀
- B. p = 0.06; reject H₀
- C. p = 0.03; fail to reject H₀
- D. p = 0.03; reject H₀ ✅
Feedback: The sample leaned the pre-stated direction, so halve: 0.06 ⁄ 2 = 0.03 ≤ 0.05 → reject. (Legitimate only because the direction was called before the data.)

Q47 (True/False · W14). A seed catalog states that a bean variety matures in 90 days on average. A gardener's 95% confidence interval from her own plots runs from 87 to 95 days, so a two-sided test at the 5% level will fail to reject the 90-day claim.
- True ✅
- False
Feedback: True — duality: μ₀ inside the 95% interval ⟺ a two-sided α = 0.05 test fails to reject it.

Q48 (MC · W15). A pub-quiz platform claims that half of all teams come back the following week. In a random sample of 100 teams, 42 returned. Using SE = the square root of (0.5 × 0.5 ⁄ 100) = 0.05, the test statistic is —
- A. z = +1.6
- B. z = −1.6 ✅
- C. z = −0.08
- D. z = −2.0
Feedback: z = (0.42 − 0.50) ⁄ 0.05 = −1.6 — the SE is built from the claimed p₀, and the sign follows the data.

Q49 (MC · W15). Security auditors compare two terminals: at Terminal 1, 66 of 300 sampled bags got extra screening (0.22); at Terminal 2, 42 of 300 (0.14). Software reports pooled proportion 0.18, z = 2.55, two-sided p = 0.011. At α = 0.05, the conclusion is —
- A. Accept H₀ — the two terminals share one true screening rate of exactly 0.18
- B. Fail to reject H₀ — the pooled proportion of 0.18 sits between the two rates
- C. Reject H₀ — convincing evidence that the terminals' true screening rates differ ✅
- D. No conclusion — two-proportion tests require equal numbers of flagged bags
Feedback: p = 0.011 ≤ 0.05 → reject, stated in context. The pooled 0.18 is H₀'s single-rate estimate, not a verdict.

Q50 (MC · W15). A cooking school compares cancellations at its two locations: downtown, 45 of 150 sign-ups canceled (0.30); lakeside, 5 of 50 canceled (0.10). For the two-proportion z-test, the pooled proportion is —
- A. 0.25 — the combined 50 cancellations over the combined 200 sign-ups ✅
- B. 0.20 — the simple average of the two locations' rates
- C. 0.30 — the cancellation rate observed at the larger of the locations
- D. 0.40 — the two locations' cancellation rates added directly together
Feedback: Pool successes over sizes: 50 ⁄ 200 = 0.25 — the single shared rate H₀ implies. Averaging rates is wrong when sizes differ.

Q51 (Select all · W15). Which of these questions call for a TWO-SAMPLE procedure on independent groups? Select all that apply.
- A. Compare mean screening times at Terminal 1 with mean screening times at Terminal 2 ✅
- B. Compare each traveler's screening time this trip with that same traveler's time last trip
- C. Compare the share of drone orders needing redelivery in District North vs. District South ✅
- D. Test whether one rink's mean rental length differs from its advertised 90 minutes
- E. Compare mean class sizes at two unrelated cooking schools ✅
Feedback: B is paired (same individual twice → subtract); D is one-sample against a claim. A, C, E compare two independent groups.

Q52 (MC · W16). An ice rink's stocking plan assumes 25% of skate rentals are figure skates. In a random sample of 240 rentals, how many figure-skate rentals does the plan predict?
- A. 25
- B. 96
- C. 60 ✅
- D. 48
Feedback: Expected count = n × claimed proportion = 240 × 0.25 = 60.

Q53 (MC · W16). A thrift store claims donations arrive evenly across its four weekly drop-off windows. A random sample of 80 donations gives counts 26, 22, 18, and 14 (expected: 20 each). With χ² = Σ(O − E)²⁄E, df = 3, and 5% critical value 7.815, the result is —
- A. χ² = 12.0 — reject the claim of an even spread across windows
- B. χ² = 0 — the positive and negative gaps cancel, so the claim stands
- C. χ² = 4.0 — reject, because 4.0 exceeds the degrees of freedom, 3
- D. χ² = 4.0 — fail to reject; the counts are consistent with an even spread ✅
Feedback: (36 + 4 + 4 + 36) ⁄ 20 = 4.0 < 7.815 → fail to reject. "Big" is defined by the critical value, never by df.

Q54 (MC · W16). To test whether donation category (five categories) is associated with drop-off location (two locations), a chi-square test of independence runs on a 5 × 2 table. Its degrees of freedom are —
- A. 10, the total number of cells in the table
- B. 4, computed from (5 − 1) × (2 − 1) ✅
- C. 9, the number of cells in the table minus one
- D. 6, computed from (5 − 1) + (2 − 1)
Feedback: df = (r − 1)(c − 1) = 4 × 1 = 4 — from the table's shape, never the sample size.

Q55 (Select all · W16). An analyst tallies skate-rental categories at one rink and plans a chi-square analysis. Which requirements must the data meet? Select all that apply.
- A. The data are counts of rentals in categories, not percentages or averages ✅
- B. Every EXPECTED count is at least 5 ✅
- C. The rentals were sampled by a random or otherwise representative method ✅
- D. Every OBSERVED count is at least 5
- E. The rental counts follow an approximately bell-shaped histogram
Feedback: The condition guards expected counts (observed zeros are legal), and χ² eats raw counts — no bell shape required anywhere.

