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## III.Completely Randomized Design (CRD)

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### III. Completely Randomized Design (CRD) III.A Design of a CRD III.B Models and estimation for a CRD III.C Hypothesis testing using the ANOVA method – PowerPoint PPT presentation

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Title: III.Completely Randomized Design (CRD)

1
III. Completely Randomized Design (CRD)
• III.A Design of a CRD
• III.B Models and estimation for a CRD
• III.C Hypothesis testing using the ANOVA method
• III.D Diagnostic checking
• III.E Treatment differences

2
III.A Design of a CRD
• Definition III.1 An experiment is set up using
a CRD when each treatment is applied a specified,
possibly unequal, number of times, the particular
units to receive a treatment being selected
completely at random.
• Example III.1 Rat experiment
• Experiment to investigate 3 rat diets with 6
rats
• Diet A, B, C will have 3, 2, 1 rats,
respectively.

3
Use R to obtain randomized layouts
• How to do this is described in Appendix B,
Randomized layouts and sample size computations
in , for all the designs that will be covered in
this course, and more besides.

4
R functions and output to produce randomized
layout
• gt Obtaining randomized layout for a CRD
• gt
• gt n lt- 6
• gt CRDRat.unit lt- list(Rat n)
• gt Diet lt- factor(rep(c("A","B","C"),
• gt times
c(3,2,1)))
• gt CRDRat.lay lt- fac.layout(unrandomizedCRDRat.uni
t,
• gt randomizedDiet,
seed695)
• gt CRDRat.lay
• fac.layout from dae package produces the
randomized layout.
• unrandomized gives the single unrandomized factor
indexing the units in the experiment.
• randomized specifies the factor, Diets, that is
to be randomized.
• seed is used so that the same randomized layout
for a particular experiment can be generated at a
later date. (01023)

5
Randomized layout
• Units Permutation Rat Diet
• 1 1 4 1 A
• 2 2 1 2 C
• 3 3 5 3 B
• 4 4 3 4 A
• 5 5 6 5 A
• 6 6 2 6 B
• remove Diet object in workspace to avoid using
it by mistake
• remove(Diet)

6
III.B Models and estimation for a CRD
• The analysis of CRD experiments uses
• least-squares or maximum likelihood estimation of
the parameters of a linear model
• hypothesis testing based on the ANOVA method or
maximum likelihood ratio testing.
• Use rat experiment to investigate linear models
and the estimation of its parameters.

7
a) Maximal model
• Definition III.2 The maximal expectation model
is the most complicated model for the expectation
that is to be considered in analysing an
experiment.
• We first consider the maximal model for the CRD.

8
Example III.1 Rat experiment (continued)
• Suppose, the experimenter measured the liver
weight as a percentage of total body weight at
the end of the experiment.
• The results of the experiment are as follows
• The analysis based on a linear model, that is
• Trick is what are X and q going to be?

9
Perhaps?
• Note numbering of Y's does not correspond to
Rats does not affect model but neater.
• This model can then be fitted using simple linear
regression techniques.

10
Using model to predict
• However, does this make sense?
• means that, for each unit increase in diet,
liver weight decreases by 0.120.
• sensible only if the diets differences are based
on equally spaced levels of some component
• For example, if the diets represent 2, 4 and 6 mg
of copper added to each 100g of food
• But, no good if diets unequally spaced (2, 4, and
10 mg Cu added) or diets differ qualitatively.

11
Regression on indicator variables
• In this method the explanatory variables are
called factors and the possible values they take
levels.
• Thus, we have a factor Diet with 3 levels A, B,
C.
• Definition III.3 Indicator variables are formed
from a factor
• create a variable for each level of the factor
• the values of a variable are either 1 or 0, 1
when the unit has the particular level for that
variable and 0 otherwise.

12
Indicator-variable model
• Hence
• EYi ak, varYi s2, covYi, Yj 0, (i ?
j)
• Can be written as
• Model suggests 3 different expected (or mean)
values for the diets.

13
General form of X for CRD
• For the general case of a set of t Treatments
suppose Y is ordered so that all observations
for
• 1st treatment occur in the first r1 rows,
• 2nd treatment occur in the next r2 rows,
• and so on with the last treatment occurring in
the last rt rows.
• i.e. order of systematic layout (prior to
randomization)
• Then XT given by the following partitioned matrix

(1 only ever vector, but 0 can be matrix)
14
Still a linear model
• In general, the model for the expected values is
still of the general form EY Xq
• and on assuming Y is
• can use standard least squares or maximum
likelihood estimation

15
Estimates of expectation parameters
• OLS equation is
• Can be shown, by examining the OLS equation, that
the estimates of the elements of a and y are the
means of the treatments.
• Example III.1 Rat experiment (continued)
• The estimates of a are

16
Example III.1 Rat experiment (continued)
• The estimates of the expected values, the fitted
values, are given by

17
Estimator of the expected values
• In general, where is the n-vector
consisting of the treatment means for each unit
and
• being least squares, this estimator can be
written as a linear combination of Y.
• that is, can be obtained as the product of an
matrix and the n-vector Y.
• let us write
• M for mean because MT is the matrix that replaces
each value of Y with the mean of the
corresponding treatment.

