I L L I N O I S UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN

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1 : Part 2 Profile Analsysis and 2 way Edps/Soc 584 and Psych 594 Applied Multivariate Statistics Carolyn J. Anderson Department of Educational Psychology I L L I N O I S UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN c Board of Trustees, University of Illinois : Part 2 Slide 1 of 75

2 Outline Extensions of 1 Way : Two groups Two or more groups Two-Way 2 way Review 2-way ANOVA 2-way Example Reading: Johnson & Wichern pages xx xx. : Part 2 Slide 2 of 75

3 Example of Profiles for Two Groups Two-Way Profile analysis is an extension of 1 way involving p response variables administered to g groups of individuals or cases. e.g., A battery of personality tests (sub-scales) where the p test scores are measured in the same units (or at least similar or commensurate units). gives you more specific hypotheses to test than the standard H o : µ 1 = µ 2 = = µ g. Consider p 1 mean vector for each group µ l = (µ l1,µ l2,...,µ lp ) and plot the means as profiles for each group. : Part 2 Slide 3 of 75

4 Example of Profiles for Two Groups 15 Example of Profiles for Two Groups Mean Value 10 5 Group 1 Group 2 Two-Way One Two Three...p Variable : Part 2 Slide 4 of 75

5 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles : The usual null hypothesis is H o : µ 1 = µ 2, but in profile analysis, we ll also consider both a more general and more restrictive hypotheses. The Three Questions (Steps): Question 1: Are the profiles parallel? or Are successive (adjacent) differences between means equal? i.e., H o1 : µ 1i µ 1,i 1 = µ 2i µ 2,i 1 for all i = 2,...,p Mean Value µ 11 µ 21 µ 12 µ Variables Group 1 Group 2 : Part 2 Slide 5 of 75

6 Question 2 Assuming that profiles are parallel, are they coincident? or Group 1 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Are the population means equal for the two groups? i.e., H o2 : µ 1i = µ 2i for all i = 1,...p; that is, H o2: µ 1 = µ 2 Mean Value µ 21 = µ 22 = µ 11 µ 12 Group Variables : Part 2 Slide 6 of 75

7 Question 3 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Assuming that the profiles are coincident, are they Level (flat)? or Are all the means equal? (over groups and variables). H o3 : µ 11 = µ 12 = = µ 1p = µ 21 = µ 22 = = µ 2p Mean Value Group 1 µ µ 1i 2i = Group Variables : Part 2 Slide 7 of 75

8 Testing Question 1: Parallel Assuming X 1j N(µ 1,Σ) and X 2j N(µ 2,Σ) : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles for j = 1,...,n l, and independent. Question 1: Are the profiles parallel? We can write the hypothesis are H o1 : Cµ 1 = Cµ 2 where C (p 1) p contrast matrix. e.g., C (p 1) p = What we re doing is linearly transforming our original p variables into (p 1) new variables. : Part 2 Slide 8 of 75

9 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Question 1 continued When we take CX = X 1 +X 2 X 2 +X 3... X p 1 +X p, which are measured on both groups (populations, etc). Random variable CX 1j N p 1 (Cµ 1,CΣC ), and Random variable CX 2j N p 1 (Cµ 2,CΣC ). To estimate the covariance matrix Σ, use S pool, S pool (n 1 1)S 1 +(n 2 1)S 2 n 1 +n 2 2 and the estimate of CΣC equals CS pool C. : Part 2 Slide 9 of 75

10 Finishing Question 1 To test H o1 : Cµ 1 Cµ 2 = C(µ 1 µ 2 ) = 0, we can just use Hotelling s T 2 for Two independent samples. : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Reject H o1 if [( T 2 = ( X 1 X 1 2 ) C + 1 ) ] 1 CS pool C C( n 1 n X 1 X 2 ) > c 2 2 where c 2 = (n 1 +n 2 2)(p 1) n 1 +n 2 p F (p 1),(n1 +n 2 p)(α) If you reject H o1 STOP.... You can do any follow-up to examine differences. If you retain H o2 Conclude profiles are parallel and PROCEED to next question. : Part 2 Slide 10 of 75

11 Testing Question 2: Coincident profiles Assuming that the the profiles are parallel, are the profiles coincident? If profiles are parallel, then one will be above the other for all i = 1,...,p; that is, : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles µ 1j > µ 2j for all i = 1,...,p or µ 1j < µ 2j for all i = 1,...,p So, profiles will be coincident only if the total heights are equal (µ 11 +µ µ 1p ) = (µ 21 +µ µ 2p ) 1 pµ 1 = 1 pµ 2 where 1 p is a (p 1) vector of ones. : Part 2 Slide 11 of 75

12 Question 2 continued The null hypothesis for question 2 is H o2 : 1 µ 1 = 1 µ 2 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles We are forming a new variable, 1 X, which is a simple sum and test whether this variable is equal for the two groups. We estimate 1 X 1j N 1 (1 µ 1,1 Σ1) j = 1,...,n 1 1 X 2j N 1 (1 µ 2,1 Σ1) j = 1,...,n 2 1 µ 1 by 1 x 1 1 µ 2 by 1 x 2 and 1 Σ1 by 1 S pool 1 : Part 2 Slide 12 of 75

