Slide set 5
Stat402 (Spring 2016)
Last update: February 3, 2016
Stat 402 (Spring 2016): slide set 5
Randomized Complete Block Design
•
Model:
yij = µ
{z τ}i +βj + ij , i = 1, . . . , a; b = 1, . . . , b
|+
µi
τi : effect of the ith treatment; a = # of treatment
i.i.d
βj : effect of the j (th) block; b = # of blocks, ij ∼ N (0, σ 2)
•
Parameter Estimates:
µ̂ = ȳ..
τ̂i = ȳi. − ȳ.. or µ̂i = ȳi.
β̂j = ȳ.j − ȳ.. and σ̂ 2 = s2e
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Stat 402 (Spring 2016): slide set 5
Randomized Complete Block Design(Cont’d)
•
Predicted (or fitted) values:
ŷij = µ̂ + τ̂i + β̂j = ȳ.. + (ȳi. − ȳ..) + (ȳ.j − ȳ..)
•
Residuals:
= yij − ŷij
= yij − [ȳ.. + (ȳi. − ȳ..) + (ȳ.j − ȳ..)]
= yij − ȳi. − ȳ.j + ȳ..
rij
•
SSRes :
Pa
i=1
Pb
2
(y
−
ȳ
−
ȳ
+
ȳ
)
ij
i.
.j
..
j=1
2
Stat 402 (Spring 2016): slide set 5
Algebraic Decomposition of the Total Variation)
•
Total SScorrected = SST =
Pa
i=1
Pb
j=1 (yij
•
Between Treatment SS = SST rt = b
•
Pb
•
Between blocks SS = SSBlk = a
Residual SS = SSRes = SSE =
Pa
2
(ȳ
−
ȳ
)
i.
..
i=1
j=1 (ȳ.j
Pa
− ȳ..)2
i=1
− ȳ..)2
Pb
2
(y
−
ȳ
−
ȳ
+
ȳ
)
ij
i.
.j
..
j=1
SST = SST rt + SSBlk + SSE
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Stat 402 (Spring 2016): slide set 5
•
ANOVA Table
SV
Treatments
Blocks
Error
Total
•
d.f.
a−1
b−1
(a − 1)(b − 1)
N − 1(= ab − 1)
SS
SST rt
SSBlk
SSE
SST
MS
F
SST rt/(a − 1) M ST rt/M SE
SSBlk /(b − 1)
M SE (= s2E )
Computational Formulae
SSBlk
2
Pa Pb
y..
2
CF = ( i=1 j=1 yij ) /N = N
Pa Pb
2
SST = i=1 j=1 yij
− CF
Pa
SST rt = ( i=1 yi.2 )/b − CF
Pb
2
= ( j=1 yij
)/a − CF , SSE = (by substraction)
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Stat 402 (Spring 2016): slide set 5
Confidence Intervals and Pairwise Comparisions
A 100(1 − α)% C.I. for τp − τq (or µp − µq ) is (ȳp. − ȳq.) ∓ t α2 ,ν · sE
ν = (a − 1)(b − 1)
q
2
b,
Declare τp, τq significantly different at α level if using the
•
LSD Procedure
p
if |ȳp. − ȳq.| > t α2 ,ν · sE 2/b, where ν = (a − 1)(b − 1)
|
{z
}
LSDα
•
Tukey HSD Procedure
if |ȳp. − ȳq.| > Tukeyα, where
√
Tukeyα = qα,p,f · sE / b and f = (a − 1)(b − 1)
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Stat 402 (Spring 2016): slide set 5
•
Single Degree of Freedom Contrasts of Treatment Effects
To test
H0:
use the t-statistic
Pa
i=1 ci τi
= 0 vs Ha :
i=1 ci τi
6= 0
Pa
i=1 ci ȳi.
t0 = q
Pa
2
i=1 ci
sE
b
or the F-statistic
F0 =
where
Pa
SSc/1
M Sc
=
M SE
M SE
a
a
X
X
SSc = (
ciȳi.)2/[
c2i /b]
i=1
i=1
with a single d.f.
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Stat 402 (Spring 2016): slide set 5
Example 4.1
A resin (PFTE) is use to produce artificial vascular grafts by extruding into
tubes. A study is performed to determine the cause of hard protrusions
called flicks. The extrusion pressure is suspected to be a factor in the
occurrence of flicks. Four levels of extrusion pressure (8500, 8700, 8900,
9100 psi) are the treatments and the response is the percentage of tubing
produced in the run that did not contain any flicks.
The experimenter also knows that there is batch-to-batch variation among
the batches of resin delivered by the supplier and thus plans on using a
RCBD wtith batches of resin as block. Six batches of resin are used in
the experiment with the four levels of extrusion pressure used with resin
samples from each batch, the order of runs within each block being randomly
determined separatetely for each batch. Thus there are a total of 24 runs
in the experiment
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Stat 402 (Spring 2016): slide set 5
Data
ANOVA Table (extracted from the JMP output)
SV
Pressure
Batch
Error
Total
d.f.
