11.5 Prediction

Step 1 — Define interpolation: Interpolation is the process of predicting inside the observed \(x\) values in the data. This is what a regression line is for, and predictions made this way are justified by the data.

Step 2 — Define extrapolation: Extrapolation is the process of predicting outside the observed \(x\) values in the data. The arithmetic still works — you substitute a number and get a number back — but the result is not justified by the data, because the line was never given a chance to observe that region.

Answer: Interpolation predicts inside the observed range of \(x\); extrapolation predicts outside it, and only interpolation is supported by the data.

Step 1 — Check the observed range first. The line was built from \(x\)-values between 3 and 20, and \(x = 12\) lies inside that range, so this prediction is interpolation and is justified.

Step 2 — Substitute \(x = 12\) into the equation.

Step 3 — Word the prediction honestly. We predict that when \(x = 12\), the mean value of \(y\) is 63, on average.

Answer: \(\hat{y} = 63\), on average (a justified interpolation).

Step 1 — Compare \(x = 25\) against the observed range. The data cover \(x\)-values from 3 to 20, and 25 is greater than 20, so the prediction falls outside the observed range.

Step 2 — Classify and judge. Predicting at \(x = 25\) is extrapolation. Even though the equation will happily return a number — \(\hat{y} = 15 + 4(25) = 115\) — that number is not reliable, because no data were collected anywhere near \(x = 25\). The real relationship may bend, flatten, or change slope there, and the line has no way of knowing.

Answer: No, the prediction at \(x = 25\) is not justified: 25 lies outside the observed range of 3 to 20, so it is an unjustified extrapolation.

Step 1 — Check 1: the scatter plot shows a linear pattern. If the pattern is not linear, a straight line is simply the wrong model. This check comes from Section 11.2.

Step 2 — Check 2: \(r\) is significant. If \(r\) is not significant, there is insufficient evidence of any linear relationship in the population, and the relationship may be nothing more than sampling accident. This check comes from Section 11.4.

Step 3 — Check 3: \(x\) is inside the observed range of the data. If \(x\) lies outside the smallest-to-largest observed values, the prediction is extrapolation and is not justified. This check comes from Sections 11.3 and 11.5.

Step 4 — Note that all three must pass. Any single failure stops the prediction; passing two of three is not enough.

Answer: (1) linear scatter plot (Section 11.2), (2) significant \(r\) (Section 11.4), (3) \(x\) within the observed range (Sections 11.3 and 11.5) — all three must hold before predicting.

Step 1 — Identify what kind of prediction this is. The data cover \(x\) from 40 to 90, and \(x = 95\) is greater than 90, so the prediction is extrapolation regardless of how the answer looks.

Step 2 — Explain why plausibility does not rescue it. A plausible-looking answer is exactly why extrapolation is dangerous: the equation returns numbers with no warning label, and nothing about the arithmetic signals that the model has left its data behind. Plausibility is not a test of validity — checking the range is.

Step 3 — State the underlying reason. The line was fitted using only points between 40 and 90. Outside that interval, the true relationship may curve or change behaviour entirely, and the fitted line carries no information about it.

Answer: The prediction is not justified because \(x = 95\) lies outside the observed range of 40 to 90; a plausible-looking extrapolation is still an extrapolation.

Step 1 — Identify the error as extrapolation carried too far. The regression was built from children aged 2 to 10. Age 40 is far outside that range, so the prediction is a massive extrapolation, not a legitimate use of the line.

Step 2 — Explain why the result is impossible. Children grow roughly linearly over ages 2 to 10, so the fitted line has a positive slope. Extending that straight line to age 40 forces it to keep climbing at the same rate, producing a "height" of 11 feet — a value no human can have.

Step 3 — Draw the general lesson. Human growth stops in the late teens, but the line knows nothing about that; it only saw children aged 2 to 10. Extrapolating a linear trend beyond its data can produce answers outside the set of possible values altogether, and the equation gives no signal that anything has gone wrong.

Answer: The prediction fails because age 40 is far outside the observed range of 2 to 10; extending a linear growth trend indefinitely produces the impossible height of 11 feet.

Step 1 — Check the range. In the third exam/final exam example, the observed \(x\)-values run from 65 to 75. Since 70 lies inside that range, this is interpolation and the prediction is justified.

Step 2 — Substitute \(x = 70\) into the equation.

Step 3 — Word the prediction honestly. We predict that statistics students who earn a grade of 70 on the third exam will earn a grade of 164.59 on the final exam, on average.

Answer: \(\hat{y} = 164.59\), on average.

Step 1 — Identify what is wrong with the sentence. Two problems: the word "proves" overstates what a regression line can do, and the sentence promises a specific outcome for an individual student, ignoring the natural variation among students who score the same on the third exam.

Step 2 — Recall what the line actually estimates. The regression line estimates the mean final exam score for all students scoring 72, not the score of any particular student. For example, two students with identical third-exam scores may earn different final scores.

Step 3 — Rewrite the sentence. We predict that students who score 72 on the third exam will score 174.2 on the final exam, on average.

Answer: "We predict that students who score 72 on the third exam will score 174.2 on the final exam, on average." — naming the group estimate and adding "on average" makes the claim honest.

Step 1 — Recall what the intercept represents. The y-intercept is the predicted value of \(y\) at \(x = 0\).

Step 2 — Note that \(x = 0\) is usually outside the observed range. Regression lines are built from data covering some range of \(x\)-values, and \(x = 0\) typically lies outside that range. So interpreting the intercept is itself an act of extrapolation, which the data do not support.

Step 3 — Give a concrete example. In the grocery-spending example, the intercept was $38 per week for a household of zero people — meaningless as a real quantity. It functions merely as a fitting constant that positions the line correctly over the observed data.

Answer: The y-intercept predicts \(y\) at \(x = 0\), which usually lies outside the observed range of the data; the intercept is then just a fitting constant, not an interpretable quantity.

Step 1 — Recall the role of significance. A non-significant \(r\) means we do not have sufficient evidence of a linear relationship in the population; the apparent pattern could be sampling accident.

Step 2 — Apply the checks. One of the three required checks — that \(r\) be significant — has failed. When \(r\) is not significant, the line CANNOT be used to model the relationship, and therefore no prediction should be made from it.

Step 3 — Conclude. Interpolation being "safe" in terms of range is irrelevant here: without a demonstrated linear relationship, there is no trustworthy line to interpolate along.

Answer: No, interpolation is not allowed. Since \(r\) is not significant, there is insufficient evidence of a linear relationship, and the regression line cannot be used for prediction at all.

Step 1 — Recall what the line estimates. A regression line estimates the mean value of \(y\) for each given \(x\), not the exact \(y\) of any single observation.

Step 2 — Explain the variation around the line. Many individuals share the same \(x\)-value but have different actual \(y\)-values — for instance, two students both scoring 69 on the third exam scored 151 and 159 on the final. A single \(x\) corresponds to a whole spread of real \(y\) values, and the line reports only the centre of that spread.

Answer: Because individuals with the same \(x\)-value have varying \(y\)-values, the regression line estimates only the mean \(y\) for that group — hence every honest prediction ends with "on average."