13 Finishing Up Question 2 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles To test H o2: 1 µ 1 = 1 µ 2, we can do a simple univariate 2 independent sample t-test. We will reject H o2 at the α-level if t = 1 ( x 1 x 2 ) ( > t n1 +n 2 2(α) 1 n n 2 )1 S pool 1 If Reject H o2 STOP and conclude that the profiles are parallel but not coincident. If Retain H o2 PROCEED and test whether profiles are flat. : Part 2 Slide 13 of 75

14 Question 3: Flat If the profiles are coincident, do they all have the same mean? i.e., : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles H o3 : µ 11 = µ 12 = = µ 1p = µ 21 = µ 22 = = µ 2p We can test this using a contrast matrix C, such as the one we used for Question 1 (testing parallel profiles) H o3 : C(µ 1 +µ 2 ) = 0 Note that we are adding rather than subtracting as we did in H o1. This tests (µ 12 µ 11 )+(µ 22 µ 21 ) (µ 13 µ 12 )+(µ 23 µ 22 ). (µ 1p µ 1,p 1 )+(µ 2p µ 2,p 1 ) Cµ 1 +Cµ 2 = 0 = 0 : Part 2 Slide 14 of 75

15 Question 3 continued Testing : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Cµ 1 +Cµ 2 = 0 (µ 12 µ 11 )+(µ 22 µ 21 ) (µ 13 µ 12 )+(µ 23 µ 22 ). = 0 (µ 1p µ 1,p 1 )+(µ 2p µ 2,p 1 ) Is the same as this, which is what we want to test, (µ 12 +µ 22 ) = (µ 11 +µ 21 ) (µ 13 +µ 23 ) = (µ 12 +µ 22 ). (µ 1p +µ 2p ) = (µ 1,p 1 +µ 2,p 1 ) Because If the profiles are coincident, then µ 1i = µ 2i for all i = 1,...,p. So if H o3 is true, then C(µ 1 +µ 2 ) = 0.. : Part 2 Slide 15 of 75

16 Question 3 continued For H o3 : C(µ 1 +µ 2 ) = 0, flat profiles, we are taking linear combinations in two ways simultaneously Over variables via the use of C. : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Over groups via an additive function. The mean C(µ 1 +µ 2 ) is estimated by taking the grand mean n1 j=1 X = X 1j + n 2 j=1 X 2j = n X 1 1 +n 2 X2 n 1 +n 2 n 1 +n 2 and the distribution is of C X is C X N p 1 ( C(µ 1 +µ 2 ),CΣC ( How do we get this? 1 n 1 +n 2 )) : Part 2 Slide 16 of 75

17 Finishing up Question 3 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Reject H o3 : C(µ 1 +µ 2 ) = 0 (i.e., that profiles are flat) at α-level if (n 1 +n 2 ) X C (CSC) 1 C X > (n 1 +n 2 1)(p 1) n 1 +n 2 p+1 where S is the total sample covariance matrix 1 n 1 S = (X 1j n 1 +n 2 1 X)(X 1j X) + j=1 n 2 j=1 Note that we use the total sample mean in computing S. F (p 1),(n1 +n 2 p+1(α) (X 2j X)(X 2j X) : Part 2 Slide 17 of 75

18 Example: WAIS data This example is from Morrison (2005). : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Fourty nine elderly men participating in an interdisciplinary study of human aging were classified into the diagnostic categories senile factor present and no senile factor on the basis of an intensive psychiatric examination. The Wechsler Adult Intelligence Scale (WAIS) was administered to all subjects by an independent investigator. Below are mean scores by group on some of the WAIS subtests. Not Senile (n = 37) Senile (n = 12) Sub-Test x std dev x std dev Information Similarities Note: Arithmetic Picture My results differ slightly from Morrison s. There is either a typo in the text or in the data set. (no way to find out which). : Part 2 Slide 18 of 75

19 WAIS Profiles 15 : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Sub-test Group Mean 10 5 Picture completion Arithmetic Similarities Information WAIS Sub-Tests No Senile Factor Senile Factor : Part 2 Slide 19 of 75

20 WAIS: Are Profiles Parallel? The within group (pooled) sample covariance matrix: : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles S pool = = 1 n 1 +n 2 2 ((n 1 1)S 1 +(n 2 1)S 2 ) Test for parallel profiles: The hypothesis is that H o : µ 1i µ 1,i 1 = µ 2,i µ 2,i 1 for i = 2,3,4; that is, simultaneously, the following three qualities hold in the population, µ 12 µ 11 = µ 22 µ 21 µ 13 µ 12 = µ 23 µ 22 µ 14 µ 13 = µ 24 µ 23 : Part 2 Slide 20 of 75

