Chapter 5

Curve Fitting: Least Squares

Linear and Polynomial Regression
We come across with practical situations wherein, given a data in terms

of the values of two variables x, y, we need to find possible relationship between

the values of x and y.
In this section we discuss the method of finding a specific relation y = t

for the data to satisfy as accurately as possible and such an equation in called
the "best fitting equation" or the "curve of best fit". Tte following procedure
Gauss, called the least square approximation is described as below.

Let y = f{x) be an approximate relation that fits into a given data (Xi, Yi)s
i = 1, 2, n, y;'sare called values and Y = f(x;) are called
the observed
the expected values. Their difference Ri yY are= called the residuals or
estimated errors. The method of least squares provides a relationship y = f(3)

such that sum of the squares of the residuals is the least.

5.1 Fitting a straight line y = ax + b or Linear approximation

It is required to find y(x) = ax + b, such that y(x) fits a set of n pairs

of given values (x, y) as nearly as possible. The residual R = y - (ax + b)

is the difference between the observed and estimated values of y. The method
of least squares is to find the parameters a and b such that the sum of squares
of the residuals is minimum (least).
n

Let S R?
1

i.e. S [y - (ax + b)]* .(1)

TreatingS as a function of two variables a and b, we choose a and b

Such that S is minimum. The necessary conditions are

da =0. Hence
2 [y - ax + b)] (- x) = 0 and 2 2 {y - (ax + b)] (- 1) = 0.

Simplyfying, we have

xy + ax + bx =0
150
Numerical and
StatisticalMMethods

and
y +
ax + b =0

n
( 1+ l+ ... n times = n)
But b =b El =bn
above equations becomes
aExb2x = Exy
ax+nb =2y 2)
These equations are called the normal equations and on solving we e
the values of a and b so that y = ax + b is the equation of the best fitting
get
straight line.
Note 1: (Direct method) : If the numbers in the given data are large

we can change the origin and scale with the substitutions X = * , Y = X

for the sake of convenience and ease in calculation work.
Note 2: (Step deviation method): This method can be preferred when
the values of x are big in magnitude and are at regular intervals. To fit a curve
y fa) in the form of a straight line or parabola we choose Y = y - y (where

i s rounded off to the nearest integer for conveniense) and X = :

n is odd, where A is the middle value of x or X =

X A+B)
if n is
h/2
even, where A and B are two middle values ofx and h is the common differene
in the values of x. We fit the curve in the new variables, (X, X) applying the
least squares method and transform X, Y into x, y.

EXAMPLE 1 Fit the best straight line to the data :

-1
y f1 0
SOLUTION Let y =
ax + b be the desired line
Here n= 4, Zx = 2, Zx' =
6, Zy =
6, Exy =
8
The normal equations are
y = alx + nb
2a+ 4b =6
Exy ax + bLx 6a+ 2b 8
Solving we get a = b = 1, giving the required straight line y x +
Ciurve

line at
EXAMPLE 2 By the method of least squares, find the straight
best
fits the following data.
2 3 5
X
y 14 27 40 55 68
SOLUTION First Method Let the straight line of best fit be

y a x +b ), Heren = 5

The nornmal equations are

Lxy =
aXx + bEx
and y = azx + nb i)
The values of 2x, 2y, Lx and Exy are calculated as shown below

Y xy
14 14
2 27 4 54
40 120
55 16 220
68 25 340
2x= 15 y 204 Ex = 55 Zxy =748

Eqs. (i) become 748 5Sa + 15b

and 204 15a + 5b
Solving these, we have a 13.6 and b =0. =

of best fit
Putting these values in Eqn. (1), we get the line 13.6x.
=
as y
Second Method Let X x - 3, Y = y - 40 and let the straight line

in new variables bee

Y = AX + B ...(i)
Now the normal equations are
EXY ALX+ BEX
=

EY = ALX + nB .ii)
y X =x - 3 Y = y -40 XY
1 14 -2 -26 4 52
2 27 -1 -13 1 13
3 40 0 0 0
4 55 1 15 1 15
5 68 2 28 4 56
EX = 0 EY = 4
X =10 LXY= 136
 c a l Met

