Read the big ideas and the method table, then try the seven questions without notes. Each question combines two or more Unit 1 topics.
These are original Marlbridge practice questions, not past exam questions, with fictional data. The rubrics are a suggested Marlbridge rubric, not official scoring.
Big ideas of the unit
- Every study starts with a question about a population. The investigative question fixes the variable, analysis and type of conclusion before data are collected (1.1, 1.10).
- Statistics estimate parameters. A sample value varies from sample to sample and rarely equals the parameter exactly (1.2).
- The type of variable decides the tools. Categorical variables are summarised with counts and proportions and shown in bar or pie charts (1.3, 1.4). Quantitative variables are shown in dotplots, stemplots, histograms and boxplots (1.5, 1.8).
- Describe distributions completely. Shape, centre, variability and unusual features, always in context (1.6).
- Shape guides the choice of summary. Skew and outliers pull the mean and standard deviation; the median and IQR resist them (1.7, 1.8).
- Compare with proportions, comparative words and relative position. Groups of different sizes need proportions; z-scores compare values from different distributions (1.9).
- Random selection protects against bias and allows generalisation. Pick the method that fits the population (1.11); biased methods push estimates one way, and a bigger sample does not help (1.12).
- Random assignment allows cause and effect. Observational studies can be confounded; a well-designed experiment with comparison, random assignment, replication and control reduces that risk (1.10, 1.13).
Key relationships and methods
| If the question asks you to… | Use… | Watch out for… |
|---|---|---|
| Summarise a categorical variable | Frequency and relative frequency tables; bar or pie chart | Comparing counts when group sizes differ |
| Display a quantitative variable | Dotplot, stem-and-leaf plot, histogram, boxplot | Bin edges; boxplots hide gaps and clusters |
| Describe or compare distributions | Shape, centre, variability, unusual features, in context | Listing values without comparative words |
| Choose a centre and spread | Median and IQR if skewed or outliers; mean and s if roughly symmetric | Saying “average” without naming the statistic |
| Check for outliers | Beyond Q1 − 1.5 × IQR or Q3 + 1.5 × IQR; or more than 2s from x̄ | Ending a whisker at the fence instead of at a data value |
| Compare positions | z = (x − mean) ÷ standard deviation | Sign: below the mean gives negative z |
| Change units | Centre and position follow the conversion; spread is only multiplied | Adding a constant to the spread |
| Generalise to a population | Random selection from that population | Volunteers represent only people like them |
| Claim cause and effect | Random assignment in a well-designed experiment | Confounding variables in observational studies |
Question 1 (multiple choice · mixed)
Daily maximum temperatures in a fictional city last summer had mean 18 °C and standard deviation 4 °C. One day reached 26 °C. A researcher converts every value to °F using F = 1.8C + 32. What are the new mean, the new standard deviation and that day’s z-score?
- (A) Mean 64.4 °F, standard deviation 39.2 °F, z = 0.37
- (B) Mean 32.4 °F, standard deviation 7.2 °F, z = 6.44
- (C) Mean 64.4 °F, standard deviation 7.2 °F, z = 2.00
- (D) Mean 64.4 °F, standard deviation 7.2 °F, z = 3.60
Answer and explanation
Answer: (C). Mean: 1.8 × 18 + 32 = 64.4 °F. Adding 32 shifts every value equally, so only the 1.8 changes the spread: 1.8 × 4 = 7.2 °F. The day is 1.8 × 26 + 32 = 78.8 °F, so z = (78.8 − 64.4) ÷ 7.2 = 2.00, the same as (26 − 18) ÷ 4. A z-score has no units, so it does not change.
- (A) adds 32 to the standard deviation.
- (B) forgets to add 32 to the mean.
- (D) multiplies the z-score by 1.8.
Topics: 1.7 (changing units), 1.9 (z-scores).
