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Plant Science 547 Biometrics for Plant Scientists

Plant Science 547 Biometrics for Plant Scientists. Association Between Characters. Effect of One Treatment on Another. Test hypothetical models for biological systems. To explain relationships (i.e. linear, quadratic, etc., orthogonal contrasts).

Samuel
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Plant Science 547 Biometrics for Plant Scientists

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  1. Plant Science 547 Biometrics for Plant Scientists Association Between Characters

  2. Effect of One Treatment on Another • Test hypothetical models for biological systems. To explain relationships (i.e. linear, quadratic, etc., orthogonal contrasts). • To predict the values of one variable according to set values of another.

  3. Possible Relationships of Interest • Predictoptimal nitrogen application to maximize seed yield. • Determine deficiencies in national supply of specific agricultural products by relating yield to weather related characters (rainfall, sunshine, etc). • Explain relationship between plant biomass and time after seeding to select for more insect tolerant cultivars.

  4. Charles Darwin • Born in England, educated in Scotland. • The father of evolution • Most famous for his travels on the Beagle to the Galapagos Islands. • “Survival of the fittest”. • Wrote The Origin of Species.

  5. History • 19th Century - Charles Darwin. • Francis Galton: In the “ law of universal regression” “ each peculiarity in a man is shared by his kinsman, but on average to a lesser degree”. • Karl Peterson & Andrew Lee (statisticians) survey 1000 fathers and sons height. • Using this data set Galton, Peterson and Lee formulated regression analyses.

  6. Regression Models • Dependant variable (of interest). Usually donated by y • One, or more, independent variable on which the dependant variable is related in a specific manner. Usually donated by x, x1, x2, etc.

  7. Common Types of Regression • Simple linear regression: y=b0+b1x • Non-linear regression: y=b0+b1x+b2x2; y=ex; y=ln(x) • Multiple regression: y=b0+b1x1+b2x2

  8. Nitrogen application v Seed yield

  9. Nitrogen application v Seed yield b1 bo Y = bo + b1x

  10. Simple Linear Regression Y = bo + b1x b1 = [SP(x,y)/SS(x)] SP(x,y) = (xi-x)(yi-y) SP(x,y) = (xy) - [(x) (y)]/n SS(x) = (xi-x)2 SS(x) = (x2) - [(x)]2/n

  11. Simple Linear Regression Y = bo + b1x bo = mean(y) = b1x mean(x)

  12. Linear regression example • Sex-linked mutations in Drosophila. • x-variable is dosage of radiation (1000’s rads). • y-variable is the percentage of mutation observed in Drosphila populations.

  13. Mutation Frequency in Drosophila

  14. Linear Regression Example

  15. Linear Regression Example SS(x) = (xi2)-[(xi)]2/n 39.25 - (11.5)2/5 12.800 Mean (y) = (yi)/n 11.5/5 2.30

  16. Linear Regression Example SS(x,y) = (xiyi)-[(xi) (yi)]/n 89.35 - (11.5 x 26.3)/5 28.860 b1 = SP(x,y)/SS(x) 28.860/12.800 = 2.255 b0 = y - b1x 5.26 - 2.255 x 2.30 = 0.735

  17. Mutation Frequency in Drosophila Y = 0.0735 + 2.255 x

  18. Analysis of Variance ~ Regression • Total variation of the dependant variable (the one of interest). • Partition into variation accountable by the regression model (linear or other): Sum of squares for regression. • Other, non-explaiable variation: Residual sum of squares.

  19. Linear Regression Example Total SS = SS(y) = (yi2)-[(yi)]2/n 204.11 - (26.3)2/5 = 65.772 Regression SS = [SP(x,y)]2/SS(x) [28.860]2/12.800 = 65.070 Residual SS [2Res] = Total SS - Regression SS 65.772 - 65.070 = 0.702

  20. Analysis of Variance Regression Residual can be tested if observations are replicated

  21. t-test and regression

  22. t-tests and Regression Is the regression slope significantly greater than zero? t = b-0/se(b) b/se(b) se(b) = {[SS(y) - [b1 SP(x,y)]]/[(n-2)SS(x)] {[65.772 - [2.255 x 28.860]]/[(3 x 12.800] = 0.134

  23. t-tests and Regression Things to Note se(b) = {[SS(y) - [b1 SP(x,y)]]/[(n-2)SS(x)] [SS(y) - [b1 SP(x,y)]] = Residual SS [SS(y) - [b1 SP(x,y)]]/(n-2) = Residual MSq se(b) = [Residual MSq/SS(x)] = [2Res/SS(x)]

  24. t-tests and Regression Is the intercept significantly different from a? t = b0-a/se(b0) se(b0) = {2Res. [1/n + [mean(x)2/SS(x)]]} {0.234 x [1/5+[(2.32/12.800]]} = 0.378

  25. Predicting the dependant variable!

  26. Predicting the Dependant Variable Y = 0.0735 + 2.255 x At x = 2.5 - 1000 Rads y = 0.073 + 2.255 x 2.5 = 5.7105 How accurate is this estimation? - se(yp) = {2 [1+1/n+(xp-x)2/SS(x)]}

  27. Linear Regression Example Predicting at x = 2.5 se(yp) = {0.234 [1+1/5+(2.5-2.3)2/12.800]} se(yp) = 0.531 yp = 5.7105 + 0.531

  28. Linear Regression Example Predicting at x = 4.5 se(yp) = {0.234 [1+1/5+(4.5-2.3)2/12.800]} se(yp) = 0.663 yp = 10.2205 + 0.663

  29. Mutation Frequency in Drosophila

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