MULTIPLE COMPARISON TESTS

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A PRIORI OR PLANNED CONTRASTS
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ANOVA is used to compare means.
However, if a difference is detected, and
more than two means are being compared,
ANOVA cannot tell you where the difference
lies.
In order to figure out which means differ, you
can do a series of tests:
 Planned or unplanned comparisons of means.
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A comparison between means identified as
being of utmost interest during the design of
a study, prior to data collection.
You can only do one or a very small number
of planned comparisons, otherwise you risk
inflating the Type 1 error rate.
You do not need to perform an ANOVA first.
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A form of “data dredging” or “data
snooping”, where you may perform
comparisons between all potential pairs of
means in order to figure out where the
difference(s) lie.
No prior justification for comparisons.
Increased risk of committing a Type 1 error.
The probability of making at least one type 1
error is not greater than α= 0.05.
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Planned comparisons may be orthogonal or
non-orthogonal.
Orthogonal: mutually non-redundant and
uncorrelated contrasts (i.e.: independent).
Non-Orthogonal: Not independent.
For example:
 4 means: Y1 ,Y2, Y3, and Y4
 Orthogonal: Y1-Y2 and Y3-Y4
 Non-Orthogonal: Y1-Y2 and Y2-Y3
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Limited number of contrasts can be made,
simultaneously.
Any set of contrasts may have k-1 number of
contrasts.
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For Example:
 k=4 means.
 Therefore, you can make 3 (i.e.: 4-1)
orthogonal contrasts at once.
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∑cijci’j=0
 where the c’s are the particular coefficients
associated with each of the means and the i indicates
the particular comparison to which you are referring.
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Multiply all of the coefficients for each particular
mean together across all comparisons.
Then add them up!
If that sum is equal to zero, then the
comparisons that you have in your set may be
considered orthogonal.
For Example:
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Set
1
cijci’j
∑cijci’j
c1
1
½
½
Coefficients
c2
c3
-1
1
½
-½
-½
-½
Contrasts
c4
-1
-½
½
Y1-Y2
Y3-Y4
(Y1+Y2)/2 –(Y3+Y4)/2
(c1)Y1 + (c2)Y2 = Y1 –Y2
Therefore c1 = 1 and c2 = -1 because
(1)Y1 + (-1)Y2 = Y1-Y2
0
(After Kirk 1982)
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There are always k-1 non-redundant
questions that can be answered.
An experimenter may not be interested in
asking all of said questions, however.
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A planned comparison addresses the null
hypothesis that all of your comparisons
between means will be equal to zero.
 Ho=Y1-Y2=0
 Ho= Y3-Y4=0
 Ho= (Y1+Y2)/2 –(Y3+Y4)/2
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These types of hypotheses can be tested
using a t statistic.
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Very similar to a two sample t-test, but the
standard error is calculated differently.
Specifically, planned comparisons use the
pooled sample variance (MSerror )based on all
k groups (and the corresponding error
degrees of freedom) rather than that based
only on the two groups being compared.
This step increases precision and power.
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Evaluate just like any other t-test.
Look up the critical value for t in the same
table.
If the absolute value of your calculated t
statistic exceeds the critical value, the null
hypothesis is rejected.
All of the t statistic calculations for all of the
comparisons in a particular set will use the same
MSerror.
 Thus, the tests themselves are not statistically
independent, even though the comparisons
that you are making are.
 However, it has been shown that, if you have a
sufficiently large number of degrees of
freedom (40+), this shouldn’t matter.
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(Norton and Bulgren, as cited by Kirk, 1982)
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You can also use an F statistic for these tests,
because t2 = F.
Different books prefer different methods.
The book I liked most used the t statistic, so
that’s what I’m going to use throughout.
SAS uses F, however.
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A confidence interval is a way of expressing
the precision of an estimate of a parameter.
Here, the parameter that we are estimating is
the value of the particular contrast that we
are making.
So, the actual value of the comparison (ψ)
should be somewhere between the two
extremes of the confidence interval.
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The values at the extremes are the 95%
confidence limits.
With them, you can say that you are 95%
confident that the true value of the
comparison lies between those two values.
If the confidence interval does not include
zero, then you can conclude that the null
hypothesis can be rejected.
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When the data are presented this way, it is
possible for the experimenter to consider all
possible null hypotheses – not just the one
that states that the comparison in question
will equal 0.
If any hypothesized value lies outside of the
95% confidence interval, it can be rejected.
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Orthogonal tests can be done in either way.
Both methods make the same assumptions
and are equally significant.
