(c) Here we focus on
ˆ
Y
as an estimator.
se
(
ˆ
Y
e
) =
s
p
(1
/n
) + (5
−
¯
x
)
2
/
(
P
(
x
i
−
¯
x
)
2
)
=
.
079.
Now,
t
8
,.
025
= 2
.
306 and
ˆ
Y
= 3
.
326. Thus, a
95% con±dence interval for
ˆ
Y
e
is (3.144, 3.508).
Thus, at 95 % con±dence, a plausible range for
the mean rate of oxygen consumption is from
3.144 to 3.508 ml/g/hr. Since the value of “3”
does not fall within this range, it is not a plau
sible value with 95% con±dence; thus the hy
pothesis that the true rate is 3 is rejected at
α
= 0
.
05.
3. There are two equivalent approaches. Let
m
= 4
n
be the total number of trials. Due to independence,
we can solve directly for
m
and choose the smallest
integer equal to or larger than
m
that is divisible
by 4.
P
(
atleast
1
success
) = 1
−
P
(0
successes
) =
1
−
.
98
m
. We wish to choose
m
so that
.
98
m
= 0
.
20.
This can be done by trial and error or, directly, using
logs.
m
∗
ln
(
.
98) =
ln
(
.
20). This results in
m
=
79
.
66. Rounding up to 80 (and dividing by 4) results
in
n
= 20. Equivalently, let
Y
be the number of
successes out of 4 trials. Then,
P
(
Y
= 0) =
.
98
4
=
.
92237. Then, with
n
as the number of ‘sets’ of trials,
.
92237
n
= 0
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 Fall '08
 Staff
 Null hypothesis, Bonferroni, null hypothesis states

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