Mock Exam 2 — Full AP Simulation (All 5 Units)

Section I — Multiple Choice (recomputed)

1. (B) Median and IQR. For a strongly skewed distribution, the resistant measures (median, IQR) best describe center and spread; the mean and SD are pulled by the right tail. (A) is for symmetric data.

2. (A) 0.067. z = (84 − 72)/8 = 1.5; P(Z > 1.5) = 1 − 0.9332 = 0.0668 (normalcdf(84, 1E99, 72, 8)). (D) 0.933 is the left tail; (C) 0.5 ignores the calculation.

3. (C) 64. IQR = 40 − 24 = 16; upper fence = Q₃ + 1.5·IQR = 40 + 24 = 64. (A) 48 forgets the 1.5 (uses +IQR... actually +Q₁); (B) 56 = Q₃ + IQR; (D) 80 = 2·Q₃.

4. (B) City has a larger median (35 vs 18) and greater variability (IQR 25 vs 16; range 60 vs 40). (A) reverses it; (C) and (D) are contradicted by the values.

5. (B) 1.17. z = (85 − 78)/6 = 1.167. (A) divides incorrectly; (C) is the raw difference; (D) is the difference times 2.

6. (C) Stratified random sample. The grade levels are strata; an SRS is drawn within each. (B) cluster would sample whole intact groups; (D) systematic uses a fixed interval.

7. (B) Nonresponse bias. Only 480 of 5,000 (under 10%) responded; respondents may systematically differ from nonrespondents, so the 70% may not reflect everyone. (A) 480 is plenty large; the issue is who responded.

8. (B) Random assignment balances confounding variables across groups so a yield difference can be causally attributed to the fertilizer. (C) "exactly equal" overstates; (A) confuses assignment with selection; (D) is false.

9. (A) Random assignment licenses a causal claim for these plots, but because the plots were not randomly selected from all gardens, the result cannot be generalized broadly. (B) over-generalizes; (C) wrongly calls a randomized experiment observational.

10. (A) Independent: P(A)·P(B) = 0.5·0.4 = 0.20 = P(A and B). (B) and (C) state irrelevant facts; disjoint ≠ independent.

11. (B) 0.40. P(Job | Sport) = 40/100 = 0.40. (D) 0.45 = 90/200 is the overall job rate; (C) 0.44 is P(Sport | Job) = 40/90.

12. (A) 1.0. E(X) = 0(0.4)+1(0.3)+2(0.2)+3(0.1) = 0+0.3+0.4+0.3 = 1.0. (D) 0.25 averages the four probabilities.

13. (B) 1.00. E(X²) = 0+0.3+0.8+0.9 = 2.0; Var = 2.0 − 1.0² = 1.0; SD = √1.0 = 1.00. (C) 1.41 = √2 forgets to subtract μ².

14. (B) 0.279. binompdf(8, 0.6, 5) = C(8,5)(0.6)⁵(0.4)³ = 56·0.07776·0.064 = 0.279. (A) 0.124 is P(X=6); (D) 0.6 is the "p itself" trap.

15. (A) mean 30, SD 5.05. μ = np = 200·0.15 = 30; σ = √(np(1−p)) = √(200·0.15·0.85) = √25.5 = 5.05. (B) reports the variance as SD; (C) miscomputes the mean.

16. (B) 0.35. Uniform on [0, 20]: P(5 < X < 12) = (12 − 5)/20 = 7/20 = 0.35. (A) 0.25 uses width 5; (D) 0.70 forgets to divide correctly.

17. (B) 0.040. σ_p̂ = √(p(1−p)/n) = √(0.6·0.4/150) = √0.0016 = 0.040. (A) 0.0016 is the variance; (C) 0.245 = √(p(1−p)) without /n.

18. (B) 2.0. σ_x̄ = σ/√n = 12/√36 = 12/6 = 2.0. (A) divides by n; (D) forgets to divide.

19. (A) (0.552, 0.648). p̂ = 240/400 = 0.60; SE = √(0.6·0.4/400) = 0.0245; 0.60 ± 1.96·0.0245 = 0.60 ± 0.048 (1-PropZInt). (B) uses too small a critical value; (D) too wide.

20. (C) Confidence is a property of the long-run method. (B) is the classic error — a single computed interval already does or does not contain the fixed p; (A) confuses the interval with the data spread; (D) describes a sampling distribution.

21. (B) 1.70. Test uses p₀: z = (0.56 − 0.50)/√(0.5·0.5/200) = 0.06/0.03536 = 1.697. (C) 2.40 wrongly uses p̂ in the SE; (D) 0.06 is the raw difference.