18
General form of mean operator
• Can be shown that the general form of MT is
• MT is a mean operator as it
• computes the treatment means from the vector to
which it is applied and
• replaces each element of this vector with its
treatment mean.

19
Estimator of the errors
• The estimator of the random errors in the
observed values of Y is, as before, the
difference from the expected values.
• That is,

20
Example III.1 Rat experimentAlternative
expression for fitted values
21
Residuals
• Fitted values for orthogonal experiments are
functions of means.
• Residuals are differences between observations
and fitted values

22
b) Alternative indicator-variable, expectation
models
• For the CRD, two expectation models are
considered
• First model is minimal expectation model
population mean response is same for all
observations, irrespective of diet.
• Second model is the maximal expectation model.

23
Minimal expectation model
• Definition III.4 The minimal expectation model
is the simplest model for the expectation that is
to be considered in analysing an experiment.
• The minimal expectation model is the same as the
intercept-only model given for the single sample
in chapter I, Statistical inference.
• Will be this for all analyses we consider.

24
Marginal models
• In regression case obtained marginal models by
zeroing some of the parameters in the full model.
• Here this is not the case.
• Here simply set ak m
• That is, intercept only model is the special case
where all ak s are equal.
• Clear for getting EYi m from EYi ak.
• What about yG from yT?
• If replace each element of a with m, then a
1tm.
• So yT XTa XT1tm XGm.
• Now marginality expressed in the relationship
between XT and XG as encapsulated in definition.

25
Marginality of models (in general)
• Definition III.5 Let C(X) denote the column
space of X.
• For two models, y1 ? X1q1 and y2 ? X2q2, the
first model is marginal to the second if C(X1) ?
C(X2) irrespective of the replication of the
levels in the columns of the Xs,
• That is if the columns of X1 can always be
written as linear combinations of the columns of
X2.
• We write y1 ? y2.
• Note marginality relationship is not symmetric
it is directional, like the less-than relation.
• So while y1 ? y2, y2 is not marginal to y1 unless
y1 ? y2.

26
Marginality of models for CRD
• yG is marginal to yT or yG ? yT because C(XG) ?
C(XT)
• in that an element from a row of XG is the sum of
the elements in the corresponding row of XT
• and this will occur irrespective of the
replication of the levels in the columns of XG
and XT.
• So while yG ? yT, yT is not marginal to yG as
C(XT) ? C(XG) so that yT ? yG.
• In geometrical terms, C(XT) is a
three-dimensional space and C(XG) is a line, the
equiangular line, that is a subspace of C(XT).

27
III.C Hypothesis testing using the ANOVA method
• Are there significant differences between the
treatment means?
• This is equivalent to deciding which of our two
expectation models best describes the data.
• We now perform the hypothesis test to do this for
the example.

28
Analysis of the rat example Example III.1 Rat
experiment (continued)
• Step 1 Set up hypotheses
• H0 aA aB aC m (yG ? XGm)
• H1 not all population Diet means are equal
• (yD ? XDa)
• Set a 0.05

29
Example III.1 Rat experiment (continued)
• Step 2 Calculate test statistic
• From table can see that (corrected) total
variation amongst the 6 Rats is partitioned into
2 parts
• variance of difference between diet means and
• the left-over (residual) rat variation.
• Step 3 Decide between hypotheses
• As probability of exceeding F of 3.60 with n1
2 and n2 3 is 0.1595 gt a 0.05, not much
evidence of a diet difference.
• Expectation model that appears to provide an
adequate description of the data is yG ? XGm.

30
b) Sums of squares for the analysis of variance
• From chapter I, Statistical inference, an SSq
• is the SSq of the elements of a vector and
• can write as the product of transpose of a column
vector with original column vector.
• Estimators of SSqs for the CRD ANOVA are SSqs of
following vectors (cf ch.I)

where Ds are n-vectors of deviations from Y
and Te is the n-vector of Treatment effects.
Definition III.6 An effect is a linear
combination of means with a set of effects
summing to zero.
31
• Want to show estimators of all SSqs can be
written as Y?QY.
• Is product of 1?n, n?n and n?1 vectors and
matrix, so is 1?1 or a scalar.
• Definition III.7 A quadratic form in a vector Y
is a scalar function of Y of the form Y?AY where
A is called the matrix of the quadratic form.

32
• Firstly write
• That is, each of the individual vectors on which
the sums of squares are based can be written as
an M matrix times Y.
• These M matrices are mean operators that are
symmetric and idempotent M' M and M2 M in
all cases.

