How to use this guide
This guide is for any student preparing for the free-response section of the Statistics exam. It shows you how to turn what you know into answers a reader can credit. Read the verb table and examples, then try the tasks before opening the model answers. Every question here is original Marlbridge practice with fictional data. None of it is past exam material. The marks shown are a suggested Marlbridge rubric, not official scoring.
The free-response section at a glance
| Feature | Detail |
|---|---|
| Questions | 4, each worth 10 points |
| Time | 90 minutes |
| Weight | 50% of the exam (each question 12.5%) |
| Question 1 | Multi-part, focused mainly on Practices 1 and 2 (formulating questions, collecting data) |
| Question 2 | Multi-part, focused mainly on Practices 3 and 4 (analysing data, interpreting results) |
| Question 3 | Inference: a hypothesis test or a confidence interval |
| Question 4 | Multi-part, mixing content areas, assessing Practices 2, 3 and 4 |
| Tools | A graphing calculator with statistical functions is expected; formulas and tables are provided |
| Format | Fully digital for May 2027 |
Questions are split into lettered parts (A, B, C…), and some parts have numbered subparts (i, ii). Label every part of your answer to match.
The task verbs and what they require
The course uses eleven task verbs in free-response questions. The verb is the instruction, so find it first.
| Verb | What your answer must do | How much to write |
|---|---|---|
| Identify / Classify | Name the thing asked for, or put it in the right category, with no reasons | A word, phrase or one short sentence |
| Calculate | Carry out the maths and reach a final value | Formula or named method, numbers substituted, answer with units |
| Estimate | Use a model, graph or table to give an approximate value | The value, plus where it came from |
| Construct | Make a graph or table from data | A labelled display: title, axes or headings, scale, units |
| Describe | Give the important features of a display or distribution | The features (for one variable: shape, centre, variability, unusual values), in context |
| Compare | Say how two or more things are alike or different; with numbers, compare the values directly | Comparative words (“greater than”, “less spread out”) and values for every group |
| Interpret | Say what a number or result means in the real situation | One or two sentences in context, with units |
| Explain | Say how or why something happens, using evidence or reasoning | A claim, the evidence and the link, usually two or three sentences |
| Justify | Give evidence for (or against) a claim, and the statistical reasoning that connects the evidence to it | Evidence with numbers, then “because…” reasoning, then the conclusion |
| Determine | Choose or decide something (a method, a question, a claim) using calculations or findings | The decision plus the work or finding that supports it |
| Complete | Carry out an inference procedure: name its parts, check its conditions, calculate its results | Parts, conditions with numbers, calculations; a conclusion if asked |
Two rules follow. A short-answer verb needs no reasons: explaining an “identify” answer wastes time and may add an error. A reasoning verb is not answered by a number alone: “justify” with only a p-value earns little.
Group 1: naming verbs (identify, classify)
Example. A fictional college records how many books each student read last term and whether the student has a library card. Classify each variable.
Answer. Books read: quantitative. Library card (yes/no): categorical.
No reason is needed.
Group 2: producing verbs (calculate, estimate, construct)
Example. Heights in a fictional class have mean 165 cm and standard deviation 8 cm. Calculate the z-score of a student who is 172 cm tall.
Answer. z = (x − mean) / SD = (172 − 165) / 8 = 0.875.
For estimate, say where the value came from (“from the fitted line at x = 30, about 52”). For construct, a display without labels or scale is incomplete.
Group 3: describing verbs (describe, compare)
Example. A scatterplot of fictional data shows weekly hours of part-time work (x) against hours of sleep per night (y) for 40 students. The points follow a straight band sloping downwards, with no points far from the rest, and r = −0.78. Describe the association.
Answer. There is a moderately strong, negative, linear association between weekly hours of part-time work and nightly hours of sleep: students who work more hours tend to sleep less. There are no unusual points.
For compare, give the same features for every group, joined by comparative words, as in Worked example 1.
Group 4: meaning and reasoning verbs (interpret, explain, justify)
Example. Interpret the z-score of 0.875 from Group 2.
Answer. The student’s height is 0.875 standard deviations above the mean height of the class.
