Math & Stat Part 2 Commerce

Obtain the two regression equations and X 1 234 5 6
estimate: Y 2 476 5 6

(i) The productivity index of a worker 12. The following sample gives the number
whose test score is 95. of hours of study (X) per day for an
examination and marks (Y) obtained by 12
(ii) The test score when productivity students.
index is 75.
X33 3 4 455 5 6 67 8
6. Compute the appropriate regression Y 45 60 55 60 75 70 80 75 90 80 75 85
equation for the following data:

X 6 8 11 Obtain the line of regression of marks on
[Independent 2 4 5 8 75 hours of study.

Veriable] 3.3 Properties of Regression Coefficients

Y The line of regression of Y on X is given by
[Dependent 18 12 10 Y = a + bYXX and the line of regression of
Veriable] X on Y is given by X = a' + bXYY.

7. The following are the marks obtained cov( X ,Y )
by the students in Economics (X) and Here, bYX var( X ) , the slope of the line
Mathematics (Y)
of regression of Y on X is called the
X 59 60 61 62 63 regression coefficient of Y on X. Simillary,
Y 78 82 82 79 81
cov( X ,Y )
Find the regression equation of Y on X. bXY = var(Y ) , the slope of the line of

8. For the following bivariate data obtain the regression of X on Y is called the regression
equations of two regression lines: coefficient of X on Y. These two regression
coefficients have the following property.
X1 2 345
Y 5 7 9 11 13 (a) bXY . bYX = r2

9. From the following data obtain the equation where r is the correlation coefficient
of two regression lines:
between X and Y,
X 6 2 10 4 8
Y 9 11 5 8 7 bXY is the regression coefficient of X on Y.
and bYX is the regression coefficient of Y
10. For the following data, find the regression on X.
line of Y on X
Proof: Note that
X1 2 3
Y2 16 bXY . bYX = cov( X ,Y ) . cov( X ,Y )
var(Y ) var( X )
Hence find the most likely value of y when
x = 4.  cov( X ,Y )  2

11. From the following data, find the regression = σ X .σY 
equation of Y on X and estimate Y when  
X = 10.
= r2.

Can it be said that the correlation coefficient
is the square root of the product of the two
regression coefficients?

42

(b) If bYX > 1, then bXY < 1. bYX + bXY = r σY +r σX
σX σY
Proof: Let, if possible, bXY > 1 and bYX > 1.
=  σσ YX +σX 
Then, using the above result, bXY . bYX > 1, σY 
implies that r2 > 1, which is impossible. 
(Can you provide the reason?)

This shows that our assumption must = r σY +σ X  (1)
be invalid. That is, both the regression  
coefficients cannot simultaneously exceed  σ X .σY 
unity.
But (σX − σY)2 > 0 and therefore

We already know that the two variances σ2X − σ2Y − 2σXσY ≥ 0

σ 2 ,σ 2 , and the correlation coefficient r σ2X + σ2Y ≥ 2σXσY
x y

satisfy the relation.

cov (X, Y) = r . σX . σY σ 2 + σ 2 ≥ 2
∴ r = y x
cov( X ,Y )
σ X .σY σ X .σY

∴ r σ 2 +σ 2 ≥ 2r. (2)
y x

The regression coefficients can also be σ X .σY
written as follows.
From (1) and (2), we have
cov( X .Y )
bYX = bYX + bXY ≥ 2r.
σ 2
X ∴ bYX + bXY ≥ r.
2
= r.σ X .σY
this result shows that the arithmetic mean
σ 2 of the two regression coefficients, namely
x bYX and bXY is greater than or equal to r.
This result, however, holds only when bYX,
= r . σY bXY and r are positive. (Can you find the
and σX reason?)

bXY = cov( X .Y )

σ 2 Consider the case where bYX = −0.8 and
y bXY = −0.45. In this case, we have r = −0.6.
(Can you find the reason?)
= r.σσX .σ Y

2
y

= r . σ X Note that bYX + bXY = −1.25, and 2r = −1.2.
σY This shows that bYX + bXY ≤ 2r.

(c) bYX + bXY ≥ |r| It may be interesting to note that
2
bYX = cov( X .Y )

Proof: We have already seen that σ 2
X

bYX = r σY and bXY = r σX , bXY = cov( X .Y )
σX σY
σ 2
y

where σX and σY are the standard deviations cov( X ,Y )
r = σ X .σ Y
of X and Y, respectively.

Therefore,

43

It is evident from the above three equations SOLVED EXAMPLES
that all the coefficients have the same
numerator and this numerator determines Ex. 1 : The table below gives the heights
their sign.As the result, all these coefficients of fathers (X) and heights of their sons (Y)
have the same sign. In other words, if r > 0, respectively.
then bYX > 0, and bXY > 0. Similary. if r < 0.
then bYX < 0, and bXY < 0. Finally. if r = 0, Heights
then bYX = bXY = 0.
of fathers 64 62 66 63 67 61 69 65 67 66
(inches)

(d) bYX and bXY are not affected by change of (X)
origin, but are affected by change of scale.
This property is known as invariance Heights of
property.
sons (inches) 67 65 67 64 68 65 67 64 70 66

(Y)

The invariance property states that bYX and (i) Find the regression line of Y on X.
bXY are invariant under change of origin,
but are not invariant under change of scale. (ii) Find the regression line of X on Y.

Proof: Let U = X −a and V = Y −b , (iii) Predict son's height if father's
h k height is 68 inches.

where a, b, h and k are constants with the (iv) Predict father's height if son's
height is 59 inches.
condition that h, k ≠ 0
Solution :
We have already proved that σ 2 = h2σ 2 , Let us use the change of origin for
X U
2 2σ 2 computations of regression coefficients.
σ Y = k V , and cov(X, Y) = hkcov(U, V).
Let ui = xi − 65 and vi = yi − 67
Therefore, cov( X .Y ) xi yi ui vi ui2 vi2 ui vi
64 67 −1 0 1 0 0
bYX = σ 2 62 65 −3 −2 9 4 6
X 66 67 1 0 1 0 0
63 64 −2 −3 4 9 6
= hk cov(U ,V ) 67 68 2 1 4 1 2
61 65 −4 −2 16 4 8
h 2σ 2 69 67 4 0 16 0 0
U 65 64 0 −3 0 9 0
67 70 2 3 4 9 6
k cov(U ,V ) 66 66 1 −1 1 1 −1
= h Total 0 −7 56 37 27
that is, σ 2
U

bYX = k bVU
h

Similatly,

bXY = k bUV
h

These two results show that regression Here, n = 10, •∑ ui = 0, •∑ vi = −7,
coefficients are invariant under change of • • • ∑ ui2 = 56, ∑ vi2 = 37, and ∑ uivi = 27
origin, but are not invariant under change
of scale. ∑ ui

u = n = 0 = 0,
10

44

∑ ( )σu2 =ui2 − 2 iii) Estimate of sons height Y for X = 68
n Y = 0.48 × 68 + 35.1
u = 67.74 inches
iv) Estimate of fathers height X for Y = 59
56 X = 0.84 × 59 + 9.31
= 10 − 02 = 58.87 inches
v) Correlation coefficient
σu2 = 5.6

∑ vi ( )∑σv2 =vi2 2
v= n , n −
v

−7 37 r = bYX .bXY
= 10 = 10 − (−0.7)2

= −0.7 = 3.21 = 0.84× 0.48

∑ cov (u, v) = uivi − u v = 0.635
We choose positive square root ! (why?)
n

27 Ex. 2: Compute regression coefficient from the
= 10 − 0 × − 0.7 following data on the variable weight (X) and
height (Y) of 8 individuals :
= 2.7
n = 8 , ∑( xi − 45) = 48
Now, you know that, regression coefficients
are independent of change of origin.

cov(u, v) 2.7 ∑ ( xi − 45)2 = 4400,
∴ bXY = bUV = σV = 3.21 = 0.84

cov(u, v) 2.7 ∑( yi −150) = 280,
and bYX = bVU = σU = 5.6 = 0.48 ∑ ( yi −150)2 = 167432,

You are also aware that mean is affected by ∑( xi − 45∑) . ( yi −150) = 21680
change of origin.

∴ x = u + 65 = 0 + 65 = 65 Solution: Let ui = xi − 45 and vi = yi − 150

and y = v + 67 = −0.7 + 67 = 66.3 • • So ∑ ui = 48, ∑ ui2 = 4400,

(i) Line of regression of Y on X is • •∑ vi = 280, ∑ vi2 = 167432,

(Y − y ) = bYX (X − x ) • ∑ uivi = 21680
∴ (Y − 66.3) = 0.48 (X − 65)

∑ 48
∴ Y = 0.48. X + 35.1 ∴ u = ui =8 = 6
n
ii) Regression line of X on Y is
∑v = vi 280
(X − x ) = bXY (Y − y ) =8 = 35
∴ (X − 65) = 0.84 (Y − 66.3) n

∴ X = 0.84. Y + 9.31 ∑ ( )σu2 = ui2 − 2
n
u

45

4400 where, bYX = cov( X ,Y )
= 8 − (6)2 = 514
σ 2
x

∑ ( )σv2 = vi2− 2 ∑( xi − x)( yi − y)
n
v n

= σ 2
x
167432
= 8 − (35)2 = 19704  1170 
=  10 
∑ cov (u, v) = uivi − u v
130
n
= 0.9
21680 and a = y − bYX x
= 8 − (6) (35) = 142 − (0.9) × (53)

= 2500

From the properties of regression = 94.3
coefficients, you know they are independent
of change of origin. Therefore, regression equation of Y on X is

cov(u, v) 2500 Y = 94.3 + 0.9 X

∴ bYX = bVU = σU = 514 = 4.86 Now, the estimate of blood pressure of
women with age 47 years is

cov(u, v) 2500 Y = 94.3 + 0.9 × 47
and bXY = bUV = σV = 19704 = 0.12
= 136.6

(Have you noticed bYX > 1 and bXY < 1?) Ex. 4: Given the following data, obtain the linear
regression & estimate of X for Y = 10
Ex. 3: The following results were obtained from
records of age (X) and systolic blood pressure x = 7.6, y = 14.8, σX = 3.5, σY = 28 and
(Y) of a group of 10 women. r = 0.8

Mean X Y Solution:
Variance 53 142 Here, we need to obtain line of regression
130 165
of X on Y which can be expressed as
( )( )∑ xi − x yi − y = 1170
X = a' − bYX Y

Find the appropriate regression equation where bXY = cov( X ,Y )
and use it to estimate the blood pressure of a
woman with age 47 years. σ 2
Y

Solution: = r σX
σY
Here, we need to find line of regression of
Y on X, which is given as (3.5)
= 0.8 (28)
Y = a + bYX X
= 0.1

46

and a' = x − bYX y Obtain
= 7.6 − (0.1) (14.8)
i) The line of regression of Y on X.

= 6.12 ii) The line of regression of X on Y

∴ Line of regression of X on Y is iii) The correlation coefficient between
X and Y.
X = 6.12 + 0.1 Y

Estimate of X for Y= 10 is 4. You are given the following information
about advertising expenditure and sales.
X = 6.12 + 0.1 × 10

X = 7.12

EXERCISE 3.2 Arithmetic Advertisment Sales
mean expenditure (Rs.in lakh)
1. For a bivariate data. (Rs.in lakh)
Standard (Y)
x = 53, y = 28, bYX = −1.2, bXY = −0.3 deviation (X) 90
Find
10 12
i) Correlation coefficient between X
and Y. 3

ii) Estimate of Y for X = 50 Correlation coefficient between X and Y is
0.8
iii) Estimate of X for Y = 25
(i) Obtain the two regression equations.
2. From the data of 20 pairs of observation on
X and Y, following results are obtained. (ii) What is the likely sales when the
advertising budget is Rs 15 lakh?
x = 199, y = 94,
(iii) What should be the advertising
2 2 budget if the company wants
to attain sales target of Rs.120 lakh?
xi − x = 1200 yi − y = 300
( ) ( )∑ ∑ 5. Bring out inconsistency if any, in the
following :
( )( )∑ xi − x yi − y = −250
(i) bYX + bXY = 1.30 and r = 0.75
Find (ii) bYX = bXY = 1.50 and r = −0.9
i) The line of regression of Y on X. (iii) bYX = 1.9 and bXY = − 0.25
ii) The line of regression of X on Y.
iii) Correlation coefficient between X 1
(iv) bYX = 2.6 and bXY = 2.6
and Y
3. From the data of 7 pairs of observations on 6. Two samples from bivariate populations
have 15 observations each. The sample
X and Y, following results are obtained. means of X and Y are 25 and 18 respectively.
The corresponding sum of squares of
∑( xi − 70) = −35 ∑( yi − 60) = −7 deviations from respective means are
∑ ( xi − 70)2 = 2989 ∑ ( yi − 60)2 = 476 136 and 150. The sum of product of
deviations from respective means is 123.
∑( xi − 70)( yi − 60) = 1064 Obtain the equation of line of regression
of X on Y.
[ Given 0.7884 = 0.8879]

47

7. For a certain bivariate data ii) What should be the advertisement
expenditure if the firm proposes a
Mean X Y sales target Rs.60 crores.
S.D. 25 20
4 3 12. For a certain bivariate data the following
information are available.
And r = 0.5. estimate y when x = 10 and
estimate x when y = 16 A.M. X Y
S.D. 13 17
8. Given the following information about the 3 2
production and demand of a commodity
obtain the two regression lines : Correlation coefficient between x and y is
0.6. estimate x when y = 15 and estimate y when
x = 10.

