2.3.10Curve Fitting

of the points (X

, Y ).It can be shown that:

i

(X i X

) (Y i Y

)

(2.17)

b

i

(X i X

)2

a

bX

(2.18)

Y

The line thus calculated is known as the line of regression of

Y on X , that is, the line indicating how

Y varies when

X is set to chosen values.

If X

is the dependent variable, the definition is modified by considering horizontal instead of

vertical deviations. In general these two definitions lead to different least square curves.

Example

13

The following data were recorded for the potential

E

of an

electrode, measured

against the saturated calomel electrode, as a function of concentration

C

(moles liter 1).

log C

E

, mV

log

C

E ,

mV

1.00

106

2.10

174

1.10

115

2.20

182

1.20

121

2.40

187

1.50

139

2.70

211

1.70

153

2.90

220

1.90

158

3.00

226

Fit the best straight line to these data;

X

i represents

log

C

, and Y i represents

E . We will perform

the calculation manually, using the following tabular lay-out.

X i

Y i

(X i X

)

(X i X )2

(Y i Y )

(X i X )(Y i Y )

1.00

106

0.975

0.951

60

58.5

1.10

115

0.875

0.766

51

44.6

1.20

121

0.775

0.600

45

34.9

1.50

139

0.475

0.226

27

12.8

1.70

153

0.275

0.076

13

3.6

1.90

158

0.075

0.006

8

0.6

2.10

174

0.125

0.016

8

1.0

2.20

182

0.225

0.051

16

3.6

2.40

187

0.425

0.181

21

8.9

2.70

211

0.725

0.526

45

32.6

2.90

220

0.925

0.856

54

50.0

3.00

226

1.025

1.051

60

61.5

X

i 23.7

Y i 1992

0

5.306

0

312.6

1.975

X

Y 166

GENERAL INFORMATION, CONVERSION TABLES, AND MATHEMATICS

2.135

and from Equation 18, and substituting the “centroid” values of the points

, the intercept(X , Y )

is:

a

166 58.91(1.975)

49.64

The best-fit equation is therefore:

E 49.64 58.91 log

C

2.3.10.2 Errors

in the Slope and

Intercept of the

Best-fit

Line.

Upon examination of the plot

of pairs of data points, the calibration line, it will be obvious that the precision involved in analyzing

an unknown sample will be considerably poorer than that indicated by replicate error alone. The

scatter of these original points about the calibration line is a good measure of the error to be expected

in analyzing an unknown sample. And this same error is considerably larger than the replication

error because it will include other sources of variability due to a variety of causes. One possible

source of variability might be the presence of different amounts of an extraneous material in the

various samples used to establish the calibration curve. While this variability causes scatter about

the calibration curve, it will not be reflected in the replication error of any one sample if the sample

is homogeneous.

The scatter of the points around the calibration line or random errors are of importance since the

best-fit line will be used to estimate the concentration of test samples by interpolation. The method

used to calculate the random errors in the values for the slope and intercept is now considered. We

must first calculate the standard deviation

s Y/X , which is given by:

i

(Y i Yˆ)2

s Y/X

q

N

2

(2.19)

Equation 19 utilizes the

Y-residuals

, Y i

Yˆ, where

Yˆi are the points on the calculated best-fit line

or the fitted

Y i values. The appropriate number of degrees of freedom is

N 2; the minus 2 arises

from the fact that linear calibration lines are derived from both a slope and an intercept which leads

to a loss of two degrees of freedom.

Now we can calculate the standard deviations for the slope and

the

intercept. These are

given by:

s b

s Y/X

(2.20)

q

i

(X i X

)2

s a sY/X q

i

X i2

(2.21)

N

i

(X i

)2

X

2.136

SECTION

2

The confidence limits for the slope are given by

b tb , where the

t-value is taken at the desired

confidence level and (

N 2) degrees of freedom. Similarly, the confidence limits for the intercept

are given by

a

ts a

. The closeness of

xˆ to

x i is answered in terms of a confidence interval for

x 0

that extends from an upper confidence (UCL) to a lower confidence (LCL) level. Let us choose 95%

for the confidence interval. Then, remembering that this is a two-tailed test (UCL and LCL), we

obtain from a table of Student’s

t

distribution the critical value of

tc (t0.975 ) and the appropriate

number of degrees of freedom.

Example 14

For the best-fit line found in Example 13, express the result in terms of confidence

intervals for the slope and intercept. We will choose 95% for the confidence interval.

The standard

deviation

s Y/X is given

by Equation 19,

but first a supplementary table must be

constructed for the

Y

residuals and other data which will be needed in subsequent equations.

Yˆ

(Y i Yˆ)

(Y i Yˆ)2

X i2

108.6

2.55

6.50

1.00

114.4

0.56

0.31

1.21

120.3

0.67

0.45

1.44

138.0

1.00

1.00

2.25

149.8

3.21

10.32

2.89

161.6

3.57

12.94

3.61

173.4

0.65

0.42

4.41

179.2

2.76

7.61

4.84

191.0

4.02

16.16

5.76

208.7

2.30

5.30

7.29

220.5

0.48

0.23

8.41

226.4

0.40

0.16

9.00

61.20

52.11

Now substitute the appropriate values into Equation 19 where there are 12

2 10 degrees of

freedom:

s X/Y

q

61.20

2.47

10

We can now calculate

s b

and

s a

from Equations 20 and 21, respectively:

s b

s Y/X

1.07

p5.31

and

s a

2.47 q

52.11

2.23

12(5.306)

Now, using a two-tailed value for Student’s

t:

b

ts b

58.91 2.23(1.07)

58.91 2.39

a

ts a

49.64 2.23(2.23)

49.64 4.97