33
• Then
• Given Ms are symmetric and idempotent, it is
relatively straightforward to show so are the
three Q matrices.
• It can also be shown that

34
• Consequently obtain the following expressions for
the SSqs

35
• Theorem III.1 For a completely randomized
design, the sums of squares in the analysis of
variance for Units, Treatments and Residual are

Proof follows the argument given above.
36
Residual SSq by difference
• That is, Residual SSq Units SSq - Treatments
SSq.

37
ANOVA table construction
• As in regression, Qs are orthogonal projection
matrices.
• QU orthogonally projects the data vector into the
n-1 dimensional part of the n-dimensional data
space that is orthogonal to equiangular line.
• QT orthogonally projects data vector into the t-1
dimensional part of the t-dimensional Treatment
space, that is orthogonal to equiangular line.
(Here the Treatment space is the column space of
XT.)
• Finally, the matrix orthogonally
projects the data vector into the n-t dimensional
Residual subspace.
• That is, Units space is divided into the two
orthogonal subspaces, the Treatments and Residual
subspaces.

38
Geometric interpretation
• Of course, the SSqs are just the squared lengths
of these vectors
• Hence, according to Pythagoras theorem, the
Treatments and Residual SSqs must sum to the
Units SSq.

39
Example III.1 Rat experiment (continued)
• Vectors for computing the SSqs are
• Total Rat deviations, Diet Effects and Residual
Rats deviations are projections into Rats, Diets
and Residual subspaces of dimension 5, 2 and 3,
respectively.
• Squared length of projection SSq
• Rats SSq is Y'QRY 0.34
• Diets SSq is Y'QDY 0.24
• Residual SSq is

Exercise III.3 is similar example for you to try
40
c) Expected mean squares
• Have an ANOVA in which we use F ( ratio of MSqs)
to decide between models.
• But why is this ratio appropriate?
• One way of answering this question is to look at
what the MSqs measure?
• Use expected values of the MSqs, i.e. EMSqs, to
do this.

41
Expected mean squares (contd)
• Remember expected value population mean
• Need EMSqs
• under the maximal model
• and the minimal model
• Similar to asking what is EYi?
• Know answer is EYi ak.
• i.e. in population, under model, average value of
Yi is ak.
• So for Treatments, what is EMSq?
• The EMSqs are the mean values of the MSqs in
populations described by the model for which they
are derived
• i.e. an EMSq is the true mean value
• it depends on the model parameters.

42
Expected mean squares (contd)
• Need EMSqs
• under the maximal model
• and the minimal model
• The EMSqs are the mean values of the MSqs in
populations described by the model for which they
are derived
• i.e. an EMSq is the true mean value and it
depends on the model parameters.
• This is analogous to saying that the expected
value of the Treatment mean vector in populations
described by the model is y.
• That is,

43
Expected mean squares (contd)
• Need EMSqs
• under the maximal model
• and the minimal model
• The EMSqs are the mean values of the MSqs in
populations described by the model for which they
are derived
• This is analogous to saying that the expected
value of the Treatment mean vector in populations
described by the model is y.
• That is,
• Put another way, what is the mean of the sampling
distribution of
• the Treatment mean vector?
• the Treatment mean square?
• (Note that both the mean vector and the mean
square are functions of Y and so are random
variables.)
• The answer is easy for the treatment mean vector
y
• But what for the Treatment mean square? Ans.
EMSq

44
EMSqs under the maximal model
• So if we had the complete populations for all
Treatments and computed the MSqs, the value of
• the Residual MSq would equal s2
• the Treatment MSq would equal s2 qT(y).
• So that the population average value of both MSqs
involves s2, the uncontrolled variation amongst
units from the same treatment.
• But what about q in Treatments EMSq.

45
The qT(y) function
• Subscript T indicates the Q matrix on which
function is based
• but no subscript on the y in qT(y),
• because we will determine expressions for it
under both the maximal (yT) and alternative
models (yG).
• That is, y in qT(y) will vary.
• Numerator is same as the SSq except that it is a
• To see what this means want expressions in terms
of individual parameters.
• Will show that under the maximal model (yT)
• and under the minimal model (yG) that

46
Example III.1 Rat experiment (continued)
• The latter is just the mean of the elements of
yT.
• Actually, the quadratic form is the SSQ of the
elements of vector

When will the SSq be zero?
47
The qT(y) function
• Now want to prove the following result
• As QT is symmetric and idempotent,
• y'QTy (QTy)'QTy
• qT(y) is the SSq of QTy, divided by (t-1).
• QTy (MT MG)y MTy MGy
• MGy replaces each element of y with the grand
mean of the elements of y
• MTy replaces each element of y with the mean of
the elements of y that received the same
treatment as the element being replaced.

48
The qT(y) function (continued)
• Under the maximal model (yT)
• Under the minimal model (yG m1n)
• MGyG MTyG yG so y'GQTyG 0 and qT(y) 0

49
Example III.1 Rat experiment (continued)
50
How qT(yT) depends on the as
• qT(yT) is a quadratic form and is basically a sum
of squares so that it must be nonnegative.
• Indeed the magnitude of depends on the size of
the differences between the population treatment
means, the aks
• if all the aks are similar they will be close to
their mean,
• whereas if they are widely scattered several will
be some distance from their mean.

