Fit a least-squares line using technology, interpret the slope and intercept in context, and predict while distinguishing interpolation from extrapolation.

How to fit the least-squares regression line with technology, interpret its slope and intercept in real units, predict with the equation, and judge interpolation against extrapolation.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

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What this dot point is asking
Fitting the line
Interpreting the coefficients
Predicting
Why extrapolation fails

What this dot point is asking

You must fit the line, write its equation, interpret the two coefficients in context with units, use the line to predict, and judge the reliability of each prediction.

Fitting the line

The least-squares line is the straight line for which the sum of the squared vertical distances from the data points is as small as possible. In this course you fit it with technology, entering the explanatory list and the response list and reading off $a$ and $b$ .

Interpreting the coefficients

Interpretation must use the real variable names and units.

Slope $b$ . For each one-unit increase in the explanatory variable, the response variable changes by $b$ units on average. State whether it rises or falls.
Intercept $a$ . The predicted response when the explanatory variable is zero. This is only meaningful if $x = 0$ is sensible in the context.

Predicting

Substitute an $x$ value into the equation to predict $y$ .

Interpolation predicts for an $x$ inside the range of the data. It is usually reliable.
Extrapolation predicts for an $x$ outside the range. It is risky, because the linear pattern may not continue.

Worked example

Fitting, interpreting and predicting

For ten cars, age ( $x$ , years) and value ( $y$ , thousands of dollars) give the least-squares line $y = 28 - 2.5x$ , with ages ranging from $1$ to $9$ years.

Slope: $b = -2.5$ , so the value falls by about $2500 for each extra year of age.

Intercept: $a = 28$ , the predicted value of a brand-new car (age $0$ ) is \ $28\,000. Age$ 0$ is just outside the data, so treat this as a borderline estimate.

Predict the value of a $5$ -year-old car: $y = 28 - 2.5 \times 5 = 28 - 12.5 = 15.5$ , that is \ $15\,500. Because$ 5 $lies inside the range$ 1 $to$ 9$, this is interpolation and is reliable.

Predict the value of a $20$ -year-old car: $y = 28 - 2.5 \times 20 = 28 - 50 = -22$ , that is minus \ $22\,000. An impossible negative value, because$ 20$ is far outside the data; this is unreliable extrapolation.

Why extrapolation fails

A least-squares line is only evidence for the pattern over the range that produced it. Pushed far beyond that range it can give impossible answers such as a negative value or a score above the maximum, which is why predictions must always be checked against the data range and against common sense.