Substituting 2X
Methods
we get 0, EY =
4, EXY =136, LX* =
and n= 5 in Eon
10 and n =

qn. (Gi),
136 = 10A
and 4 5B
Solving we get A = 13.6 and B =
0.8
From Eqn. (i), Y =
13.6X + 0.8
i.e. y 40 13.6 (x - 3) + 0.8
Or y13.6x

EXAMPLE
for the data.
3 Fit a
straight line y = a + bx in the least
square sense

6 8 99 11 14
y1 4 5 _78 9
SOLUTION The values of x are not
X. y are small in equally spaced and the values
normal magnitude. We shall solve the
problem by direct method. The
of
equations for y =
a + bx are
given by
y na + bEx
and Exy alx
=
+
bEx, Here n= 8
X
xy x2
1
1
3 1
2
9
4
16 16
6
24
8 36
5 40
9 64
63
11
8 81
88
14 9 121
126
Ex = 56 y= 40 196
Exy =
364 x
The normal =524
equations becomes
8a + 56b =
40
56a +524b =
364
Solving we get a =
0.52 and b
The
=
0.64
equation y a* bx
becomes y =
0.52 + 0.64x.
 Least Squares Linear d n a
CurveP i Fitting

Fit straight line for the data

4 a
EXAMPLE 120
50 7 0 | 100

Ly 12 1521 25
and so we can
The values of x
equally spaced are not
big enough we
SOLUTION
S O L U T I O N

Since the values of x

are
the direct imethod. integer
problem by of . y to the nearest
valucs
y taking the
he
the

n05e X
Y= x
-

X, y -
choo
for c o n v e n i e n c e .

We have

18.25
= 340 y n
= =

18.
We choose y=
Thus we have X =x 85, Y =
y - 18
aX + b are
line of fit Y
The normal equations for the straight

EY aXX + nb
=

LXY = aEX + bEX, (n 4)

given below
The relevant table is
XY X
X y X =x 85 Y =y 18 1225
12 -35 -6 210
50
-3 45 225
70 15 -15
45 225
100 21 15
7 245 1225
120 | 25 35
XY = 545 EX= 2900
EX = 0 EY =1
The normal equations becomes
545
1 and a.2900+ b.0
=

a.0+4b =

b= 0,25 and = 0.19

Solving a

aX + b, becomes y
-

18 = 0.19 (x -

85) + 0.25 i.e.

The equation Y =

y 0.19x + 2.1
EXAMPLE 5 A simply supported beam carTies a concentrated load P

S mid-point. Corresponding to various values of P the maximum deflection

measured and is given below.

P 100 120 140 160 180 200

0.70 0.80 0.85
Y 0.450.55 0.60
Fin
ind a law of the form Y = a + bP and hence estimate Y when P = 150.
 ON We have 6 equally spaced values Or hd accordi
and accorangiy 140
d60 are two middle values with h = 20.
gly 140

Hence we
choose p=
P- (140 + 160) P- 150
20/2 10
Also we have Y = 2Y = 39 -
0.65 and for convenience
y Y
6 we choose
oose
Y =Y - 0.65.
The relevant table is
given below.
P Y P 1 50
10 y Y - 0.65 PY
100
p
0.45
-0.20 1
120 0.55 25
-3 -0.10 0.3
140 0.60 -1
-0.05 0.05
160 0.70 1
1
0.05
180 0.80 3
0.05
200
0.15 0.45
0.85 5 9
5.20 1 23
Ep 0 Ey 0.05
The normal
py =
2.85 p? =70
equations for y =
a +
bp are
Zy =
na +
bZp
Epy ap +bLp*, (n 6)
6a + b.0 =0.05
a
= 0.05
6
=
0.0083
a.0b.70 2.85 b b = 2.85
The 70 0.041
equation y a+ bp bcomes
Y -