Question 2 (multiple choice · mixed)
Two fictional courier companies each delivered 120 parcels. The five-number summaries of delivery times, in days, are:
| Company | Min | Q1 | Median | Q3 | Max |
|---|---|---|---|---|---|
| R | 1 | 2 | 3 | 5 | 12 |
| S | 2 | 3 | 4 | 4.5 | 6 |
Which statement is supported by these summaries?
- (A) Every parcel from R arrived sooner than every parcel from S.
- (B) R has the smaller median, but its delivery times vary more: IQR 3 days against 1.5 days.
- (C) S has a potential high outlier at 6 days.
- (D) Because R has a high outlier, R’s mean is probably less than its median.
Answer and explanation
Answer: (B). R’s median (3 days) is below S’s (4 days). R’s IQR is 5 − 2 = 3 days; S’s is 4.5 − 3 = 1.5 days.
- (A) is impossible: R’s maximum (12) exceeds S’s minimum (2).
- (C) is wrong: S’s upper fence is 4.5 + 1.5 × 1.5 = 6.75 days, and 6 < 6.75.
- (D) has the direction backwards. R’s upper fence is 5 + 4.5 = 9.5 days, so 12 is an outlier, and it pulls R’s mean above its median.
Topics: 1.7 (outlier rule), 1.8 (boxplots and the mean), 1.9 (comparing).
Question 3 (multiple choice · mixed)
Eighty students at a fictional college each chose one revision method for a test: online quizzes or paper flashcards. The quiz group’s mean score was higher. Which variable is most likely to be a confounding variable?
- (A) The test score
- (B) The colour of the flashcards
- (C) The number of students in each group
- (D) Each student’s motivation to study
Answer and explanation
Answer: (D). A confounder is linked to both the explanatory variable and the response. More motivated students may be more likely to choose quizzes and tend to score higher anyway, so motivation could explain the difference. Randomly assigning methods would balance motivation across the groups.
- (A) is the response variable itself.
- (B) is unlikely to affect scores, and quiz users have no flashcards.
- (C) describes the groups, not a characteristic of each student.
Topics: 1.10 (observational studies, confounding), 1.13 (random assignment).
Question 4 (constructed response · mixed)
The fictional Calderbrook library service has 12,000 members. It chooses a simple random sample of 250 members and asks each one their preferred format for borrowing books. The results are: Print 105, E-book 70, Audiobook 45, No preference 30.
(a) Identify the population, N, the sample and n. Is “0.42 of the members asked prefer print” a parameter or a statistic? Explain. (b) Construct a relative frequency table. (c) Find the angle of the E-book slice in a pie chart of these data. (d) A manager says, “Most of our members prefer print.” Is this claim supported? Explain.
Worked solution and suggested Marlbridge rubric
(a) Population: all Calderbrook library members, N = 12,000. Sample: the 250 members chosen, n = 250. 105 ÷ 250 = 0.42 is a statistic, because it describes the sample, not all 12,000 members.
(b)
| Preferred format | E-book | Audiobook | No preference | Total | |
|---|---|---|---|---|---|
| Relative frequency | 0.42 | 0.28 | 0.18 | 0.12 | 1.00 |
(c) 0.28 × 360° = 100.8°.
(d) Not supported. Print is the most common choice, but 0.42 < 0.5, so it is not a majority: 58% of the sample chose something else. As a random sample, 0.42 is a reasonable estimate for all members, though another sample would give a slightly different value.
| Point | What earns it |
|---|---|
| 1 | Population, N, sample, n, and 0.42 a statistic because it describes the sample |
| 1 | All four relative frequencies correct, summing to 1 |
| 1 | Slice angle 100.8° with working |
| 1 | Not supported because 0.42 < 0.5 |
Total: 4 points. Topics: 1.1, 1.2, 1.3, 1.4.
Question 5 (constructed response · mixed)
The lengths, in minutes, of the 15 episodes of a fictional podcast are:
22, 25, 26, 28, 29, 31, 31, 33, 34, 35, 36, 38, 40, 41, 68
For these data, x̄ = 34.47 minutes and s = 10.78 minutes.