Production Demand SOLVED EXAMPLES
(X) (Y)

Mean 85 90

S.D. 5 6 Ex.1: The equations of the two lines of regression
are 3x + 2y − 26 = 0 and 6x + y − 31 = 0
Coefficient of correlation between X and Y
is 0.6. Also estimate the production when (i) Find the means of X and Y.
demand is 100.
(ii) Obtain correlation coefficient
9. Given the following data, obtain linear between X and Y.
regression estimate of X for Y = 10
(iii) Estimate Y for X = 2.
x = 7.6, y = 14.8, σX = 3.2, σY = 16 and
r = 0.7 Solution:

10. An inquiry of 50 families to study the (i) We know that the co-ordinates of the
relationship between expenditure on
accommodation (Rs. x) and expenditure point of intersection of the two lines are x
on food and entertainment (Rs. y) gave the and y , the means of X and Y.
following results. :
The regression equations are
•∑ x = 8500,•∑ y = 9600, σX = 60, σY = 20,
3x + 2y − 26 = 0
r = 0.6
and 6x + y − 31 = 0
Estimate the expenditure on food and
entertainment when expenditure on Solving these equations simultaneously,
accommodation is Rs 200. we get

11. The following data about the sales and 6x + 4y − 52 = 0
advertisement expenditure of a firms is 6x + y − 31 = 0
given below ( in Rs. Crores) (−) (−) (+)

Mean Sales Adv. Exp. 3y − 21 = 0
S.D. 40 6 ∴ 3y = 21
10 1.5 i.e. y = 7
and x = 4
i) Estimate the likely sales for a proposed
advertisement expenditure of Rs.10
crores.

48

Hence the means of X and Y are x = 4 Find
and y = 7
(i) Correlation coefficient between
(ii) Now, to find correlation coefficient, we X and Y.
have to find the regression coefficients
bYX and bXY (ii) σY2 if σX2 = 4
Solution:
For this, we have to choose one of the lines
as that of line of regression of Y on X and Here, the regression lines are specified.
other the line of regression of X on Y 41

Let 3x + 2y − 26 = 0 be the line of regression So bYX = 3 and bXY = 3
on Y on X this gives
(i) ∴ r2 = bYX . bXY
3
Y = − X + 13 41
= 3 . 3
2
4
The coefficient of X in this equation is = 9
3
22
bYX = − 2 ∴ r = + (why + only?)

Then the other equation is that of line of 33
regression of X on Y which can be written
as (ii) You know that
1 31
bYX = r . σY
X =− 6 Y + 6 σX
1
4 2 . σY
Here, the regression coefficient bXY = − 6 . ∴ 3 = 2
Now, you know that
3
r2 = bXY . bYX
∴ σY = 4
= 0.25 ∴ σY2 = 16

∴ r = ± 0.5 EXERCISE 3.3

The correlation coefficient has the sign as 1. From the two regression equations find r,
that of bYX and bXY
x and y .
∴ r = − 0.5
4y = 9x + 15 and 25x = 4y + 17
Note: We choose arbitrarily the lines as that 2. In a partially destroyed laboratory record
of regression of Y on X or X on Y. if the
product bYX . bXY is less than unity, our of an analysis of regression data, the
choice is correct. Fortunately, there are following data are legible:
only two choices. Variance of X = 9
Regression equations:
Ex 2.: The regression equation of Y on X is 8x − 10y + 66 = 0
and 40x − 18y = 214.
4
y = 3 x and the regression equation X on Y is

x= y + 5
3 3

49

Find on the basis of above information (i) Mean values of X and Y

(i) The mean values of X and Y. (ii) Standard deviation of Y

(ii) Correlation coefficient between X (iii) Coefficient of correlation between X
and Y. and Y.

(iii) Standard deviation of Y. 9. If the two regression lines for a bivariate
data are 2x = y + 15 (x on y) and
3. For 50 students of a class, the regression
equation of marks in statistics (X) on the 4y = 3x + 25 (y on x), find
marks in Accountancy (Y) is 3y − 5x + 180
= 0. The mean marks in accountancy is (i) x , (ii) y , (iii) bYX ,
44 and the variance of marks in statistics is
(iv) bXY , (v) r Given 0.375 = 0.61

 9 th of the variance of marks in 10. The two regression equations are
 16  5x − 6y + 90 = 0 and 15x − 8y − 130 = 0.

accountancy. Find the mean marks in Find x , y , r.
statistics and the correlation coefficient
between marks in two subjects. 11. Two lines of regression are 10x + 3y − 62 =
0 and 6x + 5y − 50 = 0 Identify the
4. For a bivariate data, the regression
coefficient of Y on X is 0.4 and the regression of x on y. Hence find x , y and
regression coefficient of X on Y is 0.9. Find r.
the value of variance of Y if variance of X
is 9. 12. For certain X and Y series, which are
correlated the two lines of regression are
5. The equations of two regression lines are 10y = 3x + 170 and 5x + 70 = 6y. Find the
correlation coefficient between them. Find
2x + 3y − 6 = 0 the mean values of X and Y.

and 2x + 2y − 12 = 0

Find (i) Correlation coefficient 13. Regression equations of two series are
2x − y − 15 = 0 and 3x − 4y + 25 = 0
(ii) σX Find x , y and regression coefficients.
σY
Also find coefficients of correlation.
6. For a bivariate data: x = 53, y = 28,
bYX = −1.5 and bXY = − 0.2. Estimate Y Given 0.375 = 0.61

when X = 50. 14. The two regression lines between height
(X) in inches and weight (Y) in kgs of girls
7. The equations of two regression lines are are,
x − 4y = 5 and 16y − x = 64. Find means of
X and Y. Also, find correlation coefficient 4y − 15x + 500 = 0
between X and Y.
and 20x − 3y − 900 = 0
8. In a partially destroyed record, the
following data are available variance of X Find mean height and weight of the group.
= 25. Regression equation of Y on X is 5y Also, estimate weight of a girl whose
− x = 22 and Regression equation of X on height is 70 inches.
Y is 64x − 45y = 22 Find

50

Let's Remember ∑ xi yi − nx.y
( )∑=
2

yi2 − n y

l Line of regression of Y on X is

Y = a + bX and a1 = x − b' y

where b = bYX = regression coefficient of l Line of regression of Y on X is also given
Y on X as
(Y − y ) = b(X − x )
cov( X ,Y )
bYX = var( X ) l Line of regression of X on Y is also given
as
∑( xi − x)( yi − y) (X − x ) = b' (Y − y )

n l r2 = bYX . bXY = b . b'
2 l If bYX > 1 then bXY < 1
( )∑=
xi − x l byx + bxy ≥ | r |
2
n

( )∑∑=xi yi − x.y
n
xi2 −
n 2 l Regression coefficients are independent of
change of origin but not of scale.
x
l Lines of regression have a point of
∑ xi yi − nx.y intersection( x , y )
( )∑=
2

xi2 − n x

and a = y − b x MISCELLANEOUS EXERCISE - 3

l Line of regression of X on Y is I) Choose the correct alternative.
1. Regression analysis is the theory of
X = a' + b'y a) Estimation b) Prediction
c) Both a and b d) Calculation
where b' = bXY = regression coefficient of
Y on X 2. We can estimate the value of one variable
with the help of other known variable only
cov( X ,Y ) if they are
bXY = var(Y )
a) Correlated
∑( xi − x)( yi − y) b) Positively correlated
c) Negatively correlated
n d) Uncorrelated
2
( )∑= 3. There are __________ types of regression
yi − y equations.

n a) 4 b) 2 c) 3 d) 1

∑ xi yi − x.y
n
( )∑=
yi2 − 2
n
y

51

4. In the regression equation of Y on X 12. bxy and byx are __________
a) X is independent and Y is dependent.
b) Y is independent and X is dependent. a) Independent of change of origin and
c) Both X and Y are independent. scale
d) Both X and Y are dependent.
b) Independent of change of origin but
5. In the regression equation of X on Y not of scale
a) X is independent and Y is dependent.
b) Y is independent and X is dependent. c) Independent of change of scale but not
c) Both X and Y are independent. of origin
d) Both X and Y are dependent.
d. Affected by change of origin and scale
6. bXY is __________
a) Regression coefficient of Y on X x−a y−b
b) Regression coefficient of X on Y 13. If u = c and v = d then byx =_____
c) Correlation coefficient between X and
a) d bvu b) c bvu
Y c d
d) Covariance between X and Y
c) a bvu d) b bvu
7. bYX is __________ b a
a) Regression coefficient of Y on X
b) Regression coefficient of X on Y x−a y−b
c) Correlation coefficient between X and 14. If u = c and v = d then bxy =_____

Y a) d buv b) c buv
d) Covariance between X and Y c d

c) a buv d) b buv
b a

15. Corr (x,x) = __________
a) 0 (b) 1 (c) −1 (d) can't be found

8. ‘r’ is __________ 16. Corr (x,y) = __________
a) Regression coefficient of Y on X
b) Regression coefficient of X on Y a) corr(x,x) (b) corr(y,y)
c) Correlation coefficient between X and c) corr(y,x) (d) cov(y,x)

Y 17. Corr  x−a , y−b = − corr(x,y) if,
d) Covariance between X and Y  c d 

9. bXY .bYX __________ a) c and d are opposite in sign
a) v(x) (b) σx (c) r2 (d) (σy)2
b) c and d are same in sign
10. If byx > 1 then bxy is __________
a) >1 (b) < 1 (c) > 0 (d) < 0 c) a and b are opposite in sign

11. | bxy+ byx| ≥ __________ d) a and b are same in sign
a) |r| (b) 2|r| (c) r (d) 2r
18. Regression equation of X on Y is ____

a) y − y = byx (x − x )
b) x − x = bxy (y − y )
c) y − y = bxy (x − x )
d) x − x = byx (y − y )

52

19. Regression equation of Y on X is ______ 3. Regression equation of X on Y is
a) y − y = byx (x − x ) _________
b) x − x = bxy (y − y )
c) y − y = bxy (x − x ) 4. There are __________ types of
regression equations.
d) x − x = byx (y − y )
5. Corr (x, −x) = __________

6. x−a y−b
If u=_=____c and v = then
bxy d
20. byx = __________

σ x σ y x−a y−b
a) r σ y b) r σ x 7. If u=_=____c and v = then
byx d

c) 1σy d) 1 σx 8. | bxy+ byx| ≥ __________
r σx r σy

21. bxy = __________ 9. If byx > 1 then bxy is __________

a) r σ x b) r σ y 10. bxy .byx = __________
σ y σ x

c) 1σy d) 1 σx III) State whether each of the following is
r σx r σy True or False.

22. Cov (x,y) =__________ 1. Corr (x,x) = 1

a) ∑( x − x)( y − y) 2. Regression equation of X on Y is
y − y = byx (x − x )
b) ∑( x − x)( y − y)
n 3. Regression equation of Y on X is
y − y = byx (x − x )
c) ∑ xy − xy
4. Corr (x,y) = Corr (y,x)
n

d) b and c both 5. bxy and byx are independent of change
of origin and scale.
23. If bxy < 0 and byx < 0 then 'r' is __________
a) > 0 (b) < 0 (c) >1 (d) not found 6. ‘r’ is regression coefficient of Y on X

24. If equations of regression lines are 3x + 2y 7. byx is correlation coefficient between
− 26 = 0 and 6x + y − 31 = 0 then means of X and Y
x and y are __________
8. If u = x − a and v = y − b then bxy = buv
a) (7,4) b) (4,7) c) (2,9) d) (−4,7)
9. If u = x − a and v = y − b then rxy = ruv
II) Fill in the blanks :
1. If bxy < 0 and byx < 0 then ‘r’ is 10. In the regression equation of Y on X,
__________ byx represents slope of the line.