21 WAIS: Are Profiles Parallel? : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles To test this hypothesis, we need a (3 4) contrast matrix C C = Our hypothesis can now be,expressed as H o : Cµ 1 = Cµ 2, or H o : C(µ 1 µ 2 ) = 0. This hypothesis is tested by Hotelling s T 2, and we need and CS pool C = ( x n x y ) = (3.818,4.153,3.014,3.223), (C( x n x y )) = (0.336, 1.140,0.209), : Part 2 Slide 21 of 75

22 WAIS: Are Profiles Parallel? Putting all of this into our equation for T 2 gives us : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles T 2 n 1 n 2 = ( x 1 x 2 ) C (CS pool C ) 1 C( x 1 x 2 ) n 1 +n 2 = and compare the following statistic to the F distribution, F = n 1 +n 2 p (n 1 +n 2 2)(p 1) T2 = = (47)(3) Since F 3,45 (α =.05) = is greater than our observed statistic (or the p-value of F equals.76), we do not reject the hypothesis that the profiles are parallel. Since we Retained the null, we will PROCEED to test whether the profiles are coincident. : Part 2 Slide 22 of 75

23 WAIS: Are Profiles Test for Equal (coincident) Profiles. We now test H o : µ 1i = µ 2i for i = 1,2,3,4 variables. Since we concluded that the profiles are parallel, then we can test this hypothesis by testing whether : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles (µ 11 +µ 12 +µ 13 +µ 14 ) = (µ 21 +µ 22 +µ 23 +µ 24 ) 1 µ 1 = 1 µ 2 where 1 is a (4 1) vector of ones. This is just a univariate, 2 independent sample t. To test H o : 1 (µ 1 µ 2 ) = 0, we need for each group the sums of the means over all variables, which are 1 x 1 = and 1 x 2 = We also need an estimate of the variance of the difference between means, 1 S pool 1 = : Part 2 Slide 23 of 75

24 WAIS: Are Profiles : Question 2 Question 3 Testing Question 1: Parallel Question 1 continued Finishing Question 1 Testing Question 2: Coincident profiles Question 2 continued Finishing Up Question 2 Question 3: Flat Finishing up Question 3 Example: WAIS data WAIS Profiles Putting all of these statistics together for our test statistics gives us t = ( ) = Since t 47 (α =.025) = (or the p-value of t 47 = 4.15 is <.001), we reject the null hypothesis. The profiles are not coincident. At this point we STOP and should not go on to test whether profiles are level (flat). : Part 2 Slide 24 of 75

25 Profiles Analysis for g To deal with two or more groups, we follow the same logic: 1. Using assumptions (including results from earlier tests), determine what the hypothesis implies for population parameters. This has implications for data. Profiles Analysis forg 2 Groups Two or More: Parallel GLM: Parallel and WAIS Two or More: Two or More: Level Profiles Two-Way 2. Compute a statistic that reflects the implications for data. 3. Find a transformation of the statistic with a known sampling distribution. It is easier to do profile analysis using the GLM framework. We will use X = AB+ǫ, and test hypotheses of the form H o : LBM = 0 Appropriate definitions of L and M will lead to tests for parallelism, coincidence, and level means. : Part 2 Slide 25 of 75

26 Two or More: Parallel Parallel profiles imply H o1 : (µ 1i µ 1,i 1 ) = (µ 2i µ 2,i 1 ) = = (µ gi µ g,i 1 ) Profiles Analysis forg 2 Groups Two or More: Parallel GLM: Parallel and WAIS Two or More: Two or More: Level Profiles Two-Way for all i = 2,...,p. This will be a on differences. This hypothesis takes linear combinations of the means within groups and then compare the groups. For the linear transformations of the regression parameters (i.e., the τ or µ s), define M p (p 1) = For the comparisons of groups on these new variables, define L = C (g 1) g contrast matrix. : Part 2 Slide 26 of 75

27 Profiles Analysis forg 2 Groups Two or More: Parallel GLM: Parallel and WAIS Two or More: Two or More: Level Profiles Two-Way GLM: Parallel and WAIS For the WAIS example, H o1 is LBM = (0, 1, 1) = (0,1, 1) β 01 β 02 β 03 β 04 β 11 β 12 β 13 β 14 β 21 β 22 β 23 β (β 01 β 02 ) (β 02 β 03 ) (β 03 β 04 ) (β 11 β 12 ) (β 12 β 13 ) (β 13 β 14 ) (β 21 β 22 ) (β 22 β 23 ) (β 23 β 24 ) = [(β 11 β 12 ) (β 21 β 22 ),(β 12 β 13 ) (β 22 β 23 ), So we have ( (β 11 β 12 ) (β 21 β 22 ) (β 12 β 13 ) (β 22 β 23 ) (β 13 β 14 ) (β 23 β 24 )] = (0,0) ) = ( (µ 11 µ 12 ) (µ 21 µ 22 ) (µ 12 µ 13 ) (µ 22 µ 23 ) = 0 = 0 ) = ( 0 0 ) : Part 2 Slide 27 of 75

