Interpreting Regression Output — FRQ Walkthrough
A free-response question in exam style, fully worked — then the scoring logic behind every point.
The question · original, exam-style
A gardener records, for 12 tomato plants, the hours of direct sun each plant gets per day (x) and its total fruit yield in kilograms (y). A least-squares regression gives the model ŷ = 0.4 + 0.55x, with correlation r = 0.82. (a) Interpret the slope in context. (b) Interpret r² in context. (c) Predict the yield for a plant that gets 6 hours of sun, and comment on whether predicting the yield for a plant getting 15 hours of sun would be reasonable.
Written by Marlio as an analog of the exam format. Not a College Board question.
Model response
(a) Slope: The slope 0.55 means that for each additional hour of direct sun per day, the model predicts an increase of about 0.55 kilograms in a plant's total fruit yield, on average.
(b) r²: Since r = 0.82, r² ≈ 0.67. About 67% of the variation in fruit yield among these plants is explained by the linear relationship with hours of direct sun; the remaining 33% is due to other factors.
(c) Prediction: ŷ = 0.4 + 0.55(6) = 3.7 kilograms. Predicting for 15 hours would be extrapolation — 15 hours is well outside the range of sun exposure actually observed, so the linear model may not hold there and the prediction would be unreliable.
How it’s scored — point by point
- POINT 1
Interprets the slope in context (part a).
The slope must be described as a predicted, average change in y per one-unit increase in x, in the real units (kilograms per extra hour of sun). Saying the yield "is" 0.55 kg, or dropping "predicted / on average," loses full credit.
- POINT 2
Interprets r² in context (part b).
Compute r² from r and read it as the percent of variation in the response (yield) explained by the linear model with x. Interpreting r itself, or saying "67% of the points fall on the line," does not earn the point.
- POINT 3
Correct prediction (part c).
Substitute x = 6 correctly to get ŷ = 3.7 kg. Showing the substitution keeps this point even if a later part goes wrong.
- POINT 4
Recognizes extrapolation (part c).
Identify that x = 15 is outside the observed range of x, so the model may not apply and the prediction is unreliable. Simply computing ŷ at 15 without flagging extrapolation misses the idea the question is testing.
Common point-losers
- Interpreting the slope as a fixed amount ("yield is 0.55 kg") instead of a predicted change per hour.
- Interpreting r instead of r², or saying "67% of the points lie on the line."
- Forgetting the units (kilograms, hours) in the interpretations.
- Plugging in x = 15 and reporting a number without noting it is extrapolation.
- Claiming more sun causes higher yield as if the study had proved causation.
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