AI HL · AHL 8
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Sum of squared residuals and R²
AHL 8.2
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
The dots are data. Choose a model: the orange curve is the fitted model and the teal segments are the residuals. Compare SSres and R² for all five, then look at what each model does beyond the data.
| Model | SSres | R² |
|---|
For the exponential and the power model, R² is the R² of the straight line through (x, ln y) or (ln x, ln y), as the calculator gives it. SSres is always measured on y.
Choose a model.
Drag the two orange handles to place a line among the points. Each residual is a vertical segment, and its square is drawn beside it. Make the squares add up to as little as you can.
SSres =
Drag a handle.
A coach company models the fuel used, F litres, on a journey of d km by \(F=2.1+0.34d\). The table shows six recorded journeys. Find the residual for each journey (data value minus model value), then find \(SS_{res}\).
Have a go first. Then open the working one line at a time.
\(\hat F=2.1+0.34d\): 5.5, 8.9, 14, 19.1, 25.9, 32.7
Put each d into the model. \(\hat F\) is the value the model predicts.
\(F-\hat F\): 0.4, −0.2, 0.6, −0.5, 0.1, 0.7
Data minus model. Positive means the point is above the line.
0.16, 0.04, 0.36, 0.25, 0.01, 0.49
Squares are never negative, so residuals above and below the line cannot cancel.
\(SS_{res}=\) \(0.16 + 0.04 + 0.36 + 0.25 + 0.01 + 0.49 = 1.31\)
A smaller \(SS_{res}\) means the model is closer to the data.
The table gives the value, £y, of a used laptop of age x years. Fit a linear model on the calculator and find the coefficient of determination \(R^2\) to 3 significant figures.
Have a go first. Then open the working one line at a time.
From the calculator: linear regression: a = −128, b = 1260
The model is \(y=-128x+1260\).
r = −0.985
The calculator gives the Pearson correlation coefficient r. It is negative because y falls as x rises.
\(R^2=r^2=(−0.985)^2\) = 0.971
For a linear model \(R^2=r^2\). The calculator also shows \(R^2\) directly.
About 97.1% of the variation in y is explained by the linear model
\(R^2\) is between 0 and 1. Close to 1 is a good fit.
An agricultural scientist records the crop yield, y tonnes per hectare, when 10x kg of fertiliser per hectare is used. Find \(R^2\) for a linear model and for a quadratic model. Which fits better?
Have a go first. Then open the working one line at a time.
From the calculator: linear regression: \(R^2\) = 0.391
A low value: the points rise and then fall, so a line does not fit.
From the calculator: quadratic regression: a = −0.0364, b = 0.808, c = 2.84, \(R^2\) = 0.982
The calculator shows \(R^2\) with the coefficients.
0.982 > 0.391
The quadratic (teal curve) is much closer to the points than the line (orange). The larger \(R^2\) is the better fit, and the shape agrees with the context.
The flow, y m³/s, of a river is recorded each day for eight days after a storm (x = 0 to 7). Fit a cubic model and state \(R^2\) to 3 significant figures.
Have a go first. Then open the working one line at a time.
From the calculator: cubic regression: a = 0.287, b = −3.01, c = 7.89, d = 5.64
Model: \(y = 0.287x^3 - 3.01x^2 + 7.89x + 5.64\)
\(R^2\) = 0.966
Read it from the same results screen. About 96.6% of the variation in y is explained.
Quadratic \(R^2\) = 0.000549
The data fall, then rise: two bends. A quadratic cannot do that, so the cubic is far better.
A researcher fits a linear model to data on the outdoor temperature and the amount of gas used to heat a house. Pearson's product-moment correlation coefficient is \(r=-0.934\). Find \(R^2\) and say what it means.
Have a go first. Then open the working one line at a time.
\(R^2=r^2=(-0.934)^2\) = 0.872
This works for a linear model only.
About 87.2% of the variation in gas used is explained by the linear model
The negative sign of r only shows the direction. \(R^2\) shows how much of the variation the model explains.
The table gives the distance x AU of each of the eight planets of a distant star from the star, and the orbital period T years. Find a regression model \(T=ax^3+bx^2+cx+d\), giving a, b, c and d to 3 significant figures, and state \(R^2\).
Have a go first. Then open the working one line at a time.
From the calculator: cubic regression: a = −0.00767, b = 0.255, c = 0.703, d = −0.127
a is small because x³ is large. Keep the 3 significant figures: the leading zeros do not count.
\(T = -0.00767x^3 + 0.255x^2 + 0.703x - 0.127\)
Quote each coefficient to 3 significant figures.
\(R^2\) = 1.00
A very good fit inside the data range.
Use the cubic model from the last example to predict the orbital period of a planet at 30 AU from its star. Comment on the reliability of your prediction.
Have a go first. Then open the working one line at a time.
T(30) = 43.7 years
Use the stored regression function, not the rounded coefficients.
Data range: 0.25 to 9 AU. 30 is outside it.
A prediction inside the range is interpolation. Outside it is extrapolation, and it is much less reliable.
The model reaches a maximum at x = 23.5 AU, then falls
Beyond that point the model says a planet further out takes less time to orbit. That cannot be true.
Not reliable
x = 30 is more than 3 times the largest x in the data, and the cubic turns downwards after 23.5 AU. A high \(R^2\) inside the data does not make the model reliable outside it.