= 0
.
20 using similar reasoning to the above.
Trial and error – or logs – again results in
n
= 20.
4.
(a) FALSE — The only way one can obtain a neg
ative value of F is if one has made a mistake.
Even if all data values are negative, F can
never
be less than 0.
(b) FALSE — Since you are in ‘planning’ mode,
the underlying variance should be viewed as
known. We know
V ar
(
ˆ
b
1
) =
σ
2
y.x
/
P
(
x
i
−
¯
x
)
2
.
With the 10 data points, the sum in the denom
inator is 30; for the 6 data points situation, the
sum is 70. Thus,
V ar
(
ˆ
b
1
) will be smaller for the
6 data point case. (Even if one were concerned
with CI width by considering the di²erence in
tvalues for 4df and 8df, the conclusion would
not change.)
5. The random variable of interest is
D
, the di²er
ence in glucose levels between
B
and
A
. The stated
condition requires the distribution of
D
under the
given alternative.
In this case,
D
∼
N
(
.
5
,
1
.
8).
The given condition is
P
(
¯
D
≥
0
.
3

μ
D
=
.
5) =
.
90.
We know
P
(
Z
≥ −
1
.
28) =
.
90. Thus,
−
1
.
28 =
(
.
3
−
.
5)
/
p
(1
.
8
/n
). Solving results in
√
n
= 8
.
59
and, rounding up,
n
= 74.
Grade Distribution
9099:20
8089:19
7079:37
median =
74
6069:23
5059:24
below: 5
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Staff
 Null hypothesis, Bonferroni, null hypothesis states

Click to edit the document details