Lecture 3. Experiments with a Single Factor: ANOVA Montgomery Sections 3.1 through 3.5 and Section

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1 Lecture 3. Experiments with a Single Factor: ANOVA Montgomery Sections 3.1 through 3.5 and Section Lecture 3 Page 1

2 Tensile Strength Experiment Investigate the tensile strength of a new synthetic fiber. The factor is the weight percent of cotton used in the blend of the materials for the fiber and it has five levels. percent tensile strength of cotton total average Lecture 3 Page 2

3 Data Layout for Single-Factor Experiments treatment observations totals averages 1 y 11 y 12 y 1n y 1. ȳ 1. 2 y 21 y 22 y 2n y 2. ȳ a y a1 y a2 y an y a. ȳ a Lecture 3 Page 3

4 Analysis of Variance Statistical Model (Factor Effects Model): y ij = µ+τ i +ǫ ij i = 1,2...,a j = 1,2,...,n i µ - grand mean; τ i -ith treatment effect; ǫ ij iid N(0,σ 2 ) - error Constraint: a i=1 τ i = 0 (Conceptual Approach; SAS:τ a = 0). Estimates for parameters: ˆµ = y.. ˆτ i = (y i. y.. ) ˆǫ ij = y ij y i. ( residual) Basic Hypotheses: H 0 : τ 1 = τ 2 =... = τ a = 0 vs H 1 : τ i 0 for at least onei 4 Lecture 3 Page 4

5 Analysis of Variance (ANOVA) Table If balanced: Source of Sum of Degrees of Mean F 0 Variation Squares Freedom Square Between SS Treatment a 1 MS Treatment F 0 Within SS E N a MS E Total SS T N 1 N = n a SS T = y 2 ij y2../n; SS Treatment = 1 n y 2 i. y 2../N SS E =SS T - SS Treatment If unbalanced: N = a i=1 n i SS T = y 2 ij y2../n; SS Treatment = y 2 i. n i y 2../N SS E =SS T - SS Treatment SS Treatments = a i=1 n iˆτ 2 i andss E = i j ˆǫ2 ij. 5 Lecture 3 Page 5

6 The Expected Mean Squares (EMS) are E(MS E )=σ 2 E(MS Treatment ) =σ 2 + n i τ 2 i /(a 1) Test Statistic F 0 = SS Treatments/(a 1) SS E /(N a) = MS Treatments MS E UnderH 0 : F 0 = SS Treatment/σ 2 (a 1) SS E /σ 2 (N a) = χ2 a 1/(a 1) χ 2 N a /(N a) F a 1,N a Decision Rule: IfF 0 > F α,a 1,N a then rejecth 0 Whena = 2, the square of the t-test statistict 2 0 = MS Treatment MS E = F 0. F -test and two-sample two-sided test are equivalent. 6 Lecture 3 Page 6

7 Example Twelve lambs are randomly assigned to three different diets. The weight gain (in two weeks) is recorded. Is there a difference between the diets? Diet Diet Diet N = 12, yij = 156 and ȳ.. = 156/12 = 13. n 1 = 3,y 1. = 33,ȳ 1. = 11;n 2 = 5,y 2. = 75,ȳ 2. = 15;n 3 = 4,y 3. = 48 and ȳ 3. = 12. ˆτ 1 = ȳ 1. ȳ.. = = 2; Similarly, ˆτ 2 = = 2 and ˆτ 3 = = 1. SS T = i j (y ij ȳ.. ) 2 = 246. SS Treatment = 3 ( 2) 2 +5 (2) 2 +4 ( 1) 2 = 36. SS E = = 210; MS E = ˆσ 2 = 210/(12 3) = F 0 = (36/2)/(210/9) = 0.77; P-value>0.25; Fail to rejecth 0 : τ 1 = τ 2 = τ 3 = 0. 7 Lecture 3 Page 7

8 Using SAS (lambs.sas) option nocenter ps=65 ls=80; data lambs; input diet wtgain datalines; ; symbol1 bwidth=5 i=box; axis1 offset=(5); proc gplot; plot wtgain*diet / frame haxis=axis1; run; quit; 8 Lecture 3 Page 8

