These are original Marlbridge practice questions, not past exam questions. All organisations, places and data are fictional, and the simulation results were produced for these questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: p is the true population proportion; where a test statistic z is given, the conditions for a standard normal null distribution are met; you do not need to calculate z yourself (that is Topic 3.7). A graphing calculator is assumed.
Question 1 (multiple choice · foundation)
The fictional Hollyfield Rescue says that 60% of its dogs are adopted within a month of arriving. A volunteer believes the true proportion is higher. In a random sample of 125 dogs, 85 were adopted within a month (p̂ = 0.68). For H₀: p = 0.60 and Hₐ: p > 0.60, the p-value is 0.034. Which is a correct interpretation of the p-value?
- (A) There is a 0.034 probability that the true proportion of dogs adopted within a month is 0.60.
- (B) There is a 0.034 probability that the volunteer is wrong.
- (C) Assuming the true proportion of Hollyfield dogs adopted within a month is 0.60, there is a 0.034 probability of getting a sample proportion of 0.68 or higher in a random sample of 125 dogs.
- (D) There is a 0.966 probability that more than 60% of Hollyfield dogs are adopted within a month.
Answer and explanation
Answer: (C). It contains all three parts: the assumption that p = 0.60, the probability, and the event “0.68 or higher” (the direction of Hₐ) in a random sample of 125.
- (A) treats the p-value as the probability that H₀ is true. The p-value is calculated assuming H₀ is true, so it cannot be that.
- (B) treats the p-value as the probability of a wrong conclusion. It is not a probability about the volunteer’s belief.
- (D) uses 1 − p-value as the probability that Hₐ is true. A test does not produce that probability.
Question 2 (multiple choice · core)
A test has H₀: p = 0.25 and Hₐ: p < 0.25. The test statistic is z = −1.52. What is the p-value?
- (A) 0.0643
- (B) 0.1285
- (C) 0.4357
- (D) 0.9357
Answer and explanation
Answer: (A). Hₐ is “less than”, so the p-value is the area at or below the observed z: P(Z ≤ −1.52) = normalcdf(−99, −1.52, 0, 1) ≈ 0.0643.
- (B) doubles the tail area, 2 × 0.0643. That is the p-value for a two-sided test, but this alternative is one-sided.
- (C) is the area between 0 and 1.52, a common table-reading error.
- (D) is the area at or above −1.52, the wrong tail for a “less than” alternative.
Question 3 (multiple choice · core)
A test has H₀: p = 0.50 and Hₐ: p ≠ 0.50. In a random sample of 40, the sample proportion is p̂ = 0.65. A student simulates 500 random samples of 40 from a population with p = 0.50. Of the 500 simulated sample proportions, 17 are at or above 0.65 and 21 are at or below 0.35. What is the estimated p-value?
- (A) 0.034
- (B) 0.076
- (C) 0.042
- (D) 0.924
Answer and explanation
Answer: (B). The observed p̂ is 0.15 above 0.50. For a two-sided alternative, count simulated values at least 0.15 from 0.50 on either side: at or above 0.65 and at or below 0.35. That gives 17 + 21 = 38, and 38 ÷ 500 = 0.076.
- (A) uses only the upper tail, 17 ÷ 500. That would be the p-value for Hₐ: p > 0.50.
- (C) uses only the lower tail, 21 ÷ 500, the side away from the observed value.
- (D) is the proportion of simulated values that are less extreme, (500 − 38) ÷ 500.
Question 4 (calculation · core)
A test has H₀: p = 0.45, and the test statistic is z = 1.88.
(a) Find the p-value for each alternative: (i) Hₐ: p > 0.45, (ii) Hₐ: p < 0.45, (iii) Hₐ: p ≠ 0.45. (b) Explain why the p-value in (ii) is so large.
Worked solution and suggested Marlbridge rubric
(a)
- Hₐ: p > 0.45: P(Z ≥ 1.88) = normalcdf(1.88, 99, 0, 1) ≈ 0.0301.
- Hₐ: p < 0.45: P(Z ≤ 1.88) = normalcdf(−99, 1.88, 0, 1) ≈ 0.9699.
- Hₐ: p ≠ 0.45: P(Z ≤ −1.88) + P(Z ≥ 1.88) = 2 × 0.0301 ≈ 0.0601.
(b) z = 1.88 means the sample proportion was above 0.45. A “less than” alternative counts results below 0.45 as evidence. A result above 0.45 gives no support to “less than”, so almost the whole distribution is “at least as extreme” in that direction.
| Point | What earns it |
|---|---|
| 1 | Correct upper-tail p-value, 0.0301 |
| 1 | Correct lower-tail p-value 0.9699 and two-sided p-value 0.0601 |
| 1 | Explains that the sample result is on the opposite side of p₀ from Hₐ |
Answers from a standard normal table (0.0301, 0.9699, 0.0602) are acceptable.
Question 5 (constructed response · core)
The fictional Bramwell Orchard says that 10% of its apples have skin blemishes. A buyer suspects the true proportion is higher. In a random sample of 60 apples, 11 have blemishes (p̂ ≈ 0.183). The hypotheses are H₀: p = 0.10 and Hₐ: p > 0.10.