22. (B) Fail to reject H₀. p = 0.08 > α = 0.05, so there is not convincing evidence p > 0.30. (C) "accept H₀" is never correct; (A) and (D) reject incorrectly.

23. (B) 0.375. Pooled p̂ = (45 + 30)/(100 + 100) = 75/200 = 0.375. (C) and (A) are the individual group proportions; (D) doubles incorrectly.

24. (C) 50. Expected = (row total · column total)/grand total = (100 · 150)/300 = 50. (A) 40 and (B) 45 misuse the totals; (D) 60 is the observed count.

25. (B) 4. df = (r − 1)(c − 1) = (3 − 1)(3 − 1) = 4. (D) 9 = r·c; (A) 2 uses only one dimension.

26. (C) χ² test for independence. One sample, two categorical variables measured on the same people, testing whether they are related. (B) homogeneity needs several separate samples; (A)/(D) are single-variable proportion tests.

27. (B) A higher confidence level requires a larger critical value, producing a wider interval (the price of more confidence). (A) and (D) reverse the relationship.

28. (B) (63.9, 72.1). ME = t·s/√n = 2.064·10/√25 = 2.064·2 = 4.128; 68 ± 4.13 = (63.87, 72.13). (A) rounds the margin to ~3.9 using a smaller t; (C) forgets the √n (margin ~20).*

29. (B) 2.00. t = (x̄ − μ₀)/(s/√n) = (104 − 100)/(12/6) = 4/2 = 2.00. (C) 4.00 forgets to divide by √n in the SE; (A) divides extra.

30. (B) Matched-pairs t-test. Each volunteer is measured on both layouts, so the two measurements are linked — analyze the within-person differences. (A) wrongly treats them as independent; (C)/(D) are categorical/wrong.

31. (C) 3.02. t = d̄/(s_d/√n) = 3.2/(4.1/√15) = 3.2/1.0586 = 3.02. (A) and (B) divide incorrectly; (D) forgets the √n.

32. (B) 2.01. SE = √(6²/20 + 7²/22) = √(1.8 + 2.227) = √4.027 = 2.01. (C) 4.00 is the difference in means; (A) drops a term.

33. (C) Random sample with no strong skew or outliers, so the population is plausibly Normal. (A) describes a z-procedure; (B) over-formalizes; (D) the n ≥ 30 rule is sufficient but not required — symmetric small samples are fine.

34. (B) Two-sample independent t-test. Two separate random samples of cans, one per brand, each can measured once — independent groups, quantitative response. (A) pairing would require a meaningful link between specific cans.

35. (B) Decrease the margin of error. ME ∝ s/√n; larger n shrinks s/√n (and slightly lowers t). (A)/(C) are wrong.*

36. (A) Strong, positive, linear. The points rise together, lie close to a straight line, and trend upward. (B) understates strength; (C) wrong direction.

37. (D) 0.70. r = ±√r² = ±√0.49 = ±0.70; slope is positive, so r = +0.70. (A) wrong sign; (B) ≈ r²·0.49 squared again; (C) confuses r with r².

38. (B) For each additional minute of exercise per day, predicted mood score increases by 0.85 points. (A) interprets the intercept; (C) wrongly claims causation; (D) reverses the variables.

39. (C) 29.0. ŷ = 12 + 0.85(20) = 12 + 17 = 29.0. (A) uses x = ~6; (B) forgets the intercept partly; (D) over-adds.

40. (B) 1.0. residual = actual − predicted = 30 − 29 = 1.0. (A) reverses the sign; (C) 1.5 miscomputes the prediction.

41. (B) A linear model is not appropriate — the clear curved (U-shaped) pattern in the residuals means the true relationship is nonlinear. (A) would require unstructured scatter; (C)/(D) are not what curvature shows.

42. (B) r = √0.64 = 0.80 (positive, since the slope 0.62 is positive), and R-Sq = 64% means 64% of the variation in weight is explained by the linear relationship with height. (A) confuses r with r²; (C) reads s as the slope; (D) misreads the intercept.