9 proc glm; class diet; model wtgain=diet; output out=diag r=res p=pred; run; quit; Sum of Source DF Squares Mean Square F Value Pr > F Model Error Corrected Total R-Square Coeff Var Root MSE wtgain Mean Source DF Type I SS Mean Square F Value Pr > F diet Source DF Type III SS Mean Square F Value Pr > F diet Lecture 3 Page 9

10 proc gplot; plot res*diet /frame haxis=axis1; proc sort; by pred; symbol1 v=circle i=sm50; proc gplot; plot res*pred / haxis=axis1; run; quit; 10 Lecture 3 Page 10

11 Model Checking and Diagnostics Model Assumptions 1 Model is correct 2 Independent observations 3 Errors normally distributed 4 Constant variance y ij = (y.. +(y i. y.. )) + (y ij y i. ) y ij = ŷ ij + ˆǫ ij observed = predicted + residual Note that the predicted response at treatmentiisŷ ij = ȳ i. Diagnostics use predicted responses and residuals. 11 Lecture 3 Page 11

12 Diagnostics Normality Histogram of residuals Normal probability plot / QQ plot Shapiro-Wilk Test Constant Variance Plotˆǫ ij vsŷ ij (residual plot) Bartlett s or Modified Levene s Test Independence Plotˆǫ ij vs time/space Plotˆǫ ij vs variable of interest Outliers 12 Lecture 3 Page 12

13 Diagnostics Example: Tensile Strength Experiment options ls=80 ps=60 nocenter; goptions device=win target=winprtm rotate=landscape ftext=swiss hsize=8.0in vsize=6.0in htext=1.5 htitle=1.5 hpos=60 vpos=60 horigin=0.5in vorigin=0.5in; data one; infile c:\saswork\data\tensile.dat ; input percent strength time; title1 Tensile Strength Example ; proc print data=one; run; Obs percent strength time : : : : Lecture 3 Page 13

14 symbol1 v=circle i=none; title1 Plot of Strength vs Percent Blend ; proc gplot data=one; plot strength*percent/frame; run; proc boxplot; plot strength*percent/boxstyle=skeletal pctldef=4; run; 14 Lecture 3 Page 14

15 proc glm data=one; class percent; model strength=percent; means percent / hovtest=bartlett hovtest=levene; output out=diag p=pred r=res; run; Sum of Source DF Squares Mean Square F Value Pr > F Model <.0001 Error Corrected Total Levene s Test for Homogeneity of strength Variance ANOVA of Squared Deviations from Group Means Sum of Mean Source DF Squares Square F Value Pr > F percent Error Bartlett s Test for Homogeneity of strength Variance Source DF Chi-Square Pr > ChiSq percent Lecture 3 Page 15

16 proc sort; by pred; symbol1 v=circle i=sm50; title1 Residual Plot ; proc gplot; plot res*pred/frame; run; 16 Lecture 3 Page 16

17 proc univariate data=diag normal; var res; qqplot res / normal (L=1 mu=est sigma=est); histogram res / normal; run; Moments N 25 Sum Weights 25 Mean 0 Sum Observations 0 Std Deviation Variance Skewness Kurtosis Uncorrected SS Corrected SS Coeff Variation. Std Error Mean Tests for Normality Test --Statistic p Value Shapiro-Wilk W Pr < W Kolmogorov-Smirnov D Pr > D Cramer-von Mises W-Sq Pr > W-Sq Anderson-Darling A-Sq Pr > A-Sq Lecture 3 Page 17

18 Histogram of Residuals & QQ Plot 18 Lecture 3 Page 18

19 /* Time Serial Plot */ symbol1 v=circle i=none; title1 Plot of residuals vs time ; proc gplot; plot res*time / vref=0 vaxis=-6 to 6 by 1; run; 19 Lecture 3 Page 19

