5.
(16 pts) Consider
H
0
:
p
= 0
.
32 vs the twosided
alternative
H
A
:
p
= 0
.
32, where
p
is the gerni
mation rate of oriental bittersweet seeds in tulip
poplar stands. The number of seeds that germi
nate, out of 120 seeds, has a binomial
B
(120
,
0
.
32)
distribution if the null hypothesis is true, which is
approximated by a normal
N
(38
.
4
,
5
.
11
2
) distri
bution (
np
= 38
.
4 and
nq
= 81
.
6 are both
>
5).
We observe
z
= (58

38
.
4)
/
5
.
11 = 3
.
835 (or
X
2
= 14
.
71 with a chisquare test) so that the
pvalue is
p
= 2
*
0
.
00006 = 0
.
0001. There is very
strong evidence that the germination rate in tulip
poplar stands is greater than 0
.
32.
6.(a) (6 pts) Binomial
B
(10
,
0
.
72), skewed left.
(b) (5 pts) 0
.
180
(c) (5 pts) 0
.
18 + 0
.
18

0
.
18
*
0
.
18 = 0
.
327 because
the two events are not mutually exclusive.
We
can calculate the probability of finding oriental
bittersweet in exactly 6 plots out of 10 in
both
locations (0
.
18
*
0
.
18) because these events are
independent.
(d) (10 pts) We may argue for a onesided test,
H
A
:
p
r
<
0
.
72 because of the scientist’s suspicion. We
are not told, but this suspicion could be based
upon prior data.
H
0
:
p
r
= 0
.
72. The number of
plots where oriental bittersweet is present has a
binomial distribution
X
r
∼ B
(10
, .
0
.
32) if the null
hypothesis is true, with an expected value of 3.2.
The values that are as extreme as or more extreme
than what was observed are
x
= 0 and
x
= 1, so
the pvalue is
P
(
X
r
= 0) +
P
(
X
r
= 1).
Using
the binomial formula, we get
.
000003+
.
000076 =
8
.
10

5
. (we could double this by two for a two
sided test).
There is very strong evidence that
p
r
<
0
.
32.
Frequency
20
40
60
80
100
0
10
20
30
40
Summary of grades
●
●
●
●
63
75
84
1
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 Fall '08
 Staff
 ChiSquare Test, Normal Distribution, Null hypothesis, Probability theory, Penguin

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