Biology Reference
In-Depth Information
Using the normal model, we find that
,
X
.5
3.52,
σ
X
5
1.69, and
,
Y
.5
3.94 and
σ
Y
5
2.71. To test whether the means are different, we find the probability of statistic t:
.
Þ
σ
ð
,
Y
.2,
X
t
r
(8A.13)
5
N
X
1
N
Y
N
X
N
Y
2
X
ð
N
X
2
2
Y
ð
N
Y
2
1
Þ
1
σ
1
Þ
N
X
1
N
Y
2
2
with degrees of freedom equal to (N
X
1
2). For relatively large values of N
X
and N
Y
,
the t-value will be normally distributed with a mean of zero and a standard deviation of
one, provided that the null hypothesis of equal means is true. If the absolute value of t
exceeds 1.96, we may conclude that, under the normal model, there is only a 5% chance of
the mean values being that different by chance. We can thus reject the null hypothesis at a
5% level of confidence.
The problem is that the list of lengths contained in
N
Y
2
Y
is highly non-normal. Most values
are close to 3, but there are several around 8 or 9, so
Y
appears to be rather bimodal. Also, in
a normal distribution with a mean of
2.71,
we would expect that 7.3% of the measured lengths would be less than zero. So, the distribu-
tion of
Y
3.94 and a standard deviation of
σ
Y
5
,
.5
.
To form a bootstrap version of the t-test, we simulate the null hypothesis we wish to reject.
This simple principle is the key to understanding how to form your own bootstrap tests
when asking novel statistical questions. The null hypothesis of the t-test is that the means
of the two groups are equal, which we can also phrase as the hypothesis that the two
groups in question came from a single underlying distribution that was arbitrarily subdi-
vided into two groups. If this were the case, any difference between the means would
arise simply by chance. So, to test this hypothesis, we assume that the null hypothesis is
true
i.e. that
Y
departs substantially from normality, more so than does the distribution of
X
were drawn from the same population. This means that under the
hypothesis, specimens are exchangeable between the two groups (
Anderson, 2001b
).
Therefore, we merge the two sets of observations (
X
and
Y
X
and
Y
) into a common pool of speci-
Z
Z
mens (
, one of size N
X
and one
of size N
Y
, and compute the differences in means between the two bootstrap sets. This is
repeated N
Bootstrap
times. We can then determine the number of times in which the differ-
ence between the means of paired bootstrap sets exceeds the observed difference between
the means of
) and draw (with replacement) two bootstrap sets from
. Expressed as a proportion of the total, we get an estimate of the
probability that the observed difference is due to chance; i.e. if the difference between
means of pairs of bootstrap samples exceeds the observed differences in 5% (or fewer) of
the total number of iterations, we can reject the null hypothesis that the means are equal.
This is simply another way of phrasing the statement that the observed difference is statis-
tically significant at a 5% confidence level if the observed difference between means
exceeds the 95th percentile of differences between means of the bootstrap sets.
A symbolic example of this merging and subsequent formation of two bootstrap sets
may help to develop an understanding of how the test operates. Suppose we have a set
X
and
Y
C
of five elements, and a set
D
of four elements:
C
5
f
C
1
;
C
2
;
C
3
;
C
4
;
C
5
g
(8A.14)
D
5
f
D
1
;
D
2
;
D
3
;
D
4
g
(8A.15)
