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: z* comes from the standard normal distribution (1.645 for 90%, 1.960 for 95%, 2.576 for 99%); the standard error is √[p̂1(1 − p̂1)/n1 + p̂2(1 − p̂2)/n2] with no pooling; keep p̂1 and p̂2 unrounded until the end and give endpoints to 3 decimal places. A graphing calculator is assumed.
Question 1 (multiple choice · foundation)
A regional council selects a random sample of 300 adults from the North district and an independent random sample of 250 adults from the South district. It asks each adult whether they support a new cycle lane. The council wants to estimate how much the proportion of supporters differs between the two districts. Which procedure should it use?
- (A) One-sample z-interval for a population proportion
- (B) Two-sample z-interval for a difference between population proportions
- (C) Two-sample z-test for a difference between population proportions
- (D) One-sample z-interval for each district, then subtract the endpoints
Answer and explanation
Answer: (B). There are two independent random samples, a categorical response (support or not), and the goal is to estimate a difference between two population proportions. That calls for a two-sample z-interval for pN − pS.
- (A) estimates one proportion. It cannot estimate a difference between two populations.
- (C) is a significance test. It decides whether there is evidence of a difference; it does not give an interval estimate of its size.
- (D) Subtracting the endpoints of two separate intervals does not give a correctly sized interval for the difference. The two-sample interval uses one combined standard error.
Question 2 (multiple choice · core)
In independent random samples, 54 of 120 adults in one town and 39 of 130 adults in another town own an electric bike. What is the standard error of p̂1 − p̂2 for a confidence interval?
- (A) 0.0428
- (B) 0.0606
- (C) 0.0612
- (D) 0.0856
Answer and explanation
Answer: (B). p̂1 = 54 ÷ 120 = 0.45 and p̂2 = 39 ÷ 130 = 0.30. SE = √(0.45 × 0.55 ÷ 120 + 0.30 × 0.70 ÷ 130) = √(0.0020625 + 0.0016154) ≈ 0.0606.
- (A) puts both products over the combined sample size 250. Each variance term needs its own sample size.
- (C) uses the pooled proportion 93 ÷ 250 = 0.372. Pooling is for a significance test that assumes p1 = p2, not for an interval.
- (D) adds the two separate standard errors (0.0454 + 0.0402). Add the variance terms first, then take one square root.
Question 3 (multiple choice · core)
A coach randomly assigns 100 swimmers to two training plans, 50 to each. After six weeks, 42 swimmers on Plan A and 30 on Plan B have improved their time. Which statement about a two-sample z-interval for pA − pB is correct?
- (A) All the conditions are met, so the interval can be constructed.
- (B) The 10% condition fails, because 50 is more than 10% of the 100 swimmers.
- (C) The normality condition is met, because each group has at least 30 swimmers.
- (D) The normality condition fails, because Plan A has only 8 swimmers who did not improve.
Answer and explanation
Answer: (D). The four observed counts are 42 and 8 (Plan A) and 30 and 20 (Plan B). One count, 8, is below 10, so the sampling distribution of p̂A − p̂B may not be approximately normal.
- (A) ignores the count of 8.
- (B) applies the 10% condition to an experiment. Random assignment meets the randomization condition, and the 10% condition is not needed.
- (C) uses a “30 or more” rule. For proportions, the check is at least 10 successes and 10 failures in each group.
Question 4 (calculation · core)
A report gives the 99% confidence interval (−0.052, 0.188) for p1 − p2. Find, showing your working, (a) the point estimate p̂1 − p̂2, (b) the margin of error, (c) the standard error, and (d) the interval for p2 − p1.
Worked solution
- (a) Point estimate = (−0.052 + 0.188) ÷ 2 = 0.136 ÷ 2 = 0.068.
- (b) Margin of error = (0.188 − (−0.052)) ÷ 2 = 0.240 ÷ 2 = 0.120.
- (c) For 99%, z* = 2.576, so SE = 0.120 ÷ 2.576 ≈ 0.0466.
- (d) Reverse and negate the endpoints: (−0.188, 0.052).
Suggested mark points (3): 1 for the point estimate as the midpoint; 1 for the margin of error as half the width and the standard error using z* = 2.576 (not 1.960); 1 for the reversed interval with both signs changed and the endpoints swapped.
Question 5 (constructed response · core)
Ridgeway University (18,000 students) and Calder University (24,000 students) are fictional. A survey takes a random sample of 220 Ridgeway students, of whom 77 cycle to campus, and an independent random sample of 260 Calder students, of whom 65 cycle. Construct a 95% confidence interval for the difference in the proportions of all students at the two universities who cycle to campus. Show the identify, check and calculate steps.
Worked solution and suggested Marlbridge rubric
Identify. Two-sample z-interval for a difference between population proportions. pR − pC = the proportion of all Ridgeway students who cycle to campus, minus the proportion of all Calder students who do.
Check. Randomization: independent random samples from each university. 10%: 220 ≤ 1,800 and 260 ≤ 2,400. Normality: 77 and 143 (Ridgeway), 65 and 195 (Calder), all at least 10.
