4.13 Non-linear regressionIB Maths: Applications and Interpretation HL: Revision notes

In regression we fit a model to data so that the vertical distances between the points and the curve are as small as possible. The least squares regression curve minimises the sum of the squares of these distances. The distance between an observed value $y_i$ and the model's value $\hat y_i$ is the residual. In examinations you may be asked about linear, quadratic, cubic, exponential, power and sine regression, and you evaluate the curve using technology: enter the data in lists, choose the regression type and read off the parameters (to 3 significant figures unless told otherwise). Always state which type of regression you used and write the full equation.

• Quadratic: $y=ax^2+bx+c$. Cubic: $y=ax^3+bx^2+cx+d$.
• Exponential: $y=ab^x$ (or $ae^{kx}$). If $x$ takes equally spaced values $0,1,2,\ldots$ then the $y$ values form a geometric sequence with first term $a$ and common ratio $b$ (link to SL 1.3). A value $b=1.38$ means an increase of $38\%$ per unit.
• Power: $y=ax^b$. A power model is natural where one quantity scales with another, such as period and distance in orbits.
• Sine: $y=a\sin(bx+c)+d$; amplitude $|a|$, period $\frac{2\pi}{b}$ (with $x$ in radians), principal axis $y=d$. Choose a model from the shape of the data and the context.

The sum of squared residuals, $SS_{res}=\sum(y_i-\hat y_i)^2$, measures how closely the model fits the data. A smaller $SS_{res}$ means a closer fit. It depends on the units and the size of the data, so use it to compare models fitted to the same data. The least squares curve is the one with the smallest $SS_{res}$ of its type. Example: the data $(v,d)$ in the exam question has $SS_{res}=79.0$ for the linear model and $4.67$ for the quadratic model, so the quadratic fits much more closely.

The coefficient of determination $R^2$ is evaluated using technology. It gives the proportion of the variability in the second variable ($y$) accounted for by the chosen model. For example $R^2=0.998$ means $99.8\%$ of the variation in braking distance is accounted for by the model. $R^2$ lies between $0$ and $1$; closer to 1 means a better fit. (Awareness that $R^2=1-\frac{SS_{res}}{SS_{tot}}$, and so is $1$ when $SS_{res}=0$, may help understanding but is not examined.) For a linear model, $R^2=r^2$, where $r$ is Pearson's product-moment correlation coefficient. If $r=0.982$ then $R^2=0.964$.

Many factors affect the validity of a model, and $R^2$ alone is not a good way to decide between models. Think about:

• Context: does the model make sense? A linear model for braking distance gives $-15.9$ m at $0$ km/h, which is impossible.
• Range: using the model outside the data range is extrapolation, which is less reliable than interpolation.
• Complexity: a model with more parameters (a cubic compared with a quadratic) often has a higher $R^2$ without being better.
• Residual pattern and sample size: small samples can fit by chance. In a comment, give a reason that refers to the context or the data, not just "$R^2$ is high".

Data: $t=0$ to $6$ hours, $N=3.5, 4.8, 6.7, 9.2, 12.7, 17.5, 24.2$ thousand. The ratios of successive values are about $1.38$, so an exponential model is suitable.

1. Exponential regression on the GDC: $N=3.50\times1.38^t$.
2. Predict at $t=8$: $3.50\times1.38^8=46.0$ thousand (extrapolation, so treat with care).
3. Interpret: the culture grows by about $38\%$ each hour, and the hourly values form a geometric sequence. For a periodic tide, use sine regression: with $h=1.30\sin(0.508t-1.02)+2.60$ the amplitude is $1.30$, the period is $\frac{2\pi}{0.508}=12.4$ hours and the mean depth is $2.60$.