Week11
Incomplete Block Designs
• Recall: in randomized complete block design, each of a treatments
was used once within each of b blocks.
• In some situations, it will not be possible to use each of a treatments
in each block.
• Any blocked experiment which has fewer than a units per block is
called an incomplete block design.
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Balanced Incomplete Block Design (BIBD)
• If all treatment comparisons are of equal importance and if we wish
to estimate each treatment effect with the same precision, then the
design needs to be balanced:
(1) Each block contains the same number of units.
(2) Each treatment occur the same number of times in total
(3) Each pair of treatments occurs together the same number of
times in total.
• A design that satisfies these conditions is called Balanced
Incomplete Block Design.
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Example
•
Consider a study which we wish to compares 4 treatments: A, B, C,
and D.
• Only 2 treatments can be used in each block and suppose there are 6
blocks available.
• The following is a balanced incomplete block design (BIBD):
• This design is balanced with respect to each of the 3 points above.
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• Consider the following design, which also uses 6 blocks, each with 2
units
• Although this design is balanced with regard to the first 2 items, the
same 2 treatments always appear together.
• So this would not be considered a balanced incomplete block design.
• Could this study be designed using 5 blocks? 7 blocks?
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Number of Experimental Units Required
• We will use the following notation:
a - number of treatments, b - number of blocks,
k - number of units per block ,
r - total number of times each treatment occurs
• The total number of units required in the study is therefore N = bk.
• However, the total number of units required can also be expressed as
N = ar.
• So in a balanced design, we must have bk = ar.
• Using combinatorics, we can show that the total number of times that
each pair of treatments appears together is r k 1 a 1
where λ must be an integer.
• NOTE: designs that satisfy the last two conditions may not exist for a
particular choice of a, b, k, r.
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Example
• A study was designed to compare 5 treatments.
• Although the study will use blocks to control for a nuisance factor,
there are only 3 experimental units within each block.
• Are there values of b (# of blocks) and r (# of times each treatment
is used) that would lead to a BIBD?
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Solution
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Statistical Model for BIBD
• The model equation for the BIBD is the same as for the randomized
complete block design, that is Yij = μ+τi +βj+ εij.
μ is the overall mean,
τi is the effect of the i-th treatment,
βj is the effect of the j-th block and
εij are i.i.d N(0, σ2) - the residual or random error term.
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Sums of Squares
• The total sum of squares is calculated in the usual manner:
SST Yij Y
a
b
2
i 1 j 1
• In decomposing total sum of squares, we use an adjusted treatment
sum of squares to account for the fact that each treatment did not
occur within each block.
• So treatment sum of squares sums over different blocks for each
treatment.
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Additional Notation
• Let Ti denote the sum over all experimental units which received
treatment i.
• Since each treatment does not appear in each block, the treatment
total must be adjusted to take this into account.
• Let Bi be the sum over all observations in the blocks in which
treatment i occurred.
• Then the adjusted treatment total is Qi = Ti - Bi / k
• If we were interested in testing for the equality of block effects,
we would also need to compute an adjusted block total.
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• The treatment sum of squares adjusted for blocks is
a
SSTr adj k Qi2 a
i 1
• The sum of squares for the blocking factor (not adjusted for
treatments) is
SS Bl k Y j Y
b
2
j 1
• The error sum of squares is obtained by subtraction:
SSE = SST - SSTr (adj) - SSBl
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Degrees of Freedom
• Since the same number of parameters must still be estimated in this
design (compared to the randomized complete block design), the
degrees of freedom will be the same.
• Since there are a treatments, the degrees of freedom for treatment is
a-1.
• Similarly, the degrees of freedom for blocks is b-1.
• The total degrees of freedom is N-1, where N = bk = ar
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Testing for Treatment Effects
• In order to test for treatment effects, we start by calculating the
necessary sums of squares and constructing the ANOVA Table. It is
given below
• Note that the mean square for blocks is not computed; if we wanted
to test for block effects we would have to compute an adjusted sum
of squares for blocks.
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• To test the hypotheses
H0: τ1 = τ2 = ... = τa = 0
Ha: not all τi = 0
We use the test statistic: Fobs = MSTr (adj) / MSE
• The corresponding P-value = P(F(a-1, N-a-b+1 > Fobs )
• Small values of the p-value would cause us to reject the hypothesis
that there is no treatment effect.
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Example
• A researcher planned and carried out a study to examine the inter-rater
reliability of a scale for measuring depression.
• Six examiners at a particular institution were the primary focus of the
study.
• However, since the outcome of the assessment was likely to be
influenced by the patient, it was decided to block on patients.
• It was not feasible to have all 6 examiners question every subject; each
subject was to be examined by 3 examiners.
• In order to achieve balance it is necessary to have ar = bk, which in
this case is 6r = 3b
• It is also necessary for λ = r(k-1)/(a-1) to be an integer; in this case
λ = r(2/5).
• The smallest value of r for which λ will be an integer is 5, and this
requires that the number of blocks, b, be 10.
