1. Non-Linear Regression

2. The data frame trees is made available in R with >data(trees) These record the girth in inches, height in feet and volume of timber in cubic feet of each of a sample of 31 felled black cherry trees in Allegheny National Forest, Pennsylvania. Note that girth is the diameter of the tree (in inches) measured at 4 ft 6 in above the ground.

4. We treat volume as the (continuous) response variable y and seek a reasonable model describing its distribution conditional first on the explanatory variable girth (we will call this x). This might be a first step to prediction of volume based on further observations of the explanatory variables.

6. Observation of the graph leads us to first try out whether there may be a linear dependence here. Thus the relationship is approximately y=a+bx+є, for some constants a and b We will use R to find a and b, their standard errors and the residuals.

8. The fitted model is volume = −36.9 + 5.07 × girth + residual i.e. y = −36.9 + 5.07x (+ residual) To check its validity, first look at the standard errors

9. The standard errors of both a and b are low in comparison with the actual values and the p- values associated with the coefficients show that neither of these may reasonably be taken as zero. Thus there is evidence that the model is appropriate.

11. Some measure of the success of the fitted model is also given by the residual standard error. For a good fit this should be small in relation to the variation in the response variable itself.

12. Note:  18.1 = 4.252

13. However, a full examination of the residuals, and of the nature of any further dependence they may have on the explanatory variables, is to be preferred to reliance on any single number. All this will require graphical analysis, the results of which follow.

16. There is a slight evidence of non random behaviour in the residuals with perhaps the hint of a quadratic curve. We now adapt the model.

17. The residuals from Model 1 show some further, perhaps quadratic, dependence on the explanatory variable girth, so we try introducing a nonlinear term. We consider the model volume = a + b 1 × girth + b 2 × (girth) 2 + resid The relevant R commands, and associated output, are now >model2 = lm(Volume~Girth+I(Girth^2)) > summary(model.2)

19. The fitted model is therefore volume = 10.8 − 2.09 × girth + 0.255 × (girth) 2 + residual.

20. Consider now the graphs produced by the following commands. > plot(Volume~Girth) > lines(fitted(model2)~Girth) > plot(residuals(model2)~Girth, ylab="residuals from Model 2")

23. It is clear that these residuals are both smaller than those from Model 1 and show no further obvious dependence on the explanatory variable girth. Further the very small p-value (0.00015) associated with the coefficient b 2 shows that this cannot reasonably be set equal to zero, so that Model 2 is considerably more successful than Model 1.

24. Note also that the residual standard error in Model 2 is 3.335 whilst in Model 1 it is 4.252. Further Analysis: On physical grounds, we might also consider the simpler model Volume = b 2 × (Girth) 2 + Residual For extra justification look at this R output

25. The R code to fit this model, and brief summary output, are: > model3 = lm(Volume ~ I(Girth^2) - 1) > summary(model3)