Predict from a fitted model: interpolation and extrapolation

Predicting from a fitted model

A fitted model predicts the response by substituting an explanatory-variable value into y = a + bx. The prediction is interpolation when that value is within the inclusive observed range and extrapolation when it is outside

1. Identify the variables Use x for the explanatory variable and y for the response variable in the fitted model \(\displaystyle x=\text{EV},\quad y=\text{RV}\)
2. Substitute the explanatory value Replace x with the stated explanatory value before simplifying the model \(\displaystyle \hat{y}=12+3\left(5\right)=27\)
3. Classify the range position A value inside or at an endpoint of the observed explanatory range gives interpolation; a value outside gives extrapolation \(\displaystyle 20\le x\le40\Rightarrow\text{interpolation}\)
4. Judge the prediction Interpolation uses an input within the observed explanatory range; extrapolation extends beyond it. Check output plausibility and model suitability separately. Interpolation does not guarantee an accurate individual prediction \(\displaystyle x<20\text{ or }x>40\Rightarrow\text{extrapolation}\)

Worked example

A fitted model predicts water use y in litres from garden area x in square metres: y = 24 + 0.8x. The model was fitted using x values from 10 to 40. Predict y when x = 25 and classify the prediction

1. \(\displaystyle 0.8\times25=20\)Multiply the slope by the explanatory value
2. \(\displaystyle \hat{y}=24+20=44\)Add the intercept to get a predicted water use of 44 litres
3. \(\displaystyle 10\le25\le40\Rightarrow\text{interpolation}\)The explanatory value 25 is within the observed range from 10 to 40

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