0.65 =
0.0083 +
0.041P=10
i.e. Y =
0.65 +
0.0083 +
or Y 0.0041P +
0.0041 (P -

150)
0.0433
When P =
150, Y =
0.0041 x 150 +
0.0433 =
0.66
Curve Anear and
Polynomial Regression 155
EXAMPLE 6 IfP is the
pull required to lift a load W by means of pulley
lack, find a linear law of the form P
mW + = c connecting P and W using
following data J

P: 12 15 21 25
W:500
50L 70 100 120
Where P ánd W are taken in kg-wt. Compute P when W 150kg.
[G.T.U., Sem.V(IT), Dec. 2010]
SOLUTION Heren =4
The corresponding normal equations are

P =4c + m LW

EWP c ZW +m EW2 .(1)

The values of 2P, 2w2 and EWP are calculated as shown below :
W P w2 WP
50 12 2500 600
70 15 4900 1050
100 21 10000 2100
120 25 14400 3000
LW = 340 P = 73 w2 = 31800 LWP = 6750

The equations (1) becomes

73 4 c + 340m

6750 340c + 318OOm

.e 2c + 170m = 365
(2)
34c+3180m 675 ..(3)
Multiplying (2) by 17 and subtracting from (3), we get

m = 0.1879

f r o m (2), we get c = 2.2785

Hence P = 2.2785 + 0.1879W.

When W = 150 kg. P 30.4635 kg.

 Mr
Numerical
and. Statistical
ethods
156
EXERCISES 5.1

data (1 to 8).
following
to the
straight line
Fit ne
least square
1. x
2
y 0.5
3
2. 0 27
0
y 9 11
3. x 2 9
10
2
6
y 4
2 3
4. x 0 9.35 12.05
6.50
1.00 3.85
51.0 73.2 95.7
5. 20.5 32.7 1032
765 826 873 942
8 10 12
6. 2 4 6
9.20 10.19 11.01 12.05
7.32 8.24

-3 1 0 2 4
7.
y: 0.4 -0.1 -02-0.3 -0.3 0.1 04
8. x:-1 0

y: 1 5

Find the equation of the best fitting straight line for the following data
(9 to 14).

9. x 1 2
y:14 13
10. x : 0|
3 5
9 8 24 28 26 20
11. x L 0
4
y: 2
2
6
4
12. x 62 64 65 69 70
65.7 66.8 71 72
67.2 69.3
13. x 1 2
69.8 70.5 70.5 70.9
70.9
V : 80 90 6
92 83 94 99 92
14. Year (x):_
Production (y) 19111921 1931 1941 1951
(in thousand tons) 8 10 12 10 6
Curve F i n g

the following data

15.
Obtain the least squares straight line fit to

0.2 0.4 0.6 0.8

f(x) 0.447| 0.632| 0.775| 0.894 | Summer 2013]

[G.T.U., Sem.V(IT),
Solution & Hint: Refer : Solved Ex. 1 page 168.

5.2 Fitting a parabola y = ax* + by the least soquare approximation

bx + c
Summer 2013]
[G.T.U., Sem.V(IT),
the curve
Consider a set of n pairs of the given values (x, y) for fitting between
the difference
+bx + c. The residualRy (ax* + bx + c) is = -

y ax c such that
the
and estimated values of y. We have to find a, b,
the observed
sum of the squares of the residuals is minimum (least).
...(1)
Let S E y (ax +bx + c)1
conditiOns
S function of three variables a, b, c the necessary
Treating as a
dS d S d 0. Hence
for S to be minimum are
da ob dc
(- x) 0
2 Iy -

(ax bx + c)] =

2 (y -

(ax + bx +
c) ( x) =
0

bx c)] (- 1) = 0
2 y -

(ax + +

Simplifying we have

x'y + ax+ bx + cx = 0
n
n

xy +
E ax +
bx + cx = 0
n

ax+ bx + c 0
- y*2
but c c l= Cn [ l+1+... n times=n
1
The above equations becomes
aEx+bEx3 + cEx =
Zx*y
+ bEx + c2x =
Lxy
ax
ax+ bEx + nc = 2y
These equations are called normal equations and on solving we get the

