Hout 5 Computation of estimates, ANOVA Table Orthogonal Designs In this hout e ill derive, for an arbitrary design, estimates of treatment effects in each stratum. Then the results are applied to completely romized designs, romized complete block designs Latin squares. 1. Computation of estimates in a given stratum Project the romization model (4.1) (4.2) in Hout 4 onto f, 1, e obtain the folloing linear model: ith WCœ W\ α W%, (5.1) E ÐW % œ 0 cov ÐW% œ 0 W, (5.2) here (5.2) follos from cov ÐW% œ WZ W œ WÐ 4œ! 04W4W œ 0W (since Á 4 Ê WW 4 œ!. Our goal is to compute the best linear unbiased estimator of any estimable treatment function - α in f, i.e., the estimator of - α ith minimum variance among unbiased estimators of the form +WC. When restricted to the vectors in f, the covariance matrix 0W is the same as 0M: Ð0W @ œ 0@ for all @ f. Therefore the Gauss-Markov Theorem applies, the least squares estimators are the best linear unbiased estimators. From (.) (.4) in Hout, e have Lemma 5.1 A treatment function - α is estimable in stratum f (i.e., b an unbiased estimator of the form +WC ) if only if - eð\ W\. If -α is estimable in stratum f, then its best linear unbiased estimator in f is - α^, here α^ is any solution of the normal equation \ W \ α^ œ \ WC, (5.) var Ð ^ - α œ 0 - Ð\ W \ -. Proof. The design matrix in model (5.1) is \ œ W \ ; so e have Ð\ \ œ ÐW\ W\ œ \ W\, the right-h side of the normal equation is Ð\ WC œ ÐW\ WC œ \ WC. The matrix \W\ is called the information matrix for treatment effects in stratum f. It has the folloing properties: = Lemma 5.2. The information matrix \ W \ is symmetric, nonnegative definite has zero ro sums. Proof. Symmetry nonnegative-definiteness are straightforard. To sho that it has zero ro sums, e note that by the definition of \, each of its ros has exactly one 1, all the other entries are 0's; therefore \ 1 œ 1, \ W \ 1 œ \ W 1 œ 0. R R -1-
For convenience, let G œ \ W \ (5.4) U œ \ W C. (5.5) Then (5.) becomes G α^ œ U. (5.6) Lemma 5.2 shos that rankðg Ÿ 1, that if - is estimable, then it must be a contrast. If rankðg œ 1, then e say that the design is connected in stratum f. In this case, all the treatment contrasts are estimable in. f α 2. ANOVA in a given stratum Since E ÐC œ. 1 \ α eð\ œg, under (5.1), EÐWCœWE ÐC W( g ). Let TW Ðg WC be the orthogonal projection of the data vector WC onto W( g ). Then, since W( g ) is the range of W\, from Hout, TW Ðg WC œöðw\ ÒÐW\ ÐW\ Ó ÐW\ WC œ ÐW\ Ð\ W\ \ WC. So T W C W Ðg œòðw \ Ð\ W \ \ WCÓÐW \ Ð\ W \ \ WC The residual is T f W Ðg œ CW\ Ð\ W\ \ WC (5.7) ^α œ ( ) U. (5.8) œ ( ^ α ) G α^ (5.9) W C, e have lwcl œ T WC T WC. (5.10) W Ðg f W Ðg Formula (5.10) gives the ANOVA in stratum f. The first term is the treatment sum of squares the second term is the residual sum of squares. No, EÐT W C f W Ðg œ EÐCWT WC f W Ðg œòe ÐC) ÓW T W EÐC trðw T W Z f W Ðg f W Ðg œ tr( WT W WZ ) [Since WE ÐC W( g, hich is orthogonal to f WÐg ] f Ðg -2-
œ tr( 0 WT W) (Since Z œ 0 W) f WÐg 4 4 4œ! œ tr Ð0 T (Since f W ( g ) f, e have T W œ P ) f W Ðg f W Ðg f W Ðg œ 0 Ödim( f ) dim ÒW Ðg Ó. EÐT Similarly, W C W Ðg W Ðg W Ðg œðt EÐW C trðw T W Z œ α \ W\ Ð\ W\ \ W\ α tr Ð0TW Ðg [In (5.7), replace WC ith E ÐWC, since Ðg f ] œ α Ð\ W \ α 0 dim[ W Ðg ] œ α G α 0 dim[ W Ðg ]. S = Summarizing the above discussion, e have the folloing ANOVA ithin stratum f : Sources Sums of Squares d.f. MS E(MS) Treatments ( α^ ) α^ dim[ ( )] ( α^ ) α^ G W g G 0 α G α.7òw Ðg Ó.7ÒW Ðg Ó Rresidual By subtraction 0 Total lwcl dim( f ) Note that rankðg œ rank Ð\ W \ ) œ dim[ W ( g )]. If the design is connected in f then the treatment sum of squares has 1 degrees of freedom in f.. Orthogonal designs If g Z f for some, then e say that e have an orthogonal design. In this case, for all 4Á, W4( g ) œ{ 0}, dimòw4ðg Óœ0, the treatment contrasts can only be estimated in f. We therefore drop the superscript from α^. In the folloing, e shall assume that < 4 0 for all 4 œ 1, á,, here < 4 is the number of replications of the 4 th treatment. Then dim( g Z) œ, hence all the treatment contrasts are estimable in. Furthermore, f W ( g ) œ W [ Z Š ( g Z)] œ W ( g Z) œ g Z, (5.11) --
