Resolvable BIBD: An ncomplete bloc desgn n whch each treatment appears r tmes s resolvable f the blocs can be dvded nto r groups such that each group s a complete replcaton of the treatments (.e. each group of blocs contans each treatment exactly once). Bose (194) showed b t r1 s a necessary condton. PROOF (Murty 1961): Assume t s a multple of,.e. t n, where n s an nteger. so. b rt rn b nr We have r t 1 1 r t n r r n1 r rn1 t 1 rnr t 1 br t 1 But So r r t r rn rn br t 1 All parameters are ntegers so rn s an nteger. Also, r so r 0. 1 or br t1 b rt1 NOTE: 1
1) If b rt1, the desgn s affne resolvable. ) for a resolvable desgn, SS(blocs gnorng trts)ss(reps)ss(blocs n reps), where the frst term on the rght hand sde has n1 df and the second term on the rght hand sde has r( t 1 df. General BIBD: - blocs need not be of same sze - each trt. need not appear same # of tmes - t treatments, b blocs havng plots n bloc 0, r plots for treatment r 0. - N n ; n # of tmes treatment occurs n bloc - no need to assume t for any or all - T total for treatment - B total for bloc Defntons: If n 0, 1 bnary desgn forall proper desgn r rforall equreplcate desgn Notaton: T T 1 T t, B B 1 B b, N n ncdence matrx of the desgn R txt : dagonal matrx r 1,...,r t ; K bxb : dagonal matrx 1,..., b For all desgns R1 t N1 b ( n r ) ; K1 b N1 ( n ) General model: y m e m ; m1,...,n ; 1,...,t;1,...,b m th observaton on th treatment, th bloc (f bnary desgn, omt m). Least squares: mnmze. y m
Normal equatons: 0 G y m N 0 T n r n 0 B n r Impose addtonal restrctons: r 0 G/N; r T n B r T n n B n B n Let Q T t r Q h1 n B n n h h where 1,...,t t Q r h1 n n h h Normal equatons wrtten n matrx form are: 3
G T B N 1R 1 b K R1 t R N K1 b N K Multply both sdes on the left by F N 1 0 0 0 1 t NK 1 0 NR 1 1 b NOTE: R1 t N1 b ; K1 b N1 t We get reduced equatons: G N 1 R /N1 K /N 1 Q TNK 1 B R NK 1 N B NR 1 T KNR 1 N 3 Impose sde condtons: r 1 R 0; 1 K 0. (1) becomes G/N y () Q tx1 A txt (adusted nterbloc equatons; A s sngular) (3) becomes counterparts for estmatng NOTE: A1 0 because r n h n n h Q h h r n h n h r r 0 4
KNR 1 N1 0 Confrm our study to ran(a)t-1 and ran (K-N R 1 N)t. Such a desgn s a connected desgn. Under these condtons all contrasts, are estmable. To calculate SS(regresson), get from equatons K1 N K B K 1 B K 1 N 1b Then SSregresson G B T G BK 1 B TBK 1 N G BK 1 B Q Snce A1 0 0.1, one egenvalue s zero and the correspondng egenvector s 1. Also A e e where e are egenvectors correspondng to non-zero roots and e 1 0. A 1 1 e e,a 1 1 0, A 1 Q,DQ A. Hence D( A 1 AA 1 V, where V1 0. Analyss of varance for ntrabloc source df SS total N1 y m G /N blocs (gnorng trts) b1 BK 1 B G /N Trts (adusted) t 1 Q error Nbt1 by subtracton If there are some cells wth more than 1 observaton, we can remove from the resdual SS a SS wthn cellsy m y. and attrbute the rest to nteracton between treatments and blocs. For bnary desgns, we assume no nteracton. 5
11.6 The adusted treatment and bloc totals: Q, the adusted treatment totals, are lnearly dependent Q 1 t Q 0. NOTE: Q 1 t T 1 t B K 1 N 1 t GB K 1 K1 b GG 0 Smlarly, B NR 1 T1 b 0 Calculate CovQ: Recall basc form of Nt 1 s of general lnear hypothess: XY XX wth CovY I, CovXY XI X XX. Thus, for us, the square matrx on the rght n the normal equatons s the covarance matrx of the totals G, T,B (call t XX). The covarances of G/N,Q,B NR 1 T are gven by CovFG,T, B FXXF N 1 0 0 0 R NK 1 N R NK 1 NR 1 N 0 KNR 1 NK 1 N KNR 1 N or Cov Q R NK 1 N A CovB NR 1 T KNR 1 N 11.7 Solvng the Intrabloc equatons ( methods): (reduced ntrabloc equatons Q A ) Method 1: Use H 0 where H s a row vector (e.g. H1 or 1 R) (sde condton: 1H h 0 Construct A A H H 0 B 11 B 1 B 1 B 1 then B 11 Q. 6
