Multivariate Analysis of Variance (MANOVA)

Transcription

1 Multivariate Analysis of Variance (MANOVA) Example: (Spector, 987) describes a study of two drugs on human heart rate There are 24 subjects enrolled in the study which are assigned at random to one of three groups (8 subjects to a group) Two of the groups receive the drugs under study, the other is a control group The heart rate of the subjects is measured four times, five minutes apart after administration of the drug Suppose that we are interested in understanding the similarities and differences between the three different different groups Consider the following definitions and methodology One-way Manova: We wish to compare treatment means for h different populations Population : y, y 2,, y n Population h: y h, y h2,, y hnh iid N p (µ, Σ) iid N p (µ h, Σ) Assumptions: Independent random samples from h different populations 2 Homogeneous covariance matrices 3 Normality B

2 One-way Manova, Cont Calculate the summary statistics ȳ i, S i, and the pooled estimate of the covariance matrix S Analogous to the univariate one-way Anova, it is convenient to use the effects model formulation: µ i = µ + τ i, where µ i is the population mean for population i, µ is the overall mean effect, and τ i is the treatment effect of the ith treatment For the one-way model: y ij = µ+τ i +ɛ ij, for i =,, h, j =,, n i, and ɛ ij N p (0, Σ) Note that the above model is over-parameterized; there are an infinite number of ways to define µ and the τ i s such that they add up to µ i Thus, we need to consider a constraint For example, h n i τ i = 0 or τ h = 0 i= Note that the observational equivalent of the effects model is y ij = ȳ + (ȳ i ȳ) + (y ij ȳ i ) where the terms in this model are the equivalent of the overall mean, the treatment effect, and the residual We will move ȳ to the left hand side, square both sides, and sum over all the observations B2

3 One-way Manova, Cont This gives us h n i (y ij ȳ)(y ij ȳ) = i= j= h n i (ȳ i ȳ)(ȳ i ȳ) i= + h n i (y ij ȳ i )(y ij ȳ i ) i= j= The left-hand side of this equation is the total corrected sums of squares and cross products (SSCP), denoted (W) The first term on the right-hand side is the treatment (or between subjects) sum of squares and cross products, denoted (H) The second term is the residual (or within subjects) SSCP, denoted (E) Note: E = (n )S + + (n h )S h = (n h)s Thus, we could consider the Manova table below: Source SSCP df Treatment H h Residual (error) E h i= n i h Total Corrected H + E h i= n i We want to test H 0 : τ = τ 2 = = τ h = 0 To do this, let s consider the relative sizes of E and H + E Define Wilk s Lambda, Λ by Λ = E H + E B3

4 Wilk s Lambda, Cont Properties of Wilk s Lambda: Wilk s Lambda is equivalent to the F-test in the univariate case The exact distribution of Λ can be determined for special cases For some combinations of h and p, the distribution can be shown to be an exact f distribution For large samples sizes, reject H 0 at level α if ( h ) n i p + h ( ) E log > χ 2 ( α,p(h )) 2 E + H i= Testing General Hypotheses: Consider h different treatments, with the ith treatment applied to n i subjects which are observed for p times We can consider this a p dimensional observation on a random sample from each of h different treatment populations Consider the model: y ij = µ + τ i + ɛ ij, for i =,, h and j =,, n i B4

5 Testing General Hypotheses, Cont This model can equivalently be written as the linear model Y = Xβ + ɛ, where n = h i= n i and y Y (n p) = y n, β ((h+) p) = y hn h and X (n (h+)) = µ τ τ h , ɛ (n p) = ɛ ɛ n ɛ hn h, Under this model formulation, we have ˆβ = (X X) X Y Additionally, note that the rows of Y are independent Then, to test a general hypothesis about the parameters in the model, we can write H 0 : CβM = 0 vs H a : CβM 0 Here, C is a matrix of comparisons across groups, while M is a matrix of comparisons across traits B5