28 Two or More: Assuming parallel, then coincidence implies Profiles Analysis forg 2 Groups Two or More: Parallel GLM: Parallel and WAIS Two or More: Two or More: Level Profiles Two-Way H o2 : 1 µ 1 = 1 µ 2 = = 1 µ g So we re testing equivalence of a single variable over g groups: 1-way ANOVA. The linear transformation of the variables is given by For the groups, M p 1 = See SAS for WAIS example L = I : Part 2 Slide 28 of 75

29 Two or More: Level Profiles If profiles are coincidence, then flat or level profiles means H o3 : µ 11 = µ 12 = = µ 1p = µ 21 = = µ gp Profiles Analysis forg 2 Groups Two or More: Parallel GLM: Parallel and WAIS Two or More: Two or More: Level Profiles Two-Way That is, sums of two variables are the same for pairs of groups. The required linear transformation, M (p (p 1) = The required contrasts of the groups, L = C (g 1) g which is the same one used for testing parallel profiles. : Part 2 Slide 29 of 75

30 Mini Outline: 1. Review 2 way ANOVA 2. 2 way for balanced data 3. Example: Distributed versus /instruction 4. Unbalanced designs. 5. Multivariate GLM and further extensions (MANCOVA, longitudinal) Two-Way : Part 2 Slide 30 of 75

31 2 way ANOVA 2 Factors (qualitative variables). ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests Two-Way Notation: Factor B 1 k b 1 X 11r X 1kr X 1br n observations in.... each of the gb Factor A l X l1r X lkr X lbr combinations of levels.... of the factors. g X gkr X gkr X gbr Observation: X lkr = the r th observation at level l of Factor A and level k of Factor B. Factor A: l = 1,...,g Factor B: k = 1,...,b Replications: r = 1,...,n For now, a balanced design. : Part 2 Slide 31 of 75

32 ANOVA Model for Two Factors X lkr = µ }{{} overall level + τ l }{{} fixed effect factor A at level l + β k }{{} fixed effect factor B at level k + γ lk }{{} interaction between factors A &B at levels l &,k + ǫ lkr }{{} residual ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests where g b g b τ g = β k + γ lk = γ lk = 0 l=1 k=1 l=1 k=1 and ǫ lkr N(0,σ 2 ) and all independent. Model for the mean: Two-Way E(X lkr ) = µ lk = µ+τ l +β k +γ lk TThe effects are not additive; there is an interaction. : Part 2 Slide 32 of 75

33 Examples: Interaction & No Interaction ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests Cell Mean Factor B b 3 b 1 b 2 a 1 a 2 a 3 a 4 Factor A Factor B b 1 a 1 a 2 a 3 a 4 Factor A b 2 b 3 Two-Way : Part 2 Slide 33 of 75

34 Observation as Sum of Means ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests x lkr = }{{} x +( x l x) +( x k x) }{{}}{{} ˆµ ˆτ l ˆβ k where x = overall sample mean. +( x lk x l x k x) }{{} ˆγ lk x l = mean for l th level of Factor A (or row mean). x k = mean for k th level of Factor B (or column mean). x lk = mean for the l th level of Factor A and k th level of Factor B (or the cell mean). The decomposition of the observations above shows the estimates of the population effects. +(x lkr x lk ) }{{} ˆǫ lkr Two-Way : Part 2 Slide 34 of 75

35 Decomposition of Sums of Squares x lkr = x }{{} ˆµ +( x l x) }{{} ˆτ l +( x k x) }{{} ˆβ k +( x lk x l x k x) }{{} ˆγ lk +(x lkr x lk ) }{{} ˆǫ lkr ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests Two-Way To get the sums of squares: 1. Subtract x from both sides. 2. Square both sides. 3. Sum over observations within cells, levels of Factor A, and levels of Factor B. 4. After a little algebra, g b n (x lkr x) 2 = l=1 k=1 r=1 g n( x l x) 2 + l=1 + + g l=1 g l=1 b gn( x k x) 2 k=1 b n( x lk x l x k + x) 2 k=1 b k=1 r=1 n (x lkr x lk ) 2 : Part 2 Slide 35 of 75

36 Sums of Squares Decomposition We get an orthogonal decomposition of the sums of squares: SS total = SS A +SS B +SS AB +SS residual Note: SS total is corrected for the grand mean. ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests Also, we get decomposition of degrees of freedom, (gbn 1) }{{} total = (g 1) + (b 1) }{{}}{{} Factor A Factor B +(g 1)(b 1) }{{} Interaction +gb(n 1) }{{} Residual Two-Way : Part 2 Slide 36 of 75

37 ANOVA Summary Table ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests Source of Degrees SS variation of freedom Sums of Squared Factor A (g 1) SS A = g l=1 n( x l x) 2 Factor B (b 1) SS B = b k=1 gn( x k x) 2 Interaction (g 1)(b 1) SS AB = g b l=1 k=1 n( x lk x l x k + x) 2 Residual gb(n 1) SS res = g b n l=1 k=1 r=1 (x lkr x lk ) 2 (error) Total gbn 1 SS total = g b n l=1 k=1 r=1 (x lkr x) 2 Two-Way : Part 2 Slide 37 of 75