Then A A 1 I gves (1) AB 11 H B 1 I A R NK 1 N A () AB 1 H B 0 B 11 B 11 (3) H B 11 0 (4) H B 1 1 Notng 1A 0 we get 1AB 11 1H B 1 I or 1HB 1 1. From (), smlarly, 1H B 0. Therefore B 1 1H 1 1 1 h,b 0. Tae transpose of (1),.e. B 11 A B 1 H I. Multply on left by : B 11 A B 1 H B 11 Q (snce H 0 NOTE: B 11 generalzed nverse of A.e.B 11 B 11 AB 11 If H 1R,B 1 1 N ; f H 1,B 1 1 t. Multply (1) on the left by B 11, gvng: B 11 AB 11 B 11 HB 1 B 11 B 11 AB 11 B 1 by (3) Smlarly, multply on rght of (1) by A, gvng: AB 11 A HB 1 A A AB 11 A H 1 A AB 11 A A 1 A 0. h A or AB 11 AB 11 AB 11 and B 11 AB 11 A B 11 A (.e. dempotent) Tae transpose of (1) and multply by to get B 11 A B 1 H, notng A and B 11 are symmetrc. But A Q and H 0, or B 11 Q and B 11 s a soluton matrx. (or AB 11 QH B 1 Q Q AB 11 Q Q because B 1 Q 1 Q 0 h (NOTE: Cov( CovB 11 Q B 11 AB 11 B 11 Another method: Let P 1 A HH 1. Then P 1 Q because Q A A H H. Secton 11.8 Connected desgns (due to Bose): A bloc and treatment are assocated f treatment occurs n the bloc. Two treatments are connected f t s possble to pass from one to the other by a chan of 7
treatments such that adacent treatments n the chan are n the same bloc (or two treatments A and B are connected f we can form a chan of treatments and blocs begnnng wth A and endng wth B such that every bloc n the chan s assocated wth both treatments adacent to t). A desgn s connected f all pars of treatments are connected. THEOREM: A desgn s connected f and only f all contrasts h are estmable PROOF: Wrte down the sequence and then the expected value of each term. Everythng cancels except h. ex. Bloc 1 A B C D y 11 y 1 y 13 y 14 Bloc B C E Bloc 3 D E Bloc 4 E F G Are A, G connected? ABEG, so yes. A G 1 A 1 B B E 4 E 4 Ey 11 y 1 y 1 y 5 y 45 y 47 A G s estmable (by followng the chan). THEOREM: A desgn s connected f and only f RA t 1. (to verfy ra t 1, ust chec to see f all pars of trts. are connected) ASIDE: An alternatve dervaton of average varance (v) of treatment comparsons n GIBD. (connected) ra t 1, there are t 1 non-zero egenvalues, then one egenvalue s zero. A1 0,1 s the egenvector correspondng to egenvalue 0. Let c 1,..., c t1 be egenvectors correspondng to egenvalues 1,..., t1. 8
Spectral decomposton: A t1 1 c c AA A A Generalzed nverse: A 1 t1 1 1 c c trv A 1 tr c c 1 trc c 1 PROOF: a) Suppose all elementary contrasts estmable, tae t 1 t t,...,t t1 t t (lnearly ndep. group) Normal equatons are: or G N r T r r n XY XX B n X X 1 3 Now l m, m 0, l,0 s estmable ff rxx rxx m. Now subtract n XX m. E X X A tmes (3) from () and sum over, whch does not affect ran of XX or because n 0 0 and EXX m A l or ra ra l. Hence f t 1 estmable functons (lnearly ndependent) say L we have ra r 1 0 0 A 1-1 -1 t 1 b) converse: If ra t 1, then l s estmable for every l such that l1 0. PROOF: snce ra t 1, we have c 1,...,c t1 orthogonal normal egenvectors correspondng to 1,..., t1 (nonzero). Then Ec Q c A c or c c Q. 9
Snce c s orthogonal to 1 (egenvector correspondng to 0) and l also orthogonal to 1 we have l 1 t1 c, l c E 1 t1 c Q t1 c A 1 1 t1 c l or l s estmable or desgn s connected. BIB model y ; 1,...,; 1,...,b t Q where Q T n B, Q 0 Var h t (balanced but not orthogonal) Secton 11.9: Recovery of Interbloc Informaton (nterbloc estmates): Yates (1940) assumed blocs were random. He wanted to use bloc totals..e. and ndep. of ~N0, e y, ~N0, b I Restrct ourselves to proper desgns. NOTE: The s we got above for the fxed effect model are stll unbased estmates for s under ths model (snce they were condtonally unbased, therefore uncondtonally based). Yates used lnear model theory to get another unbased estmator for s, usng bloc totals. Interbloc analyss: Consder bloc totals B n y n e Thus Ef 0 and Vf b e b e f 10