51
EMsqs in terms of parameters
• Could compute population mean of MSq if knew aks
and s2.
• Treatment MSq will on average be greater than the
Residual MSq
• as it is influenced by both uncontrolled
variation and the magnitude of treatment
differences.
• The quadratic form qT(y) will only be zero when
all the as are equal, that is when the null
hypothesis is true.
• Then the EMSqs under the minimal model are
equal so that the F value will be approximately
one.
• Not surprising if think about a particular
experiment.

52
Example III.1 Rat experiment (continued)
• So what can potentially contribute to the
difference in the observed means of 3.1 and 2.7
for diets A and C?
• Obviously, the different diets
• not so obvious that differences arising from
uncontrolled variation also contribute as 2
different groups of rats involved.
• This is then reflected in EMSq in that it
involves s2 and the "variance" of the 3 effects.

53
Justification of F-test
• Thus, the F test involves asking the question "Is
the variance in the sample treatment means gt can
be expected from uncontrolled variation alone?".
• If the variance is no greater, concluded qT(y)
0
• the minimal model is the correct model since the
expected Treatment MSq under this model is just
s2.
• Otherwise, if the variance is greater, qT(y) is
nonzero
• the maximal model is required to describe the
data.
• Similar argument to examining dotplots for
Example II.2, Paper bag experiment.

54
d) Summary of the hypothesis test
• see notes

55
d) Summary of the hypothesis test
• Summarize the ANOVA-based hypothesis test for a
CRD involving t treatments and a total of n
observed units.
• Step 1 Set up hypotheses
• H0 a1 a2 ... a3 m (yG ? XGm)
• H1 not all population Treatment means are equal
(yT ? XTa)
• Set a.

56
Summary (continued)
Step 2 Calculate test statistic
• Note that EMSqs under the maximal model are
included in this table, it being recognized that
when H0 is true qT(y) 0.
• Step 3 Decide between hypotheses
• Determine probability of observed F value that
has n1 numerator d.f. t-1 and n2
denominator d.f. n-t.
• If

then the evidence suggests that the null
hypothesis be rejected and the alternative model
be used to describe the data.
57
e) Comparison with traditional one-way ANOVA
• Our ANOVA table is essentially the same as the
• the values of the F statistic from each table are
exactly the same.
• Labelling differs and the Total would normally be
placed at the bottom of the table, not at the top.
• Difference is symbolic
• Units term explicitly represents a source of
uncontrolled variation differences between
Units.
• Our table exhibits the confounding in the
experiment.
• Indenting of Treatments under Units signifies
that treatment differences are confounded or
mixed-up with unit differences.

58
f) Computation of the ANOVA in R
• Begin with data entry (see Appendix A,
Introduction to R and, Appendix C, Analysis of
designed experiments in R)
• Next an initial graphical exploration using
boxplots defer to a second example with more
data.
• ANOVA while function lm could be used, function
aov is preferred for analysing data from a
designed experiment.
• Both use a model formula of the form
• Response variable explanatory variables (and
operators)
• So far expressions on right fairly simple one
or two explanatory variables separated by a .
• Subtlety with analysis of designed experiments
• If explanatory variable is a numeric, such as a
numeric vector, then R fits just one coefficient
for it.
• For a single explanatory variable, a
straight-line relationship fitted.
• If explanatory variable is categorical, such as a
factor, a coefficient is fit for each level of
the variable indicator variables are used.
• In analyzing a CRD important that the treatment
factor is stored in a factor object, signalling R
to use indicator variables

59
f) Computation of the ANOVA in R (continued)
• There are two ways in which this analysis can be
obtained using the aov function
• without and with an Error function in the model
formula.
• Error function used in model formula to specify a
model for the Error in the experiment (a model
for uncontrolled variation).
• summary function is used and this produces ANOVA
table.
• model.tables function used to obtain tables of
means.

60
Example III.1 Rat experiment (continued)
• Following commands to perform the two analyses of
the data
• AOV without Error
• Rat.NoError.aov lt- aov(LiverWt Diet,
CRDRat.dat)
• summary(Rat.NoError.aov)
• AOV with Error
• Rat.aov lt- aov(LiverWt Diet Error(Rat),
CRDRat.dat)
• summary(Rat.aov)
• model.tables(Rat.aov, type "means")

61
Output
• gt AOV without Error
• gt
• gt Rat.NoError.aov lt- aov(LiverWt Diet,
CRDRat.dat)
• gt summary(Rat.NoError.aov)
• Df Sum Sq Mean Sq F value Pr(gtF)
• Diet 2 0.240000 0.120000 3.6 0.1595
• Residuals 3 0.100000 0.033333
• gt
• gt AOV with Error
• gt
• gt Rat.aov lt- aov(LiverWt Diet Error(Rat),
CRDRat.dat)
• gt summary(Rat.aov)
• Error Rat
• Df Sum Sq Mean Sq F value Pr(gtF)
• Diet 2 0.240000 0.120000 3.6 0.1595
• Residuals 3 0.100000 0.033333
• Analysis without the Error function parallels the
traditional analysis and that with Error is
similar to our table.
• In this course will use Error function.