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• A balanced incomplete block design was used, and the following data were
collected:
• Do these data provide any evidence that the examiners differ with respect
to the mean depression scores that they assign to patients?
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Solution
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Contrasts
• As in other experiments, the researchers may be interested in
comparisons between subgroups of treatment means.
• Contrasts can be used to test hypotheses.
• Tests should be based on the adjusted treatment effects: Yi'
Qi
Qi
r
• To test the hypothesis
H0 : c1μ1 + c2μ2 + · · · + caμa = 0
Ha : c1μ1 + c2μ2 + · · · + caμa ≠ 0
We use Fobs = SSC / MSE, where
2
ci Qi
SS C i 1 2
a
2
ci a / kr
i 1
a
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Least Squares Estimates of the Model Parameter
• In deriving the least squares estimators, we will use nij, an indicator
for treatment I occurring in block j as follow:
1 if treatment i block j
nij
0 if not
• Start by forming the sum of the squared deviations of the observed
values from the fitted values…
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Analysis of BIBD Using SAS
• In the designs we have examined up to now, the order in which
factors were specified in the “model” statement was not important.
• However, for the BIBD case the order is important.
• Refer to the Example on slide 15-16.
• The manner in which data are input for the BIBD is similar to that
for the RBC design:
data example ;
input patient examiner score ;
cards ;
1 1 10
1 2 14
1 3 10
2 1 3
2 2 3
2 4 1
etc;
run ;
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• The glm procedure is still the appropriate choice for this design, but
is important that the block factor be entered into the model before
the treatment factor
proc glm data
class patient
model score =
title1 'block
run ;
= example ;
examiner ;
patient examiner ;
before factor' ;
• Care is needed in reading the SAS output, since Type I SS is used
for the block factor (unadjusted) while Type III SS is used for the
treatment factor (adjusted). The outputs is given below:
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• The total SS is read from the SAS output, and in this case it is
1156.66666667.
• The error SS is read from the output and it is 139.22222222.
• The unadjusted block SS is taken from the column containing the
Type I SS, and in this case is 982.00.
• The adjusted treatment SS is taken from the column containing the
Type III SS, and is 35.44444444.
• The correct value to use for the F-test is the one associated with the
Type III SS for treatment.
• For the purposes of comparison, the output generated by SAS when
the treatment factor is entered into the model before blocks is given
below
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• From this output we can get the adjusted SS for the factor (examiner),
but we cannot get the unadjusted SS for the blocks (patients).
• Because of the unbalanced nature of this design, least squares
treatment means should be used rather than simple averages.
• To obtain least squares means for the treatment factor, just add one line
to the SAS code as follows:
proc glm data = example ;
class patient examiner ;
model score = patient examiner ;
lsmeans examiner ;
title1 'block before factor' ;
run ;
• the output generated by the “lsmeans” statement is given below:
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• Compare the LS means to the treatment averages that do not take
into account block effects:
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Contrasts in SAS
• Contrasts can also be tested using SAS.
• Suppose that examiner 1 was more recently hired than the other
examiners and it is of interest to test the hypothesis that the mean for
examiner 1 is equal to the mean for the others.
• In other words, test
H0 : μ1 = (μ2 + μ3 + μ4 + μ5 + μ6)/5
Ha : μ1 ≠ (μ2 + μ3 + μ4 + μ5 + μ6)/5
• To do this test is SAS, it is still important that the block factor is entered
into the model before the treatment factor.
• Add the following statement to the glm code given above
contrast ‘1st vs others’ examiner -1 0.2 0.2 0.2 0.2 0.2
;
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• An addition line of output will be produced giving the following
information
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Additive Models and Alternatives
• Models used for randomized complete block design, BIBD, Latin
square & Graeco-Latin square design are known as additive models.
• Consider the model used for the randomized complete block design:
Yij = μ+τi +βj+ εij.
• In this model, treatment i always causes the expected response to
increase/decrease τi units from the overall mean μ.
• The effect of block j is to cause a change of βj units, regardless of
which treatment is applied.
• The effect of treatment i and block j together cause an increase or
decrease from the overall mean by τi +βj.
• In other words, the effects of the 2 factors are additive.
• This model is not always appropriate for a randomized complete block
design.
• Generally, an additive model may not be appropriate for a particular
design.
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Multiplicative Effects
• In some cases, the effect of one factor might be amplified by the
presence of another factor.
• A multiplicative model might be appropriate in this situation:
Yij = μ×τi×βj×εij.
• Taking logarithms results in an additive model:
ln{Yij} = ln{μ}+ln{τi}+ln{βj}+ln{εij}.
• This can be written as
Wij ' i' 'j ij'
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Interaction
• In some cases, there may be an element of additivity, although the
additive model may not be adequate.
• For instance, consider a randomized complete block design where
treatment i causes an extreme reaction in a particular block, but the
other treatments do not.
• Furthermore, suppose that in the other blocks treatment i does not
cause an extreme reaction.
• In this case, there is an interaction between treatments and blocks.
• The additive model would need to be modified to include an
interaction term.
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