38 Hypothesis Tests Interaction: H o : γ 11 = γ 12 = = γ gb = 0. If the null is true and all assumptions valid, F = SS AB/((g 1)(b 1)) SS res /(gb(n 1)) F (g 1)(b 1),gb(n 1) ANOVA Model for Two Factors Examples: Interaction & No Interaction Observation as Sum of Means Decomposition of Sums of Squares Sums of Squares Decomposition ANOVA Summary Table Hypothesis Tests Two-Way Factor A: H o : τ 1 = τ 2 =... = τ g. If the null is true and all assumptions valid, F = SS A /(g 1) SS res /(gb(n 1)) F (g 1),gb(n 1) Factor B: H o : β 1 = β 2 = = β b. If the null is true and all assumptions valid, = SS B /(b 1) SS res /(gb(n 1)) F (b 1),gb(n 1) We should start with the Interaction before considering whether to test the main effects. : Part 2 Slide 38 of 75

39 Two-Way For p variables, X lkr is a p 1 vector of measurements on p variables. The model X lkr = µ+τ l +β k +γ lk +ǫ lkr and ǫ lkr N p (0,Σ) and independent. where l = 1,...,g Two-Way Two-Way Decomposition of Observation Decomposition of SSCP k = 1,...,b r = 1,...,n lk Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection And for identification g τ l = l=1 b β k = k=1 g γ lk = l=1 b γ lk = 0 k=1 : Part 2 Slide 39 of 75

40 Decomposition of Observation We decompose the observation vector (x lkr x) into sums of various vectors, Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection (x lkr x) = ( x l x) +( x k x) }{{}}{{} ˆτ l ˆβ k +(x lkr x lk ) }{{} ˆǫ lkr +( x lk x l x k + x) }{{} ˆγ lk If we take the sums of squares and cross-products of this, we obtain the SSCP decomposition, ( x l x)( x l x) =... l k r } {{ } SSCP total which will take more space than is left on this slide : Part 2 Slide 40 of 75

41 Decomposition of SSCP ( x l x)( x l x) l k r } {{ } SSCP total Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection = bn( x l x)( x l x) l }{{} SSCP FactorA + gn( x k x)( x k x) k }{{} SSCP FactorB + n( x lk x l x k + x)( x lk x l x k + x) l k }{{} SSCP interaction + (x lkr x lk )(x lkr x lk ) l k r }{{} SSCP residual : Part 2 Slide 41 of 75

42 Decomposition of SSCP So, SSCP total = SSCP FactorA +SSCP FactorB +SSCP Interaction +SSCP residual which is orthogonal so long as the design is balanced (i.e., n lk = n) or proportional... what to do with unbalanced is discussed a bit later. And Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection df total = df FactorA +df FactorB +df Interaction +df residual (gbn 1) = (g 1) + (b 1) + (g 1)(b 1)+gb(n 1) : Part 2 Slide 42 of 75

43 Hypothesis Testing: Interaction First H o : γ 11 = γ 12 = = γ gp Test Statistic is Wilk s Lambda Λ = det(sscp residual ) det(sscp residual +SSCP interaction ) = det(e) det(e +H) Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection ν h = ν interaction = (g 1)(n 1) ν e = ν residual +gb(n 1) Distribution of Λ are given on the following slide for various cases. If you don t have p 2 or ν h 2, but you have large n then the approximate sampling distribution of ( ν e p+1 ν ) h ln(λ ) is χ 2 pν 2 h : Part 2 Slide 43 of 75

44 Distribution of Wilk s Lambda Λ Wilk s Λ = SSCP e SSCP e +SSCP h Number df for variables Hypothesis Sampling distribution for multivariate data Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection p = 1 ν h 1 p = 2 ν h 1 p 1 ν h = 1 p 2 ν h = 2 ( νe ( ν h )( 1 Λ Λ ) F νh,ν e )( ν e 1 1 ) Λ ν h Λ F 2νh,2(ν e 1) ( )( ) νe +ν h p 1 Λ p Λ F p,(νe +ν h p) ( νe +ν h p 1 p )( 1 Λ Λ ) F 2p,2(νe +ν h p 1) where ν h = degrees of freedom for hypothesis, and ν e = degrees of freedom for error (residual). : Part 2 Slide 44 of 75

45 Hypothesis Testing: Interaction First You should test the interaction first. If H o for the interaction is rejected, then you don t so tests for main effects because the Factor main effects do not have a clear interpretation. J&W recommend doing p-univariate 2-way ANOVAs to find out where the interactions exist. (i.e., are there interactions for just some of the p variables and not others?). Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection For variables with no interaction, you can interpret main effects. Note: In some cases you can go on to test main effects even though the interaction is significant (we ll do this in our example). In general, if you retain the null hypothesis for interaction, we continue and test main effects... : Part 2 Slide 45 of 75

