These are original Marlbridge practice questions, not past exam questions. All contexts and data are fictional. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: σ is unknown; p-values come from a t-distribution with df = n − 1; quartiles split the ordered data at the median, leaving the median out of both halves when n is odd; round t to 2 decimal places and p-values to 4. A graphing calculator is assumed.
Question 1 (multiple choice · foundation)
A fictional furniture company says its bookshelf takes 50 minutes to assemble, on average. A consumer group times a random sample of 25 buyers and finds x̄ = 52.6 minutes and s = 4.5 minutes. What is the test statistic for H₀: μ = 50?
- (A) t ≈ 0.58
- (B) t ≈ 2.83
- (C) t ≈ 2.89
- (D) t ≈ 14.44
Answer and explanation
Answer: (C). The standard error is s / √n = 4.5 / √25 = 0.9. So t = (52.6 − 50) / 0.9 = 2.6 / 0.9 ≈ 2.89, with df = 24.
- (A) divides by s = 4.5 instead of the standard error s / √n.
- (B) uses √(n − 1) = √24 in the standard error. The degrees of freedom are n − 1, but the standard error uses √n.
- (D) divides s by n = 25 instead of √25, which makes the standard error far too small.
Question 2 (multiple choice · core)
A one-sample t-test with n = 20 and Hₐ: μ ≠ μ₀ gives t = −1.85. What is the p-value?
- (A) 0.0400
- (B) 0.0643
- (C) 0.0799
- (D) 0.9600
Answer and explanation
Answer: (C). df = 19. The test is two-sided, so the p-value = 2 × P(T ≤ −1.85) ≈ 2 × 0.0400 = 0.0799 (technology, unrounded).
- (A) is only one tail. A two-sided alternative counts results this extreme in both directions.
- (B) uses the standard normal curve: 2 × P(Z ≤ −1.85) ≈ 0.0643. With σ unknown, the t-distribution has heavier tails, so the correct p-value is larger.
- (D) is 1 − 0.0400, the area on the wrong side of t.
Question 3 (multiple choice · core)
A fictional hospital canteen states that its lunchtime queue takes 6 minutes on average. A manager suspects it takes longer. In a random sample of 28 lunchtimes, the mean queue time is 6.9 minutes. A t-test of H₀: μ = 6 against Hₐ: μ > 6 gives a p-value of 0.024. Which is a correct interpretation of the p-value?
- (A) There is a 0.024 probability that the true mean queue time is 6 minutes.
- (B) Assuming the true mean queue time is 6 minutes, there is a 0.024 probability of getting a sample mean of 6.9 minutes or more in a random sample of 28 lunchtimes.
- (C) Assuming the true mean queue time is 6.9 minutes, there is a 0.024 probability of getting a sample mean of 6 minutes or less in a random sample of 28 lunchtimes.
- (D) 2.4% of lunchtime queues take 6.9 minutes or more.
Answer and explanation
Answer: (B). A p-value is calculated by assuming H₀ is true (μ = 6). It is the probability of a sample result at least as extreme as the one observed, in the direction of Hₐ (6.9 minutes or more).
- (A) treats the p-value as the probability that H₀ is true. The test assumes H₀; it cannot give the probability that H₀ is true.
- (C) assumes the sample value, not the null value, and looks in the wrong direction.
- (D) describes individual queues. The p-value is about the mean of a random sample of 28 lunchtimes.
Question 4 (multiple choice · core)
A random sample of 22 of the 400 employees at a fictional firm rated their stress (0 to 100) before and after a mindfulness course. With d = before − after, a t-test of H₀: μd = 0 against Hₐ: μd > 0 gives t = 1.75 and a p-value of 0.047. The significance level chosen in advance was α = 0.01. Which conclusion is correct?
- (A) Because 0.047 < 0.05, reject H₀. There is convincing statistical evidence that the true mean stress rating falls after the course.
- (B) Because 0.047 > 0.01, fail to reject H₀. There is not convincing statistical evidence that the true mean difference (before − after) in stress rating for all employees at the firm is greater than 0.
- (C) Because 0.047 > 0.01, fail to reject H₀. There is convincing statistical evidence that the true mean difference (before − after) in stress rating is 0.
- (D) Because 0.047 > 0.01, fail to reject H₀. The sample mean difference in stress rating is not greater than 0.