20 Non-Constant Variance: Impact and Remedy Does not affect F-test dramatically when the experiment is balanced Why concern? Comparison of treatments depends on MSE Incorrect intervals and comparison results Variance-Stabilizing Transformations Ideas for Finding Proper Transformations (E[Y] = µ, Var(Y) = σ 2 ) f(y) f(µ)+(y µ)f (µ) = Var(f(Y)) [f (µ)] 2 Var(Y) = [f (µ)] 2 σ 2 Find f such that Var(f(Y)) does not depend on µ anymore. So, Ỹ = f(y) has constant variance for different f(µ). Common transformations x,log(x),1/x,arcsin( x), and1/ x 20 Lecture 3 Page 20

21 Transformations Supposeσ 2 is a function ofµ, that isσ 2 = g(µ) Want to find transformationf such thatỹ Var(Ỹ ) does not depend onµ. = f(y)has constant variance: Have shown Var(Ỹ ) [f (µ)] 2 σ 2 [f (µ)] 2 g(µ) Want to choosef such that[f (µ)] 2 g(µ) c Examples g(µ) = µ (Poisson) f(µ) = 1 µ dµ f(x) = X g(µ) = µ(1 µ) (Binomial) f(µ) = 1 dµ f(x) = arcsin( X) µ(1 µ) g(µ) = µ 2β (Box-Cox) f(µ) = µ β dµ f(x) = X 1 β g(µ) = µ 2 (Box-Cox) f(µ) = 1 µ dµ f(x) = logx 21 Lecture 3 Page 21

22 Identify Box-Cox Transformation Using Data: Approximate Method From the previous slide, ifσ = θµ β, the transformation is Y 1 β β 1; f(y) = logy β = 1 So it is crucial to estimateβ based on datay ij,i = 1,...,a. We have logσ i = logθ +βlogµ i Lets i andȳ i. be the sample standard deviations and means. Because ˆσ i = s i and ˆµ i = ȳ i., fori = 1,...,a, logs i constant +βlogȳ i., = logs i = constant +βlogȳ i. + error i. We can plotlogs i againstlogȳ i., fit a straight line and use the slope to estimateβ. 22 Lecture 3 Page 22

23 Identify Box-Cox Transformation: Formal Method 1. For a fixedλ, perform analysis of variance on y ij (λ) = y λ ij 1 λẏ λ 1 λ 0 ẏlogy ij λ = 0 whereẏ = a n i i=1j=1 y ij 1/N. 2. Step 1 generates a transformed datay ij (λ). Apply ANOVA to the new data and obtain SS E. Because SS E depends onλ, it is denoted by SS E (λ). Repeat 1 and 2 for variousλin an interval, e.g., [-2,2], and record SS E (λ) 3 Findλ 0 which minimizes SS E (λ) and pick up a meaningfulλin the neighborhood ofλ 0. Denote it again byλ. (Maximum Likelihood Principle) 4 The transformation is: ỹ ij = y λ 0 ij ifλ 0 0; ỹ ij = log y ij ifλ 0 = Lecture 3 Page 23

24 Example: Approximate Method (trans.sas) data one; infile c:\saswork\data\boxcox.dat ; input trt resp; proc glm data=one; class trt; model resp=trt; output out=diag p=pred r=res; run; title1 Residual Plot ; symbol1 v=circle i=none; proc gplot data=diag; plot res*pred /frame; run; 24 Lecture 3 Page 24

25 proc univariate data=one noprint; var resp; by trt; output out=two mean=mu std=sigma; data three; set two; logmu = log(mu); logsig = log(sigma); proc reg; model logsig = logmu; title1 Mean vs Std Dev ; symbol1 v=circle i=rl; proc gplot; plot logsig*logmu / regeqn; run; 25 Lecture 3 Page 25

26 Example: Formal Method (trans1.sas) data one; infile c:\saswork\data\boxcox.dat ; input trt resp; logresp = log(resp); proc univariate data=one noprint; var logresp; output out=two mean=mlogresp; data three; set one; if _n_ eq 1 then set two; ydot = exp(mlogresp); do l=-2.0 to 2.0 by.25; den = l*ydot**(l-1); if abs(l) eq 0 then den = 1; yl=(resp**l -1)/den; if abs(l) < then yl=ydot*log(resp); output; end; keep trt yl l; proc sort data=three out=three; by l; 26 Lecture 3 Page 26