Values of a, b, c so that
y ax2 + bx + c is the best fitting (parabola) curve of second degree.
 and Statistical Met
Numerical
hods
158
to the data
square
parabola
AAMPLE 1 Fit the least 2
0
X
2

bx +c be the desired parabola. Here n =4 4

SOLUTION Let y =
ax +
12, Exy =16
18, Iy
=
5, Lxy =

2x= 2, Zx2 =
6, Ex' = 8, Xx =

The normal equations are

4c+ 2b + 6a = 5
2c + 6b + 8a = 12

6c+ 8b + 8a = 0

0.55, b =
2.15, a = -0.25
Solving we get c =
Hence y = 0.55 + 2.15x - 0.25x

EXAMPLE 2 Fit a parabola y = a + bx + cx for the data.

X 2 3 4

1.8 1.3 2.5 2.3

SOLUTION We shall solve the problem by direct method. The normal
equations for y = a + bx + cx are

y na + bEx + cEx*
Exy = ax + bLx +cEx3
Exy = aEx + bEx3 + cEx
(n 5)
The relevant table is given below

xy x'y
0 0
0 0
1 1.8 1.8 1.8 1 1
1.3 2.6 5.2 4 8 16
3 2.5 7.5 22.5 9 27 81
4 2.3 9.2 36.8 16 64 256
Ex =10 Ey =8.9
21.1| Exy= Exy=6.3 x=30| Ex= 100
the normnal equations becomes x=354
5a +10b + 30c = 8.9
10a + 30b + 100c i)
=21.1
30a + 100b + 354c ...(ii)
66.3 =

) x 2 (ü): -10b - 40c

- .ii)
-3.3 =

ii) x 3 (iü): -10b 54c- .iv)

=-3
(iv) (v): 14c -0.3 =

c 0.021
Curve Pfting

0.84 = -3.3
Substituting the value of c in (iv), -10b +

i.e.-10b =
4.14 b =
0.414

Finally substituting the values of b

and c in (i)
8.9
Sa+10 (0.414) + 30 (0.021) =

Sa = 5.39 a = 1.078
i.e.
0.021x*
1.078 +0.414x
-

= bx + Cx2 is y
Hence the required parabola y
a +

in the least
3 Fit a second degree parabola y ax + bx +c
EXAMPLE = 6.
estimate y at x

Square sense
for the following data and hence
2 3 4
13 16 19
10 12
y

SOLUTION Let us choose X = x -

We have y == =14
and let Y = y - y = y-14

= aX+ bX + c are
The normal equations for Y
EY = aEX+ bLX + nC
aEX + bLX2 + CLX
EXY =

n 5
EXY = aLX4 + bLX3 + CEXx
below :
The relevant table in as

xyXX=x-3|Y=y-14
XY xY X x X

4 8 -16 4 -8 16
10 -2
2 -2 -1 1
2 12 -1 -2
0 0 0
3 13 0 1
2 2 1 1
4 16 1 2
5 10 20 4 8 16
5 19 2
EY EXY XxY X EX EX
LX
= 0 = 0 22 =4 = 10 =0 = 34

becomes
The normal equations
10a 5 c = 0 (i)
. b = 2.2 .(ii)
10b = 22

34a+ 10c =4 ..ii)

4 . a = 0.29
(i), 14a
=

ii) 2 x
 nods

0 c = -0.58
Substituting in ), 10 (0.29) 5c +

The parabola of fit Y =ax + bX + c becomes

y - 14 = 0.29 (x - 3)+ 2.2 (x -

3)- 0.588
y 14 + 0.29 (x2 6 x + 9) + 2.2 (x - 3) - 0.58

or y 0.29x2 + 0.46x + 9.43

Yat x = 6 = 0.29(6) + 0.46(6) + 9.43 = 22.63

EXERCISES 5.2

Fit parabola of second degree

a
y ax + bx + c in the least square sense
for the data (1 to 4).

