1. A Bayesian random coefficient nonlinear regression for a split-plot experiment for detecting differences in the half- life of a compound Reid D. Landes Trey Spencer Ian A. Zelaya

2. Outline Background and objective Description of experiment Classical analysis plan Bayesian analysis plan –Hierarchical model (HM) Results Discussion

3. Background Shikimic acid (SA) is often studied in plant sciences to monitor effects of glyphosate (an active ingredient in herbicides) SA starts to break down after it is separated from plant tissue treated with glyphosate The SA chromophore has a half-life (  ) SA is often quantified with spectrophotometric methods –Optical density ( y ) is recorded

4. Background  is thought to (at least be) dependent upon temperature (Tmp) –Typically 2 Tmps used: 4 & 23 degrees C Maybe  is dependent upon the solution used in preparing the samples (Prep) –Typically 2 types: A and B

5. Background & Objective Lab tech can prepare and process… –many samples if  is long –not as many samples if  is short What Prep*Tmp combination prolongs  ?

6. Experiment Design Treatment structure: 3 factors @ 2 levels –Prep: A & B –Tmp: 4 & 23 degrees Celsius –Amt: 30 & 60 units of SA Laid out in a split plot design

7. Experiment Design Run i, i =1,2 Split Plot E.U. (Prep*Tmp*Amt) Sampling Unit (tube) Whole Plot E.U. (Prep*Tmp)

8. Comment If we could directly observe half-life from a tube (say g ), then use ANOVA Source of Variationdf Run 1 Prep*Tmp 3 Run*Prep*Tmp 3 Prep*Tmp*Amt 4 Run*Prep*Tmp*Amt 4 Tube(Run Prep Tmp Amt) 64 Total79

9. Observable data y

10. Model for tube data y hijkl – optical density from Prep*Tmp h in Run i with Amt j from Tube k at time l x l – time (0, 1,…, 14, 28, 50) hours a hijk – lower asymptote ( y at infinite time ) b hijk – range ( y 0 – a ) g hijk – half life  hijkl – additive mean-zero error

11. Classical Analysis Plan Since time values were consistent for each tube, and there was no missing data… 1.Estimate g for each tube via nonlinear regression 2.Fit estimated g ’s with linear mixed model having ANOVA described above 3.Make inference on Prep*Tmp treatment effects,  h with h= 1,2,3,4

12. Bayesian Analysis Plan 1.Model y data with a random coefficients nonlinear regression (a HM) 2.Assign priors to variance parameters and whole plot treatment effects 3.Use MCMC simulation in Bayes HM to obtain empirical posterior distributions 4.Make inference on  h from [  h  | y] with h= 1,2,3,4

13. g ~ Dsn(G,  g ) g Sampling Unit

14. Split Plot E.U. G G ~ Dsn(  G )

15. Whole Plot E.U.   ~ Dsn   

16. Bayes Hierarchical Model Overview

17. Bayes Hierarchical Model  a hijk b hijk g hijk y hijkl aa bb gg A hij B hij G hij xlxl

18. Bayes Hierarchical Model A hij B hij G hij AA BB GG  hi  hi    Amt j  hi hh hh hh

19. BHM Levels 1, 2, & 3

20. BHM Levels 4 & 5

21. MCMC in BHM MCMC simulation via WinBUGS 1.4 Ran 3 chains with initial values chosen from the data Burn-in = 6000 Thinning rate = every 50 th 6000 iterates to approximate posteriors Assessed posterior convergence with Brooks-Gelman-Rubin plots

22. 95% CI’s for Prep & Tmp Effects -4 -3 -2 0 1 2 3 4 5 Prep Main Effect Tmp Main Effect Prep*Tmp Interaction Bayesian Classical HoursHours

23. 95% CI’s for  0 1 2 3 4 5 6 7 8 9 A4A23B4B23 HALFLIFEHALFLIFE Bayesian Classical

24. All Pairwise Comparisons -6 -4 -2 0 2 4 6 8 10 A4- A23 A4- B4 A4- B23 A23- B4 A23- B23 B4- B23 HoursHours Bayesian Classical

25. Posterior Probabilities of Ranks Posterior prob. 1 (shortest) 234 (longest) A40.0010.0020.0470.950 A230.0190.9610.0200.001 B40.0010.0180.9320.049 B230.9790.0200.0010.000 Most Probable Ranking Sequence: P({B23, A23, B4, A4} | y ) = 0.917

26. Addressing the Objective Use Prep A at 4 degrees (A4) Given the data, the probability that A4 is the best is 0.950 From Posterior Probabilities of Rank, posterior probabilities for TMP=4 or PREP=A are easily calculated –Pr(TMP=4 > TMP=23 | y ) = [(.047+.950)+(.932+.049)]/2 = 0.989

27. Discussion Classical approach makes sense thanks to –Consistently spaced time points for each tube –No missing y data Classical approach is easy to implement Bayesian approach is flexible –Inconsistent time spacing and missing data are no problem

28. Bayesian vs. Classical Bayesian approach –has one model for all the data –produces tighter interval estimates –Allows more meaningful probability statements that address the objective Classical ANOVA approach is widely accepted, esp. by non-statistician reviewers.