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 standard deviation of p̂1 − p̂2 is √[p1(1 − p1)/n1 + p2(1 − p2)/n2]; keep at least 4 decimal places in working and round probabilities to 4 decimal places. A graphing calculator is assumed.
Question 1 (multiple choice · foundation)
In the fictional town of Oakvale, 72% of households have a garden. In the nearby town of Dunmore, 55% do. Independent random samples of 100 Oakvale households and 120 Dunmore households are taken. What is the mean of the sampling distribution of p̂O − p̂D, the difference in sample proportions (Oakvale − Dunmore) with a garden?
- (A) −0.17
- (B) 0
- (C) 0.17
- (D) 0.635
Answer and explanation
Answer: (C). The mean of p̂O − p̂D is pO − pD = 0.72 − 0.55 = 0.17. The sample sizes do not affect the mean.
- (A) subtracts in the wrong order (Dunmore − Oakvale). The question fixes the order as Oakvale − Dunmore.
- (B) confuses “unbiased” with “zero”. The difference is unbiased for pO − pD, which is 0.17, not 0.
- (D) is the average of the two proportions, (0.72 + 0.55) ÷ 2, not their difference.
Question 2 (multiple choice · core)
For two independent random samples, p1 = 0.30 with n1 = 200, and p2 = 0.20 with n2 = 250. What is the standard deviation of the sampling distribution of p̂1 − p̂2?
- (A) 0.0202
- (B) 0.0287
- (C) 0.0411
- (D) 0.0577
Answer and explanation
Answer: (C). The variances are 0.30 × 0.70 ÷ 200 = 0.00105 and 0.20 × 0.80 ÷ 250 = 0.00064. Add them: 0.00169. The square root is 0.0411.
- (A) subtracts the variances: √(0.00105 − 0.00064) = 0.0202. Variances add even when the proportions are subtracted.
- (B) puts both products over one combined sample size: √(0.37 ÷ 450) = 0.0287. Each variance needs its own sample size.
- (D) adds the two standard deviations, 0.0324 + 0.0253 = 0.0577. Add the variances first, then take one square root.
Question 3 (multiple choice · core)
A researcher has 120 volunteers. She randomly assigns 60 to a new study method and 60 to the usual method. Suppose that 45% of learners like these pass a test with the new method and 12% with the usual method. Which statement about the sampling distribution of p̂new − p̂usual is correct?
- (A) The 10% condition fails, because 60 is more than 10% of the 120 volunteers.
- (B) The randomization condition fails, because the volunteers were not a random sample.
- (C) The large-counts condition fails, because 60 × 0.12 = 7.2 is less than 10.
- (D) All the conditions are met, so a normal model is appropriate.
Answer and explanation
Answer: (C). The four expected counts are 60 × 0.45 = 27, 60 × 0.55 = 33, 60 × 0.12 = 7.2 and 60 × 0.88 = 52.8. One is below 10, so the sampling distribution may not be approximately normal.
- (A) applies the 10% condition to an experiment. It is not needed when treatments are randomly assigned.
- (B) In an experiment, random assignment of treatments meets the randomization condition. A random sample is not required.
- (D) ignores the expected count of 7.2.
Question 4 (calculation · core)
At the fictional Larchmere School (1,200 students), 40% of students walk to school. At Fenby School (1,500 students), 25% walk. A survey takes independent random samples of 80 students from Larchmere and 90 from Fenby. Find the mean and standard deviation of the sampling distribution of p̂L − p̂F, showing each step, and interpret the standard deviation in context.
Worked solution
- Mean: 0.40 − 0.25 = 0.15.
- Variance for Larchmere: 0.40 × 0.60 ÷ 80 = 0.003.
- Variance for Fenby: 0.25 × 0.75 ÷ 90 = 0.002083.
- Standard deviation: √(0.003 + 0.002083) = √0.005083 ≈ 0.0713.
- 10% check: 80 ≤ 120 and 90 ≤ 150, so the formula is accurate.
Interpretation. In repeated pairs of random samples (80 Larchmere students, 90 Fenby students), the difference in sample proportions who walk (Larchmere − Fenby) typically varies from the true difference of 0.15 by about 0.0713.
Suggested mark points (3): 1 for the mean with the order stated; 1 for adding the two variances and taking one square root; 1 for an interpretation that names both schools, the variable and repeated sampling.
Question 5 (constructed response · core)
In the fictional coastal town of Saltmarsh (30,000 residents), 38% of residents swim in the sea at least once a week in summer. In Kelby (18,000 residents), 30% do. Independent random samples of 250 Saltmarsh residents and 200 Kelby residents are taken.
(a) Find the mean and standard deviation of the sampling distribution of p̂S − p̂K. (b) Check the conditions for a normal model, in context. (c) Find the probability that the Kelby sample proportion is at least as large as the Saltmarsh sample proportion. Interpret it.
Worked solution and suggested Marlbridge rubric
(a) Mean = 0.38 − 0.30 = 0.08. Standard deviation = √(0.38 × 0.62 ÷ 250 + 0.30 × 0.70 ÷ 200) ≈ 0.0446.