i. X 0
14
2 3 5 6
18 23 29 36 40 46
22. x : 1 0 20 30 40 50 60
157 179 210 252 302 361
3. 0 2
3
Ly:1 10
22 38
x: 2 3 4
Ly:25 28 33 39 46
5. x:2 4 6 8 10
y:3.07 12.85 31,47 57.38 91.29
Fit the least
square parabola to the
6.
following data (6 to 10).
x 2
Ly:2 1
-2
7.
2
y: -1 0 5
8.x T 2
y 3.07 8 10
12.85 31.47 57.3891.29
9.
x| 2 -1 0
0.17 0.53 0.57 0.58 0.33
10. x : | 0.78
2.34 3.12
|y:2.50 1.20 1.12 2.25 3.81
4.25
Curve .

hfain (i)
Obtain (i) the least square straight line and (ii) the least square parabola
following data
to the
2 3 4
11.
y 1.7 1.8 2.3 3.2
12. x 2
0
y:
Fit the least square straight line of the form y = ox to the data
13.
3 4
2.13 2.86 3.60 4.29 4.95

Bx' thefollowing data.

14. Fit the parabola of the form y = to

x 2 4 6 8 10
y : 0.973 3.839| 8.641 15.987 23.794
in
15. Apply the. method of least squares to find the constants
y ay + a]x + a2x to fit the data
2 3 4. 6

y:3.13 3.766.94 12.62 20.86 31.53

to fit the data:
16. Find the least square straight line and parabola
0 2 3
X 16 18 21 25
11 14
and (ii) parabola which fit the data :
17. Find the best (i) straight line;
x: 0 2 3 4
-1.02-1.94| -3.04
y:0.99 0.03
ax + bx- to the following data
18, Fit a second degree curve of the form y
[G.T.U., Sem.V(IT), Summer 2013]
by the method of least squares
2 3 4 5
19.8
1.8 5.1 8.9 14.1
169.
Hint: Refer solved Ex. 2 p.
Regression
5.
SNon-polynomial Approximation or Non-linear
and engineering often we have to fit an
In certain problems of science
function to data. The most used empirical
npirical law, a non-polynomial
laws are
y ae (Exponential curve)
() y =
ax (Geometric curve)
(3) y = ab
 by taking arithms
taking logarith-
by
162 equations
equations

linear
linear

reduced
to

These formulae can be

variables.
new
and introducing
= aex
(1) y both
sides
on
logarithm
Taking equation
becomes
mes
+ the bx above

log y log a
=

a =, A,
Y and log =

Denoting log y line.

is a straight
which
Y =A + bx
= ax" yields
(2) Similarly y obtain
log = a + b log x x
=
X, we
log y a = A and log
log y
=
Y, log
Denoting line.
which is a straight
Y A + bX,
(3) y = ax" yields

log y log a + x log b obtain

log b = B,
we
A, and
Y, log
=
a
Denoting log y
=

line.
Y A + Bx, which is a straight
data:
1 Determine a and b so that y
=
ae% fits the
EXAMPLE

X 2 3 4
y 7 11 17 27
SOLUTION y = ae yields on taking logarithm,
log y log a + bx

i.e.Y A +bx, where Y

= =
log y, A =
log a
which is a straight line
Hence we have the table:
2 3 4
Y
=loge Y 1.95 2.40 2.833.30
The second line in this table is obtained
entries in the original table
by taking the logarithm of tne

Now n =4, Ex =
10, Lx* =
30, ZY =10.48, ExY =28.44 yielding uhe
normal equations
4A + 10b =
10.48
10A+ 30b = 28.44
Solving, we get A =
1.5, b =
0.45
Hence A =log a =
1.5 a =
e =
4.48, b 0.45
 Fitting.Least Squares
Curve Linear and Polynomial Regression 163
EXAMPLE 2
EXAMPLE Fit
curve of the form
y
a
= ab for the data and hence
find the estimation for y when x = 8.
2 3 4 5 6 7
y 87 97 113 129 202 195 193
[G.T.U., Sem.VIT), Jan. 2012]
sOLUTION Taking logarithm on both sides of y =ab*i
log y= log a + x log b
Denoting Y =
log y, A =
log a, B = log b, we have
Y =A + Bx, which is a
straight line.
The normal equations are
EY = nA + BEx