2. Regression equation of Y on X
is_________

53

IV) Solve the following problems. 8. Find the line of regression of X on Y for the
following data:
1. The data obtained on X, the length of time
in weeks that a promotional project has 2 2
been in progress at a small business, and
Y, the percentage increase in weekly sales xi − x = 36, yi − y = 44
over the period just prior to the beginning ( ) ( )∑ ∑ n = 8,
of the campaign.
∑( xi − x)( yi − y) = 24

X1 2 3 4 1 3 1 2 3 4 2 4 9. Find the equation of line of regression of Y
Y 10 10 18 20 11 15 12 15 17 19 13 16 on X for the following data:

Find the equation of regression line to ( )( )∑ n = 8, xi − x yi − y = 120,
predict the percentage increase in sales
if the campaign has been in progress for x = 20, y = 36, σx = 2, σy = 3.
1.5 weeks.
10. The following results were obtained from
2. The regression equation of y on x is given records of age (X) and systolic blood
by 3x + 2y − 26 = 0 Find byx. pressure (Y) of a group of 10 men.

3. If for a bivariate data x = 10, y =12, Mean X Y
v(x) = 9, σy = 4 and r = 0.6. Estimate Variance 50 140
y when x = 5. 150 165

4. The equation of the line of regression of y ( )( )∑ and xi − x yi − y = 1120

2 y7 Find the prediction of blood pressure of a
on x is y = x and x on y is x = + 6. man of age 40 years.
9 2

Find

(i) r (ii) σy2 if σx2 = 4. 11. The equations of two regression lines are
10x − 4y = 80 and 10y − 9x = −40
5. Identify the regression equations of x on y
and y on x from the following equations, Find:

2x + 3y = 6 and 5x + 7y − 12 = 0 (i) x and y

6. (i) If for a bivariate data byx = −1.2 and (ii) bYX and bXY
bxy = −0.3 then find r. (iii) If var (Y) = 36, obtain var (X)

(ii) From the two regression equations

y = 4x − 5 and 3x = 2y + 5, find x (iv) r

and y . 12. If bYX = −0.6 and bXY = −0.216 then find
correlation coefficient between X and Y.
7. The equations of the two lines of regression Comment on it.
are 3x + 2y − 26 = 0 and 6x + y − 31 = 0
Find Activities

(i) Means of X and Y

(ii) Correlation coefficient between X 1) Consider a group of 70 students of your
and Y class to take their heights in cm (x) and
weights kg (y). Hence find both the
(iii) Estimate of Y for X = 2 regression equations.

(iv) var (X) if var (Y) = 36

54

2) The age in years of 7 young couples is ∴ r = ± 4
given below: ×
10

Husband 21 25 26 24 22 30 20 r =
(x)

Wife (y) 19 20 24 20 22 24 18 If V(y) = 36 then σy = =

i) Find the equation of regression line of ∴ bxy = r σ y
age of husband on age of wife. σ x

ii) Draw the regression line of y on x ∴ = ×
σx
iii) Predict the age of wife whose husband’s
age is 27 years. ∴ σx = ( )

3) The equations of two regression lines are

10x − 4y = 80 ......... (1) ∴ V(x) = σx2 =

10y − 9x = −40 ......... (2) ( )∑ 2

∴ ( x , y ) is the point of intersection of 4) Given n = 8, xi − x = 36,
both the regression lines.
2
∴ Solve equations (i) and (ii), we get
yi − y = 40,
( )∑

x = and y = ( )( )∑ xi − x yi − y = 24

( ( )( ) )∑ ∑ ∴ byx =
Now, consider 10x − 4y = 80 xi − x yi − y

∴ a =, b = 2

xi − x

∴ slope(m1) = − a = = =
b

Consider, 10y − 9x = −40 ∑( xi − x)( yi − y)
( )∑ ∴ bxy =
∴ a = , b = 2

yi − y

∴ slope(m2) = − a = = =
b

∴ | m1| > | m2| ∴ Regression equation of Y on X :
y − y = byx (x − x )
∴ byx = 1
and bxy =

∴ 10x − 4y = 80 is the regression equations y − = (x − )

of on and

∴ 10y − 9x = −40 is the regression equations ∴ Regression equation of X on Y :
x − x = bxy (y − y )
of on .

Now, r = ± byx.bxy x − = (y − )

55

5) Consider, given bxy = cov(x, y)

( )( )∑ n = 8 xi − x yi − y = 120,

y = 36, σx = 2, σy = 3 = 9 =
∴ Regression equation of Y on X :
∴ cov(x, y) = ∑(x− x)(y− y)

n

y − = byx (x − )
y −
= = = (x − )

150 ∴ Regression equation of Y on X :

∴ byx = σ 4 = x − x = (y − y )
x

x − = (y − )

FFF

56

4 Time Series

Let's Study to analyze time series data in order to extract
meaningful statistics and understand important
l Uses of time series analysis. characteristics of the observed data.

l Components of a time series. Time Series Analysis helps us understand
the underlying forces leading to a particular
l Secular Trend pattern in the time series and helps us in
monitoring and forecasting data with help of
l Seasonal Variation appropriate statistical models.

l Cyclical Variation Analysis of time series data requires
maintaining records of values of the variable
l Irregular Variation over time.

l Mathematical Models Some examples from day-to-day life may
give a better idea of time series.
l Additive Model
1. Monthly, quarterly, or yearly production of
l Multiplicative Model an industrial product.

l Measurement of Secular Trend 2. Yearly GDP (Gross Domestic Product) of a
country.
l Graphical Method
3. Monthly sales in a departmental store.
l Method of Moving Averages
4. Weekly prices of vegetables.
l Method of Least Squares
5. Daily closing price of a share at a stock
Introduction exchange.
A manufacturing company wants to predict
6. Hourly temperature of a city recorded by
demand for its product for next year to make a the Meteorological Department.
production plan. An investor wants to know
fluctuations in share prices so that he can decide 4.1 Uses of Time Series Analysis
if he should purchase or sell certain shares. These
and many other situations involve a variable that The main objective of time series
changes with time. A variable observed over a analysis is to understand, interpret and
period of time is called a time series. Analysis assess chronological changes in values
of time series is useful in understanding the of a variable in the past, so that reliable
patterns of changes in the variable over time. Let predictions can be made about its future
us now define a time series. values. For example, the government may
be interested in predicting population
Definition growth in near future for planning its
Time Series is a sequence of observations welfare schemes, the agricultural ministry
may be interested in predicting annual crop
made on a variable at regular time intervals over yield before declaring the MSP (minimum
a specified period of time. support price) of agricultural produce or an
industrialist may be interested in predicting
Data collected arbitrarily or irregularly
does not form a time series. Time series
analysis involves the use of statistical methods

57

the weekly demand for his product for 4. It is useful in making a comparative study.
making the production schedule. Following
are considered to be some of the important A comparative study of data relating to
uses of time series ananlysis. two or more periods, regions, or industries
reveals a lot of valuable information that
1. It is useful for studying the past behaviour can guide management in taking a proper
of a variable. course of action.Atime series itself provides
a scientific basis for making comparisons
In a time series, the past observations between two or more related sets of data.
on a variable are arranged in an orderly Note that data are arranged chronologically
manner over a period of time. By simple in such a series, and the effects of its
observation of such a series, one can various components are gradually isolated,
understand the nature of changes that have analyzed, and interpreted.
taken place in values of the variable during
the course of time. Further, by applying 4.2 Components of Time Series
appropriate technique of analysis to the
series, one can study the general tendency A graphical representation of time series
of the variable in addition to seasonal data shows continuous changes in its values
changes, cyclical changes, and irregular over time, giving an impression of fluctuating
or accidental changes in values of the nature of data. A close look of the graph,
variable. however, reveals that the fluctuations are not
totally arbitrary, and a part of these fluctuations
2. It is useful for forecasting future behaviour has a steady behavior and can be related to
of a variable. time. This part is the systematic part of the time
series and the remaining part is non systematic
Analysis of a time series reveals the nature or irregular. The systematic part is further
of changes in the value of a variable during divided in the following broad categories:
the course of times. This can be useful in (i) secular trend (T), (ii) seasonal variation (S), and
forecasting the future values of the variable. (iii) cyclical variation (C). The non systematic
Thus, with the help of observations on an part is also called (iv) irregular variation (I). Every
appropriate time series, future plans can time series has some or all of these components.
be made relating to certain matters like Of course, only the systematic components of
purchase, production, sales, etc. This is a time series are useful in forecasting its future
how a planned economy makes plans for values.
the future development on the basis of time
series analysis of the relevant data. We now discuss the four components of a
time series in detail.
3. It is useful in evaluating the performance.
4.2.1. Secular Trend (T)
Evaluation of the actual performances in
comparison with predetermined targets is The secular trend is the long term pattern of
necessary to judge efficiency of the work. a time series. The secular trend can be positive
For example, the achievements of Five- or negative depending on whether the time
Year Plans are evaluated by determining series exhibits an increasing long term pattern
the annual rate of growth in the gross or a decreasing long term pattern. The secular
national product. Similarly, the national trend shows a smooth and regular long term
policy of controlling inflation and price movement of the time series. The secular trend
rises is evaluated with the help of different does not include short term fluctuations, but
price indices. All these are made possible only consists of a steady movement over a long
by analysis of time series of the relevant period of time. It is the movement that the series
variables.

58

would take if there are no seasonal, cyclical or 2. The following table shows production
irregular variations. It is the effect of factors that (in '000 tonnes) of a commodity during
are more or less constant for a long time or that years 2001-2008.
change very gradually and slowly over time.
Year 2001 2002 2003 2004
If a time series does not show an increasing Production 50.0 36.5 43.0 44.5
or decreasing pattern, then the series is stationary 2005 2006 2007 2008
around the mean. Year 38.9 38.1 32.6 33.7
Production

SOLVE EXAMPLES

1. The following table shows annual sales (in
lakh Rs.) of a departmental store for years
2011 to 2018.

Year 2011 2012 2013 2014
Sales 26.2 28.9 33.7 32.1
Year 2015 2016 2017 2018
Sales 39.8 38.7 45.4 42.6

The following graph shows the above time Fig. 4.2
series.
The above graph shows a downward
Fig. 4.1 trend.

There are ups and downs in the graph, but 4.2.2 Seasonal Variation (S)
the time series shows an upward trend in long
run. Many time series related to financial,
economic, and business activities consist of
monthly or quarterly data. It is observed very
often that these time series exhibit seasonal
variation in the sense that similar patterns are
repeated from year to year. Seasonal variation
is the component of a time series that involves
patterns of change within a year that repeat from
year to year.

Several commodities show seasonal
fluctuations in their demand. Warm clothes
and woolen products have a market during the
winter season. Fans, coolers, cold drinks and
ice creams are in great demand during summer.
Umbrellas and raincoats are in great demand
during the rainy season. Different festivals are
associated with different commodities and every
festival season is associated with an increase in
demand for related commodities. For example,
clothes and firecrackers are in great demand

59

during Diwali. Most of the seasonal variations term forecasting. Such short term forecasts are
in demand reflect changes in climatic conditions useful for a departmental store in planning its
or customs and habits of people. inventory according to months of a year. A bank
manager can use such short term forecasts in
All the above examples have one year as the managing cash flow on different days of a week
period of seasonal variation. However, the period or a month.
of seasonal variation can be a month, a week, a
day, or even an hour, depending on the nature of 3. The following table shows quarterly sales
available data. For example, cash withdrawals in (in lakh Rs.) of woolen garments in four
a bank show seasonal variation among the days consecutive years.
of a month, the number of books borrowed by
readers from a library show seasonal variation Year I II
according to days of a week, passenger traffic at
a railway station has seasonal variation during Quarter 1 2 3 4 1 2 3 4
hours of a day, and the temperature recorded
in a city exhibits seasonal variation over hours Sales 11 8 16 28 19 17 32 38
of a day, in addition to seasonal variation with
changing seasons in a year. Year III IV

Seasonal variation is measured with help Quarter 1 2 3 4 1 2 3 4
of seasonal indices, which are useful for short
Sales 33 23 39 52 41 37 44 58

Fig. 4.3

Figure 4.3 shows a pattern that is repeated 4.2.3 Cyclical Variation (C)
year after year. The values are lowest (in the
year) in second quarter and highest (in the year) Cyclical variation is a long term oscillatory
in fourth quarter of every year. Although the movement in values of a time series. Cyclical
overall graph of the time series in Figure 4.3 variation occurs over a long period, usually
shows an increasing trend, the seasonal variation several years, if seasonal variation occurs within
within every year is very clearly visible in the a year. One complete round of oscillation is
graph. called a cycle. Cyclical variations need not be
periodic in the sense that the length of a cycle
or the magnitude of variation within a cycle can
change from one cycle to another.

