Design of Experiments: III. Hasse diagrams in designed experiments

Transcription

1 Design of Experiments: III. Hasse diagrams in designed experiments R. A. Bailey 57a Reunião Anual da RBras, May 202 Outline I: Factors II: Models I. Factors Part I Factors pen. Each pen was allocated to a certain of feed. Batches of this of feed were put into the pen; calves were free to eat as much of this as they liked. Calves were weighed individually. Wine Four wines are tasted and evaluated by each of eight judges. A plot is one tasting by one judge. Experiment on cultivars and fertilizer Partial order Infimum and Supremum Orthogonality Structures defined by factors Skeleton anova. Examples Three experiments Ladybirds Three pesticides were compared for their side-effects on ladybirds. A field was divided into three areas and one pesticide applied to each area. Ladybirds were counted on three samples from each area. Calves The treatments were 4 different feeds for calves. The calves were not fed individually. They were housed in 8 pens, with 0 calves per Cropper Melba Melle Melba Cropper Melle.2 Factors Factors, levels and classes A factor F is a function for which we are more interested in knowing whether F(α) = F(β) than in knowing the value F(α).

2 So we can regard a factor as a partition into levels or classes. Let Ω = the set of observational units, and let F be a factor on Ω. We define the F-class containing α to be F[[α]] = {ω Ω : F(ω) = F(α)}. For example, if the observational units are plots in several fields, and F is the factor for fields and α is a plot, then F[[α]] consists of all the plots in the same field as the plot α. A factor is uniform if all of its classes have the same size. Cultivar example Cropper Melba Melle Melba Cropper Melle E = plot strip field U strip cultivar Aliasing We say that F is aliased with G (written F G) if every F-class is also a G-class. Some aliasing is just renaming of levels: for example, F is days with names Monday, Tuesday,..., and G is dates with names, 2,.... Some aliasing is deliberate: for example, assign each fruit-picker to pick the apples from a single row in the orchard. Two special factors The universal factor U has just one class; it has the same level everywhere. The Hasse diagram There is one point for each factor. Write F G Draw G F Convention Write n F, the number of levels of F, beside the dot for F. The equality factor E has one level per observational unit; so each class consists of a single observational unit.. Partial order The partial order We say that F is finer than G (or G is coarser than F) (written F G) if every F-class is contained in a G-class but F G. We write F G if F G or F G. So E F U for every factor F. 2

3 Cultivar example: Hasse diagram on plot factors Infimum on the Hasse diagram F G is the factor whose classes are the nonempty intersections of F-classes with G-classes. It is finer than both F and G: F G F and F G G. It has the least number of levels subject to this Cropper Melba Melle Melba Cropper Melle F G F G Infimum example on treatment factors U 2 field 6 strip cultivar fertilizer 4 treatment 2 cultivar fertilizer = treatment.4 Inf and sup 24 E Infimum of two factors Given two factors F and G, the factor F G is defined by (F G)[[ω]] = F[[ω]] G[[ω]]. Cultivar example cultivar fertilizer = treatment field cultivar = strip Supremum of two factors Given two factors F and G, the factor F G is the finest factor whose classes are unions of F-classes and unions of G-classes. If you try to fit F and G in a linear model, you will get into trouble unless you fit F G first. Cultivar example field fertilizer = U strip treatment = cultivar

4 Supremum on the Hasse diagram F G is the factor which is coarser than both F and G: F F G and G F G, and which has the largest number of levels subject to this. Its classes are the connected components of the graph whose edges are pairs with the same level of either F or G. F F G G Supremum example with mixed factors Hasse diagram for factors on the observational units U field strip E How many of each are there? Hasse diagram for factors on the treatments cultivar U = fertilizer cultivar fertilizer 4 2 T = fertlizer cultivar Factorial treatments plus control Cropper Melba Melle Melba Cropper Melle none single double Z S K M N cultivar 6 2 strip treatment strip treatment = cultivar = fumigant 2 9 U fumigant 5 treatment 4

5 Drugs at different stages of development A pharmaceutical company wants to compare 6 treatments for a certain disease. There are are different s of drug A, that has been under development for some time, and different s (not comparable with the previous ) of a new drug B, that has not been so extensively studied. A A2 A B B2 B U drug A B E U Orthogonality: general case F is orthogonal to G if, within each class of F G separately, F-classes meet G-classes proportionately. then In general, this condition means that, whenever (F G)(α) = (F G)(β), F[[α]] G[[β]] = and the following proportionality equation holds: (F G)[[α]] F[[α]] G[[α]] = (F G)[[β]] F[[β]] G[[β]]. A drug B 4 E Three different possibilities, all orthogonal Z S K M N none 6 single double Orthogonality Orthogonality: special cases Let F and G be factors on the same set. F is orthogonal to G if, within each class of F G separately, F-classes meet G-classes proportionately. Some special cases We deem F to be orthogonal to itself. If F G, then F is orthogonal to G. If every F-class intersects every G-class and F G is uniform, then F is orthogonal to G. Z S K M N none 0 single double Z S K M N none 25 single double 2 2 5

