Biology Reference
In-Depth Information
EDMA-II are as follows. Let
A
1
,
A
2
,...,
A
n
and
B
1
,
B
2
,...,
B
m
be the land-
mark coordinate matrices for individuals from the two samples.
STEP 1. Estimate the mean form matrix for sample
A
and
sample
B
.
STEP 2. Calculate the scaling factor for each mean form
matrix and use these scaling factors to standardize the mean
form matrices for each sample by dividing each entry of the
matrix by the respective scaling factor. For example, divide
each entry of
FM(A)
by the scaling factor estimated from sam-
ple
A, S
A
. The new matrix is called the shape matrix of
A
and
is denoted as
SM(A)
. Do the same for sample
B
using
S
B
to pro-
duce
SM(B)
.
STEP 3. Calculate the shape difference matrix for these sam-
ples,
SDM(B,A)
, by calculating the
difference
between like
linear distances, and sort
SDM(B,A)
written as a vector in
ascending order. The test statistic
Z
is given by either the min-
imum entry in
SDM(B,A),
or the maximum entry in
SDM(B,A)
,
whichever is most different from zero.
STEP 4. Using the estimates of the mean forms and variance-
covariance matrices, generate parametric bootstrap (Monte
Carlo) samples under the Gaussian perturbation model.
Calculate the
Z
statistic for the bootstrap samples.
STEP 5. Repeat Step 4 an adequate number of times (e.g., 200
times) to obtain
Z
1
,
Z
2
,…,
Z
200
. Sort these values in ascending
order and denote the ordered values by
Z
(1)
,
Z
(2)
,…,
Z
(200)
. The
100 (1-
)% confidence interval is delimited by (
Z
(11)
,
Z
(190)
). If
this interval contains 0, accept the null hypothesis that the
samples are similar in shape; otherwise, the test provides evi-
dence that the two forms are not similar.
The above testing procedure works well in most situations.
However there is a situation where careless application of Step 5 can
be problematic. Look carefully at the distribution of the values in
SDM
. If the maximum positive entry in
SDM
and the maximum neg-
ative entry in
SDM
have very similar magnitudes, it is necessary to
look carefully at the histogram of
Z
1
,
Z
2
,…,
Z
200
. If this histogram indi-
cates a bimodal distribution with a large dip in the number of
observations near zero, the null hypothesis should be rejected.
Careless application and interpretation of Step 5 will lead to accep-
tance of the null hypothesis in this situation. As in all statistical
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