4.
Question #1
Fig 6: Fat in Males
19.0
Δ=3.515.5
Strength of Treatment
Effect:
Signal:Noise Ratio t=
3.5
SD(1/NCtrl + 1/NAb)1/2
Are the Ns the # of
dams or # of offspring?
What is correct SD?Ctrl Ab
5.
Question #1
Group N Mean SD
Ctrl 33 18.784 1.5148
Ab 13 15.643 1.3693
Diff 3.1411 1.4766
t-test: t Value Pr > |t|
6.50 <0.0001
Paper ignores dams. Uses #s of offspring.
Simulated data gives:
Analysis assumes all 46 offspring give independent
information.
We explore the validity/necessity of that assumption.
t = 6.50 = 3.14/(1.477(1/33 +1/13)1/2
)
6.
Question #2
Does Fig 1 express biological differences
or measurement error, or both?
8.
Question #3
From Fig 1, is it possible (likely?) that litter-
mates from a mother may respond
more similarly than offspring from
different mothers (who were treated the
same)?
10.
Question #4
Suppose litter-mates do respond almost
identically.
Would an analysis, say a t-test, using
individual offspring that ignores the
mothers give about the same treatment
difference as an analysis (again, say a
t-test) using the mothers means of their
offspring?
11.
Question #5
Would the answer to question #4 change if
some litters had 3 offspring and others
had up to 8?
12.
Question #6
Continuing question #4, would the analysis
using individual offspring overstate or
understate the evidence about the
treatment difference (i.e., p-value too
low or too high)?
13.
Question #7
Suppose now that outcomes from litter-
mates differ about the same as offspring
from different mothers. Would that
justify using individual offspring, rather
than mothers, in the analysis, and
hence more power with the larger N?
14.
Question #7
Suppose now that outcomes from litter-
mates differ about the same as offspring
from different mothers. Would that
justify using individual offspring, rather
than mothers, in the analysis, and
hence more power with the larger N?
This requires the assumption of this equal
variability, an expert opinion that may
be valid, but the analysis could be faulty
if that assumption is wrong. See the
next question.
15.
Question #8
Lastly, suppose that we don’t want to
suppose as in questions #4-7.
Can we use the data itself to measure
relative intra- and inter-litter differences,
and incorporate that into the treatment
comparison?
This is what hierarchical or mixed models
accomplish.
They estimate the correlations among the
offspring so we do not have to make
assumptions as in question #7.
We now show how this is done.
16.
Basic Issue for Using Offspring as Replicates
• Dams vary.
• Overall, offspring vary.
• Do offspring from a dam vary less than offspring
from different dams (positive correlation)?
• Do offspring from a dam vary more than offspring
from different dams (negative correlation)?
What could cause this?
17.
Intra-Dam Correlation Among Offspring
Example: Four dams - A,B,C,D - with 2 offspring each:
A
A
B
B
C
C
D
D
A
A
A
B
B
B
C
C
C
D
D
D
Offspring
Fat
Dam
Means
Strong
Negative
Correlation
Strong
Positive
Correlation
Overall
Mean
No
Correlation
A
A
B
B
C
C
D
D
18.
Intra-Dam and Inter-Dam Variation
Example: Four dams - A,B,C,D - with 2 offspring each:
A
A
B
B
C
C
D
D
A
A
A
B
B
B
C
C
C
D
D
D
Offspring
Fat
Overall
Mean
Correlation = Scaled VInter - VIntra
Can be calculated from the data.
Denote correlation by r.
A
A
B
B
C
C
D
D
VInter
VIntra
19.
Correct SD Uses Both Variations
Table 6: Fat in Males
19.0
Δ=3.515.5
Strength of Treatment
Effect:
Signal/Noise Ratio t=
3.5
SD(1/NCtrl + 1/NAb)1/2
Are the Ns the # of
dams or # of offspring?
What is correct SD?Ctrl Ab
SD2
= V(1 + (n-1)r), where n=# offspring/dam
20.
Correct Analysis
Ns are #s of offspring.
Incorporate offspring correlation by using:
SD2
= V(1 + (n-1)r), where n=# offspring/dam
Signal/Noise Ratio t=
Δ
SD(1/NCtrl + 1/NAb)1/2
If r=0, then SD2
=V and same as t-test.
If r>0, then SD2
>V, so t-test overstates effect.
If r<0, then SD2
<V, so t-test understates effect.
21.
Correct Analysis
Thus, the reasoning is that the dams are clusters
of correlated outcomes (offspring).