The buyer simulates 200 random samples of 60 apples from a population with p = 0.10. The table shows how many blemished apples were in each simulated sample.
| Blemished apples in sample | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Number of simulated samples | 0 | 0 | 12 | 21 | 30 | 30 | 32 | 29 | 23 | 13 | 5 | 3 | 0 | 1 | 1 |
(a) Explain why the buyer used a simulation rather than a normal model. (b) Estimate the p-value. (c) Interpret the p-value in context. (d) Do the data give convincing evidence that more than 10% of Bramwell apples have blemishes? Explain.
Worked solution and suggested Marlbridge rubric
(a) The normality condition is not met: np₀ = 60 × 0.10 = 6, which is less than 10. The null distribution is right-skewed, so a normal model would not be appropriate. A simulation does not need that condition.
(b) Hₐ is “greater than”, so count simulated samples with 11 or more blemished apples: 3 + 0 + 1 + 1 = 5. Estimated p-value = 5 ÷ 200 = 0.025.
(c) Assuming the true proportion of Bramwell apples with blemishes is 0.10, there is about a 0.025 probability of getting 11 or more blemished apples (a sample proportion of about 0.183 or higher) in a random sample of 60 apples, by chance alone.
(d) Yes. A result this high happened in only about 2.5% of simulated samples when p = 0.10, so it would be unusual if the orchard’s claim were true. The data give convincing evidence that the true proportion of blemished apples is greater than 0.10.
| Point | What earns it |
|---|---|
| 1 | Normality condition fails, with np₀ = 6 < 10 shown |
| 1 | Counts samples with 11 or more (5) and gives 5/200 = 0.025 |
| 1 | Interpretation includes the assumption p = 0.10, the probability, and “11 or more (or p̂ ≥ 0.183) in a random sample of 60” |
| 1 | Conclusion in context, linked to the small p-value |
Counting only samples with exactly 11 (giving 0.015) does not earn point 2. An interpretation without the assumption that p = 0.10 does not earn point 3.
Question 6 (constructed response · core)
A transport authority says that 75% of buses on the fictional Route 12 run on time. A passenger group suspects the true proportion is lower. In a random sample of 80 Route 12 buses, 58 ran on time (p̂ = 0.725). For H₀: p = 0.75 and Hₐ: p < 0.75, the test statistic is z = −0.52 and the p-value is about 0.30.
(a) Interpret the p-value in context. (b) A member of the group says: “The p-value is large, so this proves that 75% of Route 12 buses run on time.” Explain what is wrong with this statement. (c) Write a correct statement of what the data show.
Model answer and suggested Marlbridge rubric
(a) Assuming the true proportion of Route 12 buses that run on time is 0.75, there is about a 0.30 probability of getting a sample proportion of 0.725 or lower in a random sample of 80 buses, by chance alone.
(b) A large p-value means that the sample result would not be unusual if H₀ were true. It does not show that H₀ is true. Many values of p, including some below 0.75, would also make a sample proportion of 0.725 quite likely. The test can fail to find evidence against the claim, but it can never prove the claim.
(c) The data do not give convincing evidence that fewer than 75% of Route 12 buses run on time.
| Point | What earns it |
|---|---|
| 1 | Interpretation with the assumption p = 0.75, the probability 0.30, and “0.725 or lower” in a sample of 80 |
| 1 | Explains that a large p-value is not evidence that H₀ is true |
| 1 | Correct non-definitive statement in context, in terms of Hₐ |
Statements such as “we accept H₀” or “the data show 75% run on time” do not earn point 3.
Question 7 (explanation · stretch)
A researcher tests H₀: p = 0.40 against Hₐ: p > 0.40 using a random sample of 150. The sample proportion is p̂ = 0.36, and the test statistic is z = −1.00.
(a) Without any calculation, explain why the p-value must be greater than 0.5. (b) Find the p-value. (c) A classmate reports the p-value as 0.1587. Explain the classmate’s error. (d) Two other random samples of 150 from the same population, tested with the same hypotheses, give p-values of 0.004 and 0.09. Which sample gives more convincing evidence that p > 0.40? Explain.
Model answer and suggested Marlbridge rubric
(a) The sample proportion, 0.36, is below 0.40, but Hₐ says the proportion is above 0.40. “At least as extreme in the direction of Hₐ” means at or above z = −1.00, which includes everything above 0, more than half of the distribution.
(b) p-value = P(Z ≥ −1.00) = normalcdf(−1, 99, 0, 1) ≈ 0.8413.
(c) 0.1587 is P(Z ≤ −1.00), the lower tail. The classmate used the tail that matches the sign of z instead of the direction of Hₐ.
(d) The sample with p-value 0.004. A smaller p-value means the observed result would be rarer if p were 0.40. A result that would happen only 0.4% of the time under H₀ is more convincing evidence for Hₐ than one that would happen 9% of the time.
| Point | What earns it |
|---|---|
| 1 | (a) links p̂ < p₀ with an upper-tail alternative, so the area includes more than half the curve |
| 1 | (b) 0.8413 |
| 1 | (c) identifies the wrong (lower) tail |
| 1 | (d) chooses 0.004 and explains that a lower p-value is more convincing evidence for Hₐ |
How did you do?
- Q1 or Q6(a) wrong: re-read “Interpreting a p-value in context” in the study guide. Start with the assumption.
- Q2, Q4 or Q7(b)–(c) wrong: revisit “Which tail? The rule for each alternative” and Worked example 2(c).
- Q3 or Q5(b) wrong: work through Worked example 1 again. Count “at least as extreme”, and use both tails for ≠.
- Q6(b) or Q7(d) wrong: revisit “What small and not-small p-values tell you”.
Then tick off the topic checklist.