More about Single Factor Experiments

Transcription

1 More about Single Factor Experiments / 23

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3 Parameter estimation Effect Model (1): Y ij = µ + A i + ɛ ij, Ji A i = 0 Estimation: µ + A i = y i. ˆµ = y..  i = y i. y.. Effect Modell (2): Y ij = µ + A i + ɛ ij, A 1 = 0 Estimation: ˆµ = y 1.  i = y i. y 1. Mean Model: Y ij = µ i + ɛ ij Estimation: ˆµ i = y i. 2 / 23

4 Remarks: To interpret parameters correctly you must know which model has been used. Coefficients have different meanings. Prediction and residuals are always the same. Prediction: { ˆµ + Âi ŷ ij = = y i. ˆµ i Residual: r ij = y ij y i. 3 / 23

5 Anova and Regression Analysis of variance models can be written as multiple regression models with indicator variables. Analysis of variance models are more intuitiv. Parameter estimators y.., y i.,... are Least Squares estimators. 4 / 23

6 Berliner Pfannkuchen 5 / 23

7 Data Response: Fat absorption of 24 Berliner [g] Type of Fat Fat Absorption Mean balanced design: equal replication 6 / 23

8 Graphical display Fat 170 Fat Type Type 7 / 23

9 R: Anova table > mod2=aov(fat~type,data=berliner) > summary(mod2) Df Sum Sq Mean Sq F value Pr(>F) type ** Residuals > coef(mod2) (Intercept) type2 type3 type Question: What do these coefficients mean? command model.matrix() can be used to see the design matrix 8 / 23

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11 Modell: Y ij = µ + A i + ɛ ij, ɛ ij N(0, σ 2 ) i.i.d. Normal plot of residuals r ij = y ij y i. To detect Outliers. Normal distribution not crucial in randomized experiments. Nonparametric test: Kruskal-Wallis Equal variances: Plot r ij vs y i. σ 2 min < 1 9 σ2 max (balanced designs), log- -transformation, weights Independent observations: Plot r ij vs time, order more complex model, analysis 10 / 23

12 Residual plots Normal Q Q Plot Sample Quantiles resid(mod2) Theoretical Quantiles fitted(mod2) 11 / 23

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14 Treatment differences F test significant = There are treatment effects. Which? How large are the effects? Treatment differences estimated by y i. y i. Fat type 2 Fat type 1: = 13 Fat type 3 Fat type 1: = 4 Fat type 4 Fat type 1: = 10 Standard error of a treatment difference: σ 2 (1/J + 1/J) = 2σ 2 /J, estimated by 2MS res /J. Example: /6 = / 23

15 Are Type 2 and 1 significantly different? t test for H 0 : A 2 = A 1 t = y 2. y 1. 2MSres /J = = > = t 0.975,20, p = Confidence interval for Type 2 - Type 1: 13 ± = 13 ± }{{} = (0.9, 25.1) LSD 14 / 23

16 Multiple pairwise comparisons Are all pairs of treatments different? Problem: α E increases. Bonferroni correction for 6 pairwise comparisons: Significance level: α T = 0.05/6 = Critical value: t /2 6,20 = Difference between Type 2 and 1 not significant. Dunnett s method for multiple comparisons with a control group. Tukey method for pairwise comparisons: critical values for the distribution of max y i. y i. 15 / 23

17 Tukey method Reject H 0 : A 2 = A 1, if t > 1 2 q 1 α,i,n I with q... the quantile of the Studentized Range distribution. Example: t > = Type 2 and 1 do not differ significantly. Tukey Confidence interval for Type 2 - Type 1: 13 ± = 13 ± }{{} = ( 3.2, 29.2) HSD 16 / 23

18 R: plot(tukeyhsd(mod2, type )) 95% family wise confidence level Differences in mean levels of type 17 / 23

19 Complex comparisons Is there a difference between fat types 1 and 4 vs 2 and 3? H 0 : A 1 + A 4 2 = A 2 + A 3 2 A 1 2 A 2 2 A A 4 2 = 0 Hypotheses can be written as linear combinations λ i A i. Question: What about the question before: is there a difference between type 1 and type 2? H 0 : A 1 A 2 = 0 18 / 23

20 Contrasts Contrast: I C = λ i A i with λ i = 0 i=1 Ex1: C 1 = ( 1 2, 1 2, 1 2, 1 2 ) Ex2: C 2 = (1, 1, 0, 0) Ex2? C can be estimated by Ĉ = λ i  i = λ i (y i. y.. ) = λ i y i. y.. λi = λ i y i.. 19 / 23

21 Testing of a contrast Reject H 0 : λ i A i = 0, if Equivalently,if Ĉ t = > t 0.975,N I λ 2 MS i res J i F = t 2 = Ĉ 2 / λ 2 i /J i MS res = SS C MS res > F 0.95,1,N I SS C denotes the sum of squares of the contrast C. 20 / 23

22 Orthogonal contrasts There are I 1 linearly independent contrasts. Two contrasts C 1 = λ i A i and C 2 = λ i A i are called orthogonal, if λ i λ i = 0. It is always possible to find I 1 orthogonal contrasts. 21 / 23

23 Partitioning of Treatment Sum of Squares (only balanced designs) orthogonal contrasts uncorrelated estimates t tests nearly independent SS C = JĈ 2 / λ 2 i sum of squares of the contrast C If C 1, C 2,..., C I 1 are orthogonal contrasts, then SS treat = SS C1 + SS C2 + + SS CI 1 22 / 23

24 Recommendation: Multiple Comparison n planned, orthogonal contrasts (n I 1) pairwise comparisons comparison with a control group complex nonorthogonal or complex unplanned comparisons Bonferroni significance level α/n Tukey method Dunnett s method Scheffé: critical value (I 1)F I 1,N I,95% 23 / 23