62
Output (continued)
• gt model.tables(Rat.aov, type"means")
• Tables of means
• Grand mean
• 3.1
• Diet
• A B C
• 3.1 3.3 2.7
• rep 3.0 2.0 1.0

63
III.D Diagnostic checking
• Assumed the following model
• Y is distributed N(y, s2In)
• where y EY Xq and
• s2 is the variance of an observation
• Maximal model is used in diagnostic checking
• yT EY XTa
• For this model to be appropriate requires that
• the response is operating additively that the
individual deviations in the response to a
treatment are similar
• the sets of units assigned the treatments are
comparable in that the amount of uncontrolled
variation exhibited by them is the same for each
treatment
• each observation is independent of other
observations and
• that the response of the units is normally
distributed.

64
Example III.2 Caffeine effects on students
• Effect of orally ingested caffeine on a physical
task was investigated (Draper and Smith, 1981,
sec.9.1).
• Thirty healthy male college students were
selected and trained in finger tapping.
• Ten men were randomly assigned to receive one of
three doses of caffeine (0, 100 or 200 mg).
• The number of finger taps after ingesting the
caffeine was recorded for each student.

65
Entering the data
• Setting up a data frame, from scratch, for data
arranged in standard order (see App C)
• factor Students with values 130,
• factor Dose with levels 0, 100 and 200 and values
depending on whether data is entered by rows or
columns (can use rep function),
• numeric vector Taps with the 30 observed values
of the response variable.

66
Entering the data (cont'd)
gt CRDCaff.dat Students Dose Taps 1 1
0 242 2 2 100 248 3 3 200
246 4 4 0 245 5 5 100 246 6
6 200 248 7 7 0 244 8
8 100 245 9 9 200 250 10 10
0 248 11 11 100 247 12 12 200
252 13 13 0 247 14 14 100
248 15 15 200 248 16 16 0
248 17 17 100 250 18 18 200
250 19 19 0 242 20 20 100
247 21 21 200 246 22 22 0
244 23 23 100 246 24 24 200
248 25 25 0 246 26 26 100
243 27 27 200 245 28 28 0
242 29 29 100 244 30 30 200 250
• set up data.frame with factors Students and Dose
and then add response variable Taps
• CRDCaff.dat lt- data.frame(Students
factor(130),
• Dose factor(rep(c(0,100,200), times10)))
• CRDCaff.datTaps lt-
• c(242,248,246,245,246,248,244,245,250,248,247,252
,247,248,248,
• 248,250,250,242,247,246,244,246,248,246,243,245,2
42,244,250)
• CRDCaff.dat

67
Boxplots for each level of Dose
• Use function
• boxplot(split(Taps, Dose), xlab"Dose",
ylab"Number of taps")
• Average number of taps increasing as dose
increases
• Some evidence dose 0 more variable than dose 100.

68
Analysis of variance for this data
• gt Caffeine.aov lt- aov(Taps Dose
Error(Students), CRDCaff.dat)
• gt summary(Caffeine.aov)
• Error Students
• Df Sum Sq Mean Sq F value Pr(gtF)
• Dose 2 61.400 30.700 6.1812 0.006163
• Residuals 27 134.100 4.967

gt model.tables(Caffeine.aov, type"means") Tables
of means Grand mean 246.5 Dose Dose
0 100 200 244.8 246.4 248.3
69
The hypothesis test
• Step 1 Set up hypotheses
• H0 a0 a100 a200 m (yG XGm)
• H1 not all population Dose means are equal
• (yD XDa)
• Set a 0.05

70
The hypothesis test (continued)
• Step 2 Calculate test statistic
•   The ANOVA table for the example is

Step 3 Decide between hypotheses P(F2,27 ?
6.18) p 0.006 lt a 0.05.   The evidence
suggests there is a dose difference and that the
expectation model that best describes the data is
yD ? XDa.
71
Examination of the residuals, eT
• Use the Residuals-versus-fitted-values plot and
the Normal Probability plot.
• In interpreting these plots
• Note ANOVA is robust to variance heterogeneity,
if treatments equally replicated, and to moderate
departures from normality.
• Most commonly find an unusually large or small
residual.
• The cause of such extreme values requires
investigation.
• May be a recording mistake or catastrophe for a
unit that can be identified.
• But, may be valid and is the result of some
unanticipated, but important effect.