60

Cyclical variations are observed in almost 4.3 Mathematical Models of Time Series
all time series related to economic or business
activities, where a cycle is known as a business Let Xt denote the value of the variable at
cycle or trade cycle. Recurring ups and downs time t. The time series is denoted by the collection
in a business are the main causes of cyclical of values, {Xt, t = 0, 1, ... , T} where T is the total
variation. duration of observation. There are two standard
mathematical models for time series based on
A typical business cycle consists of the four components mentioned earlier, namely,
the following four phases: (i) prosperity, secular trend (T), seasonal variation (S), cyclical
(ii) recession, (iii) depression, (iv) recovery. variation (C), and irregular variation (I).
Figure 4.4 depicts these four phases of a business
cycle, where every phase changes to the next 4.3.1 Additive Model
phase gradually in the order mentioned above.
The additive model assumes that the value
Fig. 4.4 Xt at time t is the sum of the four components at
Cyclical variations can consist of a period time t. Thus,
of 5 years, 10 years, or even longer duration.
The period often changes from one cycle to Xt = Tt + St + Ct + It
another. Cyclical variation may be attributed to
internal organizational factors such as purchase The additive model assumes that the four
and inventory policies or external factors such
as financial market conditions and government components of the time series are independent
policies.
of one another. It is also important to remember
4.2.4 Irregular Variation (I)
Irregular variations are unexpected that all the four components in the additive

variations in time series caused by unforeseen model must be measured in the same unit of
events that can include natural disasters like
floods or famines, political events line strikes or measurement. The magnitude of the seasonal
agitations, or international events like wars or
others conflicts. As the name suggests, irregular variation does not depend on the value of the
variations do not follow any patterns and are,
therefore, totally unpredictable. For this reason, time series in the additive model. In other words,
irregular variation are also known as unexplained
or unaccounted variations. the magnitude of the seasonal variation does no

change as the series goes up or down.

The assumption of independence of the
components is often not realistic. In such
situations, the multiplicative model can be used.

4.3.2 Multiplicative Model

The multiplicative model that the value Xt
at the time t is obtained by multiplication of the
four components at time t. That is,

Xt = Tt × St × Ct × It

The multiplicative model does not assume
independence of the four components of the
series and is, therefore, more realistic. Values of
the trend are expressed in units of measurements
and other components are expressed as
percentage or relative values, and hence are free
from units of measurements.

61

It is recommended to choose the Solution:
multiplicative model when the magnitude of
the seasonal variation in the data depends on
the magnitude of the data. In other words, the
magnitude of the seasonal variation increases
as the data values increase, and decreases as the
data values decrease.

4.4 Measurement of Secular Trend Fig. 4.5

4.4.1 Method of Freehand Curve ( Graphical 2. The publisher of a magazine wants to
Method) determine the rate of increase in the
number of subscribers. The following table
In this method, a graph is drawn for the shows the subscription information for
given time series by plotting Xt (on Y-axis) eight consecutive years.
against t (on X-axis). Then a free hand smooth
curve is plotted on the same graph to indicate the Year 1976 1977 1978 1979
general trend. 17
No. of subscribers 12 11 19 1983
This method is simple and does not (in millions) 23
require any mathematical calculation. But, in
this method, different researchers may draw Year 1980 1981 1982
different trend lines for the same set of data.
Forecasting using this method is therefore risky No. of subscribers
if the person drawing free hand curve is not (in millions)
efficient and experienced. On the other hand,
this method is quite flexible and can be used for
all types of trends, linear as well as non – linear,
and involves minimum amount of work.

SOLVED EXAMPLES 19 18 20

1. Fit a trend line to the following data using Fit a trend line by the graphical method.
the graphical method. Solution:

Year Number Year Number of
of crimes crimes
1981 1987 ('000)
1982 ('000) 1988
1983 1989 43
1984 40 1990 46
1985 42 1991 47
1986 43 45
42 46
44
44

Fig. 4.6
62

4.4.2 Method of moving Averages averages does not provide a mathematical
equation for the time series and hence cannot
The moving average of period k of a time be used for the purpose of forecasting. Another
series forms a time series of arithmetic means drawback of the method of moving averages is
of k successive observations from the original that some of the trend values at each end of the
time series. The method begins with the first given series cannot be estimated by this method.
k observations and finds the arithmetic mean
of these k observations. The next step leaves SOLVED EXAMPLES
the first observation and includes observation
number k + 1 and finds the arithmetic mean of 3. The following table shows gross capital
these k observations. This process continues till formation (in crore Rs) for years 1966 to
the average of the last k observations is found. 1975.
In other words, the method of moving averages
finds the following. Year 1966 1967 1968 1969 1970
19.3 20.9 17.8 16.1 17.6
First moving average = X1 + X 2 + ... + X k Gross 1971 1972 1973 1974 1975
k Capital 17.8 18.3 17.3 21.4 19.3
Formation
Second moving average = X 2 + X3 + ... + X k+1
k Year

Third moving average = X3 + X 4 + ... + X k+2 Gross
k Capital
Formation
and so on.
(i) Obtain trend values using 5-yearly
Each of these averages is written against moving averages.
the time point that is the middle term in the sum.
As a result, when k is an odd integer, moving (ii) Plot the original time series and
average values correspond to observed values of trend values obtained in (i) on the
the given time series. On the other hand, when same graph.
k is an even integer, the moving averages fall
mid-way between two observed values of the Solution :
given time series. In this case, a subsequent
two-unit moving average is calculated to make Table 4.3
the resulting moving average values correspond
to observed values of the given time series. 5-yearly

A moving average with an appropriate Year Xt 5-yearly moving
period smooths out cyclical variations from
the given time series and provides a good 1966 moving total averages
estimate of the trend. Cyclical fluctuations with 1967
a uniform period and a uniform amplitude can 1968 (trend value)
be completely eliminated by taking the period 1969
of moving averages that is equal to or a multiple 1970 19.3 - -
of the period of the cycles as long as the trend is 1971
linear. 1972 20.9 - -
1973
The method of moving averages is flexible 1974 17.8 91.7 18.34
in the sense that even if a few observations are 1975
added to the given series, the moving averages 16.1 90.2 18.04
calculated earlier are not affected and remain
unchanged. However, the method of moving 17.6 87.6 17.52

17.8 87.1 17.42

18.3 92.4 18.48

17.3 94.1 18.82

21.4 - -

19.3 - -

Note that 5-yearly average is not available
for the first 2 years and last 2 years.

63

The following graph shows the original Note: Entries in the third and fourth
time series and the trend values obtained in columns are between tabulated time periods,
Table 4.3. while entries in fifth and sixth columns are in
front of tabulated time periods.
Fig. 4.7
4.4.3 Method of Least Squares
4. Obtain 4-yearly centered moving averages
for the following time series. This is the most objective and perhaps the
best method of determining trend in a given
Year 1987 1988 1989 1990 1991 time series. The method begins with selection
3.6 4.3 4.3 3.4 4.4 of an appropriate form of trend equation and
Annual sales 1992 1993 1994 1995 then proceeds with estimation of the unknown
(in lakh Rs.) 5.2 3.8 4.9 5.4 constants in this equation. It is a common
practice to choose a polynomial of a suitable
Year degree and then to determine its unknown (but
constant) coefficients by the method of least
Annual sales squares. The choice of the degree of polynomial
(in lakh Rs.) is often based on the graphical representation of
the given data.
Solution:
In a linear trend, the equation is given by
4-yearly
Xt = a + bt
Year Xi 4-yearly 4-yearly 2-unit centred
moving moving moving moving The method of least squares involves
average solving the following set of linear equations,
total average total (trend commonly known as normal equations.

value) •∑ Xt = na + b•∑ t

1987 3.6 •∑ tXt = a•∑ t + b•∑ t2

1988 4.3 15.6 3.9 8 4 Where n is the number of the time periods
1989 4.3 16.4 4.1 8.425 4.2125 for which data is available, whereas,
1990 3.4 17.3 4.325 8.525 4.2625
1991 4.4 16.8 4.2 8.775 4.3875 •∑ Xt ,•∑ tXt ,•∑ t and •∑ t2 are obtained
1992 5.2 18.3 4.575 9.4
1993 3.8 19.3 4.825 4.7 from the data. The least squares estimates
1994 4.9 of a and b are obtained by solving the two
equations in the two unknowns, namely a and
1995 5.4 b. The required equation of the trend line is
the obtained by substituting these estimates in
equation Xt = a + bt

SOLVED EXAMPLES

5. Fit a trend line by the method of least
squares to the time series in Example 1
(Section 4.4.1). Also, obtain the trend value
for the number of crimes in the year 1993.

64

Solution: The normal equations are

Let the trend line be represented by the 482 = 11a' + b' (0)

equation yt = a + bt Calculations can be made 61 = a'(0) 110 b'
simpler by transforming from t to u using the
482
formula ∴ a' = = 43.8182

u = t − middle t value ......... if n is odd 11
h
61
= t − mean of two middle t values b' = 110 = 0.5545
The equation of the trend line is then given
h by

......... if n is even yt = 43.8182 + (0.5545) u,
where u = t − 1986
Where h is the difference between
successive t values. Trend values computed using this result are
given in the last column of the table.
In the given problem, n = 11 (odd), middle
t value is 1986, and h = 1. 6. Fit a trend line by the method of least
squares to the time series in Example 2
∴ we use the transformation (Section 4.4.1). Also obtain the trend value
u = t −1986 = t − 1986. for the number of subscribers in the year
1984.
1
Solution :
The equation of the trend line then becomes Let the equation of the trend line, if the n

yt = a' + b'u is even
The two normal equations are then given
yt = a + bt
by In the given problem, n = 8 (even), two

•∑ yt = na′ + b′ •∑ u , middle t – values are 1979 and 1980, and h = 1.

•∑ uyt = a′ •∑ u + b′ •∑ u2 We use the transformation

We obtain •∑ u , •∑ u2 , •∑ yt and•∑ uyt t −1979.5
u = 1
from the following table.
2
Year yt u u2 uyt Trend
t Value = 2t − 3959.

1981 40 −5 25 −200 41.0457 The equation of the trend line then becomes

1982 42 −4 16 −168 41.6002 yt = a' + b'u

1983 43 −3 9 −129 42.1547 The two normal equations are

1984 42 −2 4 −84 42.7092

1985 44 −1 1 −44 43.2637

1986 44 0 0 0 43.8182 •∑ yt = na′ + b′ •∑ u , (1)

1987 43 1 1 43 44.3727

1988 46 2 4 92 44.9272 •∑ uyt = a′•∑ u + b′ •∑ u2 (2)

1989 47 3 9 141 45.4817 We obtain •∑ u , •∑ u2 , •∑ yt and•∑ uyt

1990 45 4 16 180 46.0362 from the following table.

1991 46 5 25 230 46.5907

Total 482 0 110 61

65

Table 4.2 Fit a trend line by graphical method to the
above data.
Year yt u u2 uyt Trend
t Value 2. Use the method of least squares to fit a
12 trend line to the data in Problem 1 above.
1976 11 −7 49 −84 12.3336 Also, obtain the trend value for the year
1977 19 1975.
1978 17 −5 25 −55 13.774
1979 19
1980 18 −3 9 −57 15.2144
1981 20
1982 23 −1 1 −17 16.6548 3. Obtain the trend line for the above data
1983 using 5 yearly moving averages.
139 1 1 19 18.0952
Total
3 9 54 19.5356 4. The following table shows the index of
5 25 100 20.976 industrial production for the period from
1976 to 1985, using the year 1976 as the
7 49 161 22.4164 base year.

0 168 121

From (1) and (2), Year 1976 1977 1978 1979 1980
Index 0 23 3 2
139 = 8a' + b' (0) (3) Year
Index 1981 1982 1983 1984 1985
121 = a' (0) + b' (168) (4) 4 56 7 10

From (3), a' = 17.375.

From (4), b' = 0.7202. Fit a trend line to the above data by
graphical method.
The equation of the trend line is then given
by 5. Fit a trend line to the data in Problem 4
above by the method of least squares. Also,
yt = 17.375 + (0.7202)u, obtain the trend value for the index of
industrial production for the year 1987.
where u = 2t − 3959.
6. Obtain the trend values for the data in
Trend values computed using this result are problem 4 using 4-yearly centered moving
given in the last column of Table 4.2. averages.

Estimation of the trend value for the year 7. The following table gives the production of
1984 is obtained as shown below. steel (in millions of tonnes) for years 1976
to 1986.
For t = 1984, u = 2(1984) – 3959 = 9.