Q56 (MC · W17). For one campus-shuttle line, a least-squares fit gives ŷ = 4.0 + 2.5x, where x = the number of stops on a run and y = ride time in minutes. The slope 2.5 means —
- A. Each additional stop predicts, on average, 2.5 more minutes of ride time ✅
- B. Every single run with one more stop takes exactly 2.5 minutes longer
- C. Adding stops causes ride time to rise by 2.5 minutes, all else held fixed
- D. About 2.5% of the variation in ride time is explained by the stop count
Feedback: The four-part sentence: per one unit of x · predicted · on average · in y's units. "Exactly" and "causes" both overclaim.

Q57 (MC · W17). Using that same fitted shuttle line (ŷ = 4.0 + 2.5x), one run made 5 stops and actually took 18 minutes. Its residual is —
- A. −1.5 minutes — the run came in faster than its prediction
- B. +16.5 minutes — the predicted ride time itself is the residual
- C. +1.5 minutes — the run took longer than the line predicted ✅
- D. 0 minutes — every observed run must sit exactly on the line
Feedback: ŷ = 4 + 12.5 = 16.5; residual = actual − predicted = 18 − 16.5 = +1.5 (above the line).

Q58 (MC · W17). Across shuttle runs, the correlation between stops and ride time is r = 0.9. The correct use of r² is —
- A. 90% of the runs fall exactly on the least-squares line
- B. About 81% of the variation in ride times is explained by the number of stops ✅
- C. Predictions made from the line are correct 81% of the time
- D. Ride time grows 81% faster for each additional stop added
Feedback: r² = 0.81 = the share of variation explained — never an accuracy rate, never a slope.

Q59 (MC · W17). From n = 26 winter days, a grower regresses daily heater hours on outdoor temperature. The output shows, for the slope: Coef = −3.0, SE Coef = 1.2. Using the course t-table (df 24, 95%: t = 2.064) at α = 0.05, the conclusion is —
- A. t = −2.5, and its size 2.5 beats 2.064 — reject H₀; evidence of a real linear relationship ✅
- B. t = −2.5, and a negative t-statistic can never lead to rejecting a null hypothesis
- C. t = −0.4, and its size 0.4 is under 2.064 — fail to reject H₀; the slope may be flat
- D. The test cannot be run, because a slope's degrees of freedom equal n − 1 = 25
Feedback:* t = b ⁄ SE(b) = −3.0 ⁄ 1.2 = −2.5; slope df = n − 2 = 24; |−2.5| > 2.064 → reject.

Q60 (Matching · W17). As a course capstone, pair every scenario below with the inference tool that answers it.
| Scenario | Tool |
|---|---|
| Estimate the mean nightly rental length at one ice rink | One-sample t-interval for a mean |
| Test whether the share of shuttle riders using a transit pass differs from a claimed 40% | One-proportion z-test |
| Test whether donation category is associated with drop-off location | Chi-square test of independence |
| Test whether outdoor temperature predicts daily greenhouse heating cost | t-test for a regression slope |
Feedback: Mean → t; proportion → z; two categorical variables → chi-square independence; predict-y-from-x → slope inference. Name the answer's shape and the tool names itself.

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Answer key (quick reference)

Q	Answer		Q	Answer		Q	Answer
1	C		21	A		41	B
2	see pairs		22	D		42	D
3	A		23	B		43	C
4	D		24	C		44	A
5	B		25	B		45	D
6	B		26	A, C, D		46	D
7	A		27	D		47	True
8	C		28	A		48	B
9	D		29	C		49	C
10	A		30	C		50	A
11	C		31	B		51	A, C, E
12	False		32	D		52	C
13	B		33	A, B, D		53	D
14	D		34	A		54	B
15	D		35	B		55	A, B, C
16	A		36	D		56	A
17	B		37	False		57	C
18	True		38	C		58	B
19	C		39	A		59	A
20	A		40	B		60	see pairs

Quality gate (the four exam-QC findings, machine-enforced by tools/check_exam.py — run output: EXAM GATE CLEAN): (a) Key-sequence fit: the 50-letter MC key (C A D B B A C D A C B D D A B C A A D B C B D A C C B D A B D C A B B D C A D D B C A C D B A C B A) was designed with no ABCD cycling, no run over two, letter counts 13/13/12/12 (max 26% — under the 40% gate), a cyclic-successor rate of 12/49 ≈ 0.24 (gate ≤ 0.45), and only 3 of 25 positional coincidences with the practice exam's key (gate ≤ 50%); no stem references an option letter, so platform shuffling stays safe. (b) Option-length balance: the keyed option is uniquely longest in 14 of 50 MC items and uniquely shortest in 0 (both under the 50% gates); every item's key-to-distractor mean-length ratio sits inside 0.5–1.5. (c) Zero leakage: token-similarity screening of all 60 × 30 live-vs-practice stem pairs (< 0.45) and of every stem against all 160 weekly quiz items (< 0.50) passes; no recall-style "name-the-list" stems. (d) Consistency: every computation was re-derived by script (tools/checks/w18_math.py, PASS), each lands on the course's embedded z/t/chi-square values or clean arithmetic, and the study guide was cross-checked section-by-section against the Weeks 1–17 chapters so nothing taught this week contradicts what the exam rewards.

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Canvas placement block

canvas_object    = Quizzes::Quiz
title            = "Final Exam — Introduction to Statistics (Cumulative)"
assignment_group = "Final"
points_possible  = 120
grading_type     = points
allowed_attempts = 1
due_offset_days  = 4               # mid-week: review early, exam mid-week, breathing room after
time_limit_minutes = 75            # optional; match your section's policy and accommodations
shuffle_answers  = true
one_question_at_a_time = false     # all visible; flag-and-return encouraged
published        = true