An interpretation always returns to the context: what is measured, in what units, for which group. Explain and justify also need a claim, the evidence and a sentence linking them: claim, because evidence, which shows link.
Group 5: deciding verbs (determine, complete)
Example. In a fictional town, P(owns a bike) = 0.40, P(walks to school) = 0.35 and P(owns a bike and walks to school) = 0.12. Determine whether the two events are independent.
Answer. If independent, P(both) would equal P(bike) × P(walks) = 0.40 × 0.35 = 0.14. The actual value is 0.12 ≠ 0.14, so the events are not independent.
A bare “no” earns nothing. Complete asks for a whole inference procedure, as in Worked example 2.
Reading a multi-part question
- Read the stem slowly. Note the population, how the data were collected (random sample? random assignment?), the variables and their units.
- Read every part before you write. Later parts often show what an earlier part is for.
- Find the verb in each part and give the kind of answer it asks for.
- Look for links. “Use your answer to part B” means use it, not start again.
- If stuck, assume and move on. State a sensible value (“assuming the mean is 12 minutes”) and carry on.
- Answer the question asked. If a part asks whether a claim is supported, say yes or no, in context.
Planning your time
You have 90 minutes for four 10-point questions, about 22 minutes each. A plan that leaves time to check:
| Stage | Time |
|---|---|
| Skim all four questions and note the verbs | 3 minutes |
| Each question (about 20 minutes each) | 80 minutes |
| Final check: context, labels, every part answered | 7 minutes |
If a part is stuck after two or three minutes, move on and come back.
Showing work on calculations
A reader can only credit what they can see. For every calculation:
- Write the formula or name the method. “z = (p̂ − p₀) / √[p₀(1 − p₀)/n]” or “X is binomial with n = 12, p = 0.25”.
- Substitute the numbers.
- Give the answer with units and sensible rounding. Keep extra decimals in the middle of a calculation and round only at the end.
- Label calculator results. “binomcdf(12, 0.25, 4)” alone is not work. “P(X ≤ 4) for X ~ B(12, 0.25) = 0.8424” is.
In the sample scoring in the course description, each point is generally earned independently.
What earns and loses credit
| Earns credit | Loses credit |
|---|---|
| Context in every sentence: the variable, the units, the population | Generic answers that could fit any question |
| Answering with the verb’s kind of answer | Explaining when asked to identify; giving a number when asked to justify |
| Linking evidence to the claim (“because 0.0095 < 0.05…”) | A correct number with no conclusion drawn from it |
| Naming the parameter (the true proportion, the population mean) | Concluding about the sample statistic instead of the parameter |
| Non-definitive language (“convincing evidence that…”) | “Proves”, “accept H₀”, or causal claims from an observational study |
| One clear final answer | Two answers that disagree; the reader cannot tell which is yours |
Worked example 1: compare (Unit 1)
Question. Setting time (hours) was recorded for 11 test joints made with each of two fictional wood glues.
| Glue | Setting time (hours) |
|---|---|
| K | 14.2, 15.0, 15.6, 16.1, 16.4, 16.8, 17.0, 17.3, 17.9, 18.4, 24.5 |
| L | 11.0, 12.8, 13.9, 14.5, 15.2, 15.8, 16.0, 17.1, 18.6, 19.4, 20.3 |
Compare the distributions of setting time for the two glues.
Weak answer. “Glue L is better because it sets faster. Glue L is more spread out.”
What it misses. No values, no shape, no outlier. “Better” is a judgement, not a comparison.
Work.
| Min | Q1 | Median | Q3 | Max | IQR | 1.5 × IQR fences | |
|---|---|---|---|---|---|---|---|
| K | 14.2 | 15.6 | 16.8 | 17.9 | 24.5 | 2.3 | 12.15 to 21.35 |
| L | 11.0 | 13.9 | 15.8 | 18.6 | 20.3 | 4.7 | 6.85 to 25.65 |
Strong answer.
- Shape: both distributions are roughly symmetric, apart from one high value for Glue K.
