The Chemistry Maths Book, Second Edition

21.10 Least squares 621

Division by then gives the pair of normal equations

(21.53)

F1− 1 mE1− 1 c 1 = 10

and these have solution

(21.54)

The resulting line passes through the centroid(E, F)of the data points.

If they

i

values all have the same precision σthen the estimated parameters mand

chave standard deviations given by

(21.55)

and the linear least-squares fit with estimated errors isy 1 = 1 mx 1 + 1 cwith

(21.56)

EXAMPLE 21.15Find the linear least-squares fit for the following data points:

Table 21.7

x 0369121518

y 3.3 2.5 2.3 1.7 1.4 0.5 0.2

We have N 1 = 17 , E1= 19 , F1= 1 1.7, 1 = 1117 ,

1

= 1 64.5 271 = 1 9.21429, 1 −1E

2

1 = 136 and

1 −1EF1= 1 −42.6 271 = 1 −6.08571. Then

so that

c=.+

.×

17 ±=.±.

42 6 9

252

117

252

σσ3 2214 0 6814

m=

−.

±=−.±.

42 6

252

252

0 1691 0 0630

σ

σ

σσ σ σ σ σ

mc

==.,==.252 0 0630

117

252

0 6814

xy

x

2

x xy

2

m

xy x y

xx

cymx

mc

=

−

−

±, =− ±

22

σσ

σ

σ

σ

σ

mc

Nx x

x

Nx x

2

2

22

2

22

22

=

−

( )

,=

−

( )

m

xy x y

xx

= cymx

−

−

,=−

22

xy mx cx−−=

2

0

N

i

N

=

=

∑

1

1