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: the two samples are independent unless stated; population means and standard deviations are given; keep 4 decimal places in working and round probabilities to 4 decimal places. A graphing calculator is assumed.
Question 1 (multiple choice · foundation)
Times to run 1 km have standard deviation 12 seconds for students at the fictional School M and 9 seconds at School N. Independent random samples of 36 School M students and 25 School N students are timed. What is the standard deviation of the sampling distribution of x̄M − x̄N?
- (A) 0.87 seconds
- (B) 1.92 seconds
- (C) 2.69 seconds
- (D) 3.80 seconds
Answer and explanation
Answer: (C). √(12²/36 + 9²/25) = √(4 + 3.24) = √7.24 ≈ 2.69 seconds.
- (A) subtracts the variances: √(4 − 3.24) ≈ 0.87. Variances add even when the means are subtracted.
- (B) treats the two samples as one: √[(144 + 81) / 61] ≈ 1.92. Each variance needs its own sample size.
- (D) adds the standard deviations: 12/6 + 9/5 = 3.8. Add the variances, then take one square root.
Question 2 (multiple choice · core)
In each situation below, both population distributions are strongly skewed. In which situation is the sampling distribution of x̄₁ − x̄₂ approximately normal?
- (A) Independent random samples of 40 and 20
- (B) Independent random samples of 35 and 50
- (C) Independent random samples of 25 and 25, a total of 50
- (D) 40 people, each measured once before and once after a programme
Answer and explanation
Answer: (B). When the populations are not normal, the sampling distribution of x̄₁ − x̄₂ is approximately normal if n₁ ≥ 30 and n₂ ≥ 30. Only (B) meets this.
- (A) has n₂ = 20 < 30, so the second sample mean may still be skewed.
- (C) adds the sample sizes. The rule applies to each sample separately.
- (D) is paired data from the same people, not two independent samples, so this sampling distribution does not apply.
Question 3 (multiple choice · core)
A biologist plans independent random samples of fish from two fictional lakes. Fish lengths are normally distributed in both lakes. With her planned sample sizes, the standard deviation of x̄₁ − x̄₂ is 2.4 cm. If she multiplies both sample sizes by 4, what will the standard deviation be?
- (A) 0.6 cm
- (B) 1.2 cm
- (C) 2.4 cm
- (D) 4.8 cm
Answer and explanation
Answer: (B). Each variance term σ²/n is divided by 4, so their sum is divided by 4 and the standard deviation is divided by √4 = 2: 2.4 ÷ 2 = 1.2 cm.
- (A) divides the standard deviation by 4 instead of by √4.
- (C) assumes the spread depends only on σ₁ and σ₂. Larger samples make sample means less variable.
- (D) moves in the wrong direction: larger samples give less variability, not more.
Question 4 (multiple choice · core)
Independent random samples of 50 commuters from the fictional City F and 60 from City G record their commute times. The sampling distribution of x̄F − x̄G has mean 5 minutes and standard deviation 1.8 minutes. Which statement correctly interprets the 1.8 minutes?
- (A) Commute times of individual people in City F typically differ from those in City G by about 1.8 minutes.
- (B) In repeated pairs of random samples of 50 City F and 60 City G commuters, the difference in sample mean commute times (F − G) typically varies from 5 minutes by about 1.8 minutes.
- (C) The difference between the two sample mean commute times is 1.8 minutes.
- (D) In repeated random samples of 50 City F commuters, the sample mean commute time typically varies from the true mean by about 1.8 minutes.
Answer and explanation
Answer: (B). A standard deviation of a sampling distribution describes how the statistic (here x̄F − x̄G) varies from its mean (5 minutes) over repeated pairs of samples of the stated sizes.
- (A) describes individuals, not sample means. Individual commutes vary far more.
- (C) confuses the spread of the sampling distribution with one observed difference.
- (D) describes only one sample mean. The 1.8 minutes includes variation from both samples.
Question 5 (constructed response · core)
Egg masses at the fictional Farm R are normally distributed with mean 62 g and standard deviation 4 g. At Farm S they are normally distributed with mean 59 g and standard deviation 5 g. Each farm produces thousands of eggs a day. An inspector takes a random sample of 8 eggs from Farm R and, independently, a random sample of 10 eggs from Farm S.
(a) Find the mean and standard deviation of the sampling distribution of x̄R − x̄S. (b) The samples are small. Explain why a normal model is still appropriate, and check the other conditions. (c) Find the probability that the Farm S sample mean is greater than the Farm R sample mean. Interpret it in context.
Worked solution and suggested Marlbridge rubric
(a) Mean = 62 − 59 = 3 g. Standard deviation = √(4²/8 + 5²/10) = √(2 + 2.5) = √4.5 ≈ 2.1213 g.
(b) Both populations of egg masses are normal, so the sampling distribution of x̄R − x̄S is normal for any sample sizes. Randomization: independent random samples from the two farms ✓. 10%: 8 and 10 are far less than 10% of the thousands of eggs each farm produces ✓.
