blunt beak trait. We are assuming that birds were
randomly sampled, and that the sample sizes were
large enough for the CLT to apply. Specifically,
we can tell this last assumption is met because
50
*
0
.
20 = 10, 50

10 = 40, 80
*
0
.
35 = 28 and
80

28 = 52 are all greater than 5.
5(a) (9 pts)
H
0
: The distribution of the area colonized
by
X. perforans
beetles is the same on trees from
species 1 and on trees from species 2. In partic
ular, they have the same mean.
H
A
: these two
distributions differ.
Samples have size
n
1
= 7,
n
2
= 8.
The sum of ranks in sample 1 is 44,
which is still smaller than 7
*
(7+8+1)

44 = 68
so
T
= 44.
This is larger than 38, the criti
cal value to get
p
≤
.
05.
Therefore
p >
0
.
05.
There is no evidence that
X. perforans
beetles
colonize trees from species 1 any differently than
trees from species 2.
(b) (9 pts)
Source
df
SS
MS
F
pvalue
Tree sp.
3
681
227
2.95
.
01
< p < .
05
Residual
36
2767
76.86
Total
39
3448
(c) (5 pts)
H
0
:
μ
1
=
μ
2
=
μ
3
=
μ
4
where
μ
i
is the
mean area colonized by
X. perforans
beetles on
timber samples from trees in species
i
.
H
A
: not all
μ
i
’s are the same, which means that the average
colonized area differs in at least one tree species
versus another.
There is moderate evidence to
reject
H
0
. We conclude that
X. perforans
beetles
1
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colonize a larger average area (or show preference)
in at least one tree species versus another.
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 Fall '08
 Staff
 Normal Distribution, Standard Deviation, pts, Student's tdistribution

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