(b) Randomization: independent random samples from each town. 10%: 250 ≤ 3,000 and 200 ≤ 1,800. Large counts: 95, 155, 60 and 140 are all at least 10. So the sampling distribution is approximately normal.
(c) “Kelby at least as large” means p̂S − p̂K ≤ 0. z = (0 − 0.08) ÷ 0.0446 ≈ −1.7923, so P ≈ 0.0365. If the stated proportions are true, about 3.7% of pairs of random samples of these sizes would show Kelby’s sample proportion at least as high as Saltmarsh’s.
| Point | What earns it |
|---|---|
| 1 | Mean 0.08 and standard deviation 0.0446, with the variances added |
| 1 | All three conditions checked with numbers and linked to the context |
| 1 | Translates the event into p̂S − p̂K ≤ 0 and finds the probability 0.0365 with a normal model |
| 1 | Interpretation names both towns and repeated sampling |
Accept 0.0364 to 0.0367 from rounded working (for example, z rounded to −1.79). Accept working in the order Kelby − Saltmarsh if the event becomes p̂K − p̂S ≥ 0 with mean −0.08.
Question 6 (constructed response · core)
A fictional plant nursery tests a new fertiliser. It randomly assigns 100 tomato seedlings to the new fertiliser and 100 to the old one. Suppose that 65% of seedlings like these produce fruit within 60 days with the new fertiliser and 50% with the old one.
(a) Check the conditions for a normal model for p̂new − p̂old. (b) Find the mean and standard deviation of the sampling distribution. (c) Find the probability that the difference in sample proportions (new − old) is at least 0.25. Interpret it in context.
Worked solution and suggested Marlbridge rubric
(a) Randomization: the fertilisers were randomly assigned to the seedlings. The 10% condition is not needed in an experiment. Large counts: 65, 35, 50 and 50 are all at least 10.
(b) Mean = 0.65 − 0.50 = 0.15. Standard deviation = √(0.65 × 0.35 ÷ 100 + 0.50 × 0.50 ÷ 100) = √0.004775 ≈ 0.0691.
(c) z = (0.25 − 0.15) ÷ 0.0691 ≈ 1.4471, so P ≈ 0.0739. In about 7.4% of repetitions of this experiment, the new fertiliser group’s fruiting proportion would be at least 25 percentage points above the old fertiliser group’s.
| Point | What earns it |
|---|---|
| 1 | Random assignment named as the randomization condition, and no 10% check required (or stated as not needed) |
| 1 | Four expected counts shown, all at least 10 |
| 1 | Mean 0.15 and standard deviation 0.0691 |
| 1 | Probability 0.0739 with an interpretation about repeating the experiment |
Do not award point 1 for “the seedlings were a random sample”: they were randomly assigned.
Question 7 (explanation · stretch)
Two fictional colleges have p1 = 0.6 and p2 = 0.4 for the proportion of students who own a tablet. A student plans to take independent random samples of 150 from each college and writes:
“Standard deviation of p̂1 − p̂2 = √(0.6 × 0.4 ÷ 150) − √(0.4 × 0.6 ÷ 150) = 0.”
(a) Explain why the student’s answer cannot be right, and find the correct standard deviation. (b) The researchers want the standard deviation of p̂1 − p̂2 to be at most 0.05, using the same sample size n in each college. Find the smallest n.
Model answer and suggested Marlbridge rubric
(a) A standard deviation of 0 would mean every pair of samples gives exactly the same difference, which is impossible: each sample proportion varies. The student subtracted the standard deviations. The variances must be added: √(0.24 ÷ 150 + 0.24 ÷ 150) = √0.0032 ≈ 0.0566.
(b) With equal sizes, the standard deviation is √(0.24/n + 0.24/n) = √(0.48/n). Solve √(0.48/n) ≤ 0.05: 0.48/n ≤ 0.0025, so n ≥ 192. The smallest sample size is 192 from each college. Check: √(0.48 ÷ 192) = 0.05 exactly, while n = 191 gives about 0.0501.
| Point | What earns it |
|---|---|
| 1 | Explains that a zero standard deviation is impossible and that the variances must add, not the standard deviations subtract |
| 1 | Correct standard deviation 0.0566 |
| 1 | Sets up √(0.48/n) ≤ 0.05 and finds n = 192 in each sample |
How did you do?
- Q1 wrong: re-read “Mean and standard deviation of p̂1 − p̂2” in the study guide, and always state the order of subtraction.
- Q2, Q4 or Q7(a) wrong: add the variances, then take one square root. Work through Worked example 1 again.
- Q3 or Q6(a) wrong: revisit “The conditions and what each one is for”, especially the experiment case, and Worked example 3.
- Q5 or Q6(c) wrong: revisit Worked example 2. Turn the words into an inequality about p̂1 − p̂2 first.
- Q7(b) wrong: both variances shrink as n grows; set up the inequality and round up.
Then tick off the topic checklist.