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[
y
i
−
f
(
x
i
)]
2
S
=
=
17
.
59
S
6
σ
=
2
=
.
2
10
−
As pointed out before, this is an approximate solution of the statedproblem,since
we did not fit
y
i
, butln
y
i
. Judging by the plot, the fit seemstobe good.
Solution of Part (2)
We again fitln(
ae
bx
)
=
ln
a
+
bx
to
z
=
ln
y
,
butthistime the
weights
W
i
=
y
i
are used. FromEqs. (3.27) the weightedaverages of the dataare(recall
that we fit
z
=
ln
y
)
y
i
x
i
10
3
737
.
5
×
x
=
y
i
=
=
7
.
474
98
.
67
×
10
3
y
i
z
i
y
i
10
3
528
.
2
×
z
=
=
=
5
.
353
98
.
67
×
10
3
and Eqs. (3.28) yield for the parameters
y
i
z
i
(
x
i
−
10
3
x
)
35
.
39
×
b
=
y
i
x
i
(
x
i
−
x
)
=
=
0
.
5440
65
.
05
×
10
3
ln
a
=
z
−
b x
=
5
.
353
−
0
.
5440(7
.
474)
=
1
.
287
Therefore,
e
ln
a
e
1
.
287
a
=
=
=
3
.
622
622
e
0
.
5440
x
.Asexpected,this result issomewhat
differentfrom thatobtainedinPart(1).
The computations of the residuals and standard deviationare as follows:
so that the fitting functionis
f
(
x
)
=
3
.
y
7
.
50
16
.
10
38
.
90
67
.
00
146
.
60
266
.
20
f
(
x
)
6
.
96
16
.
61
37
.
56
68
.
33
146
.
33
266
.
20
y
−
f
(
x
)
0
.
54
−
0
.
51
1
.
34
−
1
.
33
0
.
267
0
.
00
[
y
i
f
(
x
i
)]
2
S
=
−
=
4
.
186
S
6
σ
=
2
=
.
1
023
−
Observethat the residuals and standard deviationaresmaller than in Part(1), indi-
catingabetterfit, asexpected.
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