1. Making a curved line straight Data Transformation & Regression

2. Last Class  Predicting the dependant variable and standard errors of predicted values.  Outliers.  Need to visually inspect data in graphic form.  Making a curved line straight.  Transformation.

3. Early Growth Pattern of Plants

4. y = Ln(y)

5. Early Growth Pattern of Plants y =  y

6. Homogeneity of Error Variance

7. y =Ln(y)

8. Growth Curve Y = e x

9. Growth Curve Y = Log(x)

10. Sigmoid Growth Curve 

11.  Accululative Normal Distribution

12. Sigmoid Growth Curve  Accululative Normal Distribution  T T-- T T-- ƒ (  d  d  T

13. Sigmoid Growth Curve  Accululative Normal Distribution  T T-- T T-- ƒ (  d  d  T

14. Probit Analysis Group of plants/insects exposed to different concentrations of a specific stimulant (i.e. insecticide). Group of plants/insects exposed to different concentrations of a specific stimulant (i.e. insecticide). Data are counts (or proportions), say number killed. Data are counts (or proportions), say number killed. Usually concerned or interested in concentration which causes specific event (i.e. LD 50%). Usually concerned or interested in concentration which causes specific event (i.e. LD 50%).

15. Probit Analysis ~ Example

17. Estimating the Mean  y = 50% Killed  x ~ 2.8

18. Estimating the Standard Deviation 2.8

19. 2.8 2222

20. 2.8 2222 95% values Estimating the Standard Deviation

21. 2.8 2222 95% values  = 1.2 Estimating the Standard Deviation

22. Probit Analysis

24. Probit (  ) =  + . Log 10 (concentration)  = -1.022 + 0.202  = 2.415 + 0.331 Log 10 (conc) to kill 50% (LD-50) is probit 0.5 = 0 0 = -1.022 + 2.415 x LD-50 LD-50 = 0.423 10 0.423 = 2.65%

25. Problems  Obtaining “good estimates” of the mean and standard deviation of the data.  Make a calculated guess, use iteration to get “better fit” to observed data.

26. Where Straight Lines Meet

27. Optimal Assent

28. Y 1 =a 1 +b 1 x

29. Optimal Assent Y 1 =a 1 +b 1 x Y 2 =a 2 +b 2 x

30. Optimal Assent Y 1 =a 1 +b 1 x Y 2 =a 2 +b 2 x t =[b 1 -b 2 ]/se(b) = ns = ns

31. Optimal Assent Y 1 =a 1 +b 1 x Y 3 =a 3 +b 3 x

32. Optimal Assent Y 1 =a 1 +b 1 x Y 3 =a 3 +b 3 x t =[b 1 -b 3 ]/se(b) = *** = ***

33. Optimal Assent Y 1 =a 1 +b 1 x Y 3 =a 3 +b 3 x

34. Optimal Assent Y 1 =a 1 +b 1 x Y 3 =a 3 +b 3 x t =[b 1 -b n ]/se(b) = *** = *** Y n =a n +b n x

35. Optimal Assent Y 1 =a 1 +b 1 x Y 3 =a 3 +b 3 x

36. Yield and Nitrogen

37. What application of nitrogen will result in the optimum yield response?

38. Intersecting Lines

39. Y = 2.81x + 1055.10 Y = 9.01x + 466.60

40. Intersecting Lines t = [b 11 - b 21 ]/average se(b) 6.2/0.593 = 10.45 *, With 3 df Intersect = same value of y b 10 + b 11 x = y = b 20 + b 21 x x = [b 20 - b 10 ]/[b 11 - b 21 ] = 94.92 lb N/acre with 1321.83 lb/acre seed yield

41. Intersecting Lines Y = 2.81x + 1055.10 Y = 9.01x + 466.60 94.92 lb N/acre 1321.83 lb/acre

42. Linear Y = b 0 + b 1 x Quadratic Y = b 0 + b 1 x + b 2 x 2 Cubic Y = b 0 + b 1 x + b 2 x 2 + b 3 x 3 Bi-variate Distribution Correlation