Understand · explore · practise
Practise paired data, correlation, residuals, regression coefficients, prediction reliability and causal limitations with original mixed questions and worked solutions.
Before you startComplete the correlation chapter or use the questions diagnostically.
01 / Decide what the question asks
Identify the observed pairs, the requested response and the model’s intended use.
Check units, range and context after calculating. A correct substitution does not justify an unsupported conclusion.
Try each question on paper before opening its hint or solution. These are original questions, not copied examination items. The optional task selector gives short method checks at your own pace.
Constructed pairs are (2,27), (4,20), (6,19), with fitted line ŷ = 30 − 2x. Predict y when x = 3.
Substitute: ŷ = 24. The input is inside the observed 2–6 range, but an individual response need not equal its fitted value.
02 / Read the evidence
Plot (2,27), (4,20), (6,19), with x horizontal. Describe the association.
Retain each pair and use increasing numeric scales.
Place the three points at their stated coordinates. They show negative association: larger x is associated with smaller y in this tiny constructed sample. Three observations provide limited evidence about a broader population.
Two different observations are (4,20) and (4,25). Must one be removed because x is repeated?
A scatterplot can have several responses at the same explanatory value.
No. They are distinct pairs unless record-level evidence shows a duplicate or error. Plot both points vertically above x = 4.
A U-shaped cloud has little linear correlation. Can you conclude the variables have no relationship?
Linear correlation describes straight-line association.
No. The curve can show a strong nonlinear relationship despite weak or zero linear correlation.
03 / Distinguish observation and prediction
Find the fitted values at x = 2, 4 and 6.
Apply the same line to each input.
They are 26, 22 and 18 respectively. The recorded responses remain 27, 20 and 19.
Find the three residuals and their sum of squares.
Residual = observed y − fitted y.
Residuals are 1, −2 and 1. Their squared sum is 1 + 4 + 1 = 6. Residuals can cancel in their ordinary sum, so use squares for this loss.
Does y-on-x least squares minimise the perpendicular distances from the points to the line?
The response error holds x fixed.
No. It minimises squared vertical response residuals. Perpendicular distance is a different fitting objective.
04 / Interpret levels and changes
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Suppose x is time in hours and y is temperature in °C in a fitted model ŷ = 30 − 2x. Interpret the gradient.
Name both units and the direction of change.
The fitted temperature decreases by 2°C per additional hour within the model’s intended setting. This is a modelled trend, not a guaranteed individual or causal effect.
For that model, calculate the predicted change from x = 3 to x = 5.
Multiply the input change by the gradient.
Δŷ = −2(5 − 3) = −4. Predicted levels are 24 and 20, so the response falls by 4 units; do not add the intercept to the change.
The data only cover 2–6 hours. Does the intercept 30 prove that 30°C was measured at time zero?
Zero is outside the observed explanatory range.
No. It is the fitted value at x = 0, an extrapolation here. It does not establish an actual measurement at zero.
05 / Check where the input lies
Predict at x = 3 and x = 8, and classify each use.
Substitute, then compare x with 2–6.
At 3, ŷ = 24: interpolation. At 8, ŷ = 14: extrapolation. Reliability also depends on scatter, context and whether a line is suitable.
A different model has observed x = 10–20 and observed y = 0–100. Its prediction at x = 25 is y = 50. Is this interpolation?
Check the explanatory range.
No. x = 25 lies outside 10–20. Being inside the observed response range does not make a prediction interpolation.
06 / Choose the requested response
Which x makes the stated prediction ŷ = 30 − 2x equal to 20?
The question asks for a point on this stated line.
x = 5. This is valid algebra on the supplied model, not a claim that the separate x-on-y regression must give 5.
For the same constructed pairs the x-on-y fit is x̂ = 252/19 − (8/19)y. Predict x at y = 20.
Use the equation that predicts x.
x̂ = (252 − 160)/19 = 92/19 ≈ 4.842. It differs from the inverse target value 5 because the fits minimise different response errors.
07 / Audit claims and inclusion choices
A school finds students who attend revision clubs have higher marks. Does this prove the club caused the difference?
Consider selection and other influences.
No. Motivation, prior attainment, available time and other factors may differ. The association alone does not isolate a causal effect.
A correctly recorded observation has a large residual. Is deleting it justified simply because correlation becomes stronger?
A stronger-looking result is not evidence of error.
No. Investigate the observation and context, retain raw records, justify any narrower scope and report sensitivity to the choice.
Write a careful statement for ŷ = 24 at x = 3 using the three constructed pairs.
Include the model output, range and limitations.
The fitted model predicts y = 24 at x = 3. The input lies within the observed 2–6 range, but this small constructed example does not validate a real population or guarantee an individual response.
08 / Use the chapter as a connected method
Preserve pairs, describe the pattern, choose the response, calculate carefully and explain the limitations. Revisit the linked lessons if a question exposed a gap, then retry without opening the solutions.
Section 1 of 8 · Decide what the question asks