72
The Normal Probability plot
• Should show a broadly straight line trend.

__________________
73
The Residuals-versus-fitted-values plot
• Generally, the points on scatter diagram should

_____________________________   no particular
pattern
74
Problem plots

__________________________   varianc
e increases as level increases

_________________________   systematic trend in
residuals

_________________________   variance peaks
in middle
Actually, for the CRD, this plot has a vertical
scatter of points for each treatment each
should be centred on zero and of the same width.
75
R functions used in producing these plots
• resid.errors extract the residuals from an aov
object when Error function used
• fitted.errors extract the fitted values from an
aov object when Error function used
• plot to plot the fitted values against the
residuals
• qqnorm to plot the residuals against the normal
quantiles
• qqline to add a line to the plot produced by
qqnorm.
• First 2 functions are nonstandard functions from
dae package.

76
Example III.2 Caffeine effects on students
(continued)
• A violation of the assumptions would occur if all
the students were in the same room and the
presence of other students caused anxiety to just
the students that had no caffeine.
• The response of the students is not independent.
• It may be that the inhibition of this group
resulted in less variation in their response
which would be manifest in the plot.
• Another situation that would lead to an
unacceptable pattern in the plot is if the effect
becomes more variable as the level of the
response variable increases.
• For example, caffeine increases the tapping but
at higher levels the variability of increase from
student to student is greater.
• That is there is a lack of additivity in the
response.

77
R output for getting the plots
gt res lt- resid.errors(Caffeine.aov) gt fit lt-
fitted.errors(Caffeine.aov) gt data.frame(Students,
Dose,Taps,res,fit) Students Dose Taps res
fit 1 1 0 242 -2.8 244.8 2 2
100 248 1.6 246.4 3 3 200 246 -2.3
248.3 4 4 0 245 0.2 244.8 5
5 100 246 -0.4 246.4 6 6 200 248 -0.3
248.3 7 7 0 244 -0.8 244.8 8
8 100 245 -1.4 246.4 9 9 200 250 1.7
248.3 10 10 0 248 3.2 244.8 11
11 100 247 0.6 246.4 12 12 200 252
3.7 248.3 13 13 0 247 2.2 244.8 14
14 100 248 1.6 246.4 15 15 200 248
-0.3 248.3 16 16 0 248 3.2 244.8 17
17 100 250 3.6 246.4 18 18 200 250
1.7 248.3 19 19 0 242 -2.8 244.8 20
20 100 247 0.6 246.4
• Note the use of data.frame to produce a printed
list of the original data with the residuals and
fitted values.

78
Plots for the example
21 21 200 246 -2.3 248.3 22 22 0
244 -0.8 244.8 23 23 100 246 -0.4
246.4 24 24 200 248 -0.3 248.3 25
25 0 246 1.2 244.8 26 26 100 243
-3.4 246.4 27 27 200 245 -3.3 248.3 28
28 0 242 -2.8 244.8 29 29 100 244
-2.4 246.4 30 30 200 250 1.7 248.3 gt
plot(fit, res, pch 16) gt qqnorm(res, pch
16) gt qqline(res)
• The Residuals-versus-fitted-values plot appears
to be fine.

79
Normal Probability plot
• Displaying some curvature at ends.
• Indicates data heavier in tails and flatter in
the peak than expected for a normal distribution.
• Given normality not crucial and only a few
observations involved, use analysis we have
performed.

80
The hypothesis test Summary
• The ANOVA table for the example is

P(F2,27 ? 6.18) p 0.006 lt a 0.05.   The
evidence suggests there is a dose difference and
that the expectation model that best describes
the data is yD ? XDa. Diagnostic checking
indicates model is OK
81
III.E Treatment differences
• So far all that our analysis has accomplished is
that we have decided whether or not there appears
to be a difference between the population
treatment means.
• Of greater interest to the researcher is how the
treatment means differ.
• Two alternatives available
• Multiple comparisons procedures
• used when the treatment factors are all
qualitative so that it is appropriate to test for
differences between treatments
• Fitting submodels
• When one (or more) of the treatment factors is
quantitative the fitting of smooth curves to the
trend in the means is likely to lead to a more
appropriate and concise description of the
effects of the factors.
• Often, for reasons explained in chapter II, a
low order polynomial will provide an adequate
description of the trend.

82
Note
• Multiple comparison procedures should not be used
when the test for treatment differences is not
significant.
• Submodels should be fitted irrespective of
whether the overall test for treatment
differences is significant.
• The difference in usage has to do with one being
concerned with mean differences and the other
with deciding between models.

83
a) Multiple comparisons procedures for comparing
all treatments
• Multiple comparisons for all treatments MCA
procedures.
• In general MCA procedures divide into those
based
• on family-wise error rates Type I error rate
specified and controlled for over all
comparisons, often at 0.05.
• those based on comparison-wise error rates Type
I error rate specified and controlled for each
comparison
• Problem with latter is probability of an
incorrect conclusion gets very high as the number
of comparisons increases.
• For comparison-wise error rate of 0.05,
family-wise error rate
• So recommend use MCA procedure based on
family-wise error rates use just Tukey's HSD
procedure.

