Download Intermediate Applied Statistics STAT 460

Intermediate Applied Statistics
STAT 460
Lecture 10, 10/1/2004
Instructor:
Aleksandra (Seša) Slavković
[email protected]
TA:
Wang Yu
[email protected]
Failure Times Example from Sleuth, p. 171
Boxplots of TIME by COMPOUND
(means are indicated by solid circles)
25
TIME
20
15
10
5
5
4
3
2
COMPOUND
1
0
One-way ANOVA
One-way ANOVA: TIME versus COMPOUND
Analysis of Variance for TIME
Source
DF
SS
MS
COMPOUND
4
401.3
100.3
Error
45
899.2
20.0
Total
49
1300.5
Level
1
2
3
4
5
N
10
10
10
10
10
Pooled StDev =
Mean
10.693
6.050
8.636
9.798
14.706
4.470
StDev
4.819
2.915
3.291
5.806
4.863
F
5.02
P
0.002
Individual 95% CIs For Mean
Based on Pooled StDev
--+---------+---------+---------+---(------*------)
(------*------)
(-------*------)
(------*-------)
(------*------)
--+---------+---------+---------+--4.0
8.0
12.0
16.0
ANOVA
 So far we have talked about testing the null
hypothesis
H0:
μ1 = μ2 = μ3 = . . . = μk
versus the alternative hypothesis
HA: at least one pair of groups has μi ≠ μ
j
Next, we want to know where the difference
comes from?
Approaches to Comparisons
A.
B.
C.
D.
Pairwise comparisons
Multiple comparison with a control group
Multiple comparison with the “best” group
Linear contrasts
 Each of these procedures, (e.g. pairwise
confidence intervals for the difference) are
available in addition to corresponding
tests, in most statistical packages.
All Pairwise Comparisons
 In this approach, we don’t just want to know
whether there is any significant difference, we want
to know which groups are significantly different from
which other groups.
 The most obvious approach to testing pairwise
comparisons is just to do a t-test on each pair of
groups. The problem with this is sometimes called
“compound uncertainty” or “capitalizing on chance.”
 Note: By significantly we mean systematically or demonstrably
(statistical significance), not greatly or importantly (practical
significance).
Fails 5% of time
Fails 5% of time
Fails 5% of time
Fails 5% of time
Fails 5% of time
If there are five independent parts and each fail 5% of the time,
then the whole machine fails 1-(0.95)5 = 23% of the time!
All Pairwise Comparisons
 If you have c tests, each of which has its
individual false positive rate controlled at , then
the total false positive rate may be as high as 1(1-)c  c.
 If we want to do all possible pairwise comparisons
then the number of tests will be
k * (k  1)
2
All Pairwise Comparisons
If we want to do all possible pairwise comparisons,
then the number of tests will be
k * (k  1)
c
where k is the number of groups.
If there are 5 groups then there are 10 tests!
2
All Pairwise Comparisons
 So we have to decide whether we want to
control only individual type I error risk, or
experiment-wide (or “familywise”) type I
error risk.
 We have to be much more conservative to
control the latter than we would need to be
to control the former. In fact, we might
have to make  so low that β (power) will
become very high.
All Pairwise Comparisons
 One possible answer is to control individual type I
error risk for planned comparisons (a few
comparisons which were of special interest as you
were planning the study) and control experimentwide type I error risk for unplanned comparisons
(those you are making after the study has been
done just for exploratory purposes).
 Suppose for now that there aren’t any comparisons
of special interest, we just want to look at all
possible pairwise comparisons. Then we definitely
should try to control experiment-wide error.
Method 1: Fisher’s Protected LSD
(Least Significant Difference)
 A simple approach to trying to control experimentwide error:
1. First do an overall F-test to see if there
are any differences at all
2. If this test does not reject H0 then conclude
that no differences can be found.
3. If it does reject H0 then go ahead and do all
possible two-group t-tests. (You might use a
standard deviation pooled over all groups
instead of each pair separately, but otherwise
act just as in the two-sample case.)
Method 1: Fisher’s Protected LSD
(Least Significant Difference)
 The only problem with that approach is that
it doesn’t work. At least according to some
experts, the LSD procedure does not really
control experiment-wide error. You still
have the multiple comparisons problem.
Method 2: Bonferroni Correction
 One way to do this is to use a “Bonferroni
correction” on . The idea here is that we
first set a family-wise , say .05, and then
figure out how small the individual * would
need to be in order to keep the family-wise
type I error rate at . For example, if
*=.01 and c=5, then   .05.
Method 2: Bonferroni Correction
Bonferroni Inequality:
For independent events A1, …, Ak of equal probability,
P( A1, A2 , A3 ,..., Ac )  1  (1  P( Ai ))
 c P( Ai )
c
Method 2: Bonferroni Correction
 So if we just set
 