46 Hypothesis Testing: Main Effects For Factor A: H o : τ 1 = τ 2 = = τ g = 0 versus H a : at least one τ l 0 Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection Λ A = det(sscp residual ) det(sscp residual +SSCP FactorA ) = E E +H A and ν h = g 1. For p 2 or ν A 2 (i.e., g 3), use the exact distribution given in the table. For Factor B: H o : β 1 = β 2 = = β b = 0 versus H a : at least one β k 0 Λ B = det(sscp residual ) det(sscp residual +SSCP FactorB ) = E E +H B and ν h = b 1. For p 2 or ν B 2 (i.e., v 3), use the exact distribution given in the table. For Large n, you can use ( ν e p+1 ν ) h ln(λ 2 h) χ 2 pν h : Part 2 Slide 46 of 75

47 Following Rejection Two-Way Two-Way Decomposition of Observation Decomposition of SSCP Decomposition of SSCP Hypothesis Testing: Interaction First Distribution of Wilk s Lambda Λ Hypothesis Testing: Interaction First Hypothesis Testing: Main Effects Following Rejection If a null hypothesis is rejected, you should perform additional analyses to figure out where the effects are. These could include Multivariate T 2 ; that is, generalized squared distances D 2 (X) lk l k = ( X lk X l k ) S 1 pool ( X lk X l k ) Simultaneous confidence statements/intervals. When interactions are not significant, concentrate on contrasts of levels of Factor A and of Factor B (if they re significant). These are the same as those described in 1-way notes. For example, Use Bonferroni if planned. (τ li τ mi ) or (µ li µ mi ) 2-way ANOVAs with multiple comparisons. Discriminant Analysis. : Part 2 Slide 47 of 75

48 1 Way : Data from Tatsuoka (1988), Multivariate Analysis: Techniques for Educational and Psychological Research, pp An experiment was conducted for comparing 2 methods (A & B) of teaching shorthand computer programing to 60 female seniors in a vocational high school (a dated example). Also of interest were the effects of distributed versus massed practice Two-Way The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion C 1 : C 2 : C 3 : 2 hours of instruction/day for 6 weeks 3 hours of instruction/day for 4 weeks 4 hours of instruction/day for 3 weeks So each subject received a total of 12 hours of instruction. Note: n l = 10 per cell of the design Two variables (dependent measures): X 1 = speed X 2 = accuracy : Part 2 Slide 48 of 75

49 The Design and Data Various Means: C 1 C 2 C 3 Two-Way The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion X 1 X 2 X 1 X 2 X 1 X 2 X 1 X 2 A B Notes: Cell means (black) based on n lk = 10. Means for method (green) are based on n l = 30 Means for condition (blue) are based on n k = 20. Total sample means (red) are based on n = 60. : Part 2 Slide 49 of 75

50 Cell Means & 95% Confidence Ellipses Two-Way The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion X2 (Accuracy) x AC3 x AC X 1 (Speed) x BC3 x AC1 x BC2 x BC1 : Part 2 Slide 50 of 75

51 Factor Means & 95% Confidence Ellipses Two-Way The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion X2 (Accuracy) x A X 1 (Speed) x B x C3 x C2 x C1 : Part 2 Slide 51 of 75

52 Test for Interaction H o : γ AC1 = γ AC2 = γ AC3 = γ BC1 = γ BC2 = γ BC3 = 0 H a : not all γ lk = 0 W = E = ( ) H AB = ( ) Two-Way The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion and Λ = F = det(e) det(e +H AB ) = =.6361 ( 53 2 ) ( ) = 6.73 Comparing F = 6.73 to the F 4,106 distribution, F has p <.001. : Part 2 Slide 52 of 75

53 Test for Method Main Effect... For purposes of illustration... H o : τ A = τ B = 0 versus H a : not all γ l = 0 W = E = ( ) H Method = ( ) Two-Way The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion and Λ = det(e) det(e +H Method ) = =.9077 F = ( 53 1 ) ( ) = 2.63 Comparing F = 2.63 to the F 2,106 distribution, F has p =.08. : Part 2 Slide 53 of 75

54 Test for Condition Main Effect H o : β C1 = β C2 = β C3 = 0 versus H a : not all β k = 0 W = E = ( ) H Condition = ( ) Two-Way Λ = det(e) det(e +H Condition ) = =.1341 The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion and F = ( 53 2 ) ( ) = Comparing F = to the F 4,106 distribution, F has p <.001. : Part 2 Slide 54 of 75

55 Interpretation & Conclusion Two-Way The effectiveness of the teaching methods (A and B) depends on which of the three conditions of distributed practice that a student used. (i.e., There is a significant interaction). The method by condition cell means indicate that method B is more effective in terms of both speed and accuracy under conditions C 1 and C 2, but A is more effective in terms of speed and accuracy under condition C 3. The Design and Data Cell Means &95% Confidence Ellipses Factor Means &95% Confidence Ellipses Test for Interaction Test for Method Main Effect Test for Condition Main Effect Interpretation & Conclusion Statistical tests of these conclusions should be performed in follow up analyses. Potentially useful supplement analyses include ANOVAs for each of the dependent variables, simultaneous confidence intervals for various treatment effects, and discriminant analysis. : Part 2 Slide 55 of 75