84
Tukeys Honestly Significant Difference procedure
• determines if each pair of means is significantly
different
• is based on a family-wise error rate.
• basically for equal numbers of observations for
each mean.
• A modification that is approximate will be
provided for unequal numbers.

85
The procedure
• Each application of the procedure is based on the
hypotheses
• Ho aA aB
• H1 aA ? aB
• One calculates the statistic

qt,n,a is the studentized range with t  no.
means, n  Residual df and a  significance level,
is the standard error of the
difference rA, rB are the number of replicates
for each of a pair of means being compared.
86
• Note that strictly speaking rA and rB should be
equal.
• When unequal rA and rB called the Tukey-Kramer
procedure.
• w(a) will depend on which means are being
compared.
• If the treatments are all equally replicated with
replication r, the formula for reduces to

87
In R
• aov has given Residual MSq and df
• model.tables produces tables of means.
• qtukey computes as follows
• q lt- qtukey(1 - a, t, n)

88
Example III.2 Caffeine effects on students
(continued)
• Have already concluded evidence suggests that
there is a dose difference. But which doses are
different?
• Output with means and q
• gt model.tables(Caffeine.aov, type "means")
• Tables of means
• Grand mean
• 246.5
• Dose
• Dose
• 0 100 200
• 244.8 246.4 248.3
• gt q lt- qtukey(0.95, 3, 27)
• gt q
• 1 3.506426

89
Decide on differences
• Any two means different by 2.47 or more are
significantly different.
• Our conclusion is that the mean for 0 and 200 are
different but that for 100 is somewhat
intermediate.

90
b) Fitting submodels
• For quantitative factors, like Dose, it is often
better to examine the relationship between the
response and the levels of the factor.
• This is commonly done using polynomials.
• Now, a polynomial of degree t-1 will fit exactly
t points.
• So a quadratic will fit exactly the three means.
• In practice, polynomials of order 2 often
sufficient.
• However, more than 3 points may be desirable so
that deviations from the fitted curve or lack of
fit can be tested

91
Polynomial models
• To investigate polynomial models up to order 2,
following models for expectation investigated

where xk is the value of the kth level of the
treatment factor, m is the intercept of the
fitted equation and g1 is the slope of the
fitted equation and g2 is the quadratic
coefficient of the fitted equation.
The X1 and X2 matrices made up of columns that
consist of the values of the levels of the factor
and their powers not the indicator variables of
before.
92
Example III.2 Caffeine effects on students
(continued)
• The X matrices for the example are
• The columns of each X matrix in the above list
are a linear combination of those of any of the X
matrices to its right in the list.

93
Marginality of models
• That is, the columns of each X matrix in the
above list are a linear combination of those of
any of the X matrices to its right in the list.
• Marginality is not a symmetric relationship in
that
• if a model is marginal to second model,
• the second model is not necessarily marginal to
the first.
• For example, EY X1g1 is marginal to EY
X2g2 but not vice-a-versa, except when t  2.

94
Equivalence of EY X2g2 and EY XDa
• C(X2) C(XD) as the 3 columns of one matrix can
be written as 3 linearly independent combinations
of the columns of the other matrix.
• So the two models are marginal to each other and
are equivalent
• However, while the fitted values are the same,
the estimates and interpretation of the
parameters are different
• those corresponding to X2 are interpreted as the
intercept, slope and curvature coefficient.
• those corresponding to XD are interpreted as the
expected (mean) value for that treatment.
• Also, in spite of being marginal, the estimators
of the same parameter differ depending on the
model that has been fitted.
• for example, for the model EY XGm,
• but for EY X1g1,

models not orthogonal
95
Hypothesis test incorporating submodels
• Test statistics computed in ANOVA table.
• Step 1 Set up hypotheses
•   a) H0
• H1
•   (Differences between fitted model or
•
• b) H0
• H1
• c) H0
• H1
• Set a.

96
Hypothesis test (continued)
• Step 2 Calculate test statistics
• The ANOVA table for a CRD is

Test statistics corresponds to hypothesis pairs
in Step 1.
97
Hypothesis test (continued)
• Step 3 Decide between hypotheses
• Begin with the first hypothesis pair, determine
its significance and continue down the sequence
until a significant result is obtained.
• A significant Deviations F indicates that the
description.
• Thus a model based on them would be
unsatisfactory. No point in continuing.
• If the Deviations F is not significant , then a
significant Quadratic F indicates that a 2nd
degree polynomial is required to adequately
describe the trend.
• As a linear coefficient is necessarily
incorporated in a 2nd polynomial, no point to
further testing in this case.
• If both the Deviations and Quadratic F's are not
significant, then a significant Linear F
indicates a linear relationship describes the
trend in the treatment means.

98
What if deviations is significant?
• If the Deviations are significant, then two
options are
• increase the degree of the polynomial (often not
desirable), or
• use multiple comparisons to investigate the
differences between the means.

