•
•
•
SSRes :
− ȳi. − ȳ.j + ȳ..)
2
= yij − ŷij
= yij − [ȳ.. + (ȳi. − ȳ..) + (ȳ.j − ȳ..)]
= yij − ȳi. − ȳ.j + ȳ..
j=1 (yij
b
rij
i=1
a
Residuals:
ŷij = μ̂ + τ̂i + β̂j = ȳ.. + (ȳi. − ȳ..) + (ȳ.j − ȳ..)
Predicted (or fitted) values:
2
Stat 402 (Spring 2016): slide set 5
Randomized Complete Block Design(Cont’d)
Last update: February 3, 2016
Stat402 (Spring 2016)
Slide set 5
•
•
•
•
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•
μi
i=1
a
i=1
a
− ȳi. − ȳ.j + ȳ..)2
− ȳ..)2
j=1 (yij
b
j=1 (ȳ.j
b
SST = SST rt + SSBlk + SSE
Residual SS = SSRes = SSE =
Between blocks SS = SSBlk = a
− ȳ..)2
− ȳ..)2
i=1 (ȳi.
a
j=1 (yij
b
Between Treatment SS = SST rt = b
Total SScorrected = SST =
1
3
Stat 402 (Spring 2016): slide set 5
Algebraic Decomposition of the Total Variation)
β̂j = ȳ.j − ȳ.. and σ̂ 2 = s2e
τ̂i = ȳi. − ȳ.. or μ̂i = ȳi.
μ̂ = ȳ..
Parameter Estimates:
βj : effect of the j (th) block; b = # of blocks, ij ∼ N (0, σ 2)
i.i.d
τi : effect of the ith treatment; a = # of treatment
yij = μ +
τi +βj + ij , i = 1, . . . , a; b = 1, . . . , b
Model:
Stat 402 (Spring 2016): slide set 5
Randomized Complete Block Design
•
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d.f.
a−1
b−1
(a − 1)(b − 1)
N − 1(= ab − 1)
2
j=1 yij )/a
b
2
/N =
H0 :
with a single d.f.
where
or the F-statistic
use the t-statistic
To test
SSc = (
i=1
a
ciȳi.)2/[
i=1
a
c2i /b]
SSc/1
M Sc
=
F0 =
M SE
M SE
i=1 ci τi = 0
a
a
i=1 ci ȳi.
t0 = a
2
i=1 ci
sE
b
i=1 ci τi = 0 vs Ha :
a
6
7
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
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.
Example 4.1
Stat 402 (Spring 2016): slide set 5
Stat 402 (Spring 2016): slide set 5
if |ȳp. − ȳq.| > Tukeyα, where
√
Tukeyα = qα,p,f · sE / b and f = (a − 1)(b − 1)
Tukey HSD Procedure
2/b, where ν = (a − 1)(b − 1)
LSDα
if |ȳp. − ȳq.| > t α2 ,ν · sE
LSD Procedure
2
b,
5
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Declare τp, τq significantly different at α level if using the
A 100(1 − α)% C.I. for τp − τq (or μp − μq ) is (ȳp. − ȳq.) ∓ t α2 ,ν · sE
ν = (a − 1)(b − 1)
Stat 402 (Spring 2016): slide set 5
Confidence Intervals and Pairwise Comparisions
4
− CF , SSE = (by substraction)
N
2
y..
MS
F
SST rt/(a − 1) M ST rt/M SE
SSBlk /(b − 1)
M SE (= s2E )
Stat 402 (Spring 2016): slide set 5
Single Degree of Freedom Contrasts of Treatment Effects
SSBlk = (
b
j=1 yij )
b
SS
SST rt
SSBlk
SSE
SST
2
i=1
j=1 yij − CF
a
( i=1 yi.2 )/b − CF
a
i=1
SST rt =
SST =
CF = (
a
Computational Formulae
SV
Treatments
Blocks
Error
Total
ANOVA Table
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
Stat 402 (Spring 2016): slide set 5
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C3 = [−1, 3, −3, 1]
C2 = [1, −1, −1, 1]
C1 = [−3, −1, 1, 3]
10
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
11
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.
SV
Pressure
—Linear Trend
—Lack of Fit
Batch
Error
Total
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)
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
Stat 402 (Spring 2016): slide set 5
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.
9
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•
Stat 402 (Spring 2016): slide set 5
Testing orthogonal polynomials using contrasts
8
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.
SV
Pressure
Batch
Error
Total
ANOVA Table (extracted from the JMP output)
Data
Stat 402 (Spring 2016): slide set 5
1
μ + τi + β1
μ + τ2 + β1
...
2
μ + τ1 + β2
μ + τ2 + β2
...
...
...
...
14
Interactions of type 1 above are considered non-transformable but those of
type 2 may be eliminated by transforming the data
2. Additive model does apply but not in the scale of measurement of the
response.
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.
There are two ways in which interaction could occur.
Discussion
Stat 402 (Spring 2016): slide set 5
12
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.
Treatments
1
2
3
...
Blocks
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.
Model yij = μ + τi + βj + ij
Additivity of the Model
Stat 402 (Spring 2016): slide set 5
13
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.
The block effects cancel ot. This is true for any pair of treatments in any
block.
(μ + τ1 + β1) − (μ + τ2 + β1) = τ1 − τ2
Example: Take the difference in the expected responses in block 1 for
treatments 1 and 2.
Additivity of the Model (Cont’d)