56 The assumptions are Multivariate normality. Equality of Σ s. Independence between and within groups. Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier Checks on multivariate normality: Scatter plots of residuals. QQ plots of residuals. Scatter & QQ plots of principal components. Tests of univariate normality for each residuals. Tests of multivariate normality (see text). method Checking Independence : Part 2 Slide 56 of 75 Assumption

57 Shorthand Example: Scatter Plot Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier method Checking Independence : Part 2 Slide 57 of 75 Assumption

58 Shorthand Example: QQ Plot 1 Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier method Checking Independence : Part 2 Slide 58 of 75 Assumption

59 Shorthand Example: QQ Plot 2 Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier method Checking Independence : Part 2 Slide 59 of 75 Assumption

60 Equality of Covariance Matrices When there are equal n lk s (balanced), violating equality of Σ lk s probably doesn t hurt too much. Two-Way When there are unequal n lk s and det(σ lk s differ substantially, then will tend to make more errors. Where you make more Type I or Type II errors depends on how different the n lk s and Σ lk s are from each other. Test of Equality of Σ s this gets into ideas of tests and inference for covariance matrices, which leads into topics on canonical correlation and discriminate analysis. Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier Assume: X lj N p (µ l,σ l ) for l = 1,...,g and j = 1,...,n l and independent over the g groups and with groups. The null hypothesis: H o : Σ 1 = Σ 2 = = Σ g Σ method Checking Independence : Part 2 Slide 60 of 75 Assumption

61 Testing Equality of Covariance Matrices If H o : Σ 1 = Σ 2 = = Σ g Σ is true, then ( ) g 1 S pool = S = g l=1 ν ν l ln( S i ) l where ν l = (n i 1) = df(s i ). When H o is true, then S pool is an unbiased estimate of Σ. l=1 Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier Box s test (1949ish): ( g ) M = ν l l=1 Test statistic is [ 1 2p2 +3p 1 6(p+1)(g 1) ln( S }{{} S pool ) ( g ( ) 1 l=1 g ν l ln( S l ) method Checking Independence : Part 2 Slide 61 of 75 Assumption ν l l=1 )] 1 g l=1 ν l } {{ } Correction factor i.e., asymptotically M χ 2 (g 1)p(p+1)/2

62 Example: Box s Test of Equality of Σ lk H o : Σ A,C1 = Σ A,C2 = Σ A,C3 = Σ B,C1 = Σ B,C2 = Σ B,C3 versus H a : not all Σ lk are equal. The method by condition sample covariance matrices: ( ) ( S A,C1 = S B,C1 = ) Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier S A,C2 = S A,C3 = ( ( Pooled Covariance Matrix: S pool = method Checking Independence : Part 2 Slide 62 of 75 Assumption ( ) ) S B,C2 = S B,C3 = ( ( ) ) )

63 Example of Box s Test continued M = ( l,k (n lk 1))ln S pool l,k (n lk 1) S lk Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier = 6(10 1)ln(69.14) (10 1)(ln( ln(268.82)+ln(20.30) + ln(85.07) + ln(12.73) + ln(13.04)) = = Since n lk = 10 for all K = (2)(3) = 6 combinations of method and condition (i.e., equal cell sizes), the correction factor simplifies to 1 (2p2 +3p 1)(g +1) 6(p+1)gn = 1 (2(2)2 +3(2) 1)(6+1) 6(2 + 1)10(6) =.9157 Test statistic =.9158(23.41) = with df = (K 1)p(p+1)/2 = (6 1)2(2+1)/2 = 15, which using the χ 2 distribution has p-value =.123. Retain null hypothesis; that is, the equal covariance matrix assumption is reasonable. method Checking Independence : Part 2 Slide 63 of 75 Assumption

64 Box s Test the easier method data shorthand; input speed accuracy method $ conditin $; if conditin= C1 and method= A then group=1; else if conditin= C1 and method= B then group=2; else if conditin= C2 and method= A then group=3; else if conditin= C2 and method= B then group=4; else if conditin= C3 and method= A then group=5; else if conditin= C3 and method= B then group=6; Two-Way Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier proc discrim simple pool=test Wcov Pcov list; class group; var speed accuracy; run;... and the output... method Checking Independence : Part 2 Slide 64 of 75 Assumption

65 Notation: K P N = Number of Groups = Number of Variables = Total Number of Observations - Number of Groups N(i) = Number of Observations in the i th Group - 1 N(i)/2 Within SS Matrix(i) V = N/2 Pooled SS Matrix P + 3P - 1 RHO = SUM _ N(i) N _ 6(P+1)(K-1) DF =.5(K-1)P(P+1) PN/2 N V Under the null hypothesis: -2 RHO ln PN(i)/2 _ N(i) _ is distributed approximately as Chi-Square(DF). Chi-Square DF Pr > ChiSq Since the Chi-Square value is not significant at the 0.1 level, a pooled covariance matrix will be used in the discriminant function. Reference: Morrison, D.F. (1976) Multivariate Statistical Methods p252 Note: There is a more recent edition of Morrison (I think). 64-1

