Chapter 6 Balanced Incomplete Block Design (BIBD)

The designs like CRD and RBD are the complete block designs. We now discuss the balanced incomplete block design (BIBD) and the partially balanced incomplete block design (PBIBD) which are the incomplete block designs.

A balanced incomplete block design (BIBD) is an incomplete block design in which - b blocks have the same number k of plots each and - every treatment is replicated r times in the design.

- Each treatment occurs at most once in a block, i.e., nij  0 or 1 where nij is the number of times the jth treatment occurs in ith block, ibjv1,2,..., ; 1,2,..., . - Every pair of treatments occurs together is  of the b blocks.

Five parameters denote such design as D(,bkvr ,,; ). The parameters bkvr,,, and are not chosen arbitrarily. They satisfy the following relations: (Ibkvr )  ()II ( v 1)( r k  1) (III ) b v (and hence r k ).

Hencenkij  for all i j

nrij  for all j j

n and nn nn... nn  for all jj ' 1,2,..., v . Obviously ij cannot be a constant for all 1'jij 2' jij bjj b' r j. So the design is not orthogonal.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 1 Example of BIBD In the designDbkvr ( , ; , ; ) : consider b 10 ( sayB , ,..., B ), v 6 ( sayT , ,..., T ), k 3, r 5, 2 110 16 Blocks Treatments

BTTT1 12 3

B2 TT12 T 4

B3 TT13 T 4

B4 TT14 T 6

B5 TT15 T 6

B6 TT23 T 6

B7 TT24 T 5

B8 TT25 T 6

B9 TT34 T 5

B10 TT34 T 6

Now we see how the conditions of BIBD are satisfied. (ibk )  10  3 30 and vr  6 5 30 bk vr ()ii ( v 1)2 5 10and( r k  1)5 2 10  (vrk 1) ( 1) ()iii b 106

Even if the parameters satisfy the relations, it is not always possible to arrange the treatments in blocks to get the corresponding design.

The necessary and sufficient conditions to be satisfied by the parameters for the existence of a BIBD are not known.

The conditions (I)-(III) are some necessary condition only. The construction of such design depends on the actual arrangement of the treatments into blocks and this problem is handled in combinatorial mathematics. Tables are available, giving all the designs involving at most 20 replication and their method of construction.

Theorem: ()Ibk vr ()(II v 1)( r k  1) ()III b v .

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 2 Proof: (I)

Let N ():nbvij the incidence matrix

Observing that the quantities ENE11bvand ENE 1 v ' b 1 are the scalars and the transpose of each other, we find their values. Consider

nn11 21 nb 1 1  nn n 1 ENE  (1,1,...,1)12 22b 2  11bv    nn12vv n bv1

 n1 j j   n2 j  (1,1,...,1) j     nbj j k  k  (1,1,...,1) = bk . 1b   k Similarly,

nn11 21 nb 1 1  nn n  ENE' (1,...,1)12 22b 2  11vb    nn12vv n bv1

ni i r  =(1,1,...,1) (1,1,...,1) vr  1v   r niv  i

But

ENE111bv ENE v ' b 1 as both are scalars. Thus bk vr .

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 3 Proof: (II) Consider

nn11 21 nbv 1 nn 11 12 n 1  nn n nn n NN'  12 22bv 2 21 22 2   nn12v v n bv nn b 12 b n bv 2 nnnnniiiiiv1121  ii i 2 nn n nn ii12 i 2 i 2 iv  iii   2 nniv i12 nn iv i n iv ii i r   r    . (1)    r 

2 Since nij 1 or 0 as nij  1 or 0,

2 so nij Number of times  j occurs in the design i = rj for all  1,2,..., v of times occurs in the design

and  nnij ij'' Number of blocks in which  j and j occurs together i =  for alljj '.

r  1  r 1 NNE'   v1   r 1

rv(1)  rv(1) [(1)].rvE (2)  v1  rv(1)

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 4 Also

nn11 12 n 1v 1  nn n 1 NNE'' N21 22 2v  v1    nnbb12 n bv1

 n1 j j   n2 j  N 'j    nbj j

nn11 21 nb 1 k  nn n k  12 22b 2     nniv2 v n bv k b1

 ni1 i n  i2  k i    niv i r  r  k    r

 krEv1 (3) From  2 and  3 [rvEkrE ( 1)]  vv11 or rv ( 1) kr or  (vrk 1) (  1)