If offspring were completely correlated (r=1), i.e.,
identical in a dam, then the correct analysis is the
same as using dam means. [SD2
= nV]
If there is no correlation (r=0), the analysis is the
same as ignoring dams and using offspring
results. [SD2
= V]
If there is some correlation, then SD incorporates
that correlation, i.e., relative intra- and inter-.
22.
Correct Analysis in Software
If we have the same # of offspring for every
dam, we can use repeated measures ANOVA.
Specify the dam as a “subject” and the offspring
as the repeated values.
Otherwise, use Mixed Model for Repeated
Measures.
Both of these methods consider the dams as
clusters of correlated outcomes (offspring).
23.
Numerical Illustrations
1. All Offspring for a Dam Identical
2. All Offspring for a Dam are Unique
3. Offspring for a Dam are Negatively Correlated
We will generate data that has about the
same means, but different correlations
among littermates for these 3 examples.
25.
Recall Paper Uses Offspring
Group N Mean SD
Ctrl 33 18.784 1.5148
Ab 13 15.643 1.3693
Diff 3.1411 1.4766
t-test: t Value Pr > |t|
6.50 <0.0001
Paper ignores dams. Uses #s of offspring.
Simulated data with correlation=1 gives:
Analysis assumes all 46 offspring give independent
information.
… which is wrong here.
t = 6.50 = 3.14/(1.477(1/33 +1/13)1/2
)
26.
Analysis on Dam Means
Group N Mean SD
Ctrl 9 19.000 1.500
Ab 9 15.494 1.500
Diff 3.5061 1.500
t-test: t Value Pr > |t|
4.96 <0.0001
Same data using dam means gives:
t = 4.96 = 3.51/(1.477(1/9 +1/9)1/2
)
So the previous analysis gave a signal:noise ratio t
that was 6.5/4.96=1.3 times too large. It doesn’t
matter here, but if the previous t-test gave p=0.05,
then the correct p here would be 0.13.
27.
Analysis using Calculated Correlation
Same data using mixed model gives:
CovParm Subject Estimate
CS id 2.2485
Residual 1.365E-6
Num Den
Effect DF DF F Value Pr > F
group 1 16 24.61 0.0001
group Estimate Std Err Lower Upper
Ctrl 19.0006 0.4998 17.9410 20.0602
Ab 15.4940 0.4998 14.4344 16.5536
Square root of 24.61 is t = 4.96, same as analysis on means.
R = 1 =
2.2485
(2.2485 + 0)
29.
Second Set of Simulated Data
Group N Mean SD
Ctrl 33 19.000 1.5000
Ab 13 15.500 1.5000
Diff 3.5000 1.5000
t-test: t Value Pr > |t|
7.13 <0.0001
Paper ignores dams. Uses #s of offspring.
Simulated data with correlation≈0 gives:
Analysis assumes all 46 offspring give independent
information.
… which is correct here; I generated them to be so.
30.
Analysis using Calculated Correlation
Same data using mixed model gives:
CovParm Subject Estimate
CS id -0.1860
Residual 2.4278
Num Den
Effect DF DF F Value Pr > F
group 1 16 56.20 0.0001
group Estimate Std Err Lower Upper
Ctrl 18.9690 0.2225 18.4972 19.4407
Ab 15.5102 0.4042 14.6534 16.3670
R = -0.083 =
-0.186
(-0.186+2.428)
Square root of 56.20 is t = 7.50, close to t-test ignoring dams.
31.
3. Offspring for a Dam are Negatively
Correlated
32.
Third Set of Simulated Data
Group N Mean SD
Ctrl 32 19.000 1.4756
Ab 12 15.500 1.4302
Diff 3.500 1.4639
t-test: t Value Pr > |t|
7.06 <0.0001
Use 2 offspring/dam; N=32 and 12 to be even.
Simulated data with correlation=-0.76 gives:
Analysis assumes all 46 offspring give independent
information.
… which is wrong here.
33.
Analysis using Calculated Correlation
Same data using mixed model gives:
CovParm Subject Estimate
CS id -1.5780
Residual 3.6458
Num Den
Effect DF DF F Value Pr > F
group 1 20 218.33 0.0001
group Estimate Std Err Lower Upper
Ctrl 19.0000 0.1237 18.7419 19.2580
Ab 15.5000 0.2020 15.0786 15.9214
R = -0.76 =
-1.578
(-1.578+3.646)
Square root of 218.33 is t = 14.8, twice the t-test.
But, with neg corr, probably would not have a 3.5 difference.
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