Lecture 5: Comparing Treatment Means Montgomery: Section 3-5

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1 Lecture 5: Comparing Treatment Means Montgomery: Section 3-5 Page 1

2 Linear Combination of Means ANOVA: y ij = µ + τ i + ɛ ij = µ i + ɛ ij Linear combination: L = c 1 µ 1 + c 1 µ c a µ a = a i=1 c iµ i Want to test: H 0 : L = c i µ i = L 0 pairwise comparison: µ i µ j =0 treatment vs. control: µ i µ 1 =0if treatment 1 is control comparing combinations of treatment effects: µ 1 2µ 2 + µ 3 =0. high order effects (orthogonal polynomials): linear, quadratic, etc. Page 2

3 Estimate of L: ˆL = 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 for ˆL S.E. ˆL = MSE c 2 i n i Test statistic t 0 = (ˆL L 0 ) S.E. ˆL t(n a) under H 0 Page 3

4 Example: Lambs Diet Experiment Recall there are three diets and their treatment means are denoted by µ 1, µ 2 and µ 3. Suppose one wants to consider and test H 0 : L =60. L = µ 1 +2µ 2 +3µ 3 =6µ + τ 1 +2τ 2 +3τ 3 data lambs; input diet wtgain@@; cards; ; proc glm; class diet; model wtgain=diet; means diet; estimate l1 intercept 6 diet 1 2 3; run; Page 4

5 Example: Lambs Diet Experiment SAS output Level of wtgain diet N Mean Std Dev Dependent Variable: wtgain Standard Parameter Estimate Error t Value Pr > t l <.0001 t 0 = ( )/8.89 = 1.91 P value = P (t 1.91 or t 1.91 t(12 3)) =.088 Accept H 0 : µ 1 +2µ 2 +3µ 3 =60at α =5%. Page 5

6 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 Comparing µ 1 and µ 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 Comparing µ 1 and the average of µ 2 and µ 3. Estimate for Γ: C = a i=1 c iȳ i. Page 6

7 Test: H 0 :Γ=0 t 0 = C S.E. C t(n a) t 2 0 = ( c i ȳ i. ) 2 MSE c 2 i Under H 0, t 2 0 F 1,N a. Contrast Sum of Squares n i = ( c i ȳ i. ) 2 / c 2 i /n i MSE = SS C/1 MSE SS C = ( ci y i. ) 2 / (c 2 i /n i ) SS C is the amount of variation caused by Γ. Page 7

8 SAS Code (cont.sas) Tensile Strength Example options ls=80; title1 Contrast Comparisons ; data one; infile c:\saswork\data\tensile.dat ; input percent strength time; proc glm data=one; class percent; model strength=percent; contrast C1 percent ; contrast C2 percent ; contrast C3 percent ; contrast C4 percent ; Page 8

9 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 Page 9

10 Orthogonal Contrasts Two contrasts {c i } and {d i } are Orthogonal if a i=1 c i d i n i =0 Example 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 both Γ 1 and Γ 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 and Γ 2 are orthogonal to each other. A complete set of orthogonal contrasts C = {Γ 1, Γ 2,...,Γ a 1 } if contrasts are mutually orthogonal and there does not exist a contrast orthogonal to all contrasts in C. Page 10

11 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) Page 11

12 Orthogonal Contrasts Orthoogonal contrasts (estimates) are independent with each other. Suppose C 1,C 2,...,C a 1 are the estimates of the contrasts in a complete set of comtrasts {Γ 1, Γ 2,...,Γ a 1 }, then SS Treatment = SS C1 + SS C2 + + SS Ca 1 Recall in ANOVA, F 0 = MS Treatment MSE, F 0 = SS C 1 /MSE + + SS Ca 1 /MSE a 1 = F 10 + F F (a 1)0 a 1 where F i0 is the test statistic used to test contrast Γ i. Example on Slide 9 Page 12