Section II — FRQ Rubrics (10 points each)

FRQ 1 — Model Response & Rubric

(a) Statistical question, e.g.: "Among adults having elective surgery at this hospital, does a 15-minute guided-relaxation session before surgery reduce the mean self-reported pre-operative anxiety score compared to no session?" (1 pt)

(b) Letting patients choose whether to use the session creates two self-selected groups that likely differ in important ways (e.g., patients who ask for relaxation may already be calmer, more health-conscious, or have less serious surgeries). These confounding variables are tangled up with the treatment, so any difference in anxiety cannot be attributed to the session itself. (1 pt: identifies self-selection/no random assignment; 1 pt: explains confounding makes causation unclear.) (2 pts)

(c) Completely randomized experiment: Take the 600 scheduled patients. Use a random mechanism (e.g., a random number generator, or draw names) to assign 300 patients to the treatment group (receive the 15-minute relaxation session before surgery) and 300 to the control group (no session, or a neutral 15-minute waiting period). After random assignment, measure each patient's anxiety on the 0–100 scale just before entering the OR. Compare the mean anxiety of the two groups. (1 pt: two treatments clearly described including a control; 1 pt: random assignment of patients to groups; 1 pt: groups of appropriate/equal size from the 600; 1 pt: response variable = anxiety score measured the same way for both.) (4 pts)

(d) Blocking by surgery severity: First separate the 600 patients into a major-surgery block and a minor-surgery block. Then, within each block, randomly assign half to relaxation and half to control. This controls for the known variability in anxiety due to surgery severity, so the comparison of treatment vs. control is made among similar patients — reducing unexplained variability and making it easier to detect a real effect of the session. (1 pt: correctly forms blocks by severity and randomizes within blocks; 1 pt: explains it reduces variability from the severity variable.) (2 pts)

(e) Because patients were randomly assigned to treatments, a significant difference supports a causal conclusion: the relaxation session reduced mean anxiety. However, the patients were not randomly selected from a larger population — they are the patients scheduled at this hospital this month — so the result generalizes only to patients like these at this hospital, not to all surgical patients everywhere. (1 pt: causal claim justified by random assignment AND generalization limited by lack of random selection.)

Where students lose points: vague investigative questions with no population or comparison; in (b), saying "small sample" instead of naming self-selection/confounding; in (c), forgetting the control group or the response variable; in (d), describing stratifying a sample instead of blocking in an experiment; in (e), over-generalizing to "all patients."

FRQ 2 — Model Response & Rubric

(a) The association is positive (more review hours go with higher scores), linear in form, and strong (r = 0.976, very close to 1, with points tightly clustered around the line). (1 pt: direction + form; 1 pt: strength justified by r or r².) (2 pts)

(b) Slope = 3.88: for each additional hour of review, the predicted quiz score increases by about 3.88 points. (1 pt: 3.88 tied to "per additional hour"; 1 pt: "predicted" score increase, in context.) (2 pts)

(c) r² = 0.953: about 95.3% of the variation in quiz scores is explained by the linear relationship with review hours. (1 pt: "variation in quiz scores"; 1 pt: "explained by the linear relationship with hours / the model.") (2 pts)

(d) Prediction at x = 6: ŷ = 53.29 + 3.88(6) = 53.29 + 23.28 = 76.57 ≈ 76.6. Residual = actual − predicted = 78 − 76.57 = 1.43 ≈ +1.4 (the model slightly under-predicted this student). (1 pt: correct prediction ≈ 76.6; 1 pt: correct residual ≈ +1.4 with actual − predicted.) (2 pts)

(e) No — this would be extrapolation. The data only cover review times from 1 to 8 hours; 15 hours is far outside that range, and there is no guarantee the linear pattern continues (scores cannot exceed 100, and the trend may level off). Predictions outside the observed x-range are unreliable. (1 pt: identifies 15 is outside the data range / extrapolation; 1 pt: explains why the prediction is unreliable.) (2 pts)

Where students lose points: giving direction/strength without numeric support; interpreting slope or r² without the word "predicted"/"variation" or without context; computing the residual as predicted − actual; saying "yes" to the 15-hour prediction or failing to name extrapolation.

FRQ 3 — Model Response & Rubric (PANIC)

P — Parameter & Hypotheses. Let p_A and p_B be the true click-through proportions for Version A and Version B ads, respectively (for users like these).

H₀: p_A = p_B (equivalently p_A − p_B = 0) versus Hₐ: p_A ≠ p_B.

A — Assumptions / conditions.

- Random: users were randomly assigned to see Version A or B. ✓

- Independence / 10%: the 200 users in each group are independent (random assignment); each group is plausibly less than 10% of all potential viewers. ✓

- Large Counts: using pooled p̂ = (92 + 68)/400 = 0.40, expected successes/failures are 200(0.40) = 80 and 200(0.60) = 120 in each group — all ≥ 10. ✓

N — Name the procedure. A two-proportion z-test for a difference in proportions.