- Centre: the median setting time for Glue K (16.8 hours) is greater than for Glue L (15.8 hours).
- Variability: Glue L’s setting time is more variable: its IQR is 4.7 hours compared with 2.3 hours for Glue K.
- Unusual features: Glue K has an outlier at 24.5 hours, since 24.5 > 17.9 + 1.5 × 2.3 = 21.35. Glue L has no outliers.
Suggested Marlbridge rubric (4 points).
| Point | Earned for |
|---|---|
| 1 | Centre compared with values for both glues, using a comparative word |
| 2 | Variability compared with values, using a resistant measure because of the outlier |
| 3 | Shape and the outlier correctly described, with the 1.5 × IQR check shown |
| 4 | Context throughout: setting time, hours, both glue names |
Worked example 2: complete an inference procedure (Unit 3)
Question. A fictional school district with 6,800 students says that 30% of its students walk to school. A researcher thinks the true proportion is higher. In a random sample of 250 students, 92 walk. At α = 0.05, complete the appropriate inference procedure and state a conclusion in context.
Weak answer. “z = 2.35, p = 0.0095. Reject H₀. 36.8% of students walk to school, so the district was wrong.”
What it misses. No hypotheses, procedure or conditions. The conclusion is about the sample (36.8%), not the population, and “was wrong” is too certain.
Strong answer.
- Hypotheses. Let p = the true proportion of all students in the district who walk to school. H₀: p = 0.30, Hₐ: p > 0.30, α = 0.05.
- Procedure and conditions. One-sample z-test for a population proportion. Random: the sample was random. 10%: 250 is less than 10% of 6,800 students. Large counts: np₀ = 250(0.30) = 75 ≥ 10 and n(1 − p₀) = 250(0.70) = 175 ≥ 10.
- Calculations. p̂ = 92/250 = 0.368. z = (0.368 − 0.30) / √[0.30(0.70)/250] = 0.068 / 0.02898 ≈ 2.35. p-value = P(Z ≥ 2.35) ≈ 0.0095.
- Conclusion. Because the p-value of 0.0095 is less than α = 0.05, we reject H₀. There is convincing statistical evidence that the true proportion of students in the district who walk to school is greater than 0.30.
Suggested Marlbridge rubric (4 points).
| Point | Earned for |
|---|---|
| 1 | Correct hypotheses with p defined in context |
| 2 | Procedure named and all three conditions checked with numbers |
| 3 | Correct test statistic and p-value, with the formula shown |
| 4 | Decision linked to α, conclusion about p in context, non-definitive |
Worked example 3: interpret and explain (Unit 5)
Question. A fictional hardware shop records the daily high temperature x (°C) and the number of electric fans sold y on 30 days, with temperatures from 14 °C to 31 °C. The least-squares line is ŷ = −38.6 + 4.2x. (A) Interpret the slope. (B) On a 25 °C day the shop sold 71 fans. Calculate and interpret the residual. (C) Explain whether the model should be used to predict sales on a 2 °C day.
Weak answer. “(A) The slope is 4.2. (B) 71 − 66.4 = 4.6. (C) Yes, put 2 into the equation.”
What it misses. (A) gives no meaning. (B) has no interpretation and no work for 66.4. (C) ignores the range of the data.
Strong answer.
- (A) For each additional 1 °C in the daily high temperature, the predicted number of fans sold increases by 4.2.
- (B) ŷ = −38.6 + 4.2(25) = 66.4 fans. Residual = actual − predicted = 71 − 66.4 = 4.6 fans. The shop sold 4.6 more fans than the model predicted for a 25 °C day.
- (C) No. The data cover only 14 °C to 31 °C, so 2 °C is an extrapolation and the linear pattern may not continue. The model gives ŷ = −38.6 + 4.2(2) = −30.2 fans, which is impossible.
Suggested Marlbridge rubric (3 points).
| Point | Earned for |
|---|---|
| 1 | Slope interpreted as a predicted change per 1 °C, in context |
| 2 | Residual calculated with work shown and interpreted as actual above predicted |
| 3 | Extrapolation named and linked to the 14–31 °C range |
Common mistakes
- Answering a different verb: explaining when asked to identify, or giving only a number when asked to justify.
