These are original Marlbridge practice questions, not past exam questions. All data sets 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: residual = observed value − predicted value (y − ŷ); each model was found with technology from the data described; residual plots show residuals against the explanatory variable; round final answers to 2 decimal places unless stated.
Question 1 (multiple choice · foundation)
A fictional fish farm uses the model ŷ = −180 + 18.5x to predict the mass (g) of a trout from its length x (cm). A trout 30 cm long has a mass of 390 g. What is its residual?
- (A) −15 g
- (B) 15 g
- (C) 375 g
- (D) 765 g
Answer and explanation
Answer: (B). ŷ = −180 + 18.5(30) = −180 + 555 = 375 g. Residual = observed − predicted = 390 − 375 = 15 g. The trout is 15 g heavier than the model predicts.
- (A) calculates predicted − observed, 375 − 390. The order is observed − predicted.
- (C) is the predicted mass ŷ, not the residual.
- (D) adds the observed and predicted masses, 390 + 375. A residual is a difference.
Question 2 (multiple choice · foundation)
A fictional bakery uses a linear model to predict the number of loaves it sells each day from the number of online pre-orders. On Tuesday the residual was −2.3 loaves. Which is the correct interpretation?
- (A) The bakery sold 2.3 more loaves than the model predicted, so the model underpredicted.
- (B) The bakery sold 2.3 fewer loaves than the model predicted, so the model overpredicted.
- (C) The model predicted that the bakery would sell 2.3 loaves on Tuesday.
- (D) The bakery sold 2.3 fewer loaves than the model predicted, so the model underpredicted.
Answer and explanation
Answer: (B). Residual = actual − predicted = −2.3, so the actual number sold was 2.3 loaves below the prediction. A prediction that is too high is an overprediction.
- (A) reverses the sign: “2.3 more” would be a residual of +2.3.
- (C) confuses the residual with the predicted value ŷ.
- (D) has the right size and direction for the sales but the wrong conclusion: if the actual value is below the prediction, the model predicted too much, not too little.
Question 3 (multiple choice · core)
Four linear models were fitted to four different fictional data sets. Their residual plots are shown below. Which residual plot gives the best evidence that a linear model is appropriate?
- (A) Plot W
- (B) Plot X
- (C) Plot Y
- (D) Plot Z
Answer and explanation
Answer: (C). In plot Y the residuals are scattered above and below 0 with no curve or other pattern. Apparent randomness in a residual plot supports a linear form, so a linear model is appropriate.
- (A) Plot W is U-shaped: the model underpredicts at both ends and overpredicts in the middle. This curvature shows the form is not linear.
- (B) Plot X is an arch, the reverse of W. It also shows curvature.
- (D) Plot Z has a clear wave pattern. The residuals are not random, so the straight line is missing a systematic feature of the data.
Question 4 (calculation · core)
A fictional bus company models journey time (minutes) from route distance (km) with ŷ = 4.5 + 2.6x. Five of the journeys in its data were:
| Distance (km) | 3 | 5 | 8 | 10 | 12 |
|---|---|---|---|---|---|
| Time (min) | 13 | 19 | 22 | 31 | 34 |
(a) Calculate the predicted time and the residual for each journey. (b) Which journey did the model predict best? Which did it overpredict the most? (c) Interpret the residual for the 8 km journey in context.
Worked solution
(a)
| Distance (km) | 3 | 5 | 8 | 10 | 12 |
|---|---|---|---|---|---|
| ŷ (min) | 12.30 | 17.50 | 25.30 | 30.50 | 35.70 |
| Residual (min) | 0.70 | 1.50 | −3.30 | 0.50 | −1.70 |
For example, at 8 km: ŷ = 4.5 + 2.6(8) = 25.30 minutes, and residual = 22 − 25.30 = −3.30 minutes.
(b) Best: the 10 km journey, whose residual (0.50 min) is closest to 0. Most overpredicted: the 8 km journey, with the most negative residual (−3.30 min).
(c) The 8 km journey took 3.30 minutes less than the model predicted, so the model overpredicted its journey time.
Suggested mark points (3): 1 for all five predicted values and residuals correct; 1 for 10 km (best) and 8 km (most overpredicted); 1 for an interpretation that names journey time, the direction (less than predicted, overpredicted) and the units.
Question 5 (constructed response · core)
A fictional school models its daily heating energy use (kWh) from the mean outdoor temperature (°C) with ŷ = 210 − 4.2x. The data cover days from 2 °C to 24 °C.
(a) On a day with a mean temperature of 15 °C, the residual was −9.6 kWh. Find the actual energy use that day. (b) Interpret the residual in (a) in context. (c) On a 22 °C day the school used 125 kWh. Find the residual. Did the model underpredict or overpredict? (d) Explain why a residual of 0 for one day does not show that the model predicts every day well.
Worked solution and suggested Marlbridge rubric
(a) ŷ = 210 − 4.2(15) = 210 − 63 = 147 kWh. Observed = ŷ + residual = 147 + (−9.6) = 137.4 kWh.
(b) On the 15 °C day the school used 9.6 kWh less energy than the model predicted. The model overpredicted that day’s energy use.