Proof: (III) From (I), the determinant of NN' is detNN ' [ r ( v  1)]( r  )v1

v1 rrk(1)  r  rk() r  v1  0 because since if r  from (II) that k = v. This contradicts the incompleteness of the design. Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 5

Thus NN' is a vv nonsingular matrix. Thus rank(') N N v . We know from matrix theory result rank() N rank (') N N so rank ( N )  v But rank() N b , there being b rows in N. Thus vb .

Interpretation of conditions of BIBD Interpretation of (I) bk = vr This condition is related to the total number of plots in an experiment. In our settings, there are k plots in each block and there are b blocks. So the total number of plots are bk . Further, there are v treatments and each treatment is replicated r times such that each treatment occurs atmost in one block. So total number of plots containing all the treatments is vr . Since both the statements counts the total number of plots, hence bk vr.

Interpretation of (II)

k kk(1) Each block has k plots. Thus the total pairs of plots in a block =  . 2 2 There are b blocks. Thus the total pairs of plots such that each pair consists of plots within a block = kk(1) b . 2 v vv(1) There are v treatments, thus the total number of pairs of treatment =  . 2 2 Each pair of treatment is replicated  times, i.e., each pair of treatment occurs in  blocks. vv(1) Thus the total number of pairs of plots within blocks must be   . 2 kk(1) vv 1 Hence b   22 Using bk vr in this relation, we get rk 11.   v 

Proof of (III) was given by Fisher but quite long, so not needed here.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 6 Balancing in designs: There are two types of balancing – Variance balanced and efficiency balanced. We discuss the variance balancing now and the efficiency balancing later.

Balanced Design (Variance Balanced): A connected design is said to be balanced (variance balanced) if all the elementary contrasts of the treatment effects can be estimated with the same precision. This definition does not hold for the disconnected design, as all the elementary contrasts are not estimable in this design.

Proper Design:

An incomplete block design with kk12.... kkb is called a proper design.

Symmetric BIBD:

A BIBD is called symmetrical if the number of blocks = number of treatments, i.e., bv .

Since bv , so from bk vr kr. Thus the number of pairs of treatments common between any two blocks =  .

The determinant of NN' is NN'[(1)]() r v  r v1

v1 rrk 1 r  rk(). r  v1

When BIBD is symmetric, b = v and then using bk vr, we have kr . Thus

NN'(), N2 r21 r  v so v1 Nrr().  2

Since N is an integer, hence when v is an even number, ()r   must be a perfect square. So

' NN'() r I  EEvv11 , (')NN11'1 N N

1  ' IEEvv11, rr  2

'1 1  ' NIEEvv11. rr  2 Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 7 Post-multiplying both sides by N ' , we get

' NNr'( ) IEENN vv11  '. Hence in the case of a symmetric BIBD, any two blocks have  treatment in common.

Since BIBD is an incomplete block design. So every pair of treatment can occur at most once is a block, we must have vk .

If vk , then it means that each treatment occurs once in every block which occurs in case of RBD. So in BIBD, always assume v > k .

Similarly   r. [If rvrkvk then ( 1) ( 1) which means that the design is RBD]

Resolvable design: A block design of - b blocks in which - each of v treatments is replicated r times is said to be resolvable if b blocks can be divided into r sets of br/ blocks each, such that every treatment appears in each set precisely once. Obviously, in a resolvable design, b is a multiple of r.

Theorem: If in a BIBD Dvbrk(,,, , ), b is divisible by r, then bvr1. Proof: Let bnr (where n  1 is a positive integer).

For a BIBD, (vrk 1) (  1) because vr bk (1)v  orrvrnrk or (1)k   or vnk (1)nk   (1)k  n 1 n. k 1 Sincenk 1 and 1, so n  1 is an integer. Since r has to be an integer. (1)n    is also a positive integer. k 1

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 8 Now, if possible, let bvr 1 nr v r 1 orrn ( 1) v 1 rk(1) rk (1) orrn ( 1) (because v  1 )  (1)n  1which is a contradiction as integer can not be less than one k 1 bvr1 is impossible. Thus the opposite is true. bvr1 holds correct.