*

then the
c
experiment-wide error rate will be controlled.
An advantage of this approach is that it is a simple idea
(just divide up your acceptable risk equally among all the
tests you want to do).
A disadvantage is that it can be inefficient. The
can become quite small, e.g., .05/21  .002.

c
Method 3: Tukey’s (HSD) Tests
 Tukey’s “Honest Significant Difference” test takes a
different approach that does not use the t or F
distributions but the distribution of the “studentized
range” or Q statistic.
 In a Tukey test, the Q distribution is used to find a
boundary on differences between group averages ,
 such that 95% of the time if the null hypothesis was
true, no pair of groups would have a difference larger
than this boundary.
(Ymax  Ymin )  M  SE
 So any pair of groups with a difference larger than
this critical value is declared significantly different.
Confidence Intervals
 Much like the usual t-intervals, the Bonferroni and
Tukey confidence intervals for μ1-μ2 is
yi  y j  m * SE yi  y j
best estimate
for μ1-μ2
“margin of error” or
“half-width” of interval
SE yi  y j  s p
1 1

n1 n2
where m is some multiplier. But m is chosen differently
for each procedure.
yi  y j  m * SE yi  y j
actually,
t  stands for t
2
For a t-interval or LSD t-interval,
For a Bonferroni t-interval,
m t 
m t  *
dfe,1