- Leaving out context: “the mean is 6.9” instead of “the mean sleep time of the sampled students is 6.9 hours”.
- Writing calculator syntax with no labels, formula or parameters.
- Comparing distributions with values for only one group, or with no comparative words.
- Concluding about the sample statistic rather than the population parameter.
- Saying “accept H₀” or “this proves”, or claiming cause from an observational study.
- Interpreting a slope without the word “predicted”.
- Spending too long on one question and leaving the last one unfinished.
Practise it
Task 1 (Unit 1: identify, explain). A fictional survey of 400 randomly selected adults finds that those who drink green tea daily report fewer colds per year. (i) Identify the type of study. (ii) Explain why the survey cannot show that green tea causes fewer colds.
Model answer
(i) An observational study. (ii) No treatment was assigned, so another variable may explain the link. Adults who drink green tea daily may also exercise more, and exercise may be linked to fewer colds. The survey cannot separate the effect of tea from the effect of exercise.
Task 2 (Unit 2: calculate). A fictional quiz has 12 questions, each with 4 options. A student guesses every answer at random. Calculate the probability that the student gets at least 5 correct.
Model answer
Let X = number correct. X is binomial with n = 12 and p = 0.25 (fixed number of independent guesses, same probability each time). P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.8424 = 0.1576.
Task 3 (Unit 2: calculate, interpret). A game at a fictional school fair costs 2 tokens. A player wins 10 tokens with probability 0.05, 3 tokens with probability 0.20, and nothing otherwise. Let W be the net gain per play. Calculate the mean of W and interpret it.
Model answer
Net gains: 8 (probability 0.05), 1 (probability 0.20), −2 (probability 0.75). Mean = 8(0.05) + 1(0.20) + (−2)(0.75) = 0.40 + 0.20 − 1.50 = −0.90 tokens. Over many plays, a player loses about 0.90 tokens per play on average.
Task 4 (Unit 4: interpret, justify). A random sample of 40 students at a fictional college has mean sleep 6.9 hours and standard deviation 0.88 hours. A 95% confidence interval for the population mean is 6.62 to 7.18 hours. (i) Interpret the interval. (ii) The college says students sleep 7.5 hours a night on average. Is this claim plausible? Justify.
Model answer
(i) We are 95% confident that the interval from 6.62 to 7.18 hours captures the true mean nightly sleep of all students at the college. (ii) Not plausible. 7.5 hours lies above the whole interval, so it is not a plausible value for the population mean; the true mean appears to be below 7.5 hours.
Task 5 (Unit 5: estimate, explain). For fictional used cars aged 1 to 12 years, value (thousand dollars) is modelled by ŷ = 21.5 − 1.45x, where x is age in years. (i) Estimate the value of a 6-year-old car. (ii) Explain why the model should not be used for a 20-year-old car.
Model answer
(i) ŷ = 21.5 − 1.45(6) = 12.8, so about 12.8 thousand dollars. (ii) The data only include ages 1 to 12 years, so 20 years is an extrapolation and the linear trend may not hold. The model predicts 21.5 − 1.45(20) = −7.5 thousand dollars, a negative value, which is impossible.
Where to practise next
- Comparing distributions of one quantitative variable: study guide
- Investigative questions and data collection: study guide
- The binomial distribution: practice
- Carrying out a test for a population proportion: study guide
- Justifying a claim with a confidence interval for a mean: study guide
- Least-squares regression: study guide
- Unit 3 mixed review
Quick questions
How long should an answer be?
As long as the verb needs and no longer. One phrase is enough for identify. Explain and justify need a claim, the evidence and the link between them, usually two or three sentences. Extra writing can add a wrong statement that cancels a correct one.
What if I cannot answer an early part that a later part depends on?
Write down a sensible value, say clearly that you are assuming it, and carry on. The later part can still show a correct method.
Do I need to write calculator commands?
Calculator syntax on its own is not enough. Name the procedure or distribution, give its parameters and the boundary you used, then give the result. Labelled values are clearer than key presses.