6 .6 Structures Orthogonal treatment structure An orthogonal treatment structure is a set G of factors on the set of treatments such that. U G; 2. if F G and G G then F G G;. if F G and G G then F is orthogonal to G. Factorial treatments plus control Z S K M N none single double = fumigant For each factor F in G, let W F be the set of vectors which, U are constant on every level of F and sum to zero on every level of G whenever F G and G G. Put d F = dim W F = degrees of freedom for F. Theorem. For an orthogonal treatment structure, the spaces W F are orthogonal to each other and d F = n F d G. G F Hasse diagram for factors on the treatments cultivar, 2, 2, 6 U = fertilizer cultivar fertilizer 4, T = fertlizer cultivar, 2, 9, fumigant 5, treatment Drugs at different stages of development A A2 A B B2 B U drug A B E , U Degrees of freedom calculated by subtraction A 4, 2 2, drug B 4, 2 6, 0 E 6

7 Orthogonal plot structure An orthogonal plot structure is a set F of factors on the observational units such that. every factor in F is uniform; 2. U F;. E F; 4. if F F and G F then F G F; 5. if F F and G F then F G F; 6. if F F and G F then F is orthogonal to G. Conditions 2, 4 and 6 are the same as those for an orthogonal treatment structure, so if we define W F and d F as before then we have Theorem 2. For an orthogonal plot structure, the spaces W F are orthogonal to each other and d F = n F d G. G F Hasse diagram for factors on the observational units, 2, 6, 4 24, 8 U field strip E Covariance matrix How many of each are there? Degrees of freedom calculated by subtraction Theorem. If the observational units have an orthogonal plot structure, and the covariance of Y i and Y j depends only on which factors have the same level on i and j, then the eigenspaces of the covariance matrix (that is, the strata) are the spaces W F. This implies that, if we project the data onto any one stratum W F, then estimation and testing within that stratum proceed just as they do when the covariance matrix is Iσ 2. Null analysis of variance The null analysis of variance is a table showing the strata and their degrees of freedom., 2, 6, 4 24, 8 stratum U field strip E df U field strip 4 plot 8 Total 24 Orthogonal design An orthogonal design consists of a set of treatments (which affect expectation) having an orthogonal treatment structure a set of plots (which affect covariance) having an orthogonal plot structure an allocation of treatments to plots which satisfies treatment factors remain orthogonal to each other, even if treatments are not equally replicated treatment factors are orthogonal to plot factors if F is a plot factor and G is a treatment factor then F G is a treatment factor. 7

8 .7 Skeleton anova Combined Hasse diagram Draw the combined Hasse diagram using 2 different colours of dot. Each treatment dot G is immediately above a unique plot dot, say F. Estimate the effect of W G in stratum W F. Expand the null analysis of variance to the skeleton analysis of variance by allocating each treatment subspace to the stratum in which the corresponding treatment effect is estimated. Combined Hassed diagram for cultivar example U field cultivar fertilizer strip T E Skeleton analysis of variance for cultivar example stratum df source df U mean field field strip 4 cultivar 2 residual 2 plot 8 fertilizer Total 24 cultivar fertilizer 6 residual 9 Wine example: combined Hasse diagram Four wines are tasted and evaluated by each of eight judges. A plot is one tasting by one judge. U judge 8 4 wine 2 E The main effect of cultivar is estimated in the strip stratum. The main effect of fertilizer and the cultivar-byfertilizer interaction are both estimated in the plot stratum. 8

9 Calves example: combined Hasse diagram The treatments were 4 different feeds for calves. The calves were not fed individually. They were housed in 8 pens, with 0 calves per pen. Each pen was allocated to a certain of feed. Batches of this of feed were put into the pen; calves were free to eat as much of this as they liked. Calves were weighed individually. 4 8 U feed pen Linear models A linear model for variables Y,..., Y n has the form E(Y) M and Cov(Y) = σj 2 Q j, j where M is a known subspace of R n and the matrices Q j are known. For brevity, let us call M the model. Example model with one treatment factor Plot i has treatment f (i) and E(Y i ) = τ f (i). 80 E Then M consists of all vectors v with the property that Ladybirds example: combined Hasse diagram Three pesticides were compared for their sideeffects on ladybirds. A field was divided into three areas and one pesticide applied to each area. Ladybirds were counted on three samples from each area. and v i = v j if f (i) = f (j); dim M = number of treatments. 2 II. Models area 9 U pesticide E Example of a polynomial model There is a covariate x taking value x i on plot i, and E(Y i ) = a + bx i + cx 2 i. Then M consists of all vectors which are linear combinations of the three vectors (,,,..., ), (x, x 2, x,..., x n ) and (x 2, x2 2, x2,..., x2 n); and dim M =. Part II Expectation Models Partial order Orthogonality Examples Scaled Hasse digram 2. Partial order The partial order Suppose that M and M 2 are subspaces of R n. Write M < M 2 if M is a subspace of M 2 but M = M 2. Write M M 2 if M < M 2 or M = M 2. 9