27 proc glm data=three noprint outstat=four; class trt; model yl=trt; by l; data five; set four; if _SOURCE_ eq ERROR ; keep l SS; proc print data=five; run; SS E (λ) andλ OBS L SS OBS L SS Lecture 3 Page 27

28 symbol1 v=circle i=sm50; proc gplot; plot SS*l; run; Plot of SS E (λ) vsλ 28 Lecture 3 Page 28

29 Using PROC TRANSREG proc transreg data=one; model boxcox(y/lambda=-2.0 to 2.0 by 0.1)=class(trt); run; Transformation Information for BoxCox(y) Lambda R-Square Log Like : : : * * < * * < Best Lambda : : : * Confidence Interval Convenient Lambda 29 Lecture 3 Page 29

30 Kruskal-Wallis Test: a Nonparametric Alternative for Nonnormality a treatments,h 0 : a treatments are not different. Rank the observationsy ij in ascending order Replace each observation by its rankr ij (assign average for tied observations), and apply one-way ANOVA tor ij. Test statistic (a = 2 = Wilcoxon rank-sum test) [ a ] H = 1 R 2 S 2 i. i=1 n i N(N+1)2 4 χ 2 a 1 underh 0 [ a wheres 2 = 1 ] ni N 1 i=1 j=1 R2 ij N(N+1)2 4 Decision Rule: rejecth 0 ifh > χ 2 α,a 1. LetF 0 be thef -test statistic in ANOVA based onr ij. Then F 0 = H/(a 1) (N 1 H)/(N a) 30 Lecture 3 Page 30

31 A Nonnormality Example data new; input strain nitrogen cards; ; proc glm data=new; class strain; model nitrogen=strain; output out=newres r=res; run; proc univariate data=newres normal; var res; qqplot res / normal (L=1 mu=est sigma=est); run; quit; 31 Lecture 3 Page 31

32 Tests for Normality Test Statistic p Value Shapiro-Wilk W Pr < W Kolmogorov-Smirnov D Pr > D Cramer-von Mises W-Sq Pr > W-Sq Anderson-Darling A-Sq Pr > A-Sq Lecture 3 Page 32

33 proc npar1way data=new; /* May need option: wilcoxon */ class strain; var nitrogen; run; Analysis of Variance for Variable nitrogen Classified by Variable strain Source DF Sum of Squares Mean Square F Value Pr > F Among <.0001 Within Kruskal-Wallis Test Chi-Square DF 5 Pr > Chi-Square Median One-Way Analysis Chi-Square DF 5 Pr > Chi-Square Lecture 3 Page 33

34 Linear Combinations of Treatment Means ANOVA Model: y ij = µ + τ i + ǫ ij (τ i : treatment effect) = µ i + ǫ ij (µ i : treatment mean) Linear combination with given coefficientsc 1,c 2,...,c a : L = c 1 µ 1 +c 2 µ c a µ a = a c i µ i, i=1 Want to test: H 0 : L = c i µ i = L 0 Examples: 1. Pairwise comparison: µ i µ j = 0 for all possibleiandj. 2. Compare treatment vs control: µ i µ 1 = 0 when treatment 1 is a control andi = 2,...,a are new treatments. 3. General cases such asµ 1 2µ 2 +µ 3 = 0,µ 1 +3µ 2 6µ 3 = 0, etc. 34 Lecture 3 Page 34

35 Estimate ofl: ˆL = c iˆµ i = c i ȳ i. Var(ˆL) = c 2 i Var(ȳ i.) = σ 2 c 2 i n i (= σ2 n c 2 i ) Standard Error of ˆL S.E.{ˆL} = MSE c 2 i n i Test statistic t 0 = (ˆL L 0 ) S.E.{ˆL} t(n a) underh 0 35 Lecture 3 Page 35