2
2
2
with α*= α/c
For a Tukey confidence interval,
m  (dfTreat )( FdfTrea t,df Erro r )(1   )
Method 4: Scheffe’s test
 Also available are Scheffé tests and intervals,
which use
m  (dfTreat )( FdfTreat,df Error )(1   )
but they lead to a test that may be too
conservative.
How to compare specific groups in an ANOVA
 If you have a few planned comparisons of theoretical
importance in mind, then you can just do them with a
slightly adjusted t-test
 a different value for sP and a different degrees of freedom for
the t statistic based on the fact that you pool variance over
all groups, not just two.
 If you want to do many comparisons then you may wish
to try some means of controlling compound uncertainty
(i.e., correcting for multiple comparisons).
 In the case where you want to test for all pairwise
differences, choices include Fisher’s LSD (too lenient),
Bonferroni, and Tukey (the latter also known as TukeyKramer when the sample sizes are different). There are
also various other procedures I didn’t (and won’t)
mention.
Tukey's pairwise comparisons
This is an abbreviation for,
“We are 95% confident according to
the Tukey procedure that the signed
Family error rate = 0.0500
Individual error rate = 0.00670
Critical value = 4.02
difference μ1-μ2 is between -1.040
4
hours
and +10.326 hours.” So we can’t
Intervals for (column level mean) - (row level mean)
1
2
3
conclude that μ1≠μ2.
2
-1.040
10.326
3
-3.626
7.740
-8.269
3.097
4
-4.788
6.578
-9.431
1.935
-6.845
4.521
5
-9.696
1.670
-14.339
-2.973
-11.753
-0.387
-10.591
0.775
Conclude μ2≠μ5 and
μ3≠μ5. More
specifically, μ2<μ5 and
μ3<μ5.
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.00670
Critical value = 4.02
Intervals for (column level mean) - (row level mean)
1
2
3
2
-1.040
10.326
3
-3.626
7.740
-8.269
3.097
4
-4.788
6.578
-9.431
1.935
-6.845
4.521
5
-9.696
1.670
-14.339
-2.973
-11.753
-0.387
4
-10.591
0.775
Tukey Simultaneous Tests
Response Variable TIME
All Pairwise Comparisons among Levels of COMPOUND
COMPOUND = 1 subtracted from:
Level
Difference
SE of
COMPOUND
of Means Difference
2
-4.643
1.999
3
-2.057
1.999
4
-0.895
1.999
5
4.013
1.999
T-Value
-2.322
-1.029
-0.448
2.007
Adjusted
P-Value
0.1567
0.8406
0.9914
0.2792
COMPOUND = 2 subtracted from:
Level
Difference
SE of
COMPOUND
of Means Difference
3
2.586
1.999
4
3.748
1.999
5
8.656
1.999
T-Value
1.294
1.875
4.330
Adjusted
P-Value
0.6965
0.3454
0.0008
COMPOUND = 3 subtracted from:
Level
Difference
SE of
COMPOUND
of Means Difference
4
1.162
1.999
5
6.070
1.999
T-Value
0.5812
3.0363
Adjusted
P-Value
0.9772
0.0309
COMPOUND = 4 subtracted from:
Level
Difference
SE of
COMPOUND
of Means Difference
5
4.908
1.999
T-Value
2.455
Adjusted
P-Value
0.1196
2
3
4
1
5
(“not significantly different”)
(“not significantly different”)
Multiple Comparisons with a Control
 A different situation arises when you have a
control group and are only interested in the
comparison of other groups to the control
group, not to one another.
 Then there is something like Tukey’s test,
but modified, called Dunnett’s test.
Multiple Comparisons with a Control
 Tukey confidence intervals are for μi - μJ and
there is one of them for every pair of
distinct groups, which comes out to be k*(k1)/n of them.
 Dunnett confidence intervals are for
μi - μControl and there are only k-1 of them,
one for each noncontrol group.
Multiple Comparisons with a Control
 Suppose compound 5 is thought of as a
control group because it is the compound
that has been in use previously. We want to
know if any of the other compounds have
different population mean survival times
than compound 5.
Multiple Comparisons with a Control
Dunnett's comparisons with a control
Family error rate = 0.0500
Individual error rate = 0.0149
Critical value = 2.53
Control = level (5) of COMPOUND
Intervals for treatment mean minus control mean
Level
1
2
3
4
Lower
-9.074
-13.717
-11.131
-9.969
Center
-4.013
-8.656
-6.070
-4.908
Upper -----+---------+---------+---------+-1.048
(------------*------------)
-3.595 (-----------*------------)
-1.009
(------------*-----------)
0.153
(------------*-----------)
-----+---------+---------+---------+--12.0
-8.0
-4.0
0.0
For compounds 2 and 3, we are confident that their difference
from the control is nonzero.
Multiple Comparisons with a Control
Dunnett Simultaneous Tests
Response Variable TIME
Comparisons with Control Level
COMPOUND = 5 subtracted from:
Level
COMPOUND
1
2
3
4
Difference
of Means
-4.013
-8.656
-6.070
-4.908
SE of
Difference
1.999
1.999
1.999
1.999
T-Value
-2.007
-4.330
-3.036
-2.455
Adjusted
P-Value
0.1555
0.0003
0.0141
0.0597
Again we conclude that compounds 2 and 3 are different from
the control group (actually, poorer).
Multiple comparison with the best
 There is also a kind of test called Hsu’s multiple comparison
with the best (MCB).
 This test tries to decide which groups could plausibly have
come from the population with the highest (or lowest, if you
prefer) mean.
 The group with the highest (lowest) sample average is
automatically the leading candidate, but any group that does
not significantly differ from this group is still included as a
possible best.
 So Hsu’s confidence intervals, one for each group, would be
for μi -μbest.
 This approach is harder to understand than the others,
perhaps because it seems to be doing more than one thing
at the same time.
Next Lecture
 Linear Contrasts