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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.
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Chapter 12: Considerations in Designing Experiments
Instructor’s Solutions Manual
12.8
Checking for true linearity and constant error variance cannot be performed if the data
points are spread out as far as possible.
12.9
a.
Each half of the iron ore sample should be reasonably similar, and assuming the two
methods are similar, the data pairs should be positively correlated.
b.
Either analysis compares means.
However, the paired analysis requires fewer ore
samples and reduces the sample
−
to
−
sample variability.
12.10
The sample statistics are:
d
=
−
.0217,
s
D
2
= .0008967.
a.
To test
H
0
:
μ
D
= 0 vs.
H
a
:
μ
D
≠
0, the test statistic is 
t
 =
6
0008967
.
0217
.
−
= 1.773 with 5
degrees of freedom.
Since
t
.025
= 2.571,
H
0
is not rejected.
b.
From Table 5, .10 <
p
−
value < .20.
c.
The 95% CI is
6
0008967
.
571
.
2
0217
.
±
−
=
−
.0217
±
.0314.
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 Spring '08
 Staff
 Probability

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