13 Tensile Example Try to model mean response as a function of treatments Page 13

14 Orthogonal contrasts and orthogonal polynomial model Treatments are quantitative (assume a =4) One can use general polynomial model to fit the trend (t: level or treatment). f(t) =a 0 + a 1 t + a 2 t 2 + a 3 t 3 Regression can be used to get the estimates for a 1, a 2 and a 3. We will use orthogonal polynomial model f(t) =β 0 + β 1 P 1 (t)+β 2 P 2 (t)+β 3 P 3 (t) where P 1 (t), P 2 (t) and P 3 (t) are pre-specified polynomials of order 1, 2 and 3, respectively. P 1 (t) is linear, P 2 (t) is quadratic and P 3 (t) is cubic. Let t 1, t 2,..., t a are the treatments (equally spaced), then the polynomials corresponds to the following contrasts: Page 14

15 t t 1 t 2 t a Contrasts D P 1 (t) P 1 (t 1 ) P 1 (t 2 ) P 1 (t a ) Γ 1 D 1 P 2 (t) P 2 (t 1 ) P 2 (t 2 ) P 2 (t a ) Γ 2 D 2 P 3 (t) P 3 (t 1 ) P 3 (t 2 ) P 3 (t a ) Γ 3 D 3 where D i = P i (t 1 ) 2 + P i (t 2 ) P i (t a ) 2 If Γ 1, Γ 2 and Γ 3 are orthogonal to each other, then we say P 1 (t), P 2 (t) and P 3 (t) are orthogonal polynomials. Coefficients β i can be estimated and tested by the contrasts Γ 1, Γ 2 and Γ 3. Predict f(t) when t is not a treatment used in the experiment. Page 15

16 tensile strength example: orthogonal polynomial effects Treatment levels t k : 15, 20, 25, 30, 35; Median: 25; Pace: 5 Orthogonal polynomials: let x =(t 25)/5. P 1 (t) =x P 2 (t) =x 2 2 P 3 (t) =5/6[x 3 17x/5] P 4 (t) =35/12[x 4 31x/7+72/35] Polynomial Contrasts and Effects t Contrast D Effect (Trend) P 1 (t) Γ 1 D 1 =10 linear P 2 (t) Γ 2 D 2 =14 quadratic P 3 (t) Γ 3 D 3 =10 cubic p 4 (t) Γ 4 D 4 =70 4th order The contrasts can be directly derived from Table IX or Table X. Page 16

17 Want to fit the model f(t) =β 0 + β 1 P 1 (t)+β 2 P 2 (t)+β 3 P 3 (t)+β 4 P 4 (t) Estimation and Testing β 1 :useγ 1, ˆβ 1 = c 11ȳ c 15 ȳ 5. D 1 Test: H 0 : β 1 =0, F 10 = SS C i MSE F 1,N 5. β 2 :useγ 2, ˆβ 2 = c 21ȳ c 25 ȳ 5. D 2 Test: H 0 : β 2 =0, F 20 = SS C i MSE F 1,N 5 Similar for β 3 and β 4 Question: what is the estimate for β 0? Page 17

18 General formulas for orthogonal polynomial of degrees 1-4 One factor of a levels l 1, l 2,..., l a, equally spaced. Let m be the median, δ be the difference between two consecutive levels: P 1 (t) =λ 1 ( t m ) δ P 2 (t) =λ 2 [( t m ) 2 a2 1 δ 12 ] P 4 (t) =λ 4 [( t m δ P 3 (t) =λ 3 [( t m δ ) 4 ( t m δ ) 3 ( t m δ )( 3a2 7 )] 20 ) 2 ( 3a2 13 )+ 3(a2 1)(a 2 9) (λ i ) are constants to make the polynomials have integer values at the treatment levels, they are available from Table IX or Table X. Tensile Strength Example: m=25, δ =5,(λ i )=(1, 1, 5/6, 35/12) Page 18

19 SAS tensile strength example data one; infile c:\saswork\data\tensile.dat ; input percent strength time; proc glm data=one; class percent; model strength=percent; estimate C1 percent ; estimate C2 percent ; estimate C3 percent ; estimate C4 percent ; contrast C1 percent ; contrast C2 percent ; contrast C3 percent ; contrast C4 percent ; run; Page 19

20 Output Dependent Variable: strength Sum of Source DF Squares Mean Square F Value Pr > F Model <.0001 Error Corrected Total Parameter Estimate Error t Value Pr > t C C <.0001 C C Contrast DF Contrast SS Mean Square F Value Pr > F C C <.0001 C C Page 20