Answer and explanation
Answer: (B). Compare the p-value with the α chosen in advance: 0.047 > 0.01, so fail to reject H₀. The conclusion is about the parameter and the population, in terms of Hₐ, with “not convincing evidence”.
- (A) uses α = 0.05 instead of the stated 0.01. Changing α after seeing the p-value is not allowed.
- (C) treats failing to reject as evidence for H₀. A test cannot show that μd = 0.
- (D) is about the sample, not the population, and it is false: t > 0 means the sample mean difference is greater than 0.
Question 5 (constructed response · core)
Corvo Foods (fictional) labels its oat bars “210 kcal”. A food inspector selects a random sample of 40 bars from a production run of 20,000 bars and measures the energy content of each. The sample gives x̄ = 213.4 kcal and s = 9.6 kcal.
(a) Do the data give convincing statistical evidence, at α = 0.05, that the true mean energy content of the bars in this run is different from 210 kcal? Carry out a complete test. (b) Interpret the p-value in context. (c) Would your decision change at α = 0.01? Explain.
Worked solution and suggested Marlbridge rubric
(a)
- Hypotheses. μ = the true mean energy content (kcal) of all bars in this production run. H₀: μ = 210, Hₐ: μ ≠ 210, α = 0.05.
- Method and conditions. One-sample t-test for a population mean. Random sample ✓; 40 ≤ 10% of 20,000 = 2,000 ✓; n = 40 ≥ 30 ✓.
- Calculations. Standard error = 9.6 / √40 ≈ 1.5179 kcal. t = (213.4 − 210) / 1.5179 ≈ 2.24, df = 39. p-value = 2 × P(T ≥ 2.24) ≈ 0.0309.
- Conclusion. Because 0.0309 < 0.05, reject H₀. There is convincing statistical evidence that the true mean energy content of all bars in this production run is different from 210 kcal.
(b) Assuming the true mean energy content of the bars in this run is 210 kcal, there is about a 0.0309 probability of getting a sample mean at least 3.4 kcal away from 210 kcal (213.4 or more, or 206.6 or less) in a random sample of 40 bars.
(c) Yes. 0.0309 > 0.01, so at α = 0.01 you would fail to reject H₀ and there would not be convincing evidence that the mean differs from 210 kcal.
| Point | What earns it |
|---|---|
| 1 | Parameter defined in context and correct two-sided hypotheses; procedure named with all three conditions checked |
| 1 | Correct t (≈ 2.24) with df = 39 and a two-sided p-value (≈ 0.031) |
| 1 | Decision with linkage (p-value compared with α) and a conclusion about the true mean, in context, in terms of Hₐ |
| 1 | p-value interpretation that assumes μ = 210 and describes “at least as far from 210 in either direction”; correct answer to (c) |
A one-sided p-value (0.0154) loses point 2; “proves” loses point 3.
Question 6 (constructed response · stretch)
A fictional car-rental company has 400 cars of one model. It selects a random sample of 9 of them. Each car is driven on the same test route once with standard tyres and once with low-resistance tyres, in a random order. The fuel economy (km per litre) is:
| Car | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Standard | 14.2 | 16.8 | 12.5 | 15.1 | 13.9 | 17.4 | 15.6 | 12.9 | 14.7 |
| Low-resistance | 14.9 | 17.1 | 13.4 | 14.9 | 14.8 | 17.1 | 16.5 | 13.0 | 15.0 |
(a) Explain why a matched-pairs analysis is appropriate. (b) Do the data give convincing evidence, at α = 0.05, that low-resistance tyres increase mean fuel economy for cars of this model? Carry out a complete test. (c) A classmate analyses the two rows as two independent samples and gets a p-value of about 0.30. Explain why the matched-pairs p-value is much smaller.
Worked solution and suggested Marlbridge rubric
(a) Each car was measured with both kinds of tyre, so the two values for one car belong together as a pair.
(b)
- Hypotheses. Let d = low-resistance − standard. μd = the true mean difference (low-resistance − standard) in fuel economy, in km per litre, for all 400 cars of this model. H₀: μd = 0, Hₐ: μd > 0, α = 0.05.
- Method and conditions. One-sample t-test for a population mean difference.