36 Example: Lambs Diet Experiment Denote the treatment means of three diets byµ 1,µ 2 andµ 3. Suppose one wants to testh 0 : L = 60 with L = µ 1 +2µ 2 +3µ 3 = 6µ+τ 1 +2τ 2 +3τ 3. proc glm data=lambs; class diet; model wtgain=diet; means diet; estimate l1 intercept 6 diet 1 2 3; run; Standard Parameter Estimate Error t Value Pr > t l <.0001 t 0 = ( )/8.89 = 1.91 P value = P(t 1.91 ort 1.91 t(12 3)) =.088 Fail to rejecth 0 : µ 1 +2µ 2 +3µ 3 = 60 atα = 5%. 36 Lecture 3 Page 36

37 Contrasts Γ = a i=1 c iµ i is a contrast if a i=1 c i = 0. Equivalently,Γ = a i=1 c iτ i. Examples 1. Γ 1 = µ 1 µ 2 = µ 1 µ 2 +0µ 3 +0µ 4, c 1 = 1,c 2 = 1,c 3 = 0,c 4 = 0 Comparingµ 1 andµ Γ 2 = µ 1 0.5µ 2 0.5µ 3 = µ 1 0.5µ 2 0.5µ 3 +0µ 4 c 1 = 1,c 2 = 0.5,c 3 = 0.5,c 4 = 0 Comparingµ 1 and the average ofµ 2 andµ 3. Estimate ofγ: C = a i=1 c iȳ i. 37 Lecture 3 Page 37

38 Contrast Sum of Squares SS C = ( ci y i. ) 2 / (c 2 i /n i ) SS C represents the amount of variation attributableγ. Test: H 0 : Γ = 0 t 0 = C S.E. C H 0 t(n a) t 2 0 = ( c i ȳ i. ) 2 MSE c 2 i n i = SS C /1 MSE H 0 F1,N a 38 Lecture 3 Page 38

39 Tensile Strength Example (cont.sas) proc glm data=one; class percent; model strength=percent; contrast C1 percent ; contrast C2 percent ; contrast C3 percent ; contrast C4 percent ; Dependent Variable: STRENGTH Sum of Mean Source DF Squares Square F Value Pr > F Model Error Corrected Total Source DF Type I SS Mean Square F Value Pr > F PERCENT Contrast DF Contrast SS Mean Square F Value Pr > F C C C C Lecture 3 Page 39

40 Orthogonal Contrasts Two contrasts{c i } and{d i } are Orthogonal if a i=1 c i d i n i = 0 ( a c i d i = 0 for balanced experiments) i=1 Example (Balanced Experiment) Γ 1 = µ 1 +µ 2 µ 3 µ 4, Soc 1 = 1,c 2 = 1,c 3 = 1,c 4 = 1. Γ 2 = µ 1 µ 2 +µ 3 µ 4. Sod 1 = 1,d 2 = 1,d 3 = 1,d 4 = 1 It is easy to verify that bothγ 1 andγ 2 are contrasts. Furthermore, c 1 d 1 +c 2 d 2 +c 3 d 3 +c 4 d 4 = ( 1)+( 1) 1+( 1) ( 1) = 0. Hence,Γ 1 andγ 2 are orthogonal to each other. A complete set of orthogonal contrastsc = {Γ 1,Γ 2,...,Γ a 1 } if contrasts are mutually orthogonal and there does not exist a contrast orthogonal outside ofc to all the contrasts inc. 40 Lecture 3 Page 40

41 If there are a treatments, C must contain a 1 contrasts. Complete set is not unique. For example, in the tensile strength example C 1 : includes : Γ 1 = (0, 0, 0, 1, 1) Γ 2 = (1, 0, 1, 1, 1) Γ 3 = (1, 0, 1, 0, 0) Γ 4 = (1, 4, 1, 1, 1) C 2 : includes : Γ 1 = ( 2, 1, 0, 1, 2) Γ 2 = (2, 1, 2, 1, 2) Γ 3 = ( 1, 2, 0, 2, 1) Γ 4 = (1, 4, 6, 4, 1) 41 Lecture 3 Page 41

