All 21 Chapters Covered
SOLUTIONS
,Chapter 1
Exercises
Section 1.1
1.1 From the yield data in Table 1.1 in the text, and using the given expression,
we obtain
s2A = 2.05
s2B = 7.64
from where we observe that sA2 is greater than s2B.
1.2 A table of values for di is easily generated; the histogram along with sum-
mary statistics obtained using MINITAB is shown in the Figure below.
Summary for d
Mean 3.0467
V ariance 11.0221
N 50
1st Q uartile 1.0978
3rd Q uartile 5.2501
Maximum 9.1111
Figure 1.1: Histogram for d = YA − YB data with superimposed theoretical distribution
1
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,2 CHAPTER 1.
From the data, the arithmetic average, d¯, is obtained as
d¯ = 3.05 (1.1)
And now, that this average is positive, not zero, suggests the possibility that
YA may be greater than YB. However conclusive evidence requires a measure of
intrinsic variability.
1.3 Directly from the data in Table 1.1 in the text, we obtain y¯A = 75.52; y¯B =
72.47; and sA2 = 2.05; s2B= 7.64. Also directly from the table of differences, di,
generated for Exercise 1.2, we obtain: d¯ = 3.05; however sd2 = 11.02, not 9.71.
Thus, even though for the means,
d¯= y¯A — y¯B
for the variances,
s2 /= s2 + s2
d A B
The reason for this discrepancy is that for the variance equality to hold, YA
must be completely independent of YB so that the covariance between YA and
YB is precisely zero. While this may be true of the actual random variable, it
is not always strictly the case with data. The more general expression which is
valid in all cases is as follows:
s2 = s2 + s2 — 2sAB (1.2)
d A B
where sAB is the covariance between yA and yB (see Chapters 4 and 12). In
this particular case, the covariance between the yA and yB data is computed as
sAB = —0.67
Observe that the value computed for sd2 (11.02) is obtained by adding —2sAB
to s2 + s2 , as in Eq (1.2).
A B
Section 1.2
1.4 From the data in Table 1.2 in the text, sx2 = 1.2.
1.5 In this case, with x̄ = 1.02, and variance, s2x = 1.2, even though the num-
bers are not exactly equal, within limits of random variation, they appear to be
close enough, suggesting the possibility that X may in fact be a Poisson random
variable.
Section 1.3
1.6 The histograms obtained with bin sizes of 0.75, shown below, contain 10
bins for YA versus 8 bins for the histogram of Fig 1.1 in the text, and 14 bins
for YB versus 11 bins in Fig 1.2 in the text. These new histograms show a bit
more detail but the general features displayed for the data sets are essentially
unchanged. When the bin sizes are expanded to 2.0, things are slightly different,
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, 3
Histogram of YA (Bin size 0.75)
18
16
14
12
Frequency
10
8
6
4
2
0
72.0 73.5 75.0 76.5 78.0 79.5
YA
Histogram of YB (Bin size 0.75)
6
5
4
Frequency
3
2
1
0
67.5 69.0 70.5 72.0 73.5 75.0 76.5 78.0
YB
Figure 1.2: Histogram for YA, YB data with small bin size (0.75)
Histogram of YA (Bin size 2.0)
25
20
Frequency
15
10
5
0
72 74 76 78 80
YA
Histogram of YB(Bin Size 2.0)
14
12
10
Frequency
8
6
4
2
0
67 69 71 73 75 77 79
YB
Figure 1.3: Histogram for YA, YB data with larger bin size (2.0)
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SOLUTIONS
,Chapter 1
Exercises
Section 1.1
1.1 From the yield data in Table 1.1 in the text, and using the given expression,
we obtain
s2A = 2.05
s2B = 7.64
from where we observe that sA2 is greater than s2B.
1.2 A table of values for di is easily generated; the histogram along with sum-
mary statistics obtained using MINITAB is shown in the Figure below.
Summary for d
Mean 3.0467
V ariance 11.0221
N 50
1st Q uartile 1.0978
3rd Q uartile 5.2501
Maximum 9.1111
Figure 1.1: Histogram for d = YA − YB data with superimposed theoretical distribution
1
@
@SS
eeisismm
iciicsis
oolala
titoio
nn
,2 CHAPTER 1.
From the data, the arithmetic average, d¯, is obtained as
d¯ = 3.05 (1.1)
And now, that this average is positive, not zero, suggests the possibility that
YA may be greater than YB. However conclusive evidence requires a measure of
intrinsic variability.
1.3 Directly from the data in Table 1.1 in the text, we obtain y¯A = 75.52; y¯B =
72.47; and sA2 = 2.05; s2B= 7.64. Also directly from the table of differences, di,
generated for Exercise 1.2, we obtain: d¯ = 3.05; however sd2 = 11.02, not 9.71.
Thus, even though for the means,
d¯= y¯A — y¯B
for the variances,
s2 /= s2 + s2
d A B
The reason for this discrepancy is that for the variance equality to hold, YA
must be completely independent of YB so that the covariance between YA and
YB is precisely zero. While this may be true of the actual random variable, it
is not always strictly the case with data. The more general expression which is
valid in all cases is as follows:
s2 = s2 + s2 — 2sAB (1.2)
d A B
where sAB is the covariance between yA and yB (see Chapters 4 and 12). In
this particular case, the covariance between the yA and yB data is computed as
sAB = —0.67
Observe that the value computed for sd2 (11.02) is obtained by adding —2sAB
to s2 + s2 , as in Eq (1.2).
A B
Section 1.2
1.4 From the data in Table 1.2 in the text, sx2 = 1.2.
1.5 In this case, with x̄ = 1.02, and variance, s2x = 1.2, even though the num-
bers are not exactly equal, within limits of random variation, they appear to be
close enough, suggesting the possibility that X may in fact be a Poisson random
variable.
Section 1.3
1.6 The histograms obtained with bin sizes of 0.75, shown below, contain 10
bins for YA versus 8 bins for the histogram of Fig 1.1 in the text, and 14 bins
for YB versus 11 bins in Fig 1.2 in the text. These new histograms show a bit
more detail but the general features displayed for the data sets are essentially
unchanged. When the bin sizes are expanded to 2.0, things are slightly different,
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, 3
Histogram of YA (Bin size 0.75)
18
16
14
12
Frequency
10
8
6
4
2
0
72.0 73.5 75.0 76.5 78.0 79.5
YA
Histogram of YB (Bin size 0.75)
6
5
4
Frequency
3
2
1
0
67.5 69.0 70.5 72.0 73.5 75.0 76.5 78.0
YB
Figure 1.2: Histogram for YA, YB data with small bin size (0.75)
Histogram of YA (Bin size 2.0)
25
20
Frequency
15
10
5
0
72 74 76 78 80
YA
Histogram of YB(Bin Size 2.0)
14
12
10
Frequency
8
6
4
2
0
67 69 71 73 75 77 79
YB
Figure 1.3: Histogram for YA, YB data with larger bin size (2.0)
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