- Random: random sample of cars, with the tyre order randomized. ✓
- 10%: 9 ≤ 10% of 400 = 40. ✓
- Sample data: the differences are 0.7, 0.3, 0.9, −0.2, 0.9, −0.3, 0.9, 0.1, 0.3. Ordered: −0.3, −0.2, 0.1, 0.3, 0.3, 0.7, 0.9, 0.9, 0.9. Q1 = −0.05, Q3 = 0.9, IQR = 0.95, fences −1.475 and 2.325. No outliers and no strong skewness. ✓
- Calculations. x̄d = 3.6 / 9 = 0.40 km/L, sd ≈ 0.4743 km/L. Standard error = 0.4743 / √9 ≈ 0.1581. t = 0.40 / 0.1581 ≈ 2.53, df = 8. p-value = P(T ≥ 2.53) ≈ 0.0176.
- Conclusion. Because 0.0176 < 0.05, reject H₀. There is convincing statistical evidence that the true mean fuel economy of cars of this model is greater with low-resistance tyres than with standard tyres (μd > 0).
(c) Fuel economy varies a lot from car to car (12.5 to 17.4 km/L). The two-sample analysis counts this as random error, so its standard error is large. Pairing compares each car with itself, so the differences vary much less (sd ≈ 0.47 km/L) and t is much larger.
| Point | What earns it |
|---|---|
| 1 | (a) Same car measured under both conditions, so the data are paired |
| 1 | Correct μd with the order of subtraction, one-sided hypotheses, and all conditions checked using the differences |
| 1 | Correct x̄d, sd, t ≈ 2.53 with df = 8, and p-value ≈ 0.018 |
| 1 | Decision with linkage and conclusion about the true mean difference for this model, in context; (c) explains that pairing removes car-to-car variation |
Checking the two rows for skewness instead of the differences does not earn the second point.
Question 7 (explanation · stretch)
A fictional e-reader is advertised to last 30 hours on one charge. A product tester checks a random sample of 12 e-readers from a large shipment. The battery lives show no strong skewness and no outliers; x̄ = 28.7 hours and s = 2.9 hours. A student tests H₀: μ = 30 against Hₐ: μ < 30 at α = 0.05 and writes:
z = (28.7 − 30) / (2.9 / √12) = −1.55. p-value = P(Z ≤ −1.55) = 0.0606. Since 0.0606 > 0.05, we accept H₀. The e-readers last 30 hours on average.
(a) Identify the error in the calculation and find the correct p-value. (b) Identify two problems with the conclusion, then write a correct conclusion. (c) Does correcting the calculation change the decision? Explain why the correct p-value is larger.
Model answer and suggested Marlbridge rubric
(a) σ is unknown and the standard error uses s, so the test statistic follows a t-distribution with df = 12 − 1 = 11, not the standard normal. t = −1.3 / 0.8372 ≈ −1.55, and the p-value = P(T ≤ −1.55) ≈ 0.0744.
(b) First, “accept H₀” is wrong: failing to reject is not evidence that H₀ is true. Second, “the e-readers last 30 hours on average” is definitive and not stated in terms of Hₐ. A correct conclusion: because the p-value of 0.0744 is greater than α = 0.05, we fail to reject H₀. There is not convincing statistical evidence that the true mean battery life of e-readers in this shipment is less than 30 hours.
(c) No: 0.0744 > 0.05, so the decision is still to fail to reject H₀. The t-distribution has heavier tails than the standard normal, because using s instead of σ adds extra variability, so the same test statistic gives a larger tail area.
| Point | What earns it |
|---|---|
| 1 | Names the t-distribution with df = 11 as the correct reference and gives p-value ≈ 0.074 |
| 1 | Identifies “accept H₀” and the definitive, non-parameter wording as errors, and writes a correct conclusion with linkage, about the true mean, in terms of Hₐ |
| 1 | Decision unchanged, with the heavier-tails reason |
How did you do?
- Q1 wrong: revisit “The test statistic” in the study guide. The standard error is s / √n.
- Q2 or Q7(a) wrong: revisit “Finding the p-value” and Worked example 3. Use the t-distribution and the right tail(s).
- Q3 or Q5(b) wrong: revisit “Interpreting the p-value”. Start with “Assuming the true mean is…”.
- Q4 or Q7(b) wrong: revisit “Decision and conclusion”. Compare with the stated α; never accept H₀.
- Q6 wrong: work through Worked example 2 again, especially the differences and the conditions.
Then tick off the topic checklist.