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Topic 23: Diagnostics and Remedies

Topic 23: Diagnostics and Remedies. Outline. Diagnostics residual checks ANOVA remedial measures. Diagnostics Overview. We will take the diagnostics and remedial measures that we learned for regression and adapt them to the ANOVA setting Many things are essentially the same

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Topic 23: Diagnostics and Remedies

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  1. Topic 23: Diagnostics and Remedies

  2. Outline • Diagnostics • residual checks • ANOVA remedial measures

  3. Diagnostics Overview • We will take the diagnostics and remedial measures that we learned for regression and adapt them to the ANOVA setting • Many things are essentially the same • Some things require modification

  4. Residuals • Predicted values are cell means, = • Residuals are the differences between the observed values and the cell means Yij-

  5. Basic plots • Plot the data vs the factor levels (the values of the explanatory variables) • Plot the residuals vs the factor levels • Construct a normal quantile plot and/or histogram of the residuals

  6. KNNL Example • KNNL p 777 • Compare 4 brands of rust inhibitor (X has r=4 levels) • Response variable is a measure of the effectiveness of the inhibitor • There are 10 units per brand (n=10)

  7. Plots • Data versus the factor • Residuals versus the factor • Normal quantile plot of the residuals

  8. Plots vs the factor symbol1 v=circle i=none; proc gplot data=a2; plot (eff resid)*abrand; run;

  9. Data vs the factor Means look different …common spread in Y’s

  10. Residuals vs the factor Odd dist of points

  11. QQ-plot Due to odd (lack of and large)spread Can try nonparametric analysis – last slides

  12. General Summary • Look for • Outliers • Variance that depends on level • Non-normal errors • Plot residuals vs time and other variables if available

  13. Homogeneity tests • Homogeneity of variance (homoscedasticity) • H0: σ12 = σ22 = … = σr2 • H1: not all σi2 are equal • Several significance tests are available

  14. Homogeneity tests • Text discusses Hartley, modified Levene • SAS has several including Bartlett’s (essentially the likelihood ratio test) and several versions of Levene

  15. Homogeneity tests • There is a problem with assumptions • ANOVA is robust with respect to moderate deviations from Normality • ANOVA results can be sensitive to the homogeneity of variance assumption • Some homogeneity tests are sensitive to the Normality assumption

  16. Levene’s Test • Do ANOVA on the squared residuals from the original ANOVA • Modified Levene’s test uses absolute values of the residuals • Modified Levene’s test is recommended • Another quick and dirty rule of thumb

  17. KNNL Example • KNNL p 785 • Compare the strengths of 5 types of solder flux (X has r=5 levels) • Response variable is the pull strength, force in pounds required to break the joint • There are 8 solder joints per flux (n=8)

  18. Scatterplot

  19. Levene’s Test proc glm data=a1; class type; model strength=type; means type/ hovtest=levene(type=abs); run;

  20. ANOVA Table Common variance estimated to be 2.11

  21. Output Levene's Test ANOVA of Absolute Deviations Source DF F Value Pr > F type 4 3.07 0.0288 Error 35 We reject the null hypothesis and assume nonconstant variance

  22. Means and SDs Level strength type N Mean Std Dev 1 8 15.42 1.23 2 8 18.52 1.25 3 8 15.00 2.48 4 8 9.74 0.81 5 8 12.34 0.76

  23. Remedies • Delete outliers • Is their removal important? • Use weights (weighted regression) • Transformations • Nonparametric procedures

  24. What to do here? • Not really any obvious outliers • Do not see pattern of increasing or decreasing variance or skewed dists • Will consider • Weighted ANOVA • Mixed model ANOVA

  25. Weighted least squares • We used this with regression • Obtain model for how the sd depends on the explanatory variable (plotted absolute value of residual vs x) • Then used weights inversely proportional to the estimated variance

  26. Weighted Least Squares • Here we can compute the variance for each level • Use these as weights in PROC GLM • We will illustrate with the soldering example from KNNL

  27. Obtain the variances and weights proc means data=a1; var strength; by type; output out=a2 var=s2; data a2; set a2; wt=1/s2; NOTE. Data set a2 has 5 cases

  28. Proc Means Output

  29. Merge and then use the weights in PROC GLM data a3; merge a1 a2; by type; proc glm data=a3; class type; model strength=type; weight wt; lsmeans type / cl; run;

  30. Output Data have been standardized to have a variance of 1

  31. LSMEANS Output Because of weights, standard errors simply based on sample variances of each level

  32. Mixed Model ANOVA • Relax the assumption of constant variance rather than including a “known” weight • This involves moving to a mixed model procedure • Topic will not be on exam but wanted you to be aware of these model capabilities

  33. SAS Code proc glimmix data=a1; class type; model strength=type / ddfm=kr; random residual / group=type; run; This allows the variance to differ in each level and a degrees of freedom adjustment is used to account for this

  34. GLIMMIX OUTPUT Really 3 groups of variances

  35. SAS Code proc glimmix data=a1; class type; model strength=type / ddfm=kr; random residual / group=type1; run; Type1 was created to identify Type 1 and 2, Type 3, and Type 4 and 5 as 3 groups

  36. GLIMMIX OUTPUT Better BIC but same general type conclusion

  37. Transformation Guides • When σi2 is proportional to μi, use • When σi is proportional to μi, use log(y) • When σi is proportional to μi2, use 1/y • For proportions, use arcsin( ) • arsin(sqrt(y)) in a SAS data step • Box-Cox transformation

  38. Example • Consider study on KNNL pg 790 • Y: time between computer failures • X: three locations data a3; infile 'u:\.www\datasets512\CH18TA05.txt'; input time location interval; symbol1 v=circle; proc gplot; plot time*location; run;

  39. Scatterplot Outlier or skewed distribution? Can consider transformation first

  40. Box-Cox Transformation • Can consider regression and 1-b1 is the power to raise Y by • Can try various “convenient” powers • Can use SAS directly to calculate the power

  41. E(logsig) = 0.90 + .79 logmu Power should be 1-.79 ≈ 0.20

  42. Using SAS proc transreg data=a3; model boxcox(time / lambda=-2 to 2 by .2) = class(location); run;

  43. Output

  44. Transforming data in SAS data a3; set a3; transtime = time**0.20; symbol1 v=circle i=none; proc gplot; plot transtime*location; run;

  45. Much more constant spread in data!

  46. Nonparametric approach • Based on ranks • See KNNL section 18.7, p 795 • See the SAS procedure NPAR1WAY

  47. Rust Inhibitor Analysis Highly significant F test. Even if there is a violation of Normality, the evidence is overwhelming

  48. Nonparametric Analysis

  49. Last slide • We’ve finished most of Chapters 17 and 18. • We used program topic23.sas to generate the output.

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