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Assumptions:
 The populations are approximately normally
distributed.
 Their variances are homogenous.
The t statistic is relatively robust to violations of
these assumptions when the number of
observations for each sample are equal.
 However, when the sample sizes are not equal,
the t statistic is not robust to the heterogeneity
of variances.
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When population variances are unequal, you can
replace the pooled estimator of variance, MSerror,
with individual variance estimators for the
means that you are comparing.
There are a number of possible procedures that
can be used when the variance between
populations is heterogeneous:
 Cochran and Cox
 Welch
 Dixon, Massey, Satterthwaite and Smith
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For C independent contrasts at some level of
significance (α), the probability of making
one or more Type 1 errors is equal to:
 1-(1-α)C
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As the number of independent tests
increases, so does the probability of
committing a Type 1 error.
This problem can be reduced (but not
eliminated) by restricting the use of multiple
t-tests to a priori orthogonal contrasts.
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Contrasts of interest that ARE NOT
independent.
In order to reduce the probability of making a
Type 1 error, the significance level (α) is set for
the whole family of comparisons that is being
made, as opposed to for each individual
comparison.
For Example:
 Entire value of α for all comparisons combined is
0.05.
 The value for each individual comparison would thus
be less than that.
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When contrasts are planned in advance.
They are relatively few in number.
BUT the comparisons are non-orthogonal
(they are not independent).
i.e.: When one wants to contrast a control
group mean with experimental group
means.
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A.K.A.: Bonferoni t procedure.
Involves splitting up the value of α among a
set of planned contrasts in an additive way.
For example:
 Total α = 0.05, for all contrasts.
 One is doing 2 contrasts.
 α for each contrast could be 0.025, if we wanted to
divide up the α equally.
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If the consequences of making a Type 1 error
are not equally serious for all contrasts, then
you may choose to divide α unequally across
all of the possible comparisons in order to
reflect that concern.
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This procedure also involves the calculation
of a t statistic (tD).
The calculation involved in finding tD is
identical to that for determining t for
orthogonal tests:
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However, you use a different table in order
to look up the critical value (tDα;C,v ).
 Your total α value (not the value per comparison).
 Number of comparisons (C).
 And v, the number of degrees of freedom.
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The table also only shows the critical values
for two-tailed tests.
However, you can determine the
approximate value of tDα;C,v for a onetailed test by using the following equation:
 tDα;C,v ≈ zα/C + (z3α/C + zα/C)/4(v-2)
▪ Where the value of zα/C can be looked up in yet
another table (“Areas under the Standard Normal
Distribution”).
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Instead of calculating tD for all contrasts of
interest, you can simply calculate the critical
difference (ψD) that a particular comparison
must exceed in order to be significant:
 ψD = tDα/2;C,v √(2MSerror/n).
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Then compare this critical difference value to
the absolute values of the differences
between the means that you compared.
If they exceed ψD, they are significant.
For Example:
You have
5 means that
(Y1 through
Y5).
 Those
differences
exceed
the calculated
 ΨD = 8.45
value of ΨD (8.45, in this case) are significant.
 Differences between means are:
MEANS
Y1
Y2
Y3
Y4
Y5
Y1
Y2
Y3
Y4
Y5
-
12.0
6.7
10.5
3.6
-
5.3
1.5
8.4
-
3.8
3.1
-
6.9
(After Kirk 1982)
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A modification of the Dunn procedure.
 t statistic (tDS) and critical difference (ψDS).
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There isn’t much difference between the two
procedures at α < 0.01.
However, at increased values of α, this
procedure is considered to be more powerful
and more precise.
Calculations are the same for t and ψD.
Table is different.
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However, it is not easy to allocate the total
value of α unevenly across a particular set of
comparisons.
This is because the values of α for each
individual comparison are related
multiplicatively, as opposed to additively.
Thus, you can’t simply add the α’s for each
comparison together to get the total value of
α for all contrasts combined.
For contrasts involving a control mean.
Also uses a t statistic (tD’) and critical difference
(ψD’).
 Calculations are the same for t and ψ.
 Different table.
 Instead of C, you use k, the number of means
(including the control mean).
 Note: unlike Dunn’s and Dunn-Sidak’s, Dunnet’s
procedure is limited to k-1 non-orthogonal
comparisons.
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Often, the use of more than one procedure
will appear to be appropriate.
In such cases, compute the critical
difference (ψ) necessary to reject the null
hypothesis for all of the possible procedures.
Use the one that gives the smallest critical
difference (ψ) value .
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The advantage of being able to make all
planned contrasts, not just those that are
orthogonal, is gained at the expense of an
increase in the probability of making Type 2
errors.
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When you have a large number of means,
but only comparatively very few contrasts, a
priori non-orthogonal contrasts are better
suited.
However, if you have relatively few means
and a larger number of contrasts, you may
want to consider doing an a posteriori test
instead.
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There are many kinds, all offering different
degrees of protection from Type 1 and Type 2
errors:
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Least Significant Difference (LSD) Test
Tukey’s Honestly Significant Difference (HSD) Test
Spjtotvoll and Stoline HSD Test
Tukey-Kramer HSD Test
Scheffé’s S Test
Brown-Forsythe BF Procedure
Newman-Keuls Test
Duncan’s New Multiple Range Test
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Most are good for doing all possible pairwise comparisons between means.
One (Scheffé’s method) allows you to
evaluate all possible contrasts between
means, whether they are pair-wise or not.
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Trade-off between power and the probability
of making Type 1 errors.
When a test is conservative, it is less likely
that you will make a Type 1 error.
But it also would lack power, inflating the
Type 2 error rate.
You will want to control the Type 1 error rate
without loosing too much power.
Otherwise, you might reject differences
between means that are actually significant.
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You have a data set with one dependent
variable (y) and one independent variable (a).
(a) has 4 different treatment levels (1, 2, 3,
and 4).
You want to do the following comparisons
between treatment levels:
 1)Y1-Y2
 2) Y3-Y4
 3) (Y1-Y2)/2 - (Y3-Y4)/2
SET
1
cijci’j
∑cijci’j
COEFFICIENTS
CONTRASTS
c1
c2
c3
c4
1
-1
-
-
Y1-Y2
-
-
1
-1
Y3-Y4
½
½
-1/2
-1/2
1/2
-1/2
-1/2
½
(Y2+Y2)/2 – (Y3+Y4)/2
0
YES!
...BUT THEY DON’T HAVE TO BE
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SAS INPUT:
data dataset;
Give your dataset an informative name.
input y a;
Tell SAS what you’ve inputted: column 1 is
cards;
your y variable (dependent) and column 2 is
31
your a variable (independent).
42
73
This is followed by your actual data.
74
.............
“Model” tells SAS that
Use SAS procedure proc glm.
proc glm;
class a;
“Class” tells SAS that a is categorical. you want to look at the
effects that a has on y.
model y = a;
contrast 'Compare 1 and 2' a 1 -1 0 0;
Indicate “weights” (kind of
contrast 'Compare 3 and 4' a 0 0 1 -1;
contrast 'Compare 1 and 2 with 3 and 4' a 1 1 -1 -1; like the coefficients).
run;
Then enter your “contrast”
statements.
ANOVA
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Kirk RE. 1982. Experimental design:
procedures for the behavioural sciences.
Second ed. CA: Wadsworth, Inc.
Field A, Miles J. 2010. Discovering Statistics
Using SAS. London: SAGE Publications Ltd.
Institute for Digital Research and Education
at UCLA: http://www.ats.ucla.edu/stat/
 Stata, SAS, SPSS and R.
THE END
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t = ∑cjYj/ √MSerror∑cj/nj
 Where c is the coefficient, Y is the corresponding
mean, n is the sample size, and MSerror is the Mean
Square Error.
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For Example, you want to compare 2
means:
 Y1 = 48.7 and Y2=43.4
 c1 = 1 and c2 = -1
 n=10
 MSerror=28.8
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t = (1)48.7 + (-1)43.4
= 2.21
[√28.8(12/10) + (-12/10)]
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(Y1-Y2) – tα/2,v(SE) ≤ ψ ≤ (Y1-Y2) – tα/2,v(SE)
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For Example:
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If, Y1 = 48.7, Y2 = 43.4, and SE(tα(2),df) = 4.8
(Y1-Y2) – SE(tα(2),df) = 0.5
(Y1-Y2) + SE(tα(2),df) = 10.1
Thus, you can be 95% confident that the true value of
ψ is between 0.5 and 10.1.
 Because the confidence interval does not include 0,
you can also reject the null hypothesis that Y1-Y2 = 0.
(Example after Kirk 1982)
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As the number of independent tests
increases, so does the probability of
committing a Type 1 error.
For Example, when α = 0.05:
 1-(1-0.05)3=0.14 (C=3)
 1-(1-0.05)5=0.23 (C=5)
 1-(1-0.05)10=0.40 (C=10)
(Example after Kirk 1982)

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