1. 5/13/2015330 lecture 91 STATS 330: Lecture 9

2. 5/13/2015330 lecture 92 Diagnostics Aim of today’s lecture: To give you an overview of the modelling cycle, and begin the detailed discussion of diagnostic procedures

3. 5/13/2015330 lecture 93 The modelling cycle  We have seen that the regression model describes rather specialized forms of data Data are “planar” Scatter is uniform over the plane  We have looked at some plots that help us decide if the data is suitable for regression modelling pairs reg3d coplots

4. 5/13/2015330 lecture 94 Residual analysis  Another approach is to fit the model and examine the residuals  If the model is appropriate the residuals have no pattern  A pattern in the residuals usually indicates that the model is not appropriate  If this is the case we have two options Option 1: Select another form of model eg non-linear regression – see other courses Option 2: transform the data so that the regression model fits the transformed data – see next slide

5. 5/13/2015330 lecture 95 Modelling cycle This leads us into a modelling cycle Fit Examine residuals Transform data or change model if necessary This cycle is repeated until we are happy with the fitted model Diagramatically….

6. 5/13/2015330 lecture 96 Modelling cycle Examine residuals Fit model Transform/ change Choose Model Bad fit Good fit Use model Plots, theory

7. 5/13/2015330 lecture 97 What makes a bad fit?  Non-planar data  Outliers in the data  Errors depend on covariates (non-constant scatter)  Errors not independent  Errors not normally distributed  First two points can be serious as they affect the meaning and accuracy of the estimated coefficients.  The others affect mainly standard errors, not estimated coefficients – see subsequent lectures

8. 5/13/2015330 lecture 98 Detecting non-planar data For the next 2 lectures, we look at diagnosing non-planar data and choosing a transformation. We can diagnose non-planar data (nonlinearity) by fitting the model, and plotting residuals versus fitted values, residuals against explanatory variables fitting additive models In each case, a curved plot indicates non-planar data

9. 5/13/2015330 lecture 99 Plotting residuals versus fitted values plot(cherry.lm, which=1) which = 1 selects a plot of residuals versus fitted values

10. 5/13/2015330 lecture 910 Indication of a curve: so data non-planar

11. 5/13/2015330 lecture 911 Additive models  These are models of the form Y=g 1 (x 1 ) + g 2 (x 2 ) + … + g k (x k ) +  where g 1,…,g k are transformations  Fitted using the gam function in R  The transformations are estimated by the software  Use the function to suggest good transformations

12. 5/13/2015330 lecture 912 Example: cherry trees > library(mgcv) This is mgcv 0.8-8 > cherry.gam<- gam(volume~s(height) + s(diameter),data=cherry.df) > plot(cherry.gam) Don’t forget the s if you want to estimate the transformation

13. 5/13/2015330 lecture 913 Example Strong suggestion that diameter be transformed

14. 5/13/2015330 lecture 914 Fitting polynomials  To fit model y = b 0 +b 1 x + b 2 x 2, use y ~ poly(x,2)  To fit model y = b 0 +b 1 x + b 2 x 2 + b 3 x 3, use y ~ poly(x,3) and so on

15. 5/13/2015330 lecture 915 Orthogonal polynomials The model fitted by y~poly(x,2) is of the form Y =  0 +  1 p 1 (x) +  2 p 2 (x) where p 1 is a polynomial of degree 1 ie of form ax + b and p 2 is a polynomial of degree 2 ie of form ax 2 + bx + c p 1, p 2 chosen to be uncorrelated (best possible estimation)

16. 5/13/2015330 lecture 916 Adding a quadratic term (cherry trees)  Suggestion of quadratic behaviour in diameter, so fit a quadratic model in diameter (add a term diameter^2) cherry2.lm=lm(volume~poly(diameter,2)+height,data=cherry.df) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.56553 6.72218 0.233 0.817603 poly(diameter, 2)1 80.25223 3.07346 26.111 < 2e-16 *** poly(diameter, 2)2 15.39923 2.63157 5.852 3.13e-06 *** height 0.37639 0.08823 4.266 0.000218 *** --- Residual standard error: 2.625 on 27 degrees of freedom Multiple R-Squared: 0.9771, Adjusted R-squared: 0.9745 How to add a quadratic term R2 improved from 94.8% to 97.7%

17. 5/13/2015330 lecture 917 Equation of quadratic To see which quadratic is fitted, use > cherry.quad = lm(volume ~ diameter + I(diameter^2) + height, data=cherry.df) > summary(cherry.quad) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -9.92041 10.07911 -0.984 0.333729 diameter -2.88508 1.30985 -2.203 0.036343 * I(diameter^2) 0.26862 0.04590 5.852 3.13e-06 *** height 0.37639 0.08823 4.266 0.000218 *** Residual standard error: 2.625 on 27 degrees of freedom Multiple R-Squared: 0.9771, Adjusted R-squared: 0.9745 Surface fitted fitted is -9.92041 - 2.88508 x diameter + 0.26862 x diameter 2 + 0.37639 x height Same model!