(c) ŷ = 210 − 4.2(22) = 210 − 92.4 = 117.6 kWh. Residual = 125 − 117.6 = +7.4 kWh. The model underpredicted: the school used 7.4 kWh more than predicted.
(d) A residual of 0 means only that this one day’s point lies exactly on the line. Other days can have large residuals. To judge the model as a whole, you need to look at all the residuals, for example in a residual plot.
| Point | What earns it |
|---|---|
| 1 | Correct ŷ = 147 kWh and observed = 137.4 kWh, using y = ŷ + residual |
| 1 | Interpretation: 9.6 kWh less than predicted (overpredicted), in context |
| 1 | Correct residual +7.4 kWh and “underpredicted” |
| 1 | Explains that one residual describes only one point |
A common error in (a) is 147 + 9.6 = 156.6 kWh, which ignores the negative sign; it does not earn point 1.
Question 6 (constructed response · stretch)
A fictional road-safety test measured the stopping distance (m) of one car on a wet track at nine speeds (km/h).
| Speed (km/h) | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | 110 |
|---|---|---|---|---|---|---|---|---|---|
| Stopping distance (m) | 14 | 22 | 31 | 42 | 55 | 70 | 86 | 104 | 124 |
Technology gives r = 0.99 and ŷ = −35.24 + 1.373x. The residual plot is shown.
(a) Calculate the residual at 70 km/h and check that it matches the plot. (b) Describe the pattern in the residual plot. (c) A student says: “r = 0.99, so a linear model is clearly appropriate.” Do you agree? Explain. (d) The test team wants to estimate the stopping distance at 120 km/h. Would the linear model probably give a value that is too high or too low? Explain using the residual plot.
Worked solution and suggested Marlbridge rubric
(a) ŷ = −35.24 + 1.373(70) = −35.24 + 96.11 = 60.87 m. Residual = 55 − 60.87 = −5.87 m. This matches the lowest dot in the plot, just below −5 at 70 km/h.
(b) The residuals form a clear U-shape: positive at low speeds (about +8 m at 30 km/h), negative at middle speeds (down to about −6 m at 70 km/h), and positive again at high speeds (about +8 m at 110 km/h). They are not randomly scattered.
(c) No. The value r = 0.99 shows a very strong association, but it does not show that the form is linear. The curvature in the residual plot shows that a straight line is not the most appropriate model. You can see the curve in the data too: the extra stopping distance for each 10 km/h rises from 8 m (30 to 40 km/h) to 20 m (100 to 110 km/h), so the relationship bends upwards.
(d) Too low. At the highest speeds the residuals are positive and rising, so the model already underpredicts there. The curve is getting steeper while the line keeps the same slope, so at 120 km/h (an extrapolation) the line’s prediction, ŷ = −35.24 + 1.373(120) = 129.52 m, is likely to be less than the true stopping distance.
| Point | What earns it |
|---|---|
| 1 | Correct ŷ (60.87 m) and residual (−5.87 m) |
| 1 | Describes the U-shape (positive, negative, positive) as a curved, non-random pattern |
| 1 | Disagrees, explaining that a high r does not establish a linear form and citing the residual plot |
| 1 | Says “too low”, linking positive residuals at high speeds to underprediction (and noting the extrapolation) |
In (c), “No, because the residual plot is curved” earns the point only if it also addresses why r = 0.99 is not enough.
Question 7 (explanation · stretch)
At a fictional gym, a linear model predicts the calories burned in a session from the minutes spent on a rowing machine. The residual plot shows random scatter around 0. Three students comment:
- Ali: “My residual is −6 kcal, so the model underestimated the calories I burned.”
- Bea: “My residual is 0, so this model is perfect for everyone.”
- Cem: “The residual plot shows no pattern, so every residual must be small.”
For each student, explain the mistake and give a correct statement.
Model answer and suggested Marlbridge rubric
Ali. A residual of −6 kcal means Ali’s actual calories were 6 kcal below the prediction (observed − predicted = −6). The model overestimated (overpredicted) the calories Ali burned.
Bea. A residual of 0 means only that Bea’s own point lies exactly on the line: the model predicted her calories exactly. Other people can have large positive or negative residuals, so it says nothing about how well the model fits everyone.
Cem. Random scatter in a residual plot tells you that the form is linear, so a linear model is appropriate. It does not tell you how big the residuals are. If the association is weak, the residuals can be large and still random. To judge their size, look at the values on the residual axis.
| Point | What earns it |
|---|---|
| 1 | Ali: negative means actual below predicted, so the model overestimated |
| 1 | Bea: a zero residual describes one point only |
| 1 | Cem: random scatter shows a linear form, not small residuals |
How did you do?
- Q1 or Q4(a) wrong: revisit Worked example 1 in the study guide. Always calculate observed − predicted.
- Q2, Q4(c) or Q7 (Ali) wrong: revisit “What the sign tells you”.
- Q5 wrong: work through Worked example 2. Use y = ŷ + residual and keep the sign.
- Q3 or Q6 wrong: revisit “Residual plots” and Worked example 3.
- Q7 (Bea or Cem) wrong: re-read the misconceptions list.
Then tick off the topic checklist.