ExY =
ALx + BXx', n = 7

The relevant table is given below :

X Y loge y XY x
87 4.46 4.47 1
97 4.57 9.14 4
113 4.73 14.19 9
129 4.86 19.44 16
5 202 5.31 26.55 25
6 195 5.27 31.62 36
7 193 5.26 36.82 49

2x = 28 Y = 34.47 ExY 142.23 x = 140

The normal equations become
7A + 28B = 34.47 .(i)

28A 140B = 142.23 .(11)

4.35 B =
0.1554
()-4 x ) » 28B =

ubstituting in (1), 7A + 28 (0.1554) = 34.47 . A =4.3027

e.02= 73.9
But A =
loge a i.e. a = e^ =

i.e. b = e" = e14 =

1.1681
B log. b
The = ab¥ becomes y
= 73.9 (1.1681)*
curve y

73.9 (1.1681)°
= 256.14
, when x = 8, y =
164
pressure P and the
the
EXAMPLE 3 At constant
temperature,
the
constant.
volume
Find the best e
are connected by the relation PV
=

hen PP== 4
V when
fitting
V of a gas data and
estimate

form to the following

cquation of this 1.0 1.5 2.0 2.5 3.0
0.5
Pkg. sq. cm) 460
1620 1000 750 620520
V(c.c.)
where C is a constnat
SOLUTION Consider PVY =
C,
sides.
Taking logarithm on both
C
log P + Y log V log
=

log.P log.C - y log.V

i.e.
=b, log.V x
=

Denoting log.P =
y, log.C a,
-

Y
We have y =a + bx which is a straight line
The normal equations are

2y = na + bEx
Exy = aEx + bEx, n = 6.

The relevant table is given below :

P V x = logeV y log.P xy x
0.5 1620 7.39 -0.69 -5.0991 54.6121

1.0 1000 6.91 0 47.7481

1.5 750 6.62 0.40 2.648 43.8244

| 2.0 620 6.43 0.69 4.4367 41.3449

2.5 570 6.25 0.92 5.75 39.0625

3.0 460 6.13 1.1 6.743 37.5765

2 x = 39.13| 2y 2.42 Exy = 14.4786 Ex* = 264.1o

The normal equations becomes

6a + 39.7b = 2.42
.)
39.73a+ 264.1689b = 14.478b
.i)
Solving () and (11) we obtain a = 9.8061 and b = -1.42
But log.C a. C
e = e.3061 =
=
18144
- Y b
Y-b 1.42
The curve PV C becomes
=

PVl42= 18144
 Fitting
Fiiing
: LLeast Squares Linear and Polynomial Regression
Curve

18144 Vl42=8144
P = 4, 4V
=
or
When 4
yl42 4536 V 375.94288 376.
i.e.
When P= 4, V =
376.
EXAMPLE 4 Fit a curve of the type y = C eAx for the five data points
(4, 7.5)
(0, 1.5), (1.2.5), (2, 3,.5), (3, 5.0)
and
Nov./Dec. 2011]
[G.T.U., Sem.V(IT),
.(1)
SOLUTION y C eAx
Taking logarithm on both sides
.(2)
logy logC + Ax
=

Then introduce the change of variables

..(3)
y logy, X x and B logC =

linear relation between the new variables X and Y :

This results in a

Y AX + B
.(4)
the original points and obtain
Apply the transformation (3) to
In(3.5), (3, In(5, 0)) (4, n(7.5))}
X, Y)}= {(0, In(1.5), (1, n(2.5), (2,
.(5)
1.25276), (3, 1.60944), (4, 2.01490)}
{(0, 0.40547), (1, 0.91629), (2,
Calculation of the coefficients for
the normal equations for (4) is shown in
the following table.
Y in(y X X,Yk
Yk X
0.405465 0.0 0.000000
0.0 1.5 0.0
0.916291 1.0 0.916291
1.0 2.5 1.0
1.252763 4.0 2.505526
2.0 3.5 2.0
9.0 4.828314
3.0 5.0 3.0 1.609438
2.014903 16.0 8.059612
4.0
4.07.5 X EX,Y
CX LY
= 10.0
= 6.198860 =
30.0 =
16.309743
and B is
for determining A
he resulting linear system
30A+ 1OB 16.309742
=