18. Splines  An alternative to polynomials are splines – these are piecewise cubics, which join smoothly at “knots”  Give a more flexible fit to the data  Values at one point not affected by values at distant points, unlike polynomials 5/13/2015330 lecture 918

19. Example with 4 knots 5/13/2015330 lecture 919

20. 5/13/2015330 lecture 920 Adding a spline term (cherry trees) library(splines) cherry3.lm=lm(volume~bs(diameter,6)+height,data=cherry.df) Coefficients: (Intercept) -16.3679 7.4856 -2.187 0.03921 * bs(diameter, df = 6)1 0.1941 7.9374 0.024 0.98070 bs(diameter, df = 6)2 5.5744 3.1704 1.758 0.09201. bs(diameter, df = 6)3 10.7976 3.9798 2.713 0.01240 * bs(diameter, df = 6)4 31.4053 5.5545 5.654 9.35e-06 *** bs(diameter, df = 6)5 42.2665 6.1297 6.895 4.97e-07 *** bs(diameter, df = 6)6 58.6454 4.2781 13.708 1.49e-12 *** height 0.3970 0.1050 3.780 0.00097 *** Residual standard error: 2.8 on 23 degrees of freedom Multiple R-squared: 0.9778, Adjusted R-squared: 0.971 How to add a spline term (3 knots) R2 improved from 94.8% to 97.78%

21. 5/13/2015330 lecture 921

22. 5/13/2015330 lecture 922 Example: Tyre abrasion data Data collected in an experiment to study the abrasion resistance of tyres Variables are Hardness: Hardness of rubber Tensile: Tensile strength of rubber Abrasion Loss: Amount of rubber worn away in a standard test (response)

23. 5/13/2015330 lecture 923 Tyre abrasion data > rubber.df hardness tensile abloss 1 45 162 372 2 61 232 175 3 71 231 136 4 81 224 55 5 53 203 221 6 64 210 164 7 79 196 82 8 56 200 228 9 75 188 128 10 88 119 64 11 71 151 219 … 30 observations in all

24. 5/13/2015330 lecture 924 Tyre abrasion data > rubber.lm<-lm(abloss~hardness + tensile, data=rubber.df) > summary(rubber.lm) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 885.1611 61.7516 14.334 3.84e-14 *** hardness -6.5708 0.5832 -11.267 1.03e-11 *** tensile -1.3743 0.1943 -7.073 1.32e-07 *** --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 36.49 on 27 degrees of freedom Multiple R-Squared: 0.8402, Adjusted R-squared: 0.8284 F-statistic: 71 on 2 and 27 DF, p-value: 1.767e-11

25. 5/13/2015330 lecture 925 Tyre abrasion data We will use this example to illustrate all the methods we have discussed so far to check if the data are planar, scattered about a flat regression plane i.e. Pairs Spinning plot Coplot Residual versus fitted value plot Fitting GAMS

26. 5/13/2015330 lecture 926 Tyre abrasion data (pairs)

27. Spinning 5/13/2015330 lecture 927

28. 5/13/2015330 lecture 928 Tyre abrasion data (coplot)

29. 5/13/2015330 lecture 929 Tyre abrasion data: residuals v fitted values data(rubber.df) rubber.lm = lm(abloss~., data=rubber.df) plot(rubber.lm, which=1)

30. 5/13/2015330 lecture 930 Tyre abrasion data: gams library(mgcv) rubber.gam<- gam(abloss~s(tensile) + s(hardness),data=rubber.df) plot(rubber.gam)

31. 5/13/2015330 lecture 931 Tyre abrasion data: Conclusions Pairs plot : not very informative Spinner: Hint of a “kink” Coplot: suggestion of non-linearity, a “kink” Residuals versus fitted values: weak suggestion that regression is not planar gam plots: hardness is OK, but strong suggestion that tensile needs transforming Suggested transformation: looks a bit like a cubic or 4 th degree polynomial, so try these (or a spline). Using a spline, the R 2 is increased from 84% to 94.3%.