99
Fitting polynomials in R
• Need to realize that the default contrasts for
ordered (factor) objects are polynomial
contrasts, assuming equally-spaced levels
• linear transformations of all columns of X,
except the first, such that each column
• is orthogonal to all other columns in X and
• has SSq equal to 1.
• Facilitates the computations.
• So, if a factor is quantitative
• set it up as an ordered object from the start.
• if did not make it an ordered object, then
redefine factor as an ordered.

100
Example III.2 Caffeine effects on students
(continued)
• R output for setting up ordered (factor)
• gt fit polynomials
• gt
• gt Dose.lev lt- c(0,100,200)
• gt CRDCaff.datDose lt- ordered(CRDCaff.datDose,

• levelsDose.lev)
• gt contrasts(CRDCaff.datDose) lt- contr.poly(t,

• scoresDose.lev)
• gt contrasts(CRDCaff.datDose)
• .L .Q
• 0 -7.071068e-01 0.4082483
• 100 -9.073264e-17 -0.8164966
• 200 7.071068e-01 0.4082483

101
R output for fitting polynomial submodels
• gt Caffeine.aov lt- aov(Taps Dose
Error(Students), CRDCaff.dat)
• gt summary(Caffeine.aov, split
list(Dose list(L 1, Q 2)))
• Error Students
• Df Sum Sq Mean Sq F value Pr(gtF)
• Dose 2 61.400 30.700 6.1812 0.006163
• Dose L 1 61.250 61.250 12.3322 0.001585
• Dose Q 1 0.150 0.150 0.0302 0.863331
• Residuals 27 134.100 4.967
this is reflected in C(X2) C(XD).
• Thus, Deviations line is redundant in this
example.
• When required, Deviations line is computed by
assigning extra powers to a single line named,
say, Dev in split function (e.g. Dev 36 see
notes).

102
Hypothesis test incorporating submodels
• Step 1 Set up hypotheses
•
• a) H0
• H1
• b) H0
• H1
• Set a 0.05.

103
Hypothesis test (continued)
• Step 2 Calculate test statistics
• gt Caffeine.aov lt- aov(Taps Dose
Error(Students), CRDCaff.dat)
• gt summary(Caffeine.aov, split
list(Dose list(L 1, Q 2)))
• Error Students
• Df Sum Sq Mean Sq F value Pr(gtF)
• Dose 2 61.400 30.700 6.1812 0.006163
• Dose L 1 61.250 61.250 12.3322 0.001585
• Dose Q 1 0.150 0.150 0.0302 0.863331
• Residuals 27 134.100 4.967

104
Hypothesis test (continued)
• Step 3 Decide between hypotheses
•   The Quadratic source has a probability of 0.863
gt 0.05 and so H0 is not rejected in this case.
•   The linear source has a probability of 0.002 lt
0.05 and so H0 is rejected in this case.
•   It is clear quadratic term is not significant
but linear term is highly significant so the
appropriate model for the expectation is the
linear model y X1g1 where g1 m g1.

105
Fitted equation
• Coefficients can be obtained using the coef
function on the aov object, but these are not
suitable for obtaining the fitted values.
• The fitted equation is obtained by putting the
values of the levels into a numeric vector and
using the lm function to fit a polynomial of the
order indicated by the hypothesis test.
• For the example, linear equation was adequate and
so the analysis is redone with 1 for the order of
the polynomial.

106
Fitted equation for the example
• gt D lt- as.vector(Dose)
• gt D lt- as.numeric(D)
• gt Caffeine.lm lt- lm(Taps D)
• gt coef(Caffeine.lm)
• (Intercept) D
• 244.7500 0.0175
• The fitted equation is Y 244.75 0.0175 X
where X is the number of taps
• The slope of this equation is 0.0175.
• That is, taps increase 0.0175 x 100 1.75 with
each 100 mg of caffeine.
• This conclusion seems a more satisfactory summary
of the results than that the response at 200 is
significantly greater than at 0 with 100 being
intermediate.
• The commands to fit a quadratic would be
• D2 lt- DD
• Caffeine.lm lt- lm(Taps D D2)

107
Plotting means and fitted line
• Details are in the notes
• The plot produced is as follows

108
c) Comparison of treatment parametrizations
• Two alternative bases
• Lead to different parameter estimates with
different interpretations and different
partitions of the treatment SSqs.
• The total treatment SSqs and fitted values for
treatments remain the same, while the contrasts
span the treatment space.
• That is, in this case, SSqT SSqL SSqQ

109
III.G Exercises
• Ex. III.12 look at aspects of quadratic forms
• Ex. III.3 investigates the calculations with
example that can be done with a calculator
• Ex. III.4 involves producing a randomized layout
• Ex. III.5 asks for the complete analysis of a CRD
with a qualitative treatment factor
• Ex. III.6 asks for the complete analysis of a CRD
with a quantitative treatment factor