257
Chapter 12: Considerations in Designing Experiments
12.1
(See Example 12.1) Let
n
1
=
( )
( )
90
5
3
3
2
1
1
+
σ
+
σ
σ
=
n
= 33.75 or 34 and
n
2
= 90
−
34 = 56.
12.2
(See Ex. 12.1).
If
n
1
= 34 and
n
2
= 56, then
7111
.
56
25
34
9
2
1
=
+
=
σ
−
Y
Y
In order to achieve this same bound with equal sample sizes, we must have
9
n
+
25
n
=
.7111
The solution is
n
= 47.8 or 48.
Thus, it is necessary to have
n
1
=
n
2
= 48 so that the
same amount of information is implied.
12.3
The length of a 95% CI is twice the margin of error:
2(1.96)
9
n
1
+
25
n
2
,
and this is required to be equal to two.
In Ex. 12.1, we found
n
1
= (3/8)
n
and
n
1
=
(5/8)
n
, so substituting these values into the above and equating it to two, the solution is
found to be
n
= 245.9.
Thus,
n
1
= 93 and
n
2
= 154.
12.4
(Similar to Ex. 12.3) Here, the equation to solve is
2(1.96)
9
n
1
+
25
n
1
= 2.
The solution is
n
1
= 130.6 or 131, and the total sample size required is 131 + 131 = 262.
12.5
Refer to Section 12.2.
The variance of the slope estimate is minimized (maximum
information) when
S
xx
is as large as possible.
This occurs when the data are as far away
from
x
as possible.
So, with
n
= 6, three rats should receive
x
= 2 units and three rats
should receive
x
= 5 units.
12.6
When
σ
is known, a 95% CI for
β
is given by
xx
S
z
σ
±
β
α
2
/
1
ˆ
.
Under the two methods, we calculate that
S
xx
= 13.5 for Method 1 and
S
xx
= 6.3 for
Method 2.
Thus, Method 2 will produce the longer interval.
By computing the ratio of
the margins of error for the two methods (Method 2 to Method 1), we obtain
3
.
6
5
.
13
=
1.464; thus Method 2 produces an interval that is 1.464 times as large as Method 1.
Under Method 2, suppose we take n measurements at each of the six dose levels.
It is
not difficult to show that now
S
xx
= 6.3
n
.
So, in order for the intervals to be equivalent,
we must have that 6.3
n
= 13.5, and so
n
= 2.14.
So, roughly twice as many observations
are required.
12.7
Although it was assumed that the response variable
Y
is truly linear over the range of
x
,
the experimenter has no way to verify this using Method 2.
By assigning a few points
at
x
= 3.5, the experimenter could check for curvature in the response function.