Design and analysis of the experiment

1. Balanced Incomplete Block Designs
(BIBD’s)

2. Incomplete Block Designs
Some experiments may consist of a large number of treatments
and it may not be feasible to run all the treatments in all the
blocks. Designs where only some of the treatments appear in
every block are known as incomplete block designs.
 A balanced incomplete block design (BIBD) is an incomplete
block design in which any two treatments appear together an
equal number of times.
-

3. Suppose that there are 𝑎 treatments and that each block can hold
exactly 𝑘(𝑘 < 𝑎) treatments. A BIBD may be constructed by taking
blocks and assigning
𝑎
𝑘
different combination of treatments to each
block.
In a BIBD setup, each block is selected in a balanced manner so that
any pair of treatments occur together the same number of times as
any other pair.
One way to construct a BIBD is by using
𝑎
𝑘
blocks and assigning
different combination of treatments to every block.

4. Examples
Consider three treatments, A, B, and C where two treatments are run in every
block. There are
3
2
= 3 ways of choosing 2 out of three. Thus using three blocks
 Consider five treatments, A, B, C, D, and E where 3 treatments appear per block.
We use 10 blocks

5. Statistical Analysis of the BIBD’s
Assume that there are 𝑎 treatments and 𝑏 blocks, and that each block contains 𝑘
treatments, and each treatment occurs 𝑟 times in the design (or is replicated
𝑟 times). Therefore there are 𝐍 = 𝒂𝒓 = 𝒃𝒌 total observations.
The number of times each pair of treatments appears in the same block is
The parameter 𝜆 must be an integer.
If 𝑎 = 𝑏, the design is said to be symmetric

6. In a balance incomplete block design;
-𝑏 blocks have the same number of 𝑘 treatment each and
-every treatment is replicated 𝑟 times in the design.
-Each treatment occurs at most once in block, i.e 𝑛𝑖𝑗 = 1 or 0 where nij is the
number of times the 𝑖 − 𝑡ℎ treatment occurs in 𝑗 − 𝑡ℎ block, 𝑖 =
1,2, … . . , 𝑎 𝑗 = 1,2, … … . . , 𝑏
Incomplete: cannot fit all treatment in each block
Balance: each pair of treatment occur together 𝜆 times

7. The Statistical model for the BIBD is
𝑦𝑖𝑗 = 𝜇 + 𝜏𝑖 + 𝛽𝑗 + 𝜀𝑖𝑗
𝑖 = 1,2, … … . , 𝑎
𝑗 = 1,2, … … … , 𝑏
Where 𝑦𝑖𝑗 is the 𝑖𝑡ℎ
observation in the 𝑗𝑡ℎ
block,
𝜇 is the overall mean,
𝜏𝑖 is the effect of the 𝑖𝑡ℎ treatment,
𝛽𝑗 is the effect of the 𝑗𝑡ℎblock, and
𝜀𝑖𝑗 is the 𝑁𝐼𝐷 (0, 𝜎2) random error component.
We partition the total sum of squares in the usual manner; into sum of square due to
treatments, blocks, and error.
But the difference here is that the sum of squares due to treatments needs to be
adjusted for incompleteness.

8. Total variability may be partitioned into
The total variability in the data is expressed by the total corrected sum of
squares:

9. The block sum of squares is
Where 𝑦.𝑗 is the total in the 𝑗𝑡ℎ block. 𝑆𝑆𝐵𝑙𝑜𝑐𝑘𝑠 has 𝑏 − 1 degrees of freedom.
The adjusted treatment sum of squares is
Where is the adjusted total for the 𝑖𝑡ℎ treatment, which is computed as
with 𝑛𝑖𝑗 = 1 if treatment 𝑖 appears in block 𝑗 and 𝑛𝑖𝑗=0 otherwise

10. The adjusted treatment totals will always sum to zero.
𝑆𝑆𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠(𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑) has 𝑎 − 1 degrees of freedom.
The error sum of squares is computed by subtraction as
and has 𝑁 − 𝑎 − 𝑏 − 1 degrees of freedom.
The appropriate statistic for testing the equality of the treatment effects is

11. • The ANOVA table

12. Example: Suppose that a chemical engineer thinks that the time of reaction for a
chemical process is a function of the type of catalyst employed. Four catalysts are
currently being investigated. The experimental procedure consists of selecting a
batch of raw material, loading the pilot plant, applying each catalyst in a separate
run of the pilot plant, and observing the reaction time. Because variations in the
batches of raw material may affect the performance of the catalysts, the engineer
decides to use batches of raw material as blocks. However, each batch is only large
enough to permit three catalysts to be run. Therefore, a randomized incomplete
block design must be used. The balanced incomplete block design for this
experiment, along with the observations recorded, is shown

13. The analysis of this data is as follows;
 From the data, we have 𝑎 = 4, 𝑏 = 4, 𝑘 = 3, 𝑟 = 3, and 𝑁 = 12. thus,
𝜆 =
𝑟(𝑘−1)
𝑎−1
=
3(3−1)
4−1
=2
𝜆 = 2
 The total sum of squares is
The block sum of squares is =
1
𝑘 𝑗=1
𝑏
𝑦.𝑗
2
-
𝑦..
2
𝑁

14. To compute the treatment sum of squares adjusted for blocks, we first
determine the adjusted treatment totals using
= 𝑦𝑖. −
1
𝑘 𝑗=1
𝑏
𝑛𝑖𝑗𝑦.𝑗
The adjusted sum of squares for treatments is computed from

15. The error sum of squares is obtained by subtraction as
The analysis of variance

16. We conclude that the catalyst employed has a significant effect on the
time of reaction.
Note that, in the analysis that we have described, the total sum of
squares has been partitioned into an adjusted sum of squares for
treatments, an unadjusted sum of squares for blocks, and an error
sum of squares:
That is, 𝑆𝑆𝑇 = 𝑆𝑆𝑇𝑟(𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑)+𝑆𝑆𝐵(𝑢𝑛𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑)+𝑆𝑆𝐸
Sometimes we would like to assess the block effects. To do this, we
require an alternate partitioning of SST, that is,
𝑆𝑆𝑇 = 𝑆𝑆𝑇𝑟 + 𝑆𝑆𝐵(𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑) + 𝑆𝑆𝐸
Here 𝑆𝑆𝑇𝑟 is unadjusted.

17. If the design is symmetric, that is, if 𝑎 = 𝑏, a simple formula may be obtained for
SSBlocks(adjusted). The adjusted block totals are
=𝑦.𝑗 −
1
𝑟 𝑖=1
𝑎
𝑛𝑖𝑗𝑦𝑖. 𝑗 = 1,2, … … … . 𝑏
Thus, the 𝑆𝑆𝐵 𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑 is given as

19. From
From 𝑆𝑆𝑇𝑟 =
1
𝑘 𝑖=1
𝑎
𝑦𝑖.
2
−
𝑦..
2
𝑁

20. A summary of the analysis of variance for the symmetric BIBD inclufing both
treatments and blocks
Notice that the sums of squares associated with the mean squares do not add to
the total sum of squares, that is,
This is a consequence of the nonorthogonality of treatments and blocks.

21. The estimation of the BIBD,s model parameters
Consider the previous BIBD example.
Recall: