Unformatted text preview: Linear Combinations and
Linear
Multiple Comparisons of
Means
Means
(more ANOVA stuff)
Chapter 6 1 Outline
Case Studies
Inferences About Linear Combinations of
Inferences
Group Means
Group
Simultaneous Inferences
Some Multiple Comparison Procedures
Some
Related Issues (Data Snooping)
Related 2 Case Study 1  Discrimination
randomized experiment
5 handicap conditions
– 1) Amputee
– 2) Crutches
– 3) Hearing
– 4) None
– 5) Wheelchair 3 1 HANDICAP
WHEELC HA IR NO NE HEA RING C RUTC HES A M P UTEE S C O RE Case Study 1  Discrimination
9 8 7 6 5 4 3 2 4 Research questions
A general question: do subjects systematically
general
evaluate qualifications differently depending on
the candidate’s handicap?
the
2. Followup: If so, which handicaps produce
Followup:
different evaluations from which others?
different
1. • 1. note that this question involves pairwise differences
note
(like the diet experiment), but not specific pairwise
specific
differences.
differences. Specific comparison of interest (wheelchaircrutches versus amputeehearing impaired) 5 Case Study 2 – mate
Case
preferences of fish
preferences
complicated randomized experiment
complicated
groups are male pairs of various lengths
groups
questions of interest:
– does percent of time spent with yellowsword
does
males differ among male pairs?
males
– does percent of time spent with yellowsword
does
males differ depending on length of males?
males
– Does percent of time spent with yellowsword
Does
males (averaged over male pairs) exceed 50%?
males
6 PR O PO R TIO N Case Study 2 – mate preferences
Case
of fish
of
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 28 31 33 34 35 LENGTH
7 Preview
In chapter 5 we
In
– discussed pairwise tests and confidence
discussed
intervals for comparing any 2 of the I means
based on the pooled estimate of the
standard deviation, sp, and pooled degrees of
standard
and
freedom
freedom
– mentioned the multiple comparison problem
– developed the extrasumofsquares Ftest for
developed
answering more general questions about the
means (e.g. are there any differences among
the I means?)
the
8 Preview
In chapter 6 we
– generalize the concept of pairwise
generalize
comparisons of means (a comparison of 2
means is a special case of investigation of a
linear combination of means)
linear
– suggest some solutions for the multiple
suggest
comparison problem
comparison 9 Linear Combinations
Definition:
γ = C1µ1 + C2µ2 + … + CIµI
The Ci’s are chosen to answer specific
research questions. The linear
combination is like a mathematical
expression of the research question.
10
10 A simple and familiar linear
simple
combination
combination
I want to compare the mean for crutches
want
with the mean for wheelchair – a paired
difference just like in the diet example.
difference
a set of C’s that will do this:
C1= C3 = C4 = 0, C2 = 1, C5 = 1 This means we are
examining µ5  µ2
11
11 A llinear combination for the specific
inear
research question in case study 1
research
We want to compare the average of the
We
wheelchair and crutches means to the average
of the hearing and amputee means.
of
Use C1= 1/2, C2= 1/2, C3= 1/2, C4= 0, C5= 1/2
(1=amp, 2=crut, 3=hear, 4=none, 5=wheel)
(1=amp,
Note: if the sum of the C’’s equals zero (like this)
s
the combination is called a ‘contrast’
the
12
12 Inference for Linear Combinations
Parameter : γ = C1µ1 + C2 µ 2 + ... + CI µI
Estimate : g = C1Y1 + C2Y2 + ... + CIYI
2
C1 C2
CI2
Standard Deviation : SD(g) = σ
+ 2 + ... +
n1 n2
nI Standard Error : SE(g) = sp 2
C1 C2
CI2
+ 2 + ... +
n1 n2
nI
13
13 Inference for Linear Combinations (g  γ )
t  ratio : t =
, d.f. = n  I
SE(g)
confidence interval : g ± t α
(nI)(1 )
2 SE(g) 14
14 Example – specific comparison
Example
for case study 1
for
First Perform the usual ANOVA in JMP
From this analysis find:
– The means for the groups (4.43, 5.92, 4.05,
The
4.90, 5.34)
4.90,
– The pooled standard deviation = 1.633
– The pooled degrees of freedom = degrees of
The
freedom for error = nI = 70 – 5 = 65.
freedom 15
15 Hand Calculate g and SE(g)
Remember C1= 1/2, C2= 1/2, C3= 1/2, C4= 0, C5= 1/2
1/2
1
1
1
1
g = (− )(4.43) + ( )(5.92) + (− )(4.05) + (0)(4.90) + ( )(5.34)
2
2
2
2
= 1.393
(−1 / 2) 2 (1 / 2) 2 (−1 / 2) 2 (0) 2 (1 / 2) 2
SE ( g ) = 1.633
+
+
+
+
14
14
14
14
14
= 0.436 16
16 Inference: testing γ =0
Inference:
t = (g – 0) / SE(g)
(g
= (1.3930) / 0.436
(1.3930)
= 3.19
3.19
Is this statistically significant? Compare to
a t with degrees of freedom equal to nI =
65.
65.
pvalue=.0022
The difference between the averages of
The
the specific means is real.
the
17
17 Inference: confidence interval for γ
Inference:
g +/ t(nI)(1α/2) SE(g)
=1.393 +/ 1.997(.436)
=1.393
=(0.522, 2.264)
=(0.522, 18
18 Other examples – diet study
iin diet case study, estimate rate of
n
increase in average lifetime per calorie
increase
reduction
– can be expressed as (µN/R50µN/N85)/35
– CN/R50 = 1/35
CN/N85 = 1/35
all other C’s = 0 19
19 Other examples – fish study
llinear trend in mate preference with regards to
inear
length of male pair (i.e. the change in mate
preference per mm increase in length of the
males)
males)
if Xi = length of the ith pair, and X is the
average of the lengths of all the pairs,
average
Ci=(Xi X )
fish example: C28 = 4.5, C31 = 1.5,
C33 = .5, C34 = 1.5, C34 = 1.5, C35 = 2.5
20
20 Other Examples – fish study
average mate preference
C1 = 1/6, C2 = 1/6, … , C6 = 1/6
g = (1/6)(56.41) + … + (1/6)(63.34)
= 62.38
62.38
SE(g) = (15.47) sqrt[(1/6)2/16 +…+(1/6)2/14)]
= 1.72
1.72
Test γ = 50: t = (62.3850)/1.72 = 7.20, df = 78
Test
50:
pvalue < .0001
on average, female fish think yellow swords on
on
males are cute
males
21
21 Simultaneous Inferences
consider the handicap study:
– the test of the specific planned comparison
the
planned
(wheelchaircrutches versus amputeehearing
impaired) is useful, but we have more questions.
impaired)
– the overall Ftest that there are no differences among
the
the group means is only marginally useful. It does not
tell us which handicaps produce different evaluations
from which other handicaps.
from
– the investigation of which handicaps produce different
the
evaluations from which other handicaps exemplifies
unplanned comparisons. It involves comparing all
possible pairs of means.
22
22 Simultaneous inferences
3 kinds of comparisons (ordered by how
kinds
well standard statistical tools apply):
well
– planned comparisons
– unplanned comparisons
– data snooping (look at the sample means;
data
compare only those that appear to have large
differences)
differences) 23
23 Why make this distinction?
for unplanned and data snooping
for
comparisons, we will make incorrect
inferences (e.g. call differences real when
they are not) if we don’’t worry about the
t
they
effects on our inferences of simply doing
so many comparisons.
so 24
24 The problem with multiple
The
confidence intervals
confidence
If we compute 100 independent 95% confidence
If
intervals for a population with a true mean of 0,
about how many of these intervals WILL NOT
contain 0 (even though they should)?
contain
Approximately  five
When we do many confidence intervals, we see
When
significant results when there are none.
significant
So, when we do many confidence intervals,
So,
there are 2 different confidence levels to worry
about.
about.
