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Interpret regression gradients and intercepts with units, distinguish predicted levels from changes and transform fitted equations when units or reference values change.
Before you startStraight-line equations, regression predictions and units.
01 / Read the fitted equation in context
For ŷ = a + bx, a is the predicted response at x = 0; b is the predicted response change per one unit of x.
These are model interpretations, not automatically causal effects.
Write down what x and y measure before substituting numbers. A regression prediction summarises a fitted relationship; it is not a guaranteed outcome for an individual observation.
Fictional fitted delivery model: predicted duration t = 8 + 2.5d, with d in km and t in minutes. Compare with a reference distance of 4 km.
At 0 km, the model predicts 8 minutes. Relative to 4 km, the distance change is −4 km and the predicted duration change is −10 minutes. A fitted intercept need not describe a meaningful observed trip.
02 / Attach units to the gradient
The gradient is 2.5 minutes per kilometre
It is not simply “2.5 minutes”.
An additional kilometre changes the fitted prediction by 2.5 minutes
This is a predicted difference within the model.
A 4 km delivery is predicted to take 18 minutes
The gradient alone is not the total duration.
A model predicts mass m in grams from length l in cm: m = 6 + 0.8l. Interpret 0.8.
Use grams per centimetre.
The predicted mass increases by 0.8 grams for each additional centimetre of length, according to the fitted model.
Does a fitted gradient of 0.8 prove physically increasing every object’s length by 1 cm would increase its mass by 0.8 g?
A regression may summarise observational differences between objects.
No. That intervention claim needs additional causal and physical justification. The coefficient describes the fitted prediction.
03 / Interpret zero carefully
For t = 8 + 2.5d, what is the intercept and its unit?
Set d to zero.
8 minutes: the model’s predicted duration at zero kilometres. Whether that has a practical interpretation depends on the context and observed range.
The delivery data cover distances 2–12 km. Does the fitted intercept establish an observed eight-minute zero-distance delivery?
Zero lies outside that range.
No. It is a fitted value at an unobserved distance, not a measurement or a proven fixed overhead.
04 / Subtract predictions to compare trips
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Predictions are 18 and 33 minutes
Calculate the two levels.
Difference = 33 − 18 = 15 minutes
This is not the 10 km trip’s total time.
Equivalently, 2.5×(10 − 4) = 15
The same intercept occurs in both predictions.
For m = 6 + 0.8l, estimate the predicted mass change when length rises from 12 to 17 cm.
Multiply the length change by the gradient.
0.8×5 = 4 grams. Do not add the intercept to a change.
In the delivery model, distance falls from 8 to 3 km. Find the predicted duration change.
Keep the sign of the distance change.
2.5×(3 − 8) = −12.5 minutes: a predicted decrease of 12.5 minutes.
05 / Transform the equation, not just the label
Express t = 8 + 2.5d with duration h measured in hours and d still in km.
h = t/60. Divide both coefficients.
h = 8/60 + (2.5/60)d = 2/15 + d/24. The gradient is hours per km.
Use distance D in metres instead of d in kilometres, keeping time in minutes.
d = D/1000.
t = 8 + 0.0025D. The intercept remains 8 minutes; the gradient is 0.0025 minutes per metre.
06 / Read a negative gradient as a decrease
A fictional fitted resale model is v = 1200 − 90a, with value in pounds and age in years. Interpret −90.
Name a predicted change and its units.
Each additional year of age is associated with a £90 decrease in predicted resale value under this model. It is not a guarantee for every item.
The same equation gives a negative resale value for sufficiently large ages. What does that suggest?
Do not treat the fitted line as universally valid.
The linear model should not be extended indefinitely. Check the observed range and physical context; an algebraic prediction can be inappropriate.
07 / Move the reference point
Then d = u + 4
Substitute the original variable correctly.
t = 8 + 2.5(u + 4) = 18 + 2.5u
The new intercept is the prediction at 4 km.
The gradient is still 2.5
A shift changes the reference zero, not the change per kilometre.
For m = 6 + 0.8l, let z = l − 10. Rewrite the model.
Substitute l = z + 10.
m = 14 + 0.8z. The new intercept 14 g is the predicted mass at length 10 cm.
Two equations use minutes/km and hours/metre. Can their gradient numbers be compared directly as if the units match?
Put both variables on common scales first.
No. Convert units before comparing numerical gradients. A small coefficient can reflect a unit choice rather than a weak relationship.
08 / Audit the statement
A least-squares model with an intercept has x̄ = 5, ȳ = 18 and gradient 2. Find its intercept.
The fitted line passes through the mean point.
a = 18 − 2×5 = 8. This property assumes ordinary unweighted least squares with a fitted intercept.
The line predicts y = 18 at x = 4. A recorded response is 21. Must the record be wrong?
A fitted line allows residuals.
No. The residual is 21 − 18 = 3. Investigate a value only with relevant evidence, not simply because it differs from a prediction.
09 / Say exactly what changes
Interpret the intercept at the chosen zero and the gradient in response units per explanatory unit. For a difference, multiply the x change by the gradient. Convert the equation when units change and keep predictions distinct from observations and causal claims.
Section 1 of 9 · Read the fitted equation in context