Intrablock analysis of BIBD: Consider the model

yibjvij  i  j   ij ;  1,2,..., ;  1,2,..., , where  is the general mean effect; th i is the fixed additive i block effect; th  j is the fixed additive j treatment effect and 2 ij is the i.i.d. random error with ij ~ N (0, ). We don’t need to develop the analysis of BIBD from starting. Since BIBD is also an incomplete block design and the analysis of incomplete block design has already been presented in the earlier module, so we implement those derived expressions directly under the setup and conditions of BIBD. Using

v b the same notations, we represent the blocks totals by By , treatment totals by , iij Vyjij  j1 i1

bv Gy adjusted treatment totals by Q j and grand total by  ij The normal equations are obtained ij1 by differentiating the error sum of squares. Then the block effects are eliminated from the normal equations and the normal equations are solved for the treatment effects. The resulting intrablock equations of treatment effects in matrix notations are expressible as

QC ˆ .

Now we obtain the forms ofCQ and in the case of BIBD. The diagonal elements of C are given by b 2 nij cri1 (1,2,...,) j   jj k r r . k

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 9 The off-diagonal elements of C are given by 1 b cnnjjjj(  '; , '  1,2,..., ) jj''k  ij ij i1   . k The adjusted treatment totals are obtained as 1 b QV nBj(1,2,...,)   jjk  iji i1 1 VBji k ij() where  denotes the sum over those blocks containing jth treatment. Denote ij()

TBji  ,then ij()

Tj QVjj. k The C matrix is simplified as follows: NN' CrI k 1 rI() r  I  E E' k vv11 k 1   ' rIEE()vvi1 kk

v 1  '  ()IEEvvi1 kk V EE' I vv11. kv

Since C is not as a full rank matrix, so its unique inverse does not exist. The generalized inverse of C is denoted as C  which is obtained as

' 1  EE CCvv11 . v Since

' v EEvv11 CIv kv  kC EE' orI vv11 , vvv

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 10 k the generalized inverse of C is v

1 ' 1 k  EEvv11 CC, vv

''1 EEvv11 EE vv 11 Iv  vv

 Iv . v Thus CI  . k v Thus an estimate of  is obtained from QC  as

ˆ  CQ v  Q. k

The null hypothesis of our interest is H01:...  2v against the alternative hypothesis H1 :at least one pair of  j ' s is different. Now we obtain the various sum of squares involved in the development of analysis of variance as follows.

The adjusted treatment sum of squares is

SSTreat() adj ˆ' Q k  QQ'  v k 2  Qj ,  j1

The unadjusted block sum of squares is

b 2 2 Bi G SSBlock() unadj  . i1 kbk The total sum of squares is

bv 2 2 G SSTotal y ij ij11 bk The residual sum of squares is obtained by SS SS SS  SS Error() t Total Block ( unadj ) Treat ( adj ).

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 11 A test for H01:...  2 v is then based on the statistic

SSTreat() adj /( v  1) FTr  SSError() t /( bk b v 1) v Q2 kbkbv1  j  ..j1 vv1 SSError() t

If FF then H is rejected. Tr1,1, v bk  b v 1; 0() t

This completes the analysis of variance test and is termed as intrablock analysis of variance. This analysis can be compiled into the intrablock analysis of variance table for testing the significance of the treatment effect given as follows.

Intrablock analysis of variance table of BIBD for H :...  01 2 v Source Sum of squares Degrees of Mean squares F freedom Between treatment SS v 1 SS MS Treat() adj MS  Treat() adj Treat treat MS (adjusted) v 1 E

Between blocks b - 1 SSBlock() unadj (unadjusted)

SSError() t SSError() t MSE  Intrablock error bk – b – v + 1 bk b v 1 (bysubstraction)

Total SSTotal  bk 1 2 2 G yij  ij bk

In case, the null hyperthesis is rejected, then we go for a pairwise comparison of the treatments. For that, we need an expression for the variance of the difference of two treatment effects.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 12 The variance of an elementary contrast (,') jj ' jj under the intrablock analysis is

VVar*(ˆˆjj' ) k Var() Qjj Q ' v k 2 [()Var Q Var ()2( Q Cov Q Q )] 22 jj jj' k 2 (2)cc c 2 22 jj j'' j jj ' 2 k 12 2 2221r    kk 2k   2. v This expression depends on  2 which is unknown. So it is unfit for use in the real data applications. One solution is to estimate  2 from the given data and use it in the place of  2.

