## STAT430 : BIBDReferers: Fall2007 :: (Remote :: Orphans :: Tree ) |
Dorman Wiki
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In a BIBD,

In the 2nd part of this problem on the homework, where you are asked to show that the given estimator is unbiased, it is sufficient to show unbiasedness for the example where*a* = 3, *b* =

3,*k* = 2, and lambda = 1.

However, for those of you who want to take a slightly more general approach, you will probably want to use the fact that

*r(k-1) = lambda(a-1)*, which implies that *r(k-1)* + lambda = lambda**a*

The above statement is true because for any particular treatment (say

treatment*i*), treatment *i* appears a total of *r* times, and each time it appears with

with*k-1* other treatments (because each block is of size *k*). In a BIBD, treatment *i* must also appear lambda times with the *(a-1)* other treatments, so this equality must hold.

This is actually a drawback for BIBDs, because BIBDs only exist for certain values of*a,r,k,* and *lambda*

- there are
*a*treatments, - each treatment appears a total of
*r*times, - there are
*b*blocks, - each block is of size
*k*, - each treatment appears once in a block or not at all
- all pairs of treatments appear together within a block lambda times

In the 2nd part of this problem on the homework, where you are asked to show that the given estimator is unbiased, it is sufficient to show unbiasedness for the example where

3,

However, for those of you who want to take a slightly more general approach, you will probably want to use the fact that

The above statement is true because for any particular treatment (say

treatment

with

This is actually a drawback for BIBDs, because BIBDs only exist for certain values of

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