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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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