10 The Hasse diagram on models There is one point for each model. Write M < M 2 Draw M2 M Convention Write dim M beside the dot for M. Family of models: intersection Let M be the family of expectation models that we are considering fitting to the data. Principle. If M and M 2 are both in M then M M 2 must also be in M. Why? If not, we might find that our data vector y is in M M 2 and then we cannot choose between M and M 2. (This principle is sometimes violated in fractional factorial designs.) M M 2 Family of models: sum M M 2 M + M 2 = {v + w : v M and w M 2 } Principle 2. If M and M 2 are both in M then M + M 2 must also be in M. Why? If not, we might find that v and w are allowed vectors of fitted values but v + w is not. (People who fit sparse models do not believe this.) 2.2 Orthogonality Family of models: orthogonality If M and M 2 are in M then in general they cannot be orthogonal to each other because they both contain M M 2. Principle. If M and M 2 are both in M and M 0 = M M 2 then M M0 should be orthogonal to M 2 M0. Why? If not, various paradoxes are possible in fitting the data. (Again, in practice, this is often ignored.) 2. Examples Hasse diagram of some polynomial models 4 2 cubic polynomial quadratic polynomial straight line constant Warning: the best-fitting quadratic polynomial is not usually obtained by taking the best-fitting cubic polynomial and removing the term in x. M + M 2 M M 2 Theorem 4. Two warnings If the models are defined by factors then the Hasse diagram for the models is the opposite way up from the Hasse diagram for the factors dim(m + M 2 ) = dim M + dim M 2 dim(m M 2 ). there may be more models than factors. 0

11 Drugs at different stages of development A A2 A B B2 B U drug A B tmt A 4 6 treatment B 4 2 drug constant Factorial: cultivars and fertilizer 2 treatments are all combinations of: factor levels Cultivar (C) Cropper, Melle, Melba Fertilizer (F) 0, 80, 60, 240 kg/ha 2 treatment Factorial treatments plus control none single double Z S K M N = fumigant 9 treatment fumigant constant 2.4 Scaled Hasse diagram ANOVA table Each line of the ANOVA table corresponds to a difference between models, showing the difference between the sums of squares of their fitted values, and the difference between their dimensions. C 6 C + F F 4 constant We can summarize the ANOVA table graphically by scaling the lines in the Hasse diagram.

12 An experiment on biodiversity A, B, C, D, E, F six s of freshwater shrimp. Put 2 shrimps in a jar containing stream water and alder leaf litter. Measure how much leaf litter is eaten after 28 days. Experimental unit = jar. Richness Treatment Level 6 A,..., F monoculture 2 of A 5 AB,..., EF duoculture 6 of A, 6 of B 2 20 ABC,..., DEF triculture 4 of A, 4 of B, 4 of C 4 The Hasse diagram of our family of expectation The experiment was carried out in 4 blocks of 4 models jars. (4) Treatment What models did we fit? The biologist fitted the model Richness with parameters, one for each level of richness, and found no evidence of any differences between the levels. I suggested the model Type with 6 parameters α A,..., α F : monoculture A duoculture AB triculture ABC α A α A + α B 2 α A + α B + α C In other words, if there are x i shrimps of i then E(Y) = 6 a i x i where 2a i = α i. i= ( x i = 2 always, so no need for intercept.) (8) (8) Richness + Type Richness () Type (6) Constant () (blocks included in all models) Richness Type (a i can change with each level of richness but does not depend on what else is present) (add a different constant for each level of richness) 2

13 What the data showed: lengths are mean squares Treatment Richness Type Richness + Type. Type Conclusions: The model Richness does not explain the data. The model Type explains the data well. There is no evidence that any larger model does any better. Scale: residual mean square Richness. Constant Further Reading R. A. Bailey: Design of Comparative Experiments, Cambridge Series in Statistical and Probabilistic Mathematics, Cambridge University Press, Cambridge, J. Reiss, R. A. Bailey, D. M. Perkins, A. Pluchinotta & G. Woodward: Testing effects of consumer richness, evenness and body size on ecosystem functioning. Journal of Animal Ecology, published online May 20.