(a) Construct a stem-and-leaf plot with a key. (b) Find the five-number summary. Check for potential outliers using both the 1.5 × IQR rule and the 2-standard-deviation rule. (c) Describe how to draw a boxplot that shows outliers. (d) Describe the distribution in context, and explain how the mean and median reflect its features.
Worked solution and suggested Marlbridge rubric
(a)
| Stem | Leaves |
|---|---|
| 2 | 2 5 6 8 9 |
| 3 | 1 1 3 4 5 6 8 |
| 4 | 0 1 |
| 5 | |
| 6 | 8 |
Key: 3 | 1 means 31 minutes. Keep the empty stem 5.
(b) The median is the 8th value, 33; Q1 = 28 (lower 7 values); Q3 = 38 (upper 7). Five-number summary: 22, 28, 33, 38, 68 minutes. IQR = 10, so the fences are 13 and 53 minutes: 68 is a potential outlier, and nothing is below 13. 2s rule: 34.47 + 2(10.78) = 56.03 < 68, so both rules agree.
(c) On a scale from 20 to 70 minutes, draw a box from 28 to 38 with a line at 33. Whiskers go to 22 and to 41, the largest non-outlier. Mark 68 separately.
(d) Episode lengths are roughly symmetric apart from one long episode. The median is 33 minutes and the middle half run from 28 to 38 minutes (IQR 10 minutes). A gap separates 41 from a high outlier at 68 minutes. The outlier pulls the mean (34.47 minutes) above the median; without it the mean would be 32.07, close to the new median of 32. The median and IQR describe a typical episode better.
| Point | What earns it |
|---|---|
| 1 | Correct ordered stemplot, including the empty stem, with a key |
| 1 | Five-number summary 22, 28, 33, 38, 68 |
| 1 | Both outlier checks (53 and 56.03) with a conclusion about 68 |
| 1 | Upper whisker ends at 41; 68 marked separately |
| 1 | Shape, centre, variability and outlier or gap in context, and the outlier pulling the mean up |
Total: 5 points. Topics: 1.5, 1.6, 1.7, 1.8.
Question 6 (constructed response · mixed)
A fictional athletics club has 24 adult volunteers. A random number generator assigns 12 to a dynamic warm-up (jogging and leg swings) and 12 to a static warm-up (held stretches). Each runner then runs 400 m. Times, in seconds:
- Dynamic: 58, 60, 61, 62, 63, 63, 64, 65, 66, 67, 69, 72
- Static: 61, 63, 64, 66, 67, 68, 68, 70, 71, 73, 75, 84
| Warm-up | Min | Q1 | Median | Q3 | Max | Mean | s |
|---|---|---|---|---|---|---|---|
| Dynamic | 58 | 61.5 | 63.5 | 66.5 | 72 | 64.17 | 3.93 |
| Static | 61 | 65 | 68 | 72 | 84 | 69.17 | 6.19 |
(a) Explain why this is an experiment. Identify the experimental units, the treatments and the response variable. (b) Use the 1.5 × IQR rule to check each group for outliers. (c) Compare the two distributions of 400 m times. (d) Calculate the z-score of the 84-second runner within the static group, and interpret it. (e) Assume the difference in centres is too large to be explained by chance alone. What can the club conclude, and about whom?
Worked solution and suggested Marlbridge rubric
(a) The warm-ups were imposed (assigned), so it is an experiment. Units: the 24 runners. Treatments: dynamic and static warm-ups. Response: 400 m time (s).
(b) Dynamic: IQR = 5, fences 54 and 74 s; no outliers. Static: IQR = 7, fences 54.5 and 82.5 s; 84 s is a potential outlier.
