Data transformation to linearise: VCE General Mathematics Unit 3 Data analysis

What this dot point is asking

Under "Investigating and modelling linear associations" the Study Design lists "data transformation and its use in transforming some forms of non-linear data to linearity using a square, logarithmic (base 10) or reciprocal transformation (applied to one axis only)" and "interpretation and use of the equation of the least squares line fitted to the transformed data to make predictions". You need to:

1. recognise from a scatterplot, or from a patterned residual plot, that an association is non-linear;
2. choose a transformation ($x^2$, $\log_{10} x$, $\frac{1}{x}$, $y^2$, $\log_{10} y$ or $\frac{1}{y}$) that straightens the data;
3. fit a least squares line to the transformed data with CAS and write its equation in terms of the transformed variable;
4. use that equation to predict, transforming the input first and undoing any transformation on the response at the end, and to solve backwards when the response is given;
5. compare models using $r^2$ and residual plots.

Transformation questions appear in Examination 1 every year, usually as a pair on one data set. In 2023 Examination 1 Questions 11 and 12 used plantation tree heights ($60\%$ and $76\%$ correct), and in 2025 Examination 1 Questions 11 and 12 used life expectancy and doctors ($81\%$ and $61\%$).

The answer

Why transform at all?

A least squares line assumes the association is linear. When the scatterplot curves, a straight line fits badly in a systematic way: it predicts too high in some regions and too low in others. The residual plot shows this as a pattern (an arch or a U) rather than random scatter, even when $r^2$ looks respectable.

For the 11 plantation trees in 2023 Examination 1, the raw least squares line (height $= 2.14 + 0.63 \times$ age) has $r^2 = 0.818$. Its residuals, in order of age, are negative for ages 4 and 5 (about $-1.2$ and $-1.3$), close to zero at 6, positive from 7 to 11 (up to $+1.25$), then negative again from 12 to 14: a clear arch. The reciprocal model's residuals, by contrast, are all within about 1 of zero with no pattern. The line is the wrong form, so transforming is the fix, not a better line.

A transformation re-expresses one variable on a new scale so that the points lie close to a straight line. You then fit an ordinary least squares line to the new variable.

The six moves (one axis only)

The Study Design says "applied to one axis only": in any one model you transform $x$ or $y$, not both. A log or reciprocal transformation needs every value of that variable to be positive.

Choosing a transformation: the circle of transformations

Match the shape of the scatterplot to one quarter of a circle. The transformations beside that quarter are the ones that pull the curve straight.

The logic is the same on every arc. If the curve bends so that it bulges towards large $x$ (bottom right and top right), stretch the $x$ axis with $x^2$. If it bulges towards small $x$ (top left and bottom left), compress $x$ with $\log_{10} x$ or $\frac{1}{x}$. The same rule applies to $y$: a bulge towards large $y$ calls for $y^2$, and a bulge towards small $y$ calls for $\log_{10} y$ or $\frac{1}{y}$.

The two exam data sets fit this exactly. Tree height against age rises steeply and levels off (top left), and VCAA used $\frac{1}{\text{age}}$ and $\log_{10}(\text{age})$. Life expectancy against doctors rises slowly then steeply (bottom right), and VCAA used $(\text{doctors})^2$ and $\log_{10}(\text{life})$.

Usually more than one transformation works. Choose between them by fitting each and comparing:

• the coefficient of determination $r^2$ (higher is better), and
• the residual plot (random scatter, no pattern).

Squaring age made the tree data worse ($r^2$ fell to 0.693), because it stretches the wrong end: that curve needed compressing. Choosing a transformation from the wrong arc is easy to detect this way.

Fitting the line on CAS

1. Enter the two variables as lists (for example age and height).
2. Make a new list holding the transformed values: $\frac{1}{\text{age}}$, $\log_{10}(\text{age})$ or $\text{age}^2$ (or the same for the response). On both TI-Nspire and ClassPad, log is base 10.
3. Run linear regression with the new list in place of the variable you transformed. If $x$ was transformed, the new list is the $X$ list; if $y$ was transformed, the new list is the $Y$ list.
4. Write the equation using the transformed variable name, for example height $= 13.04 - 40.22 \times \frac{1}{\text{age}}$, never height $= 13.04 - 40.22 \times$ age.

Predicting from a transformed model

The fitted equation is linear in the transformed variable, so prediction is careful substitution:

• $x$ transformed: transform the given $x$ first, then substitute. For life $= 63.12 + 2.842 \times (\text{doctors})^2$ at 2 doctors, substitute $2^2 = 4$.
• $y$ transformed: substitute $x$, which gives the transformed response, then undo it: $10^{\text{value}}$ for a log, $\sqrt{\text{value}}$ for a square, $\frac{1}{\text{value}}$ for a reciprocal.
• Solving backwards (response given, explanatory unknown): rearrange for the transformed $x$, then undo it last. For height $= -3.8 + 12.6 \log_{10}(\text{age})$ and height 8.52 m, $\log_{10}(\text{age}) = 0.9778$ and age $= 10^{0.9778} = 9.5$ years.

Interpreting the transformed equation

The slope and intercept describe the transformed variable. In height $= -3.8 + 12.6 \log_{10}(\text{age})$, height increases by 12.6 m, on average, for each increase of 1 in $\log_{10}(\text{age})$, which is a tenfold increase in age (for example from 1 to 10 years). In life $= 63.12 + 2.842 \times (\text{doctors})^2$, life increases by 2.842 years for each increase of 1 in the square of the number of doctors, so the gain from each extra doctor grows as doctors increase. You will usually be asked to use a transformed equation to predict rather than to interpret its slope, but say "for each increase of one in $\log_{10}(\text{age})$" if you are asked.

All the usual warnings about linear models still apply. A transformation is justified only over the range of the data, so extrapolation is even riskier: the reciprocal tree model predicts a negative height for a 3-year-old tree ($13.04 - \frac{40.22}{3} = -0.37$ m). And a good fit is association, not proof of causation.

Where transformations fit in the course

Transformation is the last step in the linear-modelling cycle. You describe the scatterplot (direction, form, strength), fit a least squares line, and check it with a residual plot. When the residual plot shows a pattern, the form is non-linear, and transformation lets you keep using least squares on a re-expressed variable. The same tools, $r^2$ and the residual plot, then confirm that the transformed model is better.

Logarithms also appear in the Study Design's log (base 10) scales for data spread over several orders of magnitude. A $\log_{10}$ transformation is the same idea applied to a regression: it compresses large values so that a curve spread over a wide range becomes a line.