Covf, f 0 for all so apply least squares estmaton. Mnmze b 1 B n Normal equatons: 1) wth respect to : n r and N b ) wth respect to : ( s s ) G N r B n B r s n s s n 0 n r r s s s n s s r r We get estmates: so G/N y.. n B r r because T B G tr, we have 0 h T T h r (these are alternatve estmators for h. We had h Q Q h t To show these two estmators (ntrabloc and nterbloc) are ndependent, we need only show B,Q are ndependent for all and ( cov 0 normal. b Q T s1 n s B s f treatment s not n bloc, then Q and B are ndependent. 11
If treatment s n bloc, CovB,Q CovB, T 1 VarB Vary 1 VarB T n B, VarB f b e e b Vary Var e b VarB Var n y Secton 11.10: combnng ntrabloc and nterbloc estmators: We have ndependent estmators of h : (1) OLD: (ths s a condtonal varance) h Q,Q h t V Z h t e (uner fxed model and hence under ths one V EV Z h t e () NEW: h T T h r 1
VT f CovT,T h f n r f ; n n h f V h 1 r r f f r f Usng and we construct a new estmator w 1 1 V w 1 V t e r e b w 1w w 1 w w w 1 r t e f r e b w 1 w 1 w t e w w 1 w 1 1 w r t1 b e 1 w 1 1 w 1 the above depends only on b, generally unnown varances. e w V 1 w 1 w w 1 w w 1 w 1 w 1 w w 1 V w V V,V (unformly better than ether of the earler estmators) But we are guessng!! [If we ndeed had a random model, we would have to use MLE s for the s]. The solutons are not easy. We ve shown (condtonal) E h h E h b h (uncondtonal) In general 13
c c T r, c 0 V V c f r V c c t rc c c h h f r c (condtonally unbased) combned estmator s where w 1w w 1 w w 1 t, w r f V c w 1 w We need to estmate weghts w 1 and w. s unbased usng estmated w 1 and w weghts. To get these estmates, get ANOVA tables for blocs (adusted for treatments). l l w 1 w 1 w w w 1 w l l ( w w 1 w an even functon of observatons) or E l l E l l E l l so unbased. Use w 1w w 1 w S bad Nt 1 MSE b1 Nt 14
We need estmators for b and e : 1) We get e estmate from MSE n Intrabloc analyss ANOVA (blocs elmnatng treatments (nterbloc analyss)) Source df SS E(MS) trts. (gnorng blocs) t 1 T r G N t 1 e r t b blocs (adusted for trts.) b1 R 1 R 0 e bt b1 b error (ntrabloc analyss) Nbt1 Q t s y B e total N1 y G N SSBl unad SStr ad SST unad SSbl ad Assgnment #4: 1) Show S blad. for trts. S bad B Q t T ) Also, f we defne y y, show Ey 0,y S bad y S bad y. r s R 1 R 0 for testng blocs. ~N0, e b and 3) Show ES bad b b t e b1 and hence b S badms eb1 Nt Interbloc: y n, where n s 1 f treatment occurs n bloc, and 0 otherwse. s are fxed, s are random ndependent N0, b, s are random ndependent N0, e ). where y. n n f n n, n ~N0, f 15
Treat bloc totals as observatons and get estmates of s,.e. s. Mnmze y. n 1 r n B r Here we loo at bloc totals; n ntrabloc we loo at observatons. SS error B B T Trt 1 3 4 5 6 7 8 9 10 T A 7 8 30 31 9 30 175 B 6 6 9 30 1 6 158 C 30 34 3 34 31 33 194 D 9 33 34 31 3 31 181 E 6 34 5 3 4 6 148 B 83 83 85 98 85 89 95 75 73 90 856 t 5, r 6,b 10, 3, 3 (5 detergents for dshwashng, 3 sns, observatons are # of plates washed beforee all suds dsappear from each sn. T n B Q T n B Q A 175 53 0.67 0.4489 B 158 494 6.67 44.49 C 194 56 18.67 348.57 D 181 58 5 5 E 148 497 17.67 31.3 856 568 0 Q 730.47 Q Intrabloc analyss: 16
source df SS MS E(MS) F total 9 y G /N 381.47 blocs (gnorng trts) 9 B G /N 186.13 trts (elmnatng blocs) 4 t Q 146.09 36.54 e r t 36.54/3.07 11.9 Error 16 by subtracton 49.19 3.07 e SS(Blocs ngnorng trts)ss(trts adustng for blocs)ss(trts. gnorng blocs)ss(blocs ad. for trts) SS(error)TSS-SS(col. gnorng rows)-ss(rows elmnatng columns) TSS-SS(rows gnorng columns)-ss(columns elmnatng rows) TSS-SS(fttng parameters and t Q ; V t VQ t r 1 r 1 tt t 1 tt t 1 t If bloc dfferences are large, you get a sgnfcant amount of addtonal nformaton from nterbloc (for RCB, t,b r and nterbloc dsappears). If blocng s useful, consder Var 1 Var Var 1 Var, whch wll have varance consderably less than Var and Var. Why both analyses? Degrees of freedom for error s small wth BIBD, and we need a better estmate of varance. 17