42 SupposeC 1,C 2,...,C a 1 are the estimates of the contrasts in a complete set of contrasts{γ 1,Γ 2,...,Γ a 1 }, then SS Treatment = SS C1 + SS C2 + +SS Ca 1 F 0 = MS Treatment MSE = F 10 +F F (a 1)0 a 1 wheref i0 is the test statistic used to test contrastγ i. Orthogonal contrasts (estimates) are independent with each other The results follow Cochran s Theorem, so comparisons are indepedent Example on Slide 39 Can also use orthogonal contrasts to study trend Only interesting if treatments are quantitative (ordered): X 1,,X a For equally spaced treatments andn i = n,c i in Table IX Breakdown of polynomial regressionµ(x) = β 0 +β 1 x+ +β a 1 x a 1 = µ(x) = β 0 + β 1 P 1 (x)+ + β a 1 P a 1 (x) P k (x) is of k-th order, and a i=1 P k(x i ) = 0 (constrast) 42 Lecture 3 Page 42

43 Determining Orthogonal Polynomial Coefficients Using SAS Complete set: {(P k (X 1 ),,P k (X a )) : k = 1,,a 1} Often the levels of the treatments are not equally spaced Can useproc IML to determine coefficients(p k (X 1 ),,P k (X a )) proc iml; levels={ }; /* Consider these 5 levels */ print levels; coef=orpol(levels,4); /* Gives all coefs up through 4th order */ coef=t(coef); /* Puts coefs in rows instead of cols */ coef=coef[2:5,]; /* Eliminates the first row of coef matrix*/ print coef; run; LEVELS COEF Lecture 3 Page 43

44 One contrast: Testing Multiple Contrasts (Multiple Comparisons) H 0 : Γ = c i µ i = Γ 0 vs H 1 : Γ Γ 0 atα 100(1-α) Confidence Interval (CI) forγ: CI : c i ȳ i. ±t α/2,n a MS E c 2 i n i P(CI does not containl 0 H 0 ) = α(= type I error rate) Decision Rule: RejectH 0 if CI does not containγ Lecture 3 Page 44

45 Multiple contrasts, fori = 1,2,,m, H (i) 0 : Γ (i) = Γ (i) 0, vs H(i) 1 : Γ (i) Γ (i) 0 If we construct CI 1, CI 2,..., CI m, each with 100(1-α) level, then for each CI i, P(rejectH (i) 0 H (i) 0 ) = P(CI i does not containγ (i) 0 H (i) 0 ) = α But the overall Type I error rate (probability of any type I error in testing H (i) 0 vsh (i) 1,i = 1,,m) is inflated and much larger thanα, that is, P(reject at least one of{h (i) 0 = P(at least one CI i do not containγ (i) α,i = 1,,m} H(1) 0,,H(m) 0 ) 0 H (1) 0,,H(m) 0 ) One way to achieve small overall error rate, we require much smaller error rate (α ) of each individual CI i. 45 Lecture 3 Page 45

46 Bonferroni Inequality Bonferroni Method for Testing Multiple Contrasts P(reject at least one of{h (i) 0 = P(rejectH (1) 0 or rejecth (2) 0 or... or rejecth (m),i = 1,,m} H(1) 0,,H(m) 0 ) 0 H (1) 0,,H(m) 0 ) P(rejectH (1) 0 H (1) 0 )+ +P(rejectH(m) 0 H (m) 0 ) = mα In order to control overall error rate (or, overall confidence level), let mα = α, we have,α = α/m Bonferroni CIs: CI i : c ij ȳ j. ±t α/2m (N a) MS E c 2 ij n j When m is large, Bonferroni CIs are too conservative ( overall type II error too large). 46 Lecture 3 Page 46

47 Scheffe s Method for Testing All Contrasts Consider all possible contrasts: Γ = c i µ i Estimate: C = c i ȳ i., St. Error: S.E. C = MS E c 2 i n i Critical value: (a 1)Fα,a 1,N a Scheffe s simultaneous CI:C ± (a 1)F α,a 1,N a S.E. C Overall confidence level and error rate formcontrasts P(CIs contain true parameter for any contrast) 1 α P(at least one CI does not contain true parameter) α Remark: Scheffe s method is also conservative, too conservative whenmis small 47 Lecture 3 Page 47