(c) Shape: dynamic times are roughly symmetric (mean 64.17 s, median 63.5 s); static times are roughly symmetric apart from the high outlier at 84 s, which pulls the mean (69.17 s) above the median (68 s). Centre: dynamic runners were generally faster; their median was 4.5 s lower. Variability: dynamic times varied less (IQR 5 s against 7 s; s 3.93 s against 6.19 s). Overlap: 8 of 12 dynamic times were at or below the static Q1 of 65 s.
(d) z = (84 − 69.17) ÷ 6.19 ≈ 2.40: 2.40 standard deviations slower than the static mean, unusually slow for that group.
(e) The warm-ups were randomly assigned, so the club can conclude the dynamic warm-up caused faster times. The runners were volunteers, so this applies only to adult runners similar to them.
| Point | What earns it |
|---|---|
| 1 | Treatments imposed; units, treatments and response correct |
| 1 | Correct fences for both groups; 84 s the only outlier |
| 1 | Centre and variability compared with comparative words and values |
| 1 | Shape of both groups, mentioning the outlier |
| 1 | z ≈ 2.40, interpreted as slower than the group mean |
| 1 | Causation from random assignment and generalisation only to similar volunteers |
Total: 6 points. Topics: 1.6, 1.7, 1.9, 1.10, 1.13.
Question 7 (constructed response · mixed)
The fictional Ridgeline bus company has 18,000 annual pass holders: 10,800 in the urban zone, 5,400 in the suburban zone and 1,800 in the rural zone. Buses run less often in the rural zone. The company wants to estimate the proportion of all pass holders who are satisfied with punctuality.
(a) Write an investigative question for this study that asks for an estimate, with all three components. (b) Describe how to select a stratified random sample of 300 pass holders, with the zones as strata and each stratum sampled in proportion to its size. Explain why zone is a sensible choice of strata. (c) Last year the company sent a text message with a survey link to every pass holder, and 2,100 replied. Identify the main source of bias and its likely direction. (d) A manager says, “Last year’s survey had 2,100 replies, seven times as many as 300, so it was more reliable.” Respond to this claim.
Worked solution and suggested Marlbridge rubric
(a) “Using a random sample, what is an interval of plausible values for the proportion of all 18,000 current Ridgeline annual pass holders who are satisfied with bus punctuality?” It names the variable (satisfied: yes or no), the analysis (estimating a proportion within a range) and the population the conclusion covers (all current pass holders; no cause-and-effect claim).
(b) Urban: 10,800 ÷ 18,000 = 0.60, so 0.60 × 300 = 180. Suburban: 0.30 × 300 = 90. Rural: 0.10 × 300 = 30. Within each zone, number the pass holders and use a random number generator to choose that many different numbers; combine the three groups. Zone is sensible because service differs by zone (rural buses are less frequent), so satisfaction probably differs too, and every zone is then represented.
(c) Voluntary response bias. Pass holders chose whether to reply, and those with strong feelings, often complaints about late buses, reply more. The proportion satisfied is likely to be too low. Only 2,100 of 18,000 (11.7%) replied, so it was not a census.
(d) Wrong. A biased method stays biased however many replies it gets. A random sample of 300 avoids voluntary response bias, so it should give a more trustworthy estimate.
| Point | What earns it |
|---|---|
| 1 | Question names the variable, asks for an estimate of a proportion, and names the population |
| 1 | 180, 90 and 30, with an SRS within each stratum |
| 1 | Zone justified by likely differences in satisfaction |
| 1 | Voluntary response bias, how it arises here and “too low” |
| 1 | Sample size does not remove bias, so the random sample is better |
Total: 5 points. Topics: 1.10, 1.11, 1.12.
How did you do?
Mark your answers with the rubrics and note the topics listed under each question you missed. Re-read those study guides, then tick off each checklist. If many topics went wrong, use the Unit 1 diagnostic to find the gaps.
Topic checklists: 1.1 · 1.2 · 1.3 · 1.4 · 1.5 · 1.6 · 1.7 · 1.8 · 1.9 · 1.10 · 1.11 · 1.12 · 1.13