66 Checking Independence Assumption Two-Way is generally relatively robust to violations of multivariate normality and equal covariance matrices; however, it is not robust to violations of Independence of observations within groups. Independence of observations between groups. Violation of independence generally increases the Type I error rate (i.e., it s higher than what you think it is, and possibly much larger). You can check for possible violation by computing the intra-class correlation for each of the variables using Shorthand Example: Scatter Plot Shorthand Example: QQ Plot 1 Shorthand Example: QQ Plot 2 Equality of Covariance Matrices Testing Equality of Covariance Matrices Example: Box s Test of Equality ofσ lk Example of Box s Test continued Box s Test the easier r intra = MS between MS error MS between +(n 1)MS error You can also consider the data collection procedure/method, because this is where dependence can slip in. It s the independence within that is most likely to be violated. In some cases this can be dealt with. method Checking Independence : Part 2 Slide 65 of 75 Assumption

67 Similar to 1-way ANOVA. With balanced designs (or proportional) Two-Way SSCP total = SSCP A +SSCP B +SSCP AB +SSCP residual or SSCP model = SSCP A +SSCP B +SSCP AB and interpretation is straight forward. With unbalanced designs, the partitioning of the is not longer unique. It depends on Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 66 of 75 The model. Various sub-models of model as specified by the order in which various SSCP are extracted.... These notes are based on Khattree & Naik.

68 Order in which SSCP are extracted Suppose X lkr = µ+τ l +β k +γ lk +ǫ lkr where l = 1,...,g, k = 1,...,b, and r = 1,...,n lk. We could partition total SSCP as SSCP total corrected = SSCP A µ +SSCP B µ,a +SSCP AB µ,a,b or Two-Way SSCP total corrected = SSCP B µ +SSCP A µ,b +SSCP AB µ,a,b Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 67 of 75 When the design is unbalanced, SSCP A µ SSCP A µ,b There are 4 different ways of computing SSCP s, which can lead to different results and different interpretation.

69 Type I SSCP: Sequential SSCP The partitioning of the model SSCP into component SSCP due to each variable or effect (including interactions) as it s added sequentially to the model as given in the model statement in proc glm in SAS. e.g., Two-Way Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 68 of 75 model x1 x2 = A B C A B A C B C A B C That is all main effects, 2 way interactions and a 3-way interaction. The SSCP for A C = SSCP(A C µ,a,b,c,a B); that is, SSCP is adjusted for all previously entered terms, model x1 x2 = A B C A B }{{} already entered A C B C A B C

70 Example Type I SSCP: Sequential SSCP Two-Way To demonstrate this, I deleted some observations from the shorthand data set to make it a bit unbalanced. The sample sizes are now Practice Condition Method C1 C2 C3 Total A B Total Below are the hypothesis SSCP matrices for two different orders: Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 69 of 75 Order: M C C M C M C M H m = ( ) H m = ( )

71 Type I: Sequential SSCP continued Two-Way The sum of H m +H c will be equivalent regardless of whether method order condition was entered first. With Type I SSCP, the sum of SSCP matrices of all the effects in the model will sum to the total for the model, regardless of the order in which they are entered into the model. The total SSCP is completely partitioned. Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 70 of 75

72 Type II or Partial SSCP With Type II, an SSCP matrix equals the increase in model SSCP for a particular variable (effect). e.g., Two-Way Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 71 of 75 model x1 x2 = A B C A B A C B C A B C That is all main effects, 2 way interactions and a 3-way interaction. Consider the SSCP for A C. The SSCP for A C is the increase in model SSCP by adding A C to the model that has A, B, A C and B C in it. The three-way interaction A B C contains A C in it, so the SSCP for A C is in the model. The effects A B C and A C are correlated. With Type II, the sum of the SSCP s do not equal the total model SSCP.

73 Two-Way Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 72 of 75 Example: Type II or Partial SSCP Below are the SSCP matrices for Type I and Type II and the totals: Type I: Sequential H m H c H cm Sum M C C M C M C M Type II: Partial

74 Two-Way Order in which SSCP are extracted Type I SSCP: Sequential SSCP Example Type I SSCP: Sequential SSCP Type I: Sequential SSCP continued Type II or Partial SSCP Example: Type II or Partial SSCP Type III & Type IV SSCP Type IV and When to Use Which : Part 2 Slide 73 of 75 Type IV and When to Use Which Type IV are designed for situations where there are empty cells. For balanced designs, the 4 ways to compute SSCP yield the same results. For unbalanced designs, the 4 ways can lead to different conclusions. Type III is appropriate when interest is in comparing the effects of experimental variables. For model building purposes (finding a good model for prediction), Types I and/or II might be preferable. If there are design parameters, make sure that tests do not involve these. The choice should be made on a case-by-case basis with thought given to what it is you want to test.

75 Any design that you can form for univariate case can be formed for the multivariate situation. Two-Way This can be done by creating an appropriate design matrix in the general linear model context. Models can include fixed (as we have done), but also random effects (as in the univariate case). It is possible to reformulated as a linear mixed model, and hence provides a natural way to handle missing observations (be sure there are systematic reasons for missingness).... see Goldstein. : Part 2 Slide 74 of 75