21 Estimates Hence, ˆβ 1 =8.20/10 =.82; ˆβ2 = 31/14 = ˆβ 3 = 11.4/10 = 1.14; ˆβ4 = 21.8/70 = So the fitted functional relationship between tensile strength y and cotton percent (t) is y = ˆβ P 1 (t) 2.214P 2 (t) 1.14P 3 (t) 0.311P 4 (t), where P 1 (t),...,p 4 (t) are defined on Slide 16. Page 21

22 Testing Multiple Contrasts (Multiple Comparisons) Using Confidence Intervals One contrast: H 0 :Γ= c i µ i =Γ 0 vs H 1 :Γ Γ 0 at α 100(1-α) Confidence Interval (CI) for Γ: CI : c i ȳ i. ± t α/2,n a MS E c 2 i n i P (CI not contain L 0 H 0 )=α(= type I error) Decision Rule: Reject H 0 if CI does not contain Γ 0. Page 22

23 Multiple contrasts H 0 :Γ 1 =Γ 1 0,...Γ m =Γ m 0 vs H 1 : at least one does not hold If we construct CI 1,CI 2,..., CI m, each with 100(1-α) level, then for each CI i, P (CI i not containγ i 0 H 0 )=α, for i =1,...,m But the overall error rate (probability of type I error for H 0 vs H 1 ) is inflated and much larger than α, that is, P (at least one CI i not contain Γ i 0 H 0 ) >> α One way to achieve small overall error rate, we require much smaller error rate (α ) of each individual CI i. Page 23

24 Bonferroni Inequality Bonferroni Method for Testing Multiple Contrasts P ( at least one CI i not contain Γ i 0 H 0 ) = P (CI 1 not contain..or...or CI m not contain H 0 ) P ( CI 1 not H 0 )+ + P ( CI m not H 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). Page 24

25 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 for m contrasts 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 when m is small Page 25

26 Methods for Pairwise Comparisons There are a(a 1)/2 possible pairs: µ i µ j (contrast for comparing µ i and µ j ). We may be interested in m pairs or all pairs. Standard Procedure: 1. Estimation: ȳ i. ȳ j. 2. Compute a Critical Difference (CD) (based on the method employed) 3. If ȳ i. ȳ j. > CD or equivalently if the interval (ȳ i. ȳ j. CD, ȳ i. ȳ j. +CD) does not contain zero, declare µ i µ j significant. Page 26

27 Methods for Calculating CD. Least significant difference (LSD): not control overall error rate Bonferroni method (for m pairs) 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 the m comparisons. 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 or Table VIII). Control overall error rate (exact for balanced experiments). (Example 3.7). Page 27

28 Comparing treatments with control (Dunnett s method) 1. Assume µ 1 is a control, and µ 2,...,µ a are (new) treatments 2. Only interested in a 1 pairs: 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 ) where d α (p, f) from Table IX or Table VIII: critical values for Dunnett s test. 4. Remark: control overall error rate. Read Example 3-9 (or 3-10) For pairwise comparison, which method should be preferred? LSD, Bonferroni, Tukey, Dunnett or others? Page 28

29 SAS Code 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; Page 29

30 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 Page 30

31 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 Page 31

32 NOTE: This test controls the Type I comparisonwise error rate, not t t Tests (LSD) for y 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 Page 32

33 NOTE: This test controls the Type I experimentwise error rate, but it ge Bonferroni (Dunn) t Tests for y 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 Page 33

34 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 Page 34

35 NOTE: This test controls the Type I experimentwise error rate, but it ge Tukey s Studentized Range (HSD) Test for y 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 Page 35

36 NOTE: This test controls the Type I experimentwise error for comparisons Dunnett s t Tests for y 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 *** *** *** Page 36

37 Determining Sample Size More replicates required to detect small treatment effects Operating Characteristic Curves for F tests Probability of type II error β = P ( accept H 0 H 0 is false) = P (F 0 <F α,a 1,N a H 1 is correct ) Under H 1, F 0 follows a noncentral F distribution with noncentrality λ and degrees of freedom, a 1 and N a. Let Φ 2 = n a i=1 τ 2 i aσ 2 OC curves of β vs n and Φ are included in Chart V for various α and a. Read Example 3.10 or Page 37