Calculate. p̂R = 77 ÷ 220 = 0.35 and p̂C = 65 ÷ 260 = 0.25, so the point estimate is 0.10. SE = √(0.35 × 0.65 ÷ 220 + 0.25 × 0.75 ÷ 260) ≈ 0.0419. MOE = 1.960 × 0.0419 ≈ 0.0821. Interval: 0.10 ± 0.0821 = (0.018, 0.182).
| Point | What earns it |
|---|---|
| 1 | Names the two-sample z-interval for p1 − p2 and defines the parameter with the order, the response and both populations |
| 1 | Randomization and 10% conditions checked with numbers from the context |
| 1 | All four observed counts shown and compared with 10 |
| 1 | Correct point estimate, standard error and interval (0.018, 0.182) |
Accept the order Calder − Ridgeway with interval (−0.182, −0.018) if the parameter is defined in that order. Do not award point 1 for a parameter about “the students sampled”.
Question 6 (constructed response · core)
A fictional language-learning company runs an experiment with 180 volunteers. It randomly assigns 90 to an app version with daily reminders and 90 to a version without reminders. After 30 days, 61 volunteers in the reminder group and 44 in the no-reminder group have completed the course.
(a) Construct a 99% confidence interval for the difference in completion proportions (reminders − no reminders). (b) A colleague reports the interval for (no reminders − reminders). What should it be, and does it contain different information?
Worked solution and suggested Marlbridge rubric
(a) Identify. Two-sample z-interval for p1 − p2, where pR − pN = the proportion of learners like these who would complete the course with reminders, minus the proportion who would complete it without reminders.
Check. Randomization: the app versions were randomly assigned. The 10% condition is not needed in a randomized experiment. Normality: 61 and 29, 44 and 46, all at least 10.
Calculate. p̂R = 61 ÷ 90 ≈ 0.6778 and p̂N = 44 ÷ 90 ≈ 0.4889, so the point estimate is about 0.1889. SE = √(0.6778 × 0.3222 ÷ 90 + 0.4889 × 0.5111 ÷ 90) ≈ 0.07213. For 99%, z* = 2.576, so MOE ≈ 0.1858. Interval: (0.003, 0.375).
(b) (−0.375, −0.003). It contains exactly the same information: the signs change and the endpoints swap because the order of subtraction is reversed. The width is the same.
| Point | What earns it |
|---|---|
| 1 | Parameter defined for the two treatments with the order stated |
| 1 | Random assignment named and four observed counts checked; no 10% check required |
| 1 | Correct z* = 2.576, standard error and interval (0.003, 0.375) |
| 1 | Correct reversed interval and the statement that it carries the same information |
Using z* = 1.960 gives (0.048, 0.330), which is the 95% interval and does not earn point 3.
Question 7 (explanation · stretch)
Avonbridge (24,000 adults) and Bexmoor (20,000 adults) are fictional towns. Independent random samples found that 132 of 240 Avonbridge adults and 98 of 200 Bexmoor adults attended a summer event. A student’s 95% interval is shown.
p1 − p2 = the difference between the proportions of the 240 Avonbridge adults and the 200 Bexmoor adults who attended. Conditions: n1 = 240 and n2 = 200 are both at least 30. ✓ Pooled p̂ = 230 ÷ 440 = 0.5227. SE = √[0.5227 × 0.4773 × (1/240 + 1/200)] = 0.04782. For 95%, z* = 1.645. Interval: 0.06 ± 1.645 × 0.04782 = (−0.019, 0.139).
(a) Identify four errors in the student’s work. (b) Give the correct interval.
Model answer and suggested Marlbridge rubric
(a)
- Parameter: it describes the samples. It should be the proportion of all Avonbridge adults who attended minus the proportion of all Bexmoor adults who attended.
- Conditions: “at least 30” is the wrong check. The student should check randomization (independent random samples), 10% (240 ≤ 2,400 and 200 ≤ 2,000) and the four observed counts (132, 108, 98 and 102, all at least 10).
- Standard error: pooling is for a test that assumes p1 = p2. The interval uses √(0.55 × 0.45 ÷ 240 + 0.49 × 0.51 ÷ 200) ≈ 0.04776.
- Critical value: 1.645 is for 90% confidence. For 95%, z* = 1.960.
(b) p̂1 = 0.55 and p̂2 = 0.49, so the point estimate is 0.06. MOE = 1.960 × 0.04776 ≈ 0.0936. Interval: (−0.034, 0.154).
| Point | What earns it |
|---|---|
| 1 | Identifies the parameter error and rewrites it for the populations |
| 1 | Identifies the conditions error and states the correct checks with numbers |
| 1 | Identifies both the pooling error and the z* error |
| 1 | Correct interval (−0.034, 0.154) |
How did you do?
- Q1 wrong: re-read “From one proportion to a difference” in the study guide.
- Q2 or Q7(a) wrong: revisit “Step 3: standard error, margin of error and the interval”. Do not pool; add the variance terms under one root.
- Q3 wrong: revisit “Step 2: check the three conditions” and Worked example 3(b).
- Q4 wrong: revisit Worked example 3(a) and Figure 1.
- Q5 or Q6 incomplete: follow Worked examples 1 and 2 step by step: identify, define, check, calculate.
Then tick off the topic checklist.