And hence Year 1976 1977 1978 1979 1980 1981
yt = 17.375 + (0.7202) (9) Production 0 4 4 2 6 8

= 23.8568 (in millions) Year 1982 1983 1984 1985 1986
Production 5 9 4 10 10

EXERCISE 4.1 Fit a trend line to the above data by the
graphical method.
1. The following data gives the production of
bleaching powder (in '000 tonnes) for the 8. Fit a trend line to the data in Problem 7 by
years 1962 to 1972. the method of least squares. Also, obtain
the trend value for the year 1990.
Year 1962 1963 1964 1965 1966 1972
Production 0 0 1 1 4 8 9. Obtain the trend values for the above data
using 3-yearly moving averages.
Year 1967 1968 1969 1970 1971
Production 2 4 9 7 10

66

10. The following table shows the production l A standard business cycle consists of the
of gasoline in U.S.A. for the years 1962 to following four phases:
1976.
(i) Prosperity (ii) Decline
Year Production Year Production
(million (million (iii) Depression (iv) Recovery
1962 Barrels) 1970 Barrels)
1963 1971 l Irregular variations are caused by either
1964 10 1972 6 random factors or unforeseen events like
1965 0 1973 7 floods, famines, earthquakes, strikes, wars,
1966 1 1974 8 etc.
1967 1 1975 9
1968 2 1976 8 l Additive Model:
1969 3 9
4 10 Xt = Tt + St + Ct + It
5 where Xt is the value of the variable X at

(i) Obtain trend values for the above time t and Tt , St , Ct and It are secular trend,
data using 5-yearly moving averages. seasonal variation, cyclical variation, and
irregular variation at time t, respectively.
(ii) Plot the original time series and trend
values obtained above on the same l Multiplicative Model:
graph. Xt = Tt × St × Ct × It
where the notation is same as in the
Let's Remember
additive model.
l A time series is a set of observations made
at regular intervals of time and therefore l Secular trend can be measured using
arranged chronologically (that is, according
to time). (i) Graphical method
(ii) Method of moving averages
l A time series has the following four (iii) Method of least squares
components:
l The moving averages of period k of a time
(i) Secular trend series form a new series of arithmetic
(ii) Seasonal variation means, each of k successive observations
(iii) Cyclical variation of the given time series.
(iv) Irregular variation
l Method of least squares involves solving
l Secular trend of a time series is its smooth the following two normal equations.
and regular long-term movement.
•∑ Xt = na + b•∑ t
l Seasonal variation involves patterns of
change within a year that are repeated from •∑ tXt = a•∑ t + b•∑ t2
year to year. Seasonal variations often
reflect changes in underlying climatic MISCELLANEOUS EXERCISE - 4
conditions, or customs and habits of people.
I) Choose the correct alternative.
l Cyclical variation is a long-term oscillatory
movement that occurs over a long period of 1. Which of the following can’t be a
time, mostly more than two years or more. component of a time series?

(a) Seasonality (b) Cyclical
(c) Trend (d) Mean

67

2. The first step in time series analysis is to 7. An overall upward or downward pattern in
(a) Perform regression calculations an annual time series would be contained
(b) Calculate a moving average in which component of the times series
(c) Plot the data on a graph
(d) Identify seasonal variation (a) Trend (b) Cyclical
(c) Irregular (d) Seasonal
3. Time-series analysis is based on the
assumption that 8. The following trend line equation was
developed for annual sales from 1984 to
(a) Random error terms are normally 1990 with 1984 as base or zero year.
distributed.
Y1 = 500 + 60X (in 1000 Rs). The estimated
(b) The variable to be forecast and other sales for 1984 (in 1000 Rs) is:
independent variables are correlated.
(a) Rs 500 (b) Rs 560
(c) Past patterns in the variable to be (c) Rs 1,040 (d) Rs 1,100
forecast will continue unchanged into
the future. 9. What is a disadvantage of the graphical
method of determining a trend line?
(d) The data do not exhibit a trend.
(a) Provides quick approximations
4. Moving averages are useful in identifying (b) Is subject to human error
(a) Seasonal component (c) Provides accurate forecasts
(b) Irregular component (d) Is too difficult to calculate
(c) Trend component
(d) Cyclical component 10. Which component of time series refers to
erratic time series movements that follow
5. We can use regression line for past data to no recognizable or regular pattern.
forecast future data. We then use the line
which (a) Trend (b) Seasonal
(c) Cyclical (d) Irregular
(a) Minimizes the sum of squared
deviations of past data from the line II) Fill in the blanks
1. ___ components of time series is indicated
(b) Minimizes the sum of deviations of
past data from the line. by a smooth line.
2. ___ component of time series is indicated
(c) Maximizes the sum of squared
deviations of past data from the line by periodic variation year after year.
3. ___ component of time series is indicated
(d) Maximizes the sum of deviations of
past data from the line. by a long wave spanning two or more years.
4. ___ component of time series is indicated
6. Which of the following is a major problem
for forecasting, especially when using the by up and down movements without any
method of least squares? pattern.
5. Addictive models of time series ___
(a) The past cannot be known independence of its components.
(b) The future is not entirely certain 6. Multiplicative models of time series ___
(c) The future exactly follows the patterns independence of its components.
7. The simplest method of measuring trend of
of the past time series is ____
(d) The future may not follow the patterns 8. The method of measuring trend of time
series using only averages is ___
of the past

68

9. The complicated but efficient method of 2. Fit a trend line to the data in Problem
measuring trend of time series is ____. IV (1) by the method of least squares.

10. The graph of time series clearly shows 3. Obtain trend values for data in Problem
____ of it is monotone. IV (1) using 5-yearly moving averages.

III) State whether each of the following is 4. Following table shows the amount of sugar
True or False. production (in lac tonnes) for the years
1971 to 1982.
1. The secular trend component of time series
represents irregular variations. Year Production Year Production

2. Seasonal variation can be observed over 1971 1 1977 3
several years. 1972 0 1978 6
1973 1 1979 5
3. Cyclical variation can occur several times 1974 2 1980 1
in a year. 1975 3 1981 4
1976 2 1982 10
4. Irregular variation is not a random
component of time series. Fit a trend line to the above data by
graphical method.
5. Additive model of time series does not
require the assumption of independence of 5. Fit a trend line to data in Problem 4 by the
its components. method of least squares.

6. Multiplicative model of time series does 6. Obtain trend values for data in Problem
not require the assumption of independence 4 using 4-yearly centered moving averages.
of its components.
7. The percentage of girls’ enrollment in total
7. Graphical method of finding trend is enrollment for years 1960-2005 is shown
very complicated and involves several in the following table.
calculations.
Year 1960 1965 1970 1975 1980
8. Moving average method of finding trend Production 0 3 3 4 4
is very complicated and involves several
calculations. Year 1985 1990 1995 2000 2005
Production 5 6 8 8 10
9. Least squares method of finding trend
is very simple and does not involve any Fit a trend line to the above data by
calculations. graphical method.

10. All the three methods of measuring trend 8. Fit a trend line to the data in Problem 7 by
will always give the same results. the method of least squares.

IV) Solve the following problems. 9. Obtain trend values for the data in Problem
1. The following table shows the production 7 using 4-yearly moving averages.

of pig-iron and ferro-alloys (‘000 metric 10. Following data shows the number of boxes
tonnes) of cereal sold in years 1977 to 1984.

Year 1974 1975 1976 1977 1978 Year 1977 1978 1979 1980
Production 0 4 9 9 8 103 8
No. of boxes in
Year 1979 1980 1981 1982 ten thousands 1981 1982 1983 1984
Production 5 4 8 10 10 4 5 8
Year
Fit a trend line to the above data by
graphical method. No. of boxes in
ten thousands

Fit a trend line to the above data by
graphical method.

69

11. Fit a trend line to data in Problem 10 by the Fit a trend line to the above data by the
method of least squares. method of least squares.

12. Obtain trend values for data in Problem 20. Obtain trend values for data in Problem 19
10 using 3-yearly moving averages. using 3-yearly moving averages.

13. Following table shows the number of traffic Activities
fatalities (in a state) resulting from drunken
driving for years 1975 to 1983. Note: You may change the origin and scale
in the following problems according to your
Year 1975 1976 1977 1978 1979 convenience.
06382
No. of 1. Daily SENSEX index values at opening are
deaths 1980 1981 1982 1983 given for fifty days in the following table.
9 4 5 10 Plot a graph from the data. Find the trend
Year graphically, using moving averages, and by
the method of least squares.
No. of
deaths

Fit a trend line to the above data by Date Index Date Index
graphical method.
1-Jan-19 36161.8 2-Jan-19 36198.13
14. Fit a trend line to data in Problem 13 by the 3-Jan-19 35934.5 4-Jan-19 35590.79
method of least squares. 7-Jan-19 35971.18 8-Jan-19 35964.62
9-Jan-19 36181.37 10-Jan-19
15. Obtain trend values for data in Problem 13 11-Jan-19 36191.87 14-Jan-19 36258
using 4-yearly moving averages. 15-Jan-19 35950.08 16-Jan-19 36113.27
17-Jan-19 36413.6 18-Jan-19 36370.74
16. Following table shows the all India infant 21-Jan-19 36467.12 22-Jan-19 36417.58
mortality rates (per ‘000) for years 1980 23-Jan-19 36494.12 24-Jan-19 36649.92
to 2000. 25-Jan-19 36245.77 28-Jan-19 36146.55
29-Jan-19 35716.72 30-Jan-19 36099.62
Year 1980 1985 1990 1995 31-Jan-19 35805.51 1-Feb-19 35819.67
IMR 10 7 5 4 4-Feb-19 36456.22 5-Feb-19 36311.74
Year 2000 2005 2010 6-Feb-19 36714.54 7-Feb-19 36573.04
IMR 3 1 0 8-Feb-19 36873.59 11-Feb-19 37026.56
12-Feb-19 36405.72 13-Feb-19 36585.5
Fit a trend line to the above data by 14-Feb-19 36065.08 15-Feb-19 36279.63
graphical method. 18-Feb-19 35831.18 19-Feb-19 35985.68
20-Feb-19 35564.93 21-Feb-19 35543.24
17. Fit a trend line to data in Problem 16 by the 22-Feb-19 35906.01 25-Feb-19
method of least squares. 26-Feb-19 35975.75 27-Feb-19 35837
28-Feb-19 36025.72 1-Mar-19 35983.8
18. Obtain trend values for data in Problem 5-Mar-19 36141.07 6-Mar-19 36138.83
16 using 3-yearly moving averages. 7-Mar-19 36744.02 8-Mar-19 36018.49
8-Mar-19 36753.59 12-Mar-19 6544.86
19. Following tables shows the wheat yield 36753.59
(‘000 tonnes) in India for years 1959 to 37249.65
1968.

Year Yield Year Yield

1959 0 1964 0
1960 1 1965 4
1961 2 1966 1
1962 3 1967 2
1963 1 1968 10

70

2. Onion prices (per quintal) in a market are 3. Following table gives the number of
given for fifteen days. Plot a graph of given persons injured in road accidents for
data. Find the trend graphically, using 11 years. Plot a graph from the data. Find
moving averages, and by the method of the trend graphically, using moving
least squares. averages, and by the method of least
squares.


Date Price Year 2006 2007 2008 2009 2010 2011

24/01/2019 400 No. of 29955 27464 32224 32144 25400 28366
injured
31/01/2019 650

3/2/19 650 Year 2012 2013 2014 2015 2016

7/2/19 700 No. of 27185 26314 24488 23825 22072
injured
24/02/2019 550

28/02/2019 550 4. Following table gives the number of
road accidents due to over-speeding in
5/3/19 500 Maharashtra for 9 years. Plot a graph from
the data. Find the trend graphically, using
7/3/19 600 moving averages, and by the method of
least squares.
14/03/2019 600

28/03/2019 600

7/4/19 800 Year 2008 2009 2010 2011 2012

11/4/19 801 No. of
accidents
14/04/2019 800 38680 18090 21238 28489 27054

25/04/2019 800 Year 2013 2014 2015 2016

2/5/19 800 No. of 26931 22925 24622 22071
accidents

FFF

71

5 Index Numbers

Let's Study because they are ratios. Index numbers are
usually expressed as percentages.
l Definition of Index Numbers
Maslow describes an index number as a
l Types of Index Numbers numerical value characterizing the change in a
complex economic phenomenon over a period
l Terminology and Notation of time. According to Spiegel, an index number
is a statistical measure designed to show changes
l Construction of Index Numbers in a variable or a group of related variables with
respect to time, geographical location or some
l Simple Aggregate Method other characteristic. Gregory and Ward describe
it as a measure designed to show an average
l Weighted Aggregate Method change, over time, in the price, quantity or value
of a group of items. Croxton and Cowden say
l Cost of Living Index Number that an index number is a device that measures
differences in the magnitude of a group of related
l Aggregative Expenditure Method variables. B. L. Bowley describes an index
number as a series that reflects in its trend and
l Family Budget Method fluctuations the movements of some quantity to
which it is related. Blair puts an index number as
l Uses of Cost of Living Index Number a special kind of average.