for each B, W B œ W ÐKB W ÐT B œ W ÐT B œ T B. (5.12) g Z Z Z The last equality holds because TZ B g Z g Z f. It follos from (5.12) that W\ œ T Z \, hence G œ \ W \ œ \ T Z \ (5.1) Z U œ \ WCœ ÐW\ Cœ ÐT \ C œ \ T C. (5.14) Z Consider the folloing one-ay layout model Cœ. \ α %, E Ð% œ 0 cov Ð% œ 5 M. ~ ~ ~ ~ By (.8) in Hout, estimates of α are solutions of \\ α^ œ \C ~, here \ œ T Z \, ~ ~ ~ ~. Then ~ C œ T Z C \ \ \ C are the same as G U in (5.1) (5.14), respectively. Therefore hen g Z f, the best linear unbiased estimator - α^ of a treatment contrast is the same as under the above one-ay layout model, i.e., _ - α^ œ -4 4, (5.&) _ here 4 is the average of all the observations on the 4th treatment. In the folloing, e sho that By Lemma 5.1, e have var( - α^ ) œ 0 4 <. (5.16) 4 - - œ \ W \? for some? (5.17) var( ^ - α) œ 0 - (\ W \ ) -. (5.18) By the same argument as in the proof of Lemma 5.2, W\ K? œ 0, because \ K? has constant entries. It follos that - œ \ W \ Ð? K?. So by replacing? in (5.17) ith? K?, e may assume? Z. Then \? 1 œ 0, i.e., \? Z. Since \? g, e have \? g Z. Recall that g Z f. Therefore \? f, hence W \? œ \?. Then - œ \ W \? œ \ \? œ??, here? is the diagonal matrix hose 4 th diagonal entry is < 4. Hence From (5.18), var( ^ - α) œ 0 - (\ W \ ) -? œ? -. (5.19) œ 0? (\ W \ )(\ W \ ) - [by (5.17)] -4-
œ 0?- [ CœE Dis a solution of ECœ Dif a solution exists] œ 0 (-? -), [by (5.19)] hich is (5.16). Also, the treatment sum of squares W Ðg W Ðg W Ðg W Ðg T W C œ T C [Since W ( g ) f, T W œ T ] œ lt g Z Cl [By (5.11)] œ lð KC l _ R œ < [ ( C )]. 4 4 R 6 6œ The above discussion shos that if estimators simple ANOVA: g Z f for some, then e have simple variation Sums of Squares degrees of freedom Mean square E(MS).7Ð f lwcl dim( f ) lwcl 0 ã ã ã ã ã f f _ R _ Treatments [ ( )] 1 [ < C ã 0 < ( α α) ] 4 4 R 6 4 4 6œ Residual lwcl [ _ < 4 4 R R Ð C6Ó dim( f) 1 ã 0 6œ ã ã ã ã ã = = =.7Ð = = f lwcl dim( f ) lwcl 0 Total lc KCl R 1 f = -5-
_ In the above table, α œ < α. R 4 4 No e specialize these results to three simple orthogonal designs: completely romized designs, romized complete block designs, Latin squares. In a completely romized design, the to strata are f! œ Z f œ Z. Obviously g Z Z ; so it is an orthogonal design. In a romized complete block design, each treatment appears in each block exactly once. The condition of proportional frequencies is satisfied by the treatment block factors. It follos from the Theorem in Hout 2 that ( g Z) ( U Z). Also g Z Z. Therefore g Z U ( œ f.) This establishes the orthogonality of a romized complete block design. In a Latin square (or more generally, romized ro-column designs in hich all the treatments appear equally often in each ro equally often in each column), the condition of proportional frequencies is satisfied by the treatment ro factors, also by the treatment column factors. Thus ( g Z) ( e Z) ( g Z) ( V Z). As a result, g Z ( e V) ( œ f$ ). Therefore for these designs, estimators of treatment contrasts their variances are given by (5.15) (5.16). Their ANOVA tables follo. ANOVA table for a completely romized design: variation Sums of Squares degrees of freedom Mean square E(MS) Treatments [ ] 1 [ < C. ã 0 < ( α α) ] 4 4 4 4 residual By subtraction 0 R. œ Total ( C C) R 1 ANOVA table for a romized complete block design: variation Sums of Squares d.f. Mean square E(MS),,...,... 0 œ œ Blocks C ( C ), 1 C ( C) _ Treatments [ ] 1 [, C ã 0,( α α) ] 4.. 4. Residual By subtraction 0, Total ( C C ), 1 œ 4.. -6-
ANOVA table for a Latin square design: variation Sums of Squares d.f. Mean square E(MS)...... 0 œ œ Ros C ( C ) 1 C ( C). 4... 4.. 0 Columns C ( C ) 1 C ( C) _ Treatments [ ] 1 [ C ã 0 ( α α) ] 6.. $ 6. 6œ 6œ Residual By subtraction 0 $ Total ( C C ) 1 œ 4.. -7-