10A+ 5B
= 6.198860 .(6)
0.3912023 andB =0457367.
ne solution is A
=

Then from (3) C =e = 1.579910

C exp (0.457367)
166
in (1), we get
of A and C
Substituting values data
linearization method)
1.579910 03912023x
(fit by
y =
fit.
1S required exponential EXERCISES 5.3

data and hence estimate

te yy when
whe
ab* for the
1. Fit a curve of the form y =
x = 8.
3 4 5 6
x 0 2
132 190 275
Ly:32 47 65 92 L
form y = ax" for the data:
2. Fit a curve of the
6
|x: 2 3
5.21 6.1 6.8 7.5
y:2.98 4.26
of the best fitting curve in the form y = ae" for the
3. Find the equation
data
10
5 6 7 8 9

y:133 55 23 7 2 2
Determine the constants in y = ae fitting the data

1 2 3

y:60 30 20 15
5. Determine the constants in y =
ax fitting the
:2.2 2.7 4.1 3.5
65 605350
6. Find a and b so that y =
ae"* fits the data
x 0.2 0.3 04 0.5 0.6 0.7 0.8
y : | 3.16 2.38 1.75 1.34 1.00 0.74 0.56
7. Find the constants in acx
y =
fitting the data
x 77 100 185 239 285
y:2.4 3.4 7.0 11.1 19.6
8. Fit a curve of the type y =
ab" to the
following data
X
50 450
780 1200 4400 4800 5300
5300
y 28 30 32 36 51 58 69

Hint: Proceed as
[G.T.U., Sem.V(IT), Summer 2014)
per solved Ex. 2 page
174.]
 mial Kegre
ression 167

ANSWERS
EXERCISES 5.1

2. y =-4.2 8.8x
3. y = 0.9431x - 0.0244
4. y = 1.03 + 2.76x
5. y = 702 + 3.39x
6. y =
6.37 + 0.47x
7. y0
8. y =
0.4 -

1.8x
9. y-3.2x + 18.2
10. y =
3.23x + 11.096
11. y = 0.5x + 1.336
12. y =
0.52x + 33.46
13. y =2x + 82 14. y = 0.16x - 302.56

EXERCISES 5.2

1. y =
0.083x + 4.96x + 13.46 2. y =
0.046x- 0.84x + 143.67

3. y =
2.23x + 0.18x + 146 4. 0.64x
y = + 1.46x + 2.78

5. y =
0.9925x2 0.86x 0.7 6. 1.15
-
+ y -

1.05x 0.25x
7. y =
1.3x + 0.5x2 8. y 0.696 0.855x
= -

+ 0.992x
9. y 0.66314 +0.037x 0.10357x
10. y =
5.045 4.043x + 1.009x
. 2 0.5x
(i) y = 1 + 0.5x (ii) y = +
0.2x
22x 29
12. i) y = 22x-29 Gi) y =

26 26

13. y =
0.713x
 Statistical
Numerical
and
Method
168

14. y = 0.24125x2

15. y = 4.98 3.13x + 1.26x*

= 11.499 1.755x +
0.181x
l. y 10.857 + 2.657x, y

1.00299x (ü) y
= 0.98856- 0.96013x - 0.010712
17. (i) y = 1.00999 -

EXERCISES 5.3

1. y = 32.15 (1.43), 562 2. y 2.98x031

3. y = 12882 e-0.92x 4. a = 84.8, b = -0.456

5. a =
91.8, b =
-0.434 6. a =
5.63, b =
-2.89
7. a =
1.2, b =0.0096