Transcript The Random Effects Model – Introduction

The Random Effects Model – Introduction
•
Sometimes, treatments included in experiment are randomly chosen
from set of all possible treatments.
• Conclusions from such experiment can then be generalized to other
treatments.
• When the treatments are random sample, the treatment effects, τi,
are random variables.
• This type of model is called a random effects model or a
components of variance model.
• Note it is important to determine at the planning phase of an
experiment whether treatment effects are fixed or random.
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Example - Weaving Loom Homogeneity
•
Textile Company uses large number of looms to produce fabric. The
looms function in uniform manner so fabric produced by company is
of uniform strength.
• To investigate whether there are any difference between looms, 4
looms will be selected at random from all those owned by company.
• Four samples will be obtained from each loom.
• Fixed or Random Effects? Since looms are randomly selected, a
random effects model must be used. Results obtained from these 4
looms can be generalized to the set of all looms owned by company.
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The Random Effect Model
• The equation for the statistical model remains the same as for fixed
effects model is: Yij = μ + τi + εij .
• As in the fixed effects model, the εij are assumed to be i.i.d. N(0, σ2).
• However, unlike the fixed effects model, random effects model has
treatment effects, τi, which are random variables.
• Therefore, instead of assuming that ,  i  0 random effects model
requires that E(τi) = 0.
• Under this assumption, the overall mean is still μ. To see that this is the
case, consider…
• It is also assumed that:
1. the τi are statistically independent of the εij
2. that the τi are i.i.d N 0,  2 .


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Expected Mean Squares for Random Effects Model
• To construct hypothesis test concerning treatment effects, we need to
find expected mean squares.
• Since τi are random, E(MSTreat) is not the same as in fixed effects
model.
• To find E(MSTreat) start by expressing the treatment sample means in
terms of model parameters as follows:….
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Hypothesis Testing
• In a random effect model, testing for no treatment effect is in fact
testing whether the variance of treatment effects is 0.
• The Hypothesis are therefore: H 0 :  2  0
H a :  2  0
• When H0 is true, then it follows that E(MSTreat) = σ2 = E(MSE).
• Otherwise, if H0 is false, then E(MSTreat) > σ2.
• So the ratio of MSTreat to MSE provides a test statistic which is given by
Fobs
MSTreat

.
MS E
• The calculation of the P-value is then the same as for fixed effect
model. That is, P-value = P(F(a-1, n-a) > Fobs).
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Analysis of Variance Table
• The only aspect of ANOVA table that is different from fixed effects
case is E(MSTreat).
• The ANOVA table for random effect model is given below
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Estimating the Variance Components
• In a random effects model, both σ2 and  2 are known as the
components of variation.
• Point estimates of these quantities can be obtained by using
information contained in the ANOVA table.
• As we can see the MSE is unbiased estimate of σ2.
• Exercise: Show that an unbiased estimate of  2 is:
ˆ2 
a  1MSTreat  MSE 
ri2

n  i1
a
n
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Proportion of Variance Explained
• The total variation in Yij is    therefore, the proportion of
variance explained by the treatment is:
2
2
 2
 2   2
• An estimate of this proportion can be obtained by substituting
ˆ 2 and ˆ2 into the equation above.
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Confidence Interval for Variance Component
• Confidence interval is useful for assessing amount of treatment
variability relative to error variability.
• That is, we want a CI for 
2
2.
• It can be shown that:
SSTreat
SSE
2
2
~

and
~

a 1
na
 2  c 2
2
• Further, to use Cochrane’s theorem, we divide each of the above by its
degrees of freedom and take ratio. We get…
• This can be used to derive a confidence interval for by taking the
following steps…
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Example Weaving Looms Continued
• Recall, 4 looms randomly selected from all those owned by the
company.
• Four cloth samples obtained from each loom & strength is recorded.
• The data is given in the following table:
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Using SAS to Conduct ANOVA
• To conduct ANOVA in SAS we need to add a RANDOM statement.
The code is then,
proc glm data = looms ;
class loom ;
model strength = loom ;
RANDOM loom ;
run ;
• The extra line in the code generated information regarding E(MSTreat).
• The output is given in the following slide (slide 12).
• Exercise, verify that E(MSTreat) is as stated in output…
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Factorial Designs – Introduction
• In a factorial design experiments, at least 2 factors are studied for
their simultaneous impact on the response.
• Suppose there are 2 factors, A and B where A has levels 1, 2, . . . , a
and B has levels 1, 2, . . . , b.
• Every level of A will be used with every level of B. That is, there are
a × b possible treatment combinations.
• The experimental units are randomly assigned to the treatments.
• In this type of design factors are said to be crossed.
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Advantages and Disadvantages of Factorial Designs
Advantage s
• Factorial designs are more efficient than doing multiple studies to
examine one factor at a time.
• They provide information about joint impact of factors on response.
Disadvantage
• Studies with many factors can be complex and costly to design and
execute.
• Further, if too many factors are included, it may result in insufficient
degrees of freedom for estimating variability due to experimental
error.
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Example - Toxins and Their Antidotes
• Three toxic agents are thought to have varying impacts on survival time.
• Four potential poison antidotes are also to be studied to determine if they
also had an impact on survival, in the presence of a toxic agent.
• Four different animals were randomly allocated to each of the 12
treatments.
• The mean survival times for 4 animals exposed to each treatment are
given in the Table below:
Antidote
Toxin
B1
B2
B3
B4
A1
41.25
88.00
56.75
61.00
A2
32.00
81.50
37.50
66.75
A3
21.00
33.50
23.50
32.50
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• To see relationship between toxic agent, antidote, and survival time, it
is useful to plot mean survival times. The plot is given below:
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• As we can see, for antidotes B1, B2, and B3, toxic agent A1 has the
greatest mean survival time, followed by A2, and then A3.
• For antidote B4 there is a different pattern.
• Also notice that the plots for antidotes B1 and B3 are almost parallel
but those for B2 and B4 are very different.
• It appears that there is an interaction between toxic agents and
antidotes.
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Understanding Interaction
• If there is no interaction between factors A and B then plot should
contain approximately parallel curves.
• Variability in data can be explained as a contribution due to factor A
and a contribution due to factor B.
• However, in presence of interaction, contributions by factors A and
B don’t explain all of the differences seen in the data.
• There is also a contribution associated with each particular
combination of A and B.
• The following figures illustrate what might be expected when there
is or isn’t an effect of Factor A, Factor B, or an interaction effect.
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