48 Methods for Pairwise Comparisons There area(a 1)/2 possible pairs: µ i µ j (contrast for comparingµ i andµ j ). We may be interested inmpairs or all pairs. Standard Procedure: 1. Estimation: ȳ i. ȳ j. 2. Compute a Critical Difference (CD) (based on the method employed) 3. If or equivalently if the interval ȳ i. ȳ j. > CD (ȳ i. ȳ j. CD, ȳ i. ȳ j. +CD) does not contain zero, declareµ i µ j significant. 48 Lecture 3 Page 48

49 Least significant difference (LSD): not control overall error rate Bonferroni method (for m pairs) Methods for Calculating CD CD = t α/2,n a MS E (1/n i +1/n j ) CD = t α/2m,n a MS E (1/n i +1/n j ) control overall error rate for themcomparisons. Tukey s method (for all possible pairs) CD = q α(a,n a) 2 MS E (1/n i +1/n j ) q α (a,n a) from studentized range distribution (Table VII). Control overall error rate (exact for balanced experiments). (Example 3.7). 49 Lecture 3 Page 49

50 Comparing Treatments with Control (Dunnett s Method) 1. Assumeµ 1 is a control, andµ 2,...,µ a are (new) treatments 2. Only interested ina 1pairs: µ 2 µ 1,...,µ a µ 1 3. Compare ȳ i. ȳ 1. to CD = d α (a 1,N a) MS E (1/n i +1/n 1 ) whered α (p,f) from Table VIII: critical values for Dunnett s test. 4. Remark: control overall error rate. Read Example 3.9 For pairwise comparison, which method should be preferred? LSD, Bonferroni, Tukey, Dunnett or others? 50 Lecture 3 Page 50

51 Tensile Strength Example data one; infile c:\saswork\data\tensile.dat ; input percent strength time; proc glm data=one; class percent; model strength=precent; /* Construct CI for Treatment Means*/ means percent /alpha=.05 lsd clm; means percent / alpha=.05 bon clm; /* Pairwise Comparison*/ means percent /alpha=.05 lines lsd; means percent /alpha=.05 lines bon; means percent /alpha=.05 lines scheffe; means percent /alpha=.05 lines tukey; means percent /dunnett; run; 51 Lecture 3 Page 51

52 The GLM Procedure t Confidence Intervals for y Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of t Half Width of Confidence Interval % Confidence trt N Mean Limits Lecture 3 Page 52

53 The GLM Procedure Bonferroni t Confidence Intervals for y Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of t Half Width of Confidence Interval Simultaneous 95% trt N Mean Confidence Limits Lecture 3 Page 53

54 t Tests (LSD) for y NOTE: This test controls the Type I comparisonwise error rate, not the experimentwise error rate. Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of t Least Significant Difference Means with the same letter are not significantly different. t Grouping Mean N trt A B B C C Lecture 3 Page 54

55 Bonferroni (Dunn) t Tests for y NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of t Minimum Significant Difference Means with the same letter are not significantly different. Bon Grouping Mean N trt A B A B C C C Lecture 3 Page 55

56 Scheffe s Test for y NOTE: This test controls the Type I experimentwise error rate. Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of F Minimum Significant Difference Means with the same letter are not significantly different. Scheffe Grouping Mean N trt A A B A B B C C C C C Lecture 3 Page 56

57 Tukey s Studentized Range (HSD) Test for y NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of Studentized Range Minimum Significant Difference Means with the same letter are not significantly different. Tukey Grouping Mean N trt A A B A B B C C D C D D Lecture 3 Page 57

58 Dunnett s t Tests for y NOTE: This test controls the Type I experimentwise error for comparisons of all treatments against a control. Alpha 0.05 Error Degrees of Freedom 20 Error Mean Square 8.06 Critical Value of Dunnett s t Minimum Significant Difference Comparisons significant at the 0.05 level are indicated by ***. Difference trt Between Simultaneous 95% Comparison Means Confidence Limits *** *** *** Lecture 3 Page 58