(c) “S greater than R” means x̄R − x̄S < 0. z = (0 − 3) ÷ 2.1213 ≈ −1.4142, so P ≈ 0.0786. If the stated distributions are correct, about 7.9% of pairs of random samples (8 Farm R eggs, 10 Farm S eggs) would give a larger sample mean mass for Farm S than for Farm R.
| Point | What earns it |
|---|---|
| 1 | Mean 3 g with the order R − S, and standard deviation ≈ 2.12 g from added variances |
| 1 | Normal shape justified by both populations being normal (not by sample size) |
| 1 | Randomization and 10% conditions checked in context |
| 1 | Correct inequality (x̄R − x̄S < 0), probability ≈ 0.079 and an interpretation naming both farms and the sample sizes |
Claiming the normal model fails because n < 30 loses point 2.
Question 6 (constructed response · stretch)
Call durations at two fictional call centres are skewed to the right. Last month Centre X handled 12,000 calls, with mean 8.5 minutes and standard deviation 5.2 minutes. Centre Y handled 15,000 calls, with mean 7.0 minutes and standard deviation 4.0 minutes. An analyst takes independent random samples of 45 calls from X and 60 calls from Y.
(a) Find the mean and standard deviation of the sampling distribution of x̄X − x̄Y. (b) Justify the use of a normal model. Explain why the sample sizes matter here. (c) The analyst’s samples give x̄X = 10.4 minutes and x̄Y = 7.1 minutes. Find the probability of a difference at least this large if the stated values are correct. Does this cast doubt on them?
Worked solution and suggested Marlbridge rubric
(a) Mean = 8.5 − 7.0 = 1.5 minutes. Standard deviation = √(5.2²/45 + 4.0²/60) = √(0.6009 + 0.2667) = √0.8676 ≈ 0.9314 minutes.
(b) Randomization: independent random samples ✓. 10%: 45 ≤ 1,200 and 60 ≤ 1,500 ✓. The populations are skewed, so the shape of each sample mean’s distribution depends on sample size. Because nX = 45 ≥ 30 and nY = 60 ≥ 30, the sampling distribution of x̄X − x̄Y is approximately normal. With small samples from these skewed populations, a normal model would not be justified.
(c) Observed difference = 10.4 − 7.1 = 3.3 minutes. z = (3.3 − 1.5) ÷ 0.9314 ≈ 1.9325, so P(x̄X − x̄Y ≥ 3.3) ≈ 0.0266. A difference this large would happen in only about 2.7% of pairs of samples if the stated means were correct. That is fairly unusual, so it gives some reason to doubt the stated values, but it does not prove them wrong.
| Point | What earns it |
|---|---|
| 1 | Mean 1.5 minutes and standard deviation ≈ 0.93 minutes |
| 1 | All three conditions in context, with both n ≥ 30 linked to the skewed populations |
| 1 | Correct observed difference, z and probability ≈ 0.027 |
| 1 | Judgement (“reason to doubt”, not “proves”) linked to the small probability |
Question 7 (explanation · stretch)
Use the call centres in Question 6.
(a) A student writes: “σ(x̄X − x̄Y) = 5.2/√45 + 4.0/√60 ≈ 1.29 minutes.” Explain the error and why the correct method adds variances. (b) A manager wants to compare the mean durations of two kinds of call for the same 45 agents, using each agent’s average for each kind. Can the formula from this topic be used? Explain. (c) The analyst increases the Centre X sample to 180 calls but keeps 60 from Centre Y. Does the standard deviation halve? Show your working.
Model answer and suggested Marlbridge rubric
(a) The student added the two standard deviations. Each sample mean varies independently, and the variation from both ends up in the difference. For independent variables, it is the variances that add, so σ = √(5.2²/45 + 4.0²/60) ≈ 0.93 minutes. Adding standard deviations overstates the spread.
(b) No. The two values for each agent come from the same person, so the samples are paired, not independent. The manager should find each agent’s difference and use one-sample methods on the differences.
(c) No. Only the first term changes: √(5.2²/180 + 4.0²/60) = √(0.1502 + 0.2667) ≈ 0.6457 minutes, which is more than half of 0.9314 (0.4657). The standard deviation halves only if both sample sizes are multiplied by 4.
| Point | What earns it |
|---|---|
| 1 | Identifies adding the standard deviations as the error and explains that the variances of independent sample means add |
| 1 | Recognises the paired design and says the formula needs independent samples |
| 1 | Correct new standard deviation (≈ 0.646) and explanation that only one variance term shrank |
How did you do?
- Q1 or Q7(a) wrong: revisit “Mean and standard deviation of x̄₁ − x̄₂” in the study guide.
- Q2, Q5(b) or Q6(b) wrong: revisit “The conditions and what each one is for”.
- Q3 or Q7(c) wrong: revisit Worked example 3(c). Each variance term has its own sample size.
- Q4 or Q5(c) wrong: revisit the interpretations in Worked examples 1 and 2.
- Q6(c) wrong: revisit Worked example 2(c).
Then tick off the topic checklist.