25
25 Two Different Confidence
Two
Levels
Levels
INDIVIDUAL
INDIVIDUAL
confidence level is the
frequency with which
a single interval
captures the
parameter.
parameter.
e.g.  95% confidence
e.g.
that the interval
the
includes the true
value
value FAMILYWISE
FAMILYWISE
confidence level is the
frequency with which
all intervals
all
simultaneously capture
their parameters.
their
e.g. – 95% confidence
e.g.
that all intervals include
all
the true values
the
26
26 Confidence Intervals that control
Confidence
the familywise confidence level
the
usual confidence interval procedure:
g +/ t(nI)(1α/2) SE(g)
it’’s like:
s
estimate +/ multiplier x SE(estimate)
estimate
multiplier SE(estimate)
where the multiplier comes from the ttable
comes
confidence intervals that control the
confidence
familywise confidence level are of exactly
the same form, but they use different
multipliers
multipliers
27
27 Multipliers for different methods
Procedure Multiplier LSD t Tukey − Kramer q( I ,n − I )(1−α ) / 2 α
( n − I )(1− )
2 Scheffe′
Bonferroni ( I − 1) F( I −1,n − I )(1−α )
t α
( n − I )(1− )
2k Table
Table
A.2 A.5
A.4
software
28
28 Comments on methods
LSD
LSD
– the usual method
– doesn’t control familywise confidence level TukeyKramer
– good for unplanned comparisons or data
good
snooping
snooping
– only works for comparisons like µ1 – µ2
only Scheffe
– good for unplanned comparisons or data
good
snooping, all kinds of comparisons
snooping,
– price: very conservative 29
29 Comments on methods
Bonferroni
– not for data snooping (OK for unplanned
not
comparisons)
comparisons)
– works for all kinds of comparisons
– table A.2 doesn’’t have enough values for this
t
method – need to use JMP
method
– excellent for small number of comparisons;
excellent
not so good for large number
not 30
30 NonParametric FamilyWise Tests
There are familywise tests associated with
There
the KruskalWallis test.
the
This is called Dunn’s Test
It is not available in JMP 31
31 Com paris on of Multipliers
5 Treatm ents and 65 Degrees of Freedom
S ch e f f e M u l ti p l i e r 3.0 B o n fe rro n i
T u ke y 2.5 Ne wm a n K e u l s L SD 2.0
0 5 10 15 20 25 Nu m b e r o f Co n tra sts
32
32 Data Snooping
DNA example
2,436 mononucleotides along a DNA
2,436
molecule. (page 166)
molecule.
11 breaks occur.
6 of these breaks occur within four
of
mononucleotides of the TGG
trinucleotides.
trinucleotides. 33
33 Does TGG seem to cause the
Does
break?
That depends.
That
If TGG was hypothesized before the data was
If
observed, then the chances of 6 of the 11
breaks occurring within four units after a TGG is
only 0.000243. So it looks significant.
only
Note, the arbitrary number of four units also
Note,
needs to be specified before observing the data
needs
However, TGG was hypothesized after looking
However,
at the data and observing the patterns before the
break.
break.
34
34 Computer Simulation
This pattern of 2,436 mononucleotides was put
This
into a computer simulation.
into
11 breaks were placed in the sequence at
11
random. This was repeated 1000 times.
random.
The most frequent trinucleotide within four units
The
upstream from the breaks was observed.
upstream
How often did the most frequent occur six or
How
more times out of the 11? (highly significant
ignoring multiple comparison problem)
ignoring
35
35 Computer Simulation
Computer
Continued.
Continued.
Over 300 out of 1000 therefore the
Over
presence of six TGG’’s upstream from the
s
presence
breaks is not statistically significant.
breaks
Data snooping went on here – and data
Data
snooping is OK, if we do not quote psnooping
values.
Do you have any examples of data
Do
snooping from your field?
snooping 36
36 The End 37
37 ...
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This note was uploaded on 11/30/2011 for the course STAT 380 taught by Professor Stevens during the Spring '11 term at Brigham Young University, Hawaii.
 Spring '11
 Stevens

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