I — Test computation.

`

p̂_A = 92/200 = 0.46 p̂_B = 68/200 = 0.34

pooled p̂ = (92 + 68)/(200 + 200) = 160/400 = 0.40

SE = √[ p̂(1−p̂)(1/n_A + 1/n_B) ] = √[ 0.40·0.60·(1/200 + 1/200) ]

= √(0.24 · 0.01) = √0.0024 = 0.04899

z = (0.46 − 0.34) / 0.04899 = 0.12 / 0.04899 = 2.449

p-value = 2 · P(Z > 2.449) ≈ 0.0143

`

TI-84: STAT → TESTS → 2-PropZTest, x1=92, n1=200, x2=68, n2=200, p1 ≠ p2 → z = 2.449, p = 0.0143.

C — Conclusion in context. Because p ≈ 0.0143 < α = 0.05, we reject H₀. There is convincing evidence that the true click-through rates of Version A and Version B ads differ (Version A's rate appears higher).

10-point rubric:

| Pts | Awarded for |

|---|---|

| 1 | Parameters p_A, p_B defined in context (true click-through rates) |

| 1 | Correct hypotheses H₀: p_A = p_B, Hₐ: p_A ≠ p_B |

| 1 | Random/assignment condition checked |

| 1 | Independence / 10% condition checked |

| 1 | Large Counts checked using pooled p̂ (all expected counts ≥ 10) |

| 1 | Procedure named (two-proportion z-test) |

| 1 | Correct pooled p̂ = 0.40 and SE = 0.049 |

| 1 | Correct test statistic z ≈ 2.45 |

| 1 | Correct p-value ≈ 0.014 |

| 1 | Decision linked to α AND conclusion in context (rates differ) |

Where students lose points: writing hypotheses about p̂ instead of p; using unpooled SE for the test; checking Large Counts with the separate proportions instead of pooled; "reject H₀" with no context; forgetting to double the tail for the two-sided p-value.

FRQ 4 — Model Response & Rubric

(a) z = (450 − 500)/40 = −1.25; P(X < 450) = P(Z < −1.25) = 0.1056 ≈ 0.106 (normalcdf(−1E99, 450, 500, 40)). (1 pt: correct z = −1.25; 1 pt: correct probability ≈ 0.106.) (2 pts)

(b) The sampling distribution of x̄ is approximately Normal (because the population of bulb lifetimes is Normal, x̄ is exactly Normal for any n), centered at μ_x̄ = 500 hours, with standard deviation σ_x̄ = σ/√n = 40/√16 = 10 hours. (1 pt: shape Normal with justification; 1 pt: mean = 500; 1 pt: SD = 10 computed correctly.) (3 pts)

(c) z = (480 − 500)/10 = −2.0; P(x̄ < 480) = P(Z < −2.0) = 0.0228 ≈ 0.023 (normalcdf(−1E99, 480, 500, 10)). (1 pt: uses σ_x̄ = 10 to get z = −2.0; 1 pt: correct probability ≈ 0.023.) (2 pts)

(d) PANIC (one-sample t-test).

- P: Let μ = the true mean lifetime (hours) of all the company's bulbs. H₀: μ = 500 vs. Hₐ: μ < 500.

- A: Random sample (stated); 16 bulbs < 10% of all bulbs produced; lifetimes are stated to be approximately Normal, so the t-procedure is valid for n = 16.

- N: One-sample t-test for a mean (df = 15).

- I: SE = s/√n = 38/√16 = 9.5; t = (486 − 500)/9.5 = −14/9.5 = −1.47; df = 15; p-value = P(t₁₅ < −1.47) ≈ 0.081 (T-Test: μ₀=500, x̄=486, Sx=38, n=16, μ<μ₀ → t = −1.47, p = 0.081).

- C: Because p ≈ 0.081 > α = 0.05, we fail to reject H₀. There is not convincing evidence that the true mean bulb lifetime is less than 500 hours.

(1 pt: parameter + correct one-sided hypotheses; 1 pt: correct t ≈ −1.47 with df = 15 and p ≈ 0.081; 1 pt: correct decision linked to α AND conclusion in context.) (3 pts)

Where students lose points: in (a)/(c), mixing up σ (=40) with σ_x̄ (=10); in (b), forgetting to justify Normality or miscomputing 40/√16; in (d), using a z-test instead of t (σ unknown), running a two-sided test, or writing "accept H₀."