3
5
15
23
SS
178.17125
192.25208
109.88625
480.30958
MS
59.3904
38.4504
7.3257
F
8.1071
p-value
.0019
Since the p-value is < .05 reject H0 : µ1 = µ2 = µ3 = µ4 at α = .05.
The treatment means suggest that the percentage of tubing with no flicks
decreases with increasing level of extrusion pressure.
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Stat 402 (Spring 2016): slide set 5
Testing orthogonal polynomials using contrasts
•
•
•
•
•
In this experiment, the experimenter may be interested in determining
whether the observed mean percentage is some function of the actual
temperature levels.
A test can be made to check if this is a linear a relationship i.e, mean
percentage is a linear function of temperature.
Orthogonal polynomials is a method of partitioning the treatment sum
of squares into components that allows the experimenter to construct
F -statistics with 1 degree of freedom each for the numerators.
Each of these can then be used to test whether the relationship can
be represented by a linear, quadratic, cubic, etc. function of the factor
levels, in a sequential fashion.
To make the computations easier, this partitioning of the treatment sum
of squares can be performed using appropriate orthogonal contrasts.
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Stat 402 (Spring 2016): slide set 5
•
•
•
•
A suggested procedure is to start with a test of a linear trend of the mean
response on the factor levels, and then test to check if the remaining
treatment sum of squares is significant (lack of fit) or noy.
If there is lack of fit, proceed by successively increasing the order of the
polynomial and performing lack of fit tests on the remaining part of the
treatment Sum of squares.
The required contrast coefficents can be obtained from a standard table
of orthogonal polynomials.
For an experiment with a = 4, the contrasts corresponding to the linear,
quadratic, and cubic orthogonal polynomials have the coefficients given
by
C10 = [−3, −1, 1, 3]
C20 = [1, −1, −1, 1]
C30 = [−1, 3, −3, 1]
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Stat 402 (Spring 2016): slide set 5
In the vascular grafts experiment let us test if there is a linear trend in the
decrease of percentage of usable tubing as the temperature increases. Test
the contrast −3 − 1 + 1 − 3 in JMP to obtain the part of the Pressure
SS due to a linear trend
ANOVA Table (extracted from the second JMP output)
SV
Pressure
—Linear Trend
—Lack of Fit
Batch
Error
Total
d.f.
3
1
2
5
15
23
SS
178.17125
171.6
6.57
192.25208
109.88625
480.30958
MS
59.3904
171.6
3.28
38.4504
7.3257
F
8.1071
23.4245
.45
p-value
.0019
.0002
The p-value for this F-statistic is .0002; thus it is significant, so the mean
percentage has a straight line relationship with the temperature levels. The
part of the Pressure SS labeled Lack of Fit (LoF) is the what is left over
when the SS due to linear trend is removed from the Pressure SS. The
F-statistic for LoF is small and so not significant. Thus we know that there
there is no LoF and thus the relationsship is linear.
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Stat 402 (Spring 2016): slide set 5
Additivity of the Model
Model yij = µ + τi + βj + ij
The expected response in the ij th cell is E(yij ) = µ + τi + βj . This
model is called additive because the overall mean µ is incremented by
adding quantities τi and βj to obtain the expected response in block j for
treatment i.
Blocks
Treatments
1
2
3
...
1
µ + τi + β1
µ + τ2 + β1
...
2
µ + τ1 + β2
µ + τ2 + β2
...
...
...
...
This model implies that the difference in the expected response for any
pair of treatments in the same block will contain no block effects i.e., that
difference will depend only on treatment effects.
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Stat 402 (Spring 2016): slide set 5
Additivity of the Model (Cont’d)
Example: Take the difference in the expected responses in block 1 for
treatments 1 and 2.
(µ + τ1 + β1) − (µ + τ2 + β1) = τ1 − τ2
The block effects cancel ot. This is true for any pair of treatments in any
block.
If the data or the experiment is such that the block effects and the treatment
effects cannot be considered additive, then an interaction is said to occur
between blocks and treatments. In this case the above model is not
appropriate.
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Stat 402 (Spring 2016): slide set 5
Discussion
There are two ways in which interaction could occur.
1. The response to a treatment changes from block to block. e.g.,
comparing 4 chemical processes. Batches of raw material are considered
as blocks. An impurity in batch 2 affects one of the processes adversely
resulting in lower yields.
2. Additive model does apply but not in the scale of measurement of the
response.
Interactions of type 1 above are considered non-transformable but those of
type 2 may be eliminated by transforming the data
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