Introduction Let's Learn
The value of money does not remain the
same for all the time. It cannot be observed 5.1 Definition of Index Numbers.
directly, but can be understood by observing
the general level of prices. A rise in the price Index Numbers are defined in different
level indicates a fall in the value of money ways by different experts. Some of the most
and a fall in the price level indicates a rise in popular definitions of Index Numbers are given
the value of money. Changes in the value of below.
money are reflected in changes in general level
of prices over a period of time. Changes in the 1. An Index Number is a statistical measure
value of money are found to be inversely related of changes in a variable or a group of
to changes in price levels. So, changes in the variables with respect to time, geographical
value of money can be understood by observing location, or some other characteristic such
changes in the general level of prices over a as production, income, etc.
specified time period. Changes in the general
level of prices are measured using a statistical 2. An Index Number is used for measuring
tool known as index numbers. Index numbers changes in some quantity that can not be
provide one of the most popular statistical tools measured directly.
used in economics.
3. An Index Number is a single ratio, usually
Index numbers cannot be measured expressed as percentage, that measures
directly, but are constructed with help of some aggregate (or average) change in several
mathematical formula. Index numbers are not
expressed in terms of any units of measurement

72

variables between two different times, a group of commodities) in the current year
places, or situations. and its value in the base year. A value index
number combines prices and quantities by
After reading the above definitions, we can taking the product of price and quantity as
conclude that an Index Number is an 'economic the value. The value index number thus
indicator' of business activities. measures the percentage change in the value
of a commodity or a group of commodities
Examples of index numbers. during the current year in comparison to its
NIFTY: value during the base year.

The NIFTY 50 index is National Stock 5.3 (a) : Terminology.
Exchange of India's benchmark broad based
stock market index for the Indian equity market. Base Period: The base period of an
It represents the weighted average of 50 Indian index number is the period against which
company stocks in 13 sectors and is one of the comparisons are made. For example, the
two main stock indices used in India, the other Central Statistical Organisation (CSO) is
being the BSE Sensex. constructing the Consumer Price Index by
taking 2010 as the base year. It means that
SENSEX: the prices in 2015 are compared with 2010
The BSE SENSEX (also known as the prices by taking them as 100. The base
period is indicated by subscript Zero.
S&P Bombay Stock Exchange Sensitive Index
or simply the SENSEX) is a free-float market- Current Period : The present period
weighted stock market index of 30 well- is called the current period of an index
established and financially sound companies number. An index number measures the
listed on Bombay Stock Ex change changes between the base period and
the current period. The current period is
5.2 Types of Index Numbers indicated by subscript 1.
Following are three major types ot index
Note:
numbers.
The period used in index numbers can be a
1. Price Index Number day, a month, or a year. We shall use a year as the
Price index numbers measure changes period in our study.

in the level of prices in the economy. It 5.3 (b) : Notation.
compares the price of the current year,
with that of the base year to indicate the p0 : Price of a commodity in the base year.
relative variation. It is a very good measure
of inflation in the economy. q0 : Quantity (produced, purchased, or
consumed) of a commodity in the base
2. Quantity Index Number year.
As the name suggests, quantity index
p1 : Price of a commodity in the current year.
numbers measure changes in the quantities
of goods between the two specified years. q1 : Quantity (produced, purchased, or
This can be the number of goods produced, consumed) of a commodity in the current
sold, consumed, etc. It is a good indication year.
of the output of an economy.
w : Weight assigned to a commodity according
3. Value Index Number to its relative importance in the group.
A value index number is the ratio of the
I : Simple index number. It is also called the
aggregate value of a given commodity (or price relative. It is given by

73

p1 (1) (b) Simple Aggregate Method to find
I = p0 × 100 Quantity Index Number

P01 : Price index for the current year with respect Quantity Index Number can be calculated
to the base year. by the same procedure as above, only replacing
prices by quantities.
Q01: Quantity index for the current year with
respect to the base year. Step I : Quantities of all commodities are added
for the base year. This total is denoted by
V01 : Value index for the current year with
respect to the base year. ∑ q0

5.4 Construction of Index Numbers Step II : Quantities of all commodities are added
Index number are constructed by the for the current year. This total is denoted by

following two methods ∑ q1

1. Simple Aggregate Method. Step III : The total obtained in Step II is divided
by the total obtained in Step I. The ratio is
2. Weighted Aggregate Method. then multiplied by 100.

Let us now learn how index numbers are Thus, the required quantity index number
constructed by these two methods.
is given by

5.4.1 Method 1: Simple Aggregate Method ∑∑ Q01 = q1 × 100.
q0
This is the simplest method of constructing
index numbers. This method assumes that every (c) Simple Aggregate Method to find Value
commodity is equally important. Index Number

(a) Simple Aggregate Method to find Price Value of a commodity is defined as the
Index Number product of its price and quantity. Value Index
Number is then calculated using the same
The procedure of calculating Price Index procedure as above, where price or quantity is
Number by the Simple Aggregate Method is as replaced by value.
follows.
Step I : Values (that is, products of prices and
Step I : Prices of all commodities are added quantities) of all commodities are added
for the base year. This total is denoted by for the base year. This total is denoted by

∑ p0 ∑ p0q0

Step II : Prices of all commodities are added for Step II : Values (that is, products of prices and
the current year. This total is denoted by quantities) of all commodities are added
for the current year. This total is denoted
∑ p1.
∑ by p1q1
Step III : The total obtained in Step II is divided
by the total obtained in Step I. The ratio is Step III : The total obtained in Step II is divided
then multiplied by 100. by the total obtained in Step I. The ratio is
then multiplied by 100.
Thus, the required price index number is
given by Thus, the required value index number is

∑∑ P01 = p1 × 100. given by p1q1 × 100.
p0 p0q0
∑∑ V01 =

74

SOLVED EXAMPLES 2. Calculate the Quantity Index Number for
the following data using Simple Aggregate
1. Calculate the price index number for the Method. take year 2000 as the base year.
following data using the Simple Aggregate
Method. Take 2000 as the base year. Commodity I II III IV V VI

Commodities A B C D E Quantity 30 55 65 70 40 90
in 2000

Price (in Rs.) 30 35 45 55 25 Quantity 40 60 70 90 55 95
for 2000 in 2004

Price (in Rs.) 30 50 70 75 40 Solution:
for 2003 We first tabulate the data in the following

Solution : tabular form.

Let us first tabulate the data in the following Table 5.2
tabular form.
Commodities Quantity in Quantity in
Table 5.1 2000 (Base 2004 (Current
I
Commodities Price in 2000 Price in 2003 II year) q0 year) q1
(Base year) p0 (Current year) III
A IV 30 40
B 30 p1 V 55 60
C 35 VI 65 70
D 45 40 70 90
E 55 50 Total 40 55
Total 25 70 90 95
75
∑ p0 = 190 40 ∑ q0 = 350 ∑ q1 = 410

∑ p1 = 275 Quantity Index Number is then given by

Price Index Number is then given by ∑∑ Q01 = q1 × 100
q0
∑∑ P01 = p1
p0 × 100 410
= 350 × 100

275 = 117.14.
= 190 × 100

= 144.74 This means that the output in terms of
quantity rose by approximately 17% in year
Interpretation: 2004 from year 2000.

If the price of a commodity was Rs.100
in the year 2000, then the price of the same
commodity is approximately Rs.145 in the year
2003. Hence, the overall increase in the price
level is 45% in three years.

75

3. Calculate the Value Index Number for the Solution:
following data using the Simple Aggregate
Method. Table 5.4

Commodities Base Year Current Year Commodity I II III IV Total
Price in 1995 6 15 x 4 25 + x
p Price Quantity Price Quantity Price in 1998 8 18 28 6 60
Q Rs. p0 (units) Rs. p1 (units)
R From the above table, we have
S 10 q0 q1
T 6 ∑ ∑ p0 = 25 + x, p1 = 60, and
60 7

20 4 70 6 P01 = 120.

30 7 80 8

40 8 90 9 The value of x is then found from the

50 3 100 5 formula ∑∑P01 = p1
p0
Solution : First, prepare the following table. × 100
Table 5.3
So that we obtain

Commodity Base Year Current 60
Year ∴ 120 = 25 + x × 100

p0 q0 p1 q1 p0 q0 p1 q1 60
12 = 25 + x × 10
10 6 60 7 60 420
p 20 4 70 6 80 420 ∴ 12(25 + x) = 600
Q 30 7 80 8 210 640
R 40 8 90 9 320 810 ∴ 300 + 12x = 600
S 50 3 100 5 150 500
T ∴ 12x = 600 − 300
820 2790
Total ∴ 12x = 300

∑∑Note : that p0q0 = 820, p1q1 = 2790, ∴ x = 25.
Hence, x = 25.
and, therefore, Value Index Number is

given by p1q1 × 100 5. The Price Index Number for year 2004,
p0q0 with respect to year 2000 as base year, is
∑∑ V01 = known to be 130. Find the missing
numbers in the following table if
2790
= 820 × 100 ∑ p0 = 320

= 340.24 Commodity A B C D E F

4. Find x in the following table if theAggregate Price 40 50 30 x 60 100
Price Index Number for year 1998 with (in Rs.)
respect to Base Year 1995 is 120. in 2000

Commodity I II III IV Price
Price in 1995 6 15 x 4
Price in 1998 8 18 28 6 (in Rs.) 50 70 30 85 y 115

in 2005

76

Solution: 2. Use 1995 as base year in the following
problem.
We first tabulate the given data as shown in
the following table. Commodity A B C D E

Table 5.5 Price
(in Rs.)
Commodities Price in 2000 Price in 2005 in 1995 42 30 54 70 120
(Base year) p0 (Current year)
A Price
B 40 p1
C 50 (in Rs.) 60 55 74 110 140
D 30 50
E x 70 in 2005
F 60 30
100 85 3. Unit Base Year Current
y Price Year
115 Commodity kg (in Rs.) Price
kg
From the above table, we have Wheat litre 28 (in Rs.)
Rice meter 40
∑ ∑ p0 = 280 + x, p1 = 350 + y, Milk litre 35 36
Clothing 82 56
∑ But it is given that p0 = 320, so that Fuel 58 45
104
280 + x = 320 72

∴ x = 40 4. Use 2000 as base year in the following
problem.
Further, using the formula

∑∑ P01 = p1 × 100 Price Price
p0
Commodity (in Rs.) for (in Rs.)for
We have
year 2000 year 2006
350 + y
130 = 320 × 100 Watch 900 1475
Shoes 1760 2300
∴ 416 = 350 + y Sunglasses 600 1040
Mobile 4500 8500
∴ y = 66.

EXERCISE 5.1 5. Use 1990 as base year in the following
problem.

Find the Price Index Number using Simple Commodity Unit Price Price
Aggregate Method in each of the following (in Rs.) (in Rs.)
examples. Butter kg for 1990 for 1997
Cheese kg
1. Use 1995 as base year in the following Milk litre 27 33
problem. Bread loaf 30 36
Eggs doz 25 29
Commodity P Q R S T Ghee tin 10 14
Price (in Rs.) 24 36
250 320
in 1995 15 20 24 22 28

Price (in Rs.) 27 38 32 40 45
in 2000

77

6. Assume 2000 to be base year in the 10.
following problem.
Commodity Base Year Current Year
Fruit Unit Price Price
(in Rs.) (in Rs.) Price Quantity Price Quantity
Mango doz in 2000 in 2007
Banana doz A 30 22 40 18
Apple kg 250 300 B 40 16 60 12
Peach kg 12 24 C 10 38 15 24
Orange doz 80 110 D 50 12 60 16
Sweet Lime doz 75 90 E 20 28 25 36
36 65
30 45 11.

7. Use 2005 as base year in the following Commodity Base Year Current Year
problem.
Price Quantity Price Quantity
Vegetable Unit Price Price
(in Rs.) (in Rs.) A 50 22 70 14
Ladies Finger kg in 2005 in 2012 B 70 16 90 22
Capsicum kg C 60 18 105 14
Brinjal kg 32 38 D 120 12 140 15
Tomato kg 30 36 E 100 22 155 28
Potato kg 40 60
40 62 12. Find x if the Price Index Number by Simple
16 28 Aggregate Method is 125.

Commodity P Q R S T

Find the Quantity Index Number using Base Year 8 12 16 22 18
Simple Aggregate Method in each of the Price (in Rs.)
following examples.
Current Year 12 18 x 28 22
8. Price (in Rs.)