An unbiased estimator of σ2 is SS ˆ 2  Error() t . bk b  1

Thus an unbiased estimator of V * can be obtained by substituting ˆ 2 in it as 2k SS Vˆ  ..Error() t * vbkb1

If H0 is rejected, then we make pairwise comparison and use the multiple comparison test. To test

Hjj0':('), jj ' a suitable statistic is

kbk(1) b v QQ t  . jj' v SSError() t which follows a t-distribution with (1)bk b v degrees of freedom under H0 . A question arises that how a BIBD compares to an RBD. Note that BIBD is an incomplete block design whereas RBD is a complete block design. This point should be kept in mind while making such restrictive comparison.

We now compare the efficiency of BIBD with a randomized block (complete) design with r replicates. The variance of an elementary contrast under a randomized block design (RBD) is 2 2 VVar*2()ˆˆ * R j j' RBD r

2 where Var() yij   * under RBD.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 13 Thus the relative efficiency of BIBD relative to RBD is

2 2*  Var()ˆˆ r j j' RBD   2 Var()ˆˆj j' BIBD 2k  v 2 v *  2 . rk  v The factor  E (say) is termed as the efficiency factor of BIBD and rk vvk1 E  rk k v 1 1 11 11  kv 1(sincevk ). The actual efficiency of BIBD over RBD not only depends on the efficiency factor but also on the

22 2 2 ratio of variances  * /  . So BIBD can be more efficient than RBD as  * can be more than  because kv .

Efficiency balanced design: A block design is said to be efficiency balanced if every contrast of the treatment effects is estimated through the design with the same efficiency factor.

If a block design satisfies any two of the following properties: (i) efficiency balanced, (ii) variance balanced and (iii) an equal number of replications, then the third property also holds true.

Missing observations in BIBD:

th The intrablock estimate of missing (i, j) observation yij is

vr(1)(1)(1) k B k v Q v Q' y  ijj ij kk(1)( bkb v 1)

' th Q j : the sum of Q value for all other treatment (but not the j one) which are present in the ith block. All other procedures remain the same. Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 14 Interblock analysis and recovery of interblock information in BIBD In the intrablock analysis of variance of an incomplete block design or BIBD, the treatment effects were estimated after eliminating the block effects from the normal equations. In a way, the block effects were assumed to be not marked enough and so they were eliminated. It is possible in many situations that the block effects are influential and marked. In such situations, the block totals may carry information about the treatment combinations also. This information can be used in estimating the treatment effects which may provide more efficient results. This is accomplished by an interblock analysis of BIBD and used further through the recovery of interblock information. So we first conduct the interblock analysis of BIBD. We do not derive the expressions a fresh but we use the assumptions and results from the interblock analysis of an incomplete block design. We additionally assume that

2 the block effects are random with variance   .

After estimating the treatment effects under interblock analysis, we use the results for the pooled estimation and recovery of interblock information in a BIBD.

In case of BIBD, 2 nnnnniiiiiv1121  ii i 2 nn n  nn ii12 i 2  i 2 iv NN'  ii i   2 nniv i12 nn iv i  n iv ii i r   r      r ' ()rIEEvvv  11

' 1 1 EEvv11 (')NN Iv rrk  

The interblock estimate of  can be obtained by substituting the expression on NN' 1 in the earlier obtained interblock estimate. GE  (')NN1 NB 'v1 . bk

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 15 Our next objective is to use the intrablock and interblock estimates of treatment effects together to find an improved estimate of treatment effects.

In order to use the interblock and intrablock estimates of  together through pooled estimate, we consider the interblock and intrablock estimates of the treatment contrast.

