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STATS 330: Lecture 9

STATS 330: Lecture 9. Diagnostics. Diagnostics. Aim of today’s lecture: To give you an overview of the modelling cycle, and begin the detailed discussion of diagnostic procedures. The modelling cycle. We have seen that the regression model describes rather specialized forms of data

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STATS 330: Lecture 9

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  1. STATS 330: Lecture 9 Diagnostics 330 lecture 9

  2. Diagnostics Aim of today’s lecture: To give you an overview of the modelling cycle, and begin the detailed discussion of diagnostic procedures 330 lecture 9

  3. 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 330 lecture 9

  4. 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 330 lecture 9

  5. 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…. 330 lecture 9

  6. Choose Model Plots, theory Fit model Examine residuals Transform/ change Good fit Bad fit Use model Modelling cycle 330 lecture 9

  7. 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 330 lecture 9

  8. 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 330 lecture 9

  9. Plotting residuals versus fitted values plot(cherry.lm, which=1) which = 1 selects a plot of residuals versus fitted values 330 lecture 9

  10. Indication of a curve: so data non-planar 330 lecture 9

  11. Additive models • These are models of the form Y=g1(x1) + g2(x2) + … + gk(xk) + e where g1,… ,gk are transformations • Fitted using the gam function in R • The transformations are estimated by the software • Use the function to suggest good transformations 330 lecture 9

  12. Example: cherry trees Don’t forget the s if you want to estimate the transformation > library(mgcv) This is mgcv 0.8-8 > cherry.gam<- gam(volume~s(height) + s(diameter),data=cherry.df) > plot(cherry.gam) 330 lecture 9

  13. Example Strong suggestion that diameter be transformed 330 lecture 9

  14. Fitting polynomials • To fit model y = b0 +b1x + b2 x2, use y ~ poly(x,2) • To fit model y = b0 +b1x + b2 x2 + b3 x3, use y ~ poly(x,3) and so on 330 lecture 9

  15. Orthogonal polynomials The model fitted by y~poly(x,2) is of the form Y = b0 + b1p1(x) + b2p2(x) where p1 is a polynomial of degree 1 ie of form ax + b and p2 is a polynomial of degree 2 ie of form ax2 + bx + c p1, p2 chosen to be uncorrelated (best possible estimation) 330 lecture 9

  16. 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% 330 lecture 9

  17. 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 diameter2 + 0.37639 x height Same model! 330 lecture 9

  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 330 lecture 9

  19. Example with 4 knots 330 lecture 9

  20. 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% 330 lecture 9

  21. 330 lecture 9

  22. 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) 330 lecture 9

  23. 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 330 lecture 9

  24. 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-11 330 lecture 9

  25. 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 330 lecture 9

  26. Tyre abrasion data (pairs) 330 lecture 9

  27. Spinning 330 lecture 9

  28. Tyre abrasion data (coplot) 330 lecture 9

  29. Tyre abrasion data: residuals v fitted values data(rubber.df) rubber.lm = lm(abloss~., data=rubber.df) plot(rubber.lm, which=1) 330 lecture 9

  30. Tyre abrasion data: gams library(mgcv) rubber.gam<- gam(abloss~s(tensile) + s(hardness),data=rubber.df) plot(rubber.gam) 330 lecture 9

  31. 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 4th degree polynomial, so try these (or a spline). Using a spline, the R2 is increased from 84% to 94.3%. 330 lecture 9

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