Commodity I II III IV V 13. Find y if the Price Index Number by Simple
Aggregate Method is 120, taking 1995 as
Base Year 140 120 100 200 225 base year
Quantities
Commodity A B C D

Current Year 100 80 70 150 185 Price (in Rs.) 95 y 80 35
Quantities for 1995

9. Price (in Rs.) 116 74 92 42
for 2003
Commodity A B C D E

Base Year 360 280 340 160 260 5.4.2 Method 2: Weighted Aggregate Method
Quantities
This method assigns suitable weights
Current Year 440 320 470 210 300 to different commodities before aggregating
Quantities their prices, quantities, or values. These
weights indicate relative importance of various
Find the Value Index Number using Simple commodities in the group. If w denotes the
Aggregate Method in each of the following weight attached to a commodity, then the Price
examples. Index Number is given by

78

∑∑P01 = p1w (f) Walsh’s Price Index Number
p0 w
× 100 ∑∑ P01(W) = p1 q0q1
p0 q0q1
× 100

Weights are usually defined in terms of SOLVED EXAMPLES
quantities in the weighted aggregate method.
Index numbers constructed by the weighted 1. Calculate (a) Laspeyre’s, (b) Paasche’s,
aggregate method are known by names of the (c) Dorbish-Bowley’s and Marshall-
developers of these index numbers. Following are Edgeworth’s Price Index Number for the
most popular price index numbers constructed following data.
by the weighted aggregate method.
Commodity Base Year Current Year
(a) Laspeyre’s Price Index Number Price Quantity Price Quantity

∑∑P01(L) = p1q0 × 100 p 12 20 18 24
p0q0 Q 14 12 21 16
R 8 10 12 18
Note:This construction uses base year quantities S 16 15 20 25
as weights.

(b) Paasche’s Price Index Number

∑ p1q1 Solution:
∑P01(P) = p0q1 × 100 Let us first prepare the following table.

Note: This construction uses current year Table 5.6
quantities as weights.
Base Current
(c) Dorbish-Bowley’s Price Index Number Commodity Year Year p0 q0 p1 q0 p0 q1 p1 q1

∑ ∑p1q0 + p1q1 p0 q0 p1 q1
∑ ∑p0q0 p0q1
P 12 20 18 24 240 360 288 432

P01(D − B) = 2 × 100 Q 14 12 21 16 168 252 224 336

R 8 10 12 18 80 120 144 216

(d) Fisher’s Ideal Price Index Number S 16 15 20 25 240 300 400 500

∑∑ ∑∑ P01(F) = p1q0 × p1q1 Total 728 1032 1056 1484
p0q0 p0q1 × 100
From the above table, we have
Question:
∑ ∑ p0q0 = 728, p1q0 = 1032
Can you find any relation among ∑ ∑ p0q1 = 1056 p1q1 = 1484
Laspeyre’s, Paasche’s, Dorbish-Bowley’s and
Fisher’s Price Index Number? (a) Laspeyre’s Price Index Number is
then
(e) Marshall-Edgeworth’s Price Index
Number ∑ p1q0
∑= p0q0 × 100
∑∑ p1 ( q0 + q1 ) P01(L)
p0 ( q0 + q1 )
P01(M − E) = × 100

∑∑ ∑∑ p1q0 + p1q1 1032
p0q0 + p0q1 = × 100

728

= × 100 ∴ P01(L) = 141.76

79

(b) Paasche’s Price Index Number is given by Table 5.7

∑ p1q1 Comm- Base Current q0q1 p0 q0q1 P1 q0q1
∑ P01(P) = p0q1 × 100 odity Year Year

1484 p0 q0 p1 q1
= 1056 × 100
A 20 9 30 4 6 120 180
5 50 250
B 10 5 50 5 4 160 40
2 60 40
∴ P01(P) = 140.53 C 40 8 10 2
390 510
D 30 4 20 1

(c) Dorbish-Bowley’s Price Index Number is Total
given by

P01(L) + P01(P) From the above table, we get
2
P01(D − B) = ∑ p0 q0q1 = 390

141.76 +140.53 ∑ p1 q0q1 = 510
= 2

= 141.15 Walsh’s Price Index Number is given by

(d) Marchall-Edgeworth’s Price Index Number ∑∑ P01(W) = p1 q0q1 × 100
is given by p0 q0q1

(∑ p1q0 + ∑ p1q1 ) 510
(∑ ∑ ) P01(M − E) = = 390 × 100
p0q0 + p0q1 × 100

(1032 +1484) 5100
= (728 +1056) × 100 = 39

∴ P01(W) = 130.77

2516 3. If P01(L) = 225, P01(P) = 144, then calculate
= 1784 × 100 = 141.03 P01(F) and P01(D − B)

∴ P01(M − E) = 141.03 Solution :
Given P01(L) = 225, P01(P) = 144, we obtain
2. Calculate Walsh’s Price Index Number for
the following data. P01(F) = P01(L) × P01(P)

Commodity Base Year Current Year = 225×144
Price Quantity Price Quantity = 15 × 12

A 20 9 30 4 ∴ P01(F) = 180
B 10 5 50 5 Next,
C 40 8 10 2 = P01(L) + P01(P)
D 30 4 20 1 P01(D − B) 2

Solution: Let us prepare the following table. 225 +144
= 2


∴ P01(D − B) = 184.50

80

Example 4: ∑∑5. If p0q0 = 120, p0q1 = 200

Find the missing price in the following ∑ p1q1 = 300, and P01(L) = 150, find
table if Laspeyre’s and Paasche’s Price Index
Numbers are the same. P01(M − E).
Solution: Note that
Commodity Base Year Current Year
∑ p1q0
Price Quantity Price Quantity ∑ P01(L) = p0q0 × 100

A 1 10 2 5

B 15-2

Solution: ∑ p1q0
Let us denote the missing value by x, and ∴ 150 = 120 × 100

reconstruct the table as follows. ∑∴ p0q1 = 15 × 12

Table 5.8 ∑ ∴ p0q1 = 180.

Comm- Base Current

odity Year Year p0 q0 p1 q0 p1 q1 p0 q1 Now, (∑ p1q0 + ∑ p1q1 )
p0 q0 p1 q1
10 5 p0q0 + p0q1 × 100
A 1 10 2 5 10 20 2x 2 (∑ ∑ ) P01(M − E) =
B 1 5 x 2 5 5x
180 + 300
Total 15 20+5x 10+2x 7 = 120 + 200 × 100

The above table gives 480
= 320 × 100
∑ p0q0 = 15, ∑ p1q0 = 20+5x

∑ p0q1 = 7 ∑ p1q1 = 10+2x ∴ P01(M − E) = 150.

It is given that ∑ ∑6. If p0q0 = 180, p0q1 = 200

P01(L) = P01(P) ∑ p1q1 = 280, and P01(M − E) = 150,

∑ p1q0 ∑ p1q1 find P01(P).
∑ ∑ p0q0 × 100 = p0q1 × 100
Solution:
5x + 20 2x +10
∴ 15 = 7 Let us denote ∑ p0q1 by x. Then, using the

fact that

5(x + 4) 2x + 10
15 = 7
∴ (∑ p1q0 + ∑ p1q1 )

(∑ ∑ ) P01(M − E) =
∴ 7(x + 4) = 3(2x + 10) p0q0 + p0q1 × 100

∴ 7x + 28 = 6x + 30 200 + 280
∴ 150 = 180 + x × 100
∴ x = 2. ∴ 15(180 + x) = 4800

The missing price is 2.

81

4800 4.
∴ 180 + x = 15
Commodity Base Year Current Year

∴ 180 + x = 320 Price Quantity Price Quantity

∴ x = 140 I 10 12 20 9

∑ ∴ p0q1 = 140. II 20 4 25 8

Now, ∑ p1q1 III 30 13 40 27

∑ P01(P) = p0q1 × 100 IV 60 29 75 36

280 5. If P01(L) = 90, and P01(P) = 40,
= 140 × 100 find P01(D − B) and P01(F)

∴ P01(P) = 200 ∑∑6. If p0q0 = 140, p0q1 = 200,

EXERCISE 5.2 ∑ ∑ p0q1 = 350, p1q1 = 460, find

Calculate Laspeyre’s, Paasche’s, Dorbish- Laspeyre’s, Paasche’s, Dorbish-Bowley’s
Bowley’s and Marshall-Edgeworth’s Price Index and Marshall-Edgeworth’s Price Index
Numbers in Problems 1 and 2 Numbers.

1. 7. Given that Laspeyre’s and Dorbish-
Bowley’s Price Index Numbers are 160.32
Commodity Base Year Current Year and 164.18 respectively. Find Paasche’s
Price Index Number.
A Price Quantity Price Quantity
B 8. Given that ∑ p0q0 ∑= 220, p0q1 = 380,
C 8 20 11 15
D 7 10 12 10 ∑ p1q1 = 350 and Marshall-Edgeworth’s
3 30 5 25
2 50 4 35 Price Index Number is 150, find Laspeyre’s
Price Index Number.

2. 9. Find x in the following table if Laspeyre’s
and Paasche’s Price Index Numbers are
Commodity Base Year Current Year equal.

Price Quantity Price Quantity

I 10 9 20 8 Commodity Base Year Current Year
II 20 5 30 4
III 30 7 50 5 Price Quantity Price Quantity
IV 40 8 60 6
A 2 10 2 5

B 25x2

Calculate Walsh’s Price Index Number in 10. If Laspeyre's Price Index Number is four
Problem 3 and 4. times Paasche's Price Index Number, then
find the relation between Dorbish-Bowley's
3. and Fisher's Price Index Numbers.

Commodity Base Year Current Year 11. If Dorbish-Bowley's and Fisher's Price
Price Quantity Price Quantity Index Numbers are 5 and 4, respectively,
then find
L 4 16 3 19
M 6 16 8 14 Laspeyre's and Paasche's Price Index
N 8 28 7 32 Numbers.

82

5.5 Cost of Living Index Number commodities should be chosen according
to the purpose of the index number.
Cost of Living Index Number, also known
as Consumer Price Index Number, is an index In choosing commodities, the following
number of the cost of buying goods and services points must be kept in mind:
in day-to-day life for a specific consumer class.
Different classes of consumers show different (a) The commodities must represent
patterns of consumption of goods and services. the tastes, habits and customs of the
As a result, a general index number cannot reflect people.
changes in cost of living for a specific consumer
class. For example, cost of living index numbers (b) Commodities should be recognizable.
for rural population are different from cost of
living index numbers for urban population. The (c) Commodities should have the same
goods and services consumed by members of quality over different periods and
different consumer classes can be different and places.
therefore cost of living index numbers calculated
for different consumer classes can be based on (d) The economic and social importance
costs of different sets of goods and services. of different commodities should be
taken in consideration.
Steps in Construction of Cost of Living
Index Numbers (e) The commodities should be
sufficiently large in number,
Construction of cost of living index
numbers involves the following steps: (f) All varieties of a commodity should
be included that are in common use
1. Choice of Base Year: and are stable in nature.

The first step in preparing cost of living 3. Collection of Prices:
index numbers is choice of base year. Base
years defined as that year with reference After choosing the commodities, the
to which price changes in other years are next step is collection of their prices.
compared and expressed as percentages. The following points are important while
The base year should be a normal year. It collecting prices of commodities chosen for
should be free from abnormal conditions constructing cost of living index numbers.
like wars, famines, floods, political
instability, etc. (a) From where prices are to be collected.

Base year can be chosen in two ways: (b) Whether to collect wholesale prices
or retail prices.
(a) Using fixed base method, where the
base year remains fixed; and (c) Whether to include taxes in prices.

(b) Using chain base method, where Following points are to be noted while
the base year goes on changing. For collecting prices:
example, 1979 will be the base year
for 1980, it will be 1978 for 1979, and (a) Prices must be collected from places
so on. where a particular commodity is
traded in large quantities
2. Choice of Commodities:
(b) If published information on prices is
The second step in construction of cost available, it must be used,
of living index numbers is choosing the
commodities. Since all commodities (c) Care should be taken while collecting
cannot be included, only representative price quotations from individuals or
institutions that they provide correct
information.

