Math 558 Lecture #28
Balanced incomplete block designs
Theorem .1
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(Fisher’s Inequality) In a balanced incomplete-block design,b ≥ t.
The other two requirements of a BIBD are
λ(t − 1) = r(k − 1)
We can use these relations to find minimum number of blocks for a BIBD. From the last equation we can see that k − 1 must divide
λ(t − 1). In our catalysts example t = 4, k = 3 λ(t − 1) = λ × 3 must be divided by k − 1 = 2. Therefore the smallest integer for which the above relation is satisfied is λ = 2. Using the last equation again we get r = 2.
Finding a BIBD in R
The function BIBsize in the R package daewr provides a quick way of finding values of λ and r to satisfy Equations given in the last slide when given the number of the treatments t and the block size k. Examples: 1. library(daewr)
BIBsize(6, 3)
Posible1 BIB design with b= 10 and r= 5 λ = 2 2. BIBsize(4, 3)
Posible BIB design with b= 4 and r= 3 λ = 2
1These are the spellings used in R output
Finding a BIBD in R
To find a field plan for BIBD in R use the R package AlgDesign and the function optBlock. The function optBlock will take the arguments optBlock( withinData = factor(1:7),blocksizes = rep(3, 7))
Where“ withinData= (1:t)” stores information on the number of treatments (factor levels) and “blocksizes = rep(k, b))” stores block size and number of blocks. After creating a BIBD by using the optBlock function we can print out the plan as follows.
BIBD <- optBlock( withinData = factor(1:7),blocksizes = rep(3, 7)) design <- BIBDrows
design <- matrix(design, nrow = 7, ncol = 3, byrow = TRUE, dimnames = list(c( "B1", "B2", "B3", "B4", "B5", "B6", "B7"), c("unit1", "unit2", "unit3")))
Finding a BIBD in R
unit1 unit2 unit3 B1 1 2 4 B2 3 4 6 B3 1 3 7 B4 2 6 7 B5 4 5 7 B6 1 5 6 B7 2 3 5
According to this plan the three experimental units in the frst block would receive treatments 1,2, 4, the experimental units in the second block would receive 3,4,6 and so forth. By inspection, it can be seen that each treatment occurs r = 3 times in this design, and that every treatment pair occurs within the same block with every other treatment level λ = 1 times.
yij = μ + bi + τj + εij
which is identical to the model for a randomized complete block design given . However, the analysis is slightly different due to the missing observations as the blocks are incomplete.
Example (Lawson chapter 7
Consider the data from a taste panel experiment reported by Moskowitz (1988), in the next slide. This experiment is conducted according to BIBD with t=4 treatments ( recipes), and block size k=2. Each panelist in the experiment tastes only two of the four recipes in a random order, and assigns a category scale score (score from 1-10 with description of each category)to the recipe. We can have (42) = 6 panelists to conduct the experiment. Panelists are taken as blocks in this experiment. However, the experimentalist had enough resources to double the number of panelists (blocks) to increase the power of the experiment. . Thus the first six blocks and the last six are a repeat of the same BIB design.Block numbers were randomly assigned to the panelists and the order of the two recipes tasted by each panelist was randomized as well.
Data for the Taste example
Recipes PanelistA B CD
155-- 27-6- 35--4 4-67- 5-6-4 6--86 767-- 85-8- 94--5
10 - 7 7 - 11 - 6 - 5 12 - - 7 4
Design the Taste example
Recipes PanelistA B CD
1AB-- 2A-C- 3A--D 4-BC- 5-B-D 6--CD 7AB-- 8A-C- 9A--D
10 - B C - 11 - B - D 12 - - CD
Statistical Analysis
model <- aov( score panelist + recipe, data = taste) summary(model)
panelist recipe Residuals
Df Sum Sq 11 19.333 3 9.125 9 6.875
Mean Sq 1.7576 3.0417 0.7639
F value 2.301 3.982
0.1106 0.0465 *
Further Investigation
install.packages("lsmeans")
library(lsmeans)
lsmeans(model, pairwise recipe, adjust = ("tukey"))
recipe lsmean SE df
lower.CL upper.CL A 5.46 0.418 9 4.51 6.40
B 6.21 0.418 9 5.26 7.15
C 6.83 0.418 9 5.89 7.78
D 4.83 0.418 9 3.89 5.78
Results are averaged over the levels of: panelist Confidence level used: 0.95
Further investigation
$contrasts
contrast estimate SE df
t.ratio p.value -1.214 0.6342 -2.225 0.1882 1.011 0.7472 -1.011 0.7472 2.225 0.1882 3.236 0.0421
A-B -0.750 A-C -1.375 A-D 0.625 B-C -0.625 B-D 1.375 C-D 2.000
0.618 9 0.618 9 0.618 9 0.618 9 0.618 9 0.618 9
Results are averaged over the levels of: panelist P value adjustment: tukey method for comparing a family of 4 estimates
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