The intrablock estimate of treatment contrast l ' is llCQ''ˆ   k  lQ' v k  lQj j v j ˆ  l jj ,say. j

The interblock estimate of treatment contrast l ' is lNB'' llE'  (since ' 0) r   v1 1 vb  lnBjiji r   ji11

1 v  lTjj r   j1 v   l jj . j1

The variance of l 'ˆ is obtained as

k  ˆ Var(' l )  Var ljj Q v j

2 k 2 ljj Var()2 Q l jjjj l'' Cov (, Q Q ). v jjjj'( ) Since

1 2 Var() Qj  r  1 , k   Cov(, Q Q ) 2 ,( j  j '), jj' k so

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 16 2 2 k 1    ˆ 22 22 Varlrlll(' ) 1 jjj  vk  jjj k  2 krk(1) 222 lljj (since  j  0 being contrast) vkjj k j

2 k 1 2 (1)vl j (using(1)(1)) rkv  vk j

k 22   l j . v j Similarly, the variance of  'ˆ is obtained as

1 2    2 Var(' l ) ljj Var ( T ) 2 l jjjj l'' Cov ( T , T ) r   jjjj'( ) 2 1 2  22 2 2 rlfj f l j l j r   jjj  2  f 2  l j . r   j The information on 'ˆ and 'ˆ can be used together to obtain a more efficient estimator of ' by considering the weighted arithmetic mean of  'ˆ and  ' . This will be the minimum variance unbiased and estimator of  ' when the weights of the corresponding estimates are chosen such that they are inversely proportional to the respective variances of the estimators. Thus the weights to be assigned to intrablock and interblock estimates are reciprocal to their variances as vk/( 2 ) and

2 ()/,r    f respectively. Then the pooled mean of these two estimators is vr v r  ll''ˆˆ l ˆ l  kk22  2jj  2 jj L* ffjj vr  vr 22 22 kkff  v 1 lrˆ () l k jj2 jj  jj v ()r k 12 ˆ  vlkr12jj ()  l jj   jj vkr12() 

vkr12ˆjj()   l j  j vkr12()  *  l jj j

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 17

* 12ˆˆjjkr()    11 where  j  , 1222,. 12kr()     f

* Now we simplify the expression of  j so that it becomes more compatible in further analysis.

* Since ˆ jjjj(/kQ ) and  Tr /(  ),so the numerator of  j can be expressed as

12ˆ j  kr()   jjj  12 kQ  kT

Similarly, the denominator of  * can be expressed as j

12vkr()  vr(1) k r (1) k   kr(using ( v 1) rk ( 1)) 12   vv11  1 vr(1)( k kr v k ). v 1 12 Let

* WvkVvTkGjjj()(1)(1)  

* where Wj  0. Using these results we have j

(1)vkQkT * 12jj  j  12rv(1)( k kr v  k ) (1)(vkVTkT ) 12jj j Tj = (usingQVjj ) rvk12(1)( kvk  ) k kv(1)( V k  )(1) v  T = 121jj rvk12(1)( kvk  ) kv(1)( V k ) W*  ( v kV ) (1) k G  112jjj  rvk12(1)( kvk  ) kv(1)(  k )() v  k V  (  k ) W*  (1) k  G  11212jj  rvk12(1)( kvk  )

1 12 k * VWkGjj  ( 1) rvkkvk12(1)(  ) 1 VWkG *  ( 1) jj r where

12 k 11 ,,.1222 12vk(1)( kv  k ) f

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 18 Thus the pooled estimate of the contrast l ' is

* ll'*  jj j

1 * lVjj( W j ) (since l j  0 being contrast) r jj

The variance of l '* is

k 2 Var('*) l   l j vkr12()  j

kv(1) 2 lvrkj (using ( 1) ( 1) rvk(112 kv ( k ) j 2 l j   2 j E r where

2 kv(1)  E  vk(1)(12 kv k ) is called as the effective variance.