(d) Choice of wholesale or retail prices
depends on the purpose of preparing

83

index numbers. Wholesale prices are Total expendutre in current year
used in the construction of general CLI = Total expendutre in base year
price index, while retail prices are
used in the construction of cost of ∑ p1q0
living index. ∑= p0q0 × 100

(e) Prices must be averaged if collected The above formula is similar to that of a
from several sources. weighted Index Number. Do you recognize that
Index Number?
4. Choice of Average:
5.5.2 Family Budget Method
Since the index numbers are a specialized
average, it is important to choose a suitable (Weighted Relative Method)
average. Geometric mean is theoretically
the best, but arithmetic mean is used in Cost of Living Index Number is defined as
practice because it is easier to calculate.
follows.
5. Choice of Weights:
∑ IW
Generally, all the commodities included CLI = ∑W
in the construction of index numbers
are not equally important. Therefore, where
proper weights must be assigned to the
commodities according to their relative I = p1 × 100
importance. For example, cost of living p0
index for teachers will assign higher
weightage to prices of books than cost of = price relative for current year
living index for workers. Weights should
be chosen rationally and not arbitrarily. and

6. Purpose of Index Numbers: W = p0q0

The most important consideration in = base year weightage.
the construction of index numbers is
their objective. All other steps are to be Do you find the above two methods of
viewed in light of the purpose for which a calculating the Cost of Living Index Numbers
particular index number is being prepared. to be same?
Since every index number is prepared with
a specific purpose, no single index number SOLVED EXAMPLES
can be 'all purpose' index number. It is
important to have a clear idea about the 1. Construct the Cost of Living Index Number
purpose of the index number before it is for the following data.
constructed.
Group Base Year Current
Methods of constructing Cost of Living Price Quantity Year
Index Numbers Food & Price
Clothing 40 3 70
5.5.1 Aggregative Expenditure Method Fuel &
(Weighted Aggregate Method) Lighting 30 5 60
House Rent
This method uses quantities consumed Miscellaneous 50 2 50
in base year as weights, so that Cost of Living 60 3 90
Index Number is defined as follows.

84

Solution: Table 5.10
We shall begin by preparing the following table.
Group I W IW

Group Base Current Food 120 3 360

Year Year p1q0 p0q0 Clothing 100 6 600
p0 q0 p1
Fuel &

Food & 40 3 70 210 120 Lighting 140 5 700
Clothing
House Rent 160 2 320

Fuel & 30 5 60 300 150 Miscellaneous 150 4 600

Lighting Total - 20 2580

House Rent 50 2 50 100 100

Miscellaneous 60 3 90 270 180 ∑ The above table shows that W =20 and

Total 880 550 ∑ IW = 2580, and
∑ IW
We shall use Aggregative Expenditure
Method since p0, q0 and p1 are given. ∴CLI = ∑W

∑ p1q0 2580
CLI = p0q0 × 100 = 20
∑

880 ∴ CLI = 129.
= 550 × 100
3. Find x in the following table if the Cost of
∴ CLI = 160. Living Index Number is 121.

Interpretation. A person earning Rs 100 Group Food Cloth- Fuel House Misce-
in the base year, should earn Rs 160 in the current ing & Light- Rent llane-
year to maintain the same standard of living. ous
ing

2. The following table gives the base year I 100 125 174 x 90
weightage (W) and current year price
relative (I) for five commodities. Calculate W 13 12 10 8 7
the Cost of Living Index Number.
Solution:
First, we prepare the following table.

Group Food Cloth- Fuel House Misce- Group I W IW
ing & Light- Rent llane- 100 13 1300
ous Food 125 12 1500
ing Clothing
Fuel & 174 10 1740
I 120 100 140 160 150 Lighting x 8 8x
House Rent 90 7 630
W36 5 2 4 Miscellaneous - 50 5170+8x

Solution: Total

We use Family Budget Method since I and It can be found from the above table that
W are given. For this, we prepare the following
table.

85

∑ ∑ W = 50 and IW = 5170 + 8x the expenditure of a person in year 1999 if
his expenditure in year 1995 was 800.
∑ IW
∴ CLI = ∑W Group Price Price W
in year in year
∴ 121 = 5170 + 8x Food 1995 1999 6
50 Clothing 12
Fuel & 8 24
∴ 6050 = 5170 + 8x Lighting 18 36 8
House Rent 4
∴ x = 110. Miscellaneous 20 40 10
15 30
4. Cost of Living Index Numbers for the years 10 22
2000 and 2005 are 120 and 200, respectively.
If a person has monthly earnings of Rs Solution:
10800 in year 2000, what should be his Let us first prepare the following table.
monthly earnings in year 2005 in order to
maintain same standard of living? Table 5.12

Solution: Price Price I=
For the year 2000, it is given that in year in year
Group p1 × W IW
CLI = 120, and Income = Rs 10800. These two p0
give us real income as follows. 1995 1999 100

Income Food 8 24 300 6 1800
Real Income = CLI × 100
Clothing 18 36 200 12 2400
10800
= 120 × 100 Fuel &

∴ Real Income = 9000. Lighting 20 40 200 8 1600

This shows that the real income is Rs 9000. House Rent 15 30 200 4 800

The CLI for year 2005 is 220. Miscellaneous 10 22 220 10 2200

Income Total - - - 40 8800
Real Income = CLI × 100
By Family Budget Method,
Income
∴ 9000 = 220 × 100 ∑ IW
∴ CLI = ∑W
∴ Income = 19800.
8800
This shows that the monthly income of the = 40
person should be Rs 19800 in year of 2005 in
order to maintain the same standard of living as ∴ CLI = 220
in year 2000
Now, the expenditure in 1995 was
5. Calculate the Cost of Living Index Number Rs 800. In other words, the expenditure is
for the year 1999 by Family Budget Rs 800 when CLI is 100. The question is to find
Method from the following data. Also, find expenditure when CLI is 220 in 1999.

220
∴ Expenditure in 1999 = 100 × 800

= 1760

Thus, the expenditure in 1999 is Rs 1760.

86

EXERCISE 5.3 4.

Calculate the cost of living index in Group Food Cloth- Fuel House Misce-
problems 1 to 3. ing & Light- Rent llane-
ous
1. ing

I 70 90 100 60 80

Current W53 2 4 6
Year
Group Base Year Price 5.
170
Food Price Quantity 190 Group Food Cloth- Fuel House Misce-
Clothing ing & Light- Rent llane-
Fuel & 120 15 220 ous
Lighting 150 20 180 ing
House Rent 200
Miscellaneous I 400 300 150 120 100

130 30 W33 4 5 2
160 10
200 12 6.

2. Group Food Cloth- Fuel House Misce-
ing & Light- Rent llane-
Current ous
Group Base Year Year ing
Price
Food 45 I 200 150 120 180 160
Clothing 35
Fuel & Price Quantity W 30 20 10 40 50
Lighting 40 15 25
House Rent 30 10 70 7. Find x if the cost of living index is 150.
Miscellaneous 80
20 17 Group Food Cloth- Fuel House Misce-
60 22 ing & Light- Rent llane-
70 25 ous
ing

I 180 120 300 100 160

W45 6 x 3

3. 8. Find y if the cost of living index is 200

Group Base Year Current Cloth- Fuel House Misce-
Year ing & Light- Rent llane-
Food Price Quantity Price Group Food ous
Clothing 132 10 170 ing
Fuel & 154 12 160
Lighting I 180 120 160 300 200
House Rent 180
Miscellaneous 195 W45 3 y 2
120
164 20 9. The Cost of Living Index Number for
175 18 years 1995 and 1999 are 140 and 200
128 5 respectively. A person earns Rs. 11,200 per
month in the year 1995. What should be his
Base year weights (W) and current year monthly earnings in the year 1999 in order
price relatives (I) are given in Problems to maintain his standard of living as in the
4 to 8. Calculate the cost of living index in year 1995 ?
each case

87

5.6 Uses of Cost of Living Index Number Where

1. Cost of Living Index Number is used to P01 : Price index Number for the current
regulate the dearness allowance or the grant year with respect to base year
of bonus to employees in order to enable
them bear the increased cost of living. P1 : Price of the commodity in current year

2. Cost of Living Index Number is used for P0 : Price of the commodity in base year
settling dispute related to salaries and
wages. l Price Index Number using Weighted
Aggregate Method is calculated by the
3. Cost of Living Index Number is used in following formula.
calculating purchasing power of money.
∑∑ P01 = p1w × 100
Purchasing power of money p0 w

1 Where
= Cost of Living Index Number
w : Weight assigned to a commodity
4. Cost of Living Index Number is used in
determining real wages. l Laspeyre’s Price Index Number

Real Wages ∑ p1q0
∑ P01(L) = p0q0 × 100
Money wages
= Cost of Living Index Number × 100 l Paasche’s Price Index Number

5. Cost of Living Index Numbers are widely ∑ p1q1
used in negotiations of wages in wage ∑ P01(P) = p0q1 × 100
contracts.
l Dorbish-Bowley’s Price Index Number

Let's Remember ∑ ∑p1q0 + p1q1

∑ ∑ p0q1 × 100
P01(D−B) = p0q0
l There are three types of index numbers. 2
(i) Price Index Number
(ii) Quantity Index Number l Fisher’s Price Index Number
(iii) Value Index Number
∑∑ ∑∑ P01(F) = p1q0 × p1q1
p0q0 p0q1 × 100

l There are two methods of constructing l Marshall-Edgeworth’s Price Index
index numbers. Number

(i) Simple Aggregate Method ∑∑ p1 ( q0 + q1 ) 100
P01(M − E) = p0 ( q0 + q1 ) ×
(ii) Weighted Aggregate Method

l Price Index Number using Simple l Walsh’s Price Index Number
aggregate method is calculated by the
following formula. ∑∑ p1 q0q1
p0 q0q1
∑∑ P01 = p1 P01(W) = × 100
p0
× 100

88

l Cost of Living Index Number using 3. Value Index Number by Simple Aggregate
Aggregate Expenditure Method Method is given by

∑ p1q0 ∑ (a) p1q0 × 100
∑ CLI = p0q0 × 100 p0q1

l Cost of Living Index Number using ∑ (b) pp00qq10 × 100
Weighted Relative Method
∑∑ (c) pp11qq10 × 100
∑ IW
CLI = ∑W

where I = p1 × 100 ∑∑ (d) p1q1 × 100
p0 p 0 q0

and w = p0q0

4. Price Index Number by Weighted
Aggregate Method is given by
MISCELLANEOUS EXERCISE - 5
p1w
I) Choose the correct alternative. ∑ (a) p0 w × 100

1. Price Index Number by Simple Aggregate ∑ (b) pp10ww × 100
Method is given by
∑ p1w
(a) ∑∑∑ pp10 × 100 (c) ∑ p 0 w × 100

(b) ∑∑∑ pp10 × 100 (d) ∑ p0w × 100
∑ p1w
∑ p1 × 100
(c) ∑ p0
5. Quantity Index Number by Weighted
∑ p0 Aggregate Method is given by
(d) ∑ p1 × 100
∑ (a) q1w × 100
q0 w
2. Quantity Index Number by Simple
Aggregate Method is given by
∑ (b) q0 w × 100
∑ q1 q1w
(a) q0 × 100

∑ q0 × 100 (c) ∑ q1w × 100
(b) q1 ∑q0w

(c) ∑∑qq10 × 100 ∑ q0 × 100
(d) ∑ q1

∑ q0 × 100
(d) ∑ q1

89

6. Value Index Number by Weighted ∑ ∑p1q1 + p0q0
Aggregate Method is given by ∑ ∑p0q0 p1q1
(b) × 100
∑ (a) p1q0w × 100 2

p0 q0 w ∑∑ ∑∑ (c) p1q0 + p1q1
p0q0 p0q1 × 100
∑ (b) p0q1w × 100
2
p0 q0 w
∑ ∑p0q0 + p0q1
∑ p1q1w ∑ ∑ (d) p1q1 × 100
∑ (c) p0q1w × 100 p1q0
2
∑ p1q1w
∑ (d) p0q0w × 100 10. Fisher’s Price Number is given by

7. Laspeyre’s Price Index Number is given by ∑∑ ∑∑ (a) p1q0 × p1q1
p0q0 p0q1
∑∑ (a) p0q0 × 100 × 100
p1q0
∑∑ ∑∑ (b)
∑∑ (b) p0q1 × 100 p0q0 × p0q1 × 100
p1q1 p1q0 p1q1

∑∑ (c) p1q0 × 100 ∑∑ ∑∑ (c) p0q1 × p1q1 × 100
p0q0 p0q0 p1q0

∑∑ (d) p1q1 × 100 ∑∑ ∑∑ (d) p1q0 × p0q0 × 100
p0q1 p1q1 p0q1

8. Paasche’s Price Index Number is given by 11. Marshall-Edgeworth’s Price Index Number
is given by
∑∑ (a) p0q0 × 100
p1q0 ∑ p1 (q0 + q1 )
∑ (a) p0 (q0 + q1 ) × 100
∑∑ (b) p0q1 × 100
p1q1
∑∑ (b) p0 (q0 + q1 )
∑∑ (c) p1q0 p1 (q0 + q1 ) × 100
p0q0
× 100

∑∑ (d) p1q1 ∑∑ (c) q1 ( p0 + p1 ) × 100
p0q1 q0 ( p0 + p1 )
× 100

9. Dorbish-Bowley’s Price Index Number is ∑∑ (d) q1 