6 Testing General Hypotheses, Cont We can compute the general treatment corrected sums of squares and cross products by H = M Y X(X X) C [C(X X) C ] C(X X) X Y M or, for the null hypothesis H 0 : CβM = D, H = (C ˆβM D) [C(X X) C ] (C ˆβM D) Similarly, we can compute the general matrix of residual sums of squares and cross products as E = M Y [I X(X X) X ]Y M = M [Y Y ˆβ (X X) ˆβ]M Then, we can compute the following statistics, all based upon the eigenvalues of HE Wilk s Criterion: Λ = E The degrees of freedom depend upon the rank of C, M, and H+E X Lawley-Hotelling Trace: U = tr(he ) Pillai Trace: V = tr( H H+E ) Roy s Maximum Root: largest eigenvalue of HE B6

7 Profile Analysis We wish to explore possible similarities between the treatment effects This can be particularly useful for longitudinal analysis and for clinical trials Recall that the test of equal means, H 0 : τ = τ 2 = = τ h implies that all treatments have the same average effect Acceptance or rejection of this hypothesis may not provide adequate insight into the nature of the similarities and differences between the treatments Instead, we can break the test into three sequential steps: Are the profiles parallel? (Equivalently, is there no interaction between treatment and time?) If so, are the profiles coincidental? (Are the profiles identical?) If so, are the profiles horizontal? (Are there no differences between any time points?) Alternatively, if we reject the null hypothesis that the profiles are parallel, we could test any of the following: Are there differences among groups within some subset of the total time points? B7

8 Profile Analysis, Cont Are there differences among time points in a particular group (or groups)? Are there differences within some subset of the total time points in a particular group of groups? Parallel Profile: Are the profiles for each population identical except for a mean shift? H 0 : µ µ 2 = µ 2 µ 22 = = µ t µ 2t µ µ 3 = µ 2 µ 32 = = µ t µ 3t for a total of h equations We could write this hypothesis equivalently as [ ] 0 µ µ 4 µ 0 2 µ µ 3 µ = 0 Let s use the space below to show why this is true: Note that this is not the only set of contrasts that we could use for this test B8

9 Profile Analysis, Cont We could also write H 0 as [ ] µ 0 0 τ 0 0 τ 2 τ 3 or as H 0 : LβM = 0 rank(m) = p = 0 Note that rank(l) = h and Note that the choice of L and M are not unique could have chosen [ ] L = and M = and gotten the same test statistic We Coincidental Profiles: Given that the profiles are parallel, are they identical? If we know that the profiles are parallel, then the profiles are identical if the sums of the components of µ i are identical for all of the treatments We can write this test as H 0 : p µ = p µ 2 = = p µ h B9

10 Profile Analysis, Cont Equivalently, we could formulate this test as H 0 : LβM = 0, where [ ] 0 0 L = and M = 0 0 Once again, matrix multiplication will show that this choice of L and M yields the correct comparison Again as before, these choices of L and M are not unique, but other choices will continue to yield the same test Horizontal Profiles: Assuming that we cannot reject the null hypothesis that all h profiles are the same, we can next ask the question, Are all of the elements of the common profile equal? Similarly to previous tests, there are many ways to formulate this test One requires testing H 0 : LβM = 0 where L = [ ] and M = A quick note: if we fail to reject all three hypotheses, then we have failed to reject the null hypotheses of both no difference between treatments and no differences between traits B0

11 Orthogonal Contrasts We might wish to test other contrasts in the treatment mean vectors, as well For example, we might wish to test for linear, quadratic, and cubic effects for the four time periods We could do this by using the contrasts C l = ( 3,,, 3) C q = (,,, ) C c = (, 3, 3, ) We would test these hypotheses using the formulation H 0 : C βm = 0 where 0 0 M = We will now spend some time talking about how to do this using SAS software Please also see the notes in Microsoft Word about how to run this in SAS B