Note that the variance of any elementary contrast based on the pooled estimates of the treatment effects is 2 Var() ** 2 . ijr E The effective variance can be approximately estimated by

2 ˆ E MSE1( v k )*  where MSE is the mean square due to error obtained from the intrablock analysis as SS MSE  Error() t bk b v 1 and    *  12 . vk(1)(12 kv k )

2 2 The quantity  * depends upon the unknown  and   . To obtain an estimate of  * , we can obtain the unbiased estimates of  2 and  2 and then substitute them back in place of  2 and  2 in    * . To do this, we proceed as follows.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 19 2 An estimate of 1 can be obtained by estimating  from the intrablock analysis of variance as

1 1 ˆ1 []MSE . ˆ2

22 2 The estimate of 2 depends on ˆˆand   . To obtain an unbiased estimator of   , consider

SSBlock() adj SS Treat () adj SS Block ( unadj )  SS Treat ( unadj ) for which

22 ESS()()(1).Block() adj  bk v   b 

2 Thus an unbiased estimator of   is 1 ˆˆ22SS(1) b  bk v Block() adj 1 SS(1) b MSE bk v Block() adj

b 1 MS MSE bk v Block() adj b 1 MS MSE vr(1) Block() adj where SS MS  Block() adj . Block() adj b 1 Thus 1 ˆ 2  22 kˆˆ 

1  .  vr(1)(1) kb SSBlock() adj  ( v kSS ) Error () t

Recall that our main objective is to develop a test of hypothesis for H01:...  2v and we now want to develop it using the information based on both interblock and intrablock analysis.

To test the hypothesis related to treatment effects based on the pooled estimate, we proceed as follows.

Consider the adjusted treatment totals based on the intrablock and the interblock estimates as TT** * Wj ;  1,2,..., v jj j and use it as usual treatment total as in earlier cases.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 20 * The sum of squares due to Tj is

2 v T * v  j 2*2j1 STTj*  . j1 v

Note that in the usual analysis of variance technique, the test statistic for such hull hypothesis is developed by taking the ratio of the sum of squares due to treatment divided by its degrees of freedom and the sum of squares due to error divided by its degrees of freedom. Following the same idea, we define the statistics Svr2 /[( 1) ] F*  T* MSE[1 ( v k )ˆ *] where ˆ * is an estimator of *. It may be noted that F * depends on ˆ *. The value of ˆ * itself

22 depends on the estimated variances ˆˆand  f . So it cannot be ascertained that the statistic F * necessary follow the F distribution. Since the construction of F * is based on the earlier approaches where the statistic was found to follow the exact F -distribution, so based on this idea, the distribution of F * can be considered to be approximately F distributed. Thus the approximate distribution of F * is considered as F distribution with (1)v  and (1)bk b v degrees of freedom. Also, ˆ * is an estimator of * which is obtained by substituting the unbiased estimators of 1 and 2 .

An approximate best pooled estimator of l j j is j1

v ˆ VWj  j l j j1 r and its variance is approximately estimated by

2 kl j j . vrkˆˆ12()

2 In case of the resolvable BIBD, ˆ  can be obtained by using the adjusted block with replications sum

* of squares from the intrablock analysis of variance. If sum of squares due to such block total is SSBlock and corresponding mean square is SS * MS *  Block Block br then

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 21 *2()(1)vkr 2 EMS()Block    br (1)rk 22 r  and kb()() r rv  k for a resolvable design. Thus

*22 ErMSBlock  MSE(1)( r k  ) and hence

* 1 rMSblock  MSE ˆ2  , r 1 1 ˆ1  MSE .

The analysis of variance table for the recovery of interblock information in BIBD is described in the following table:

Source Sum of squares Degrees of Mean square F * freedom Between treatment S 2 v - 1 MS T * F*  Blocks() adj (unadjusted) MSE

Between blocks

(adjusted) b - 1 SSBlock() adj SSBlock() adj  MSBlocks() adj  b 1 SSTreat() adj 

SSBlock() unadj 

SSTreat() unadj Intrablock error

SSError() t SS MSE  Error() t bk – b – v + 1 bk b v 1 (by substraction)

Total SS bk - 1 Total

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 22 The increase in the precision using interblock analysis as compared to intrablock analysis is Var()ˆ 1 Var(*) vkr ()  12 1 v 1 ()rk  2 . v 1 Such an increase may be estimated by ˆ ()rk  2 . v ˆ1

Although 12  but this may not hold true for ˆˆ12and  . The estimates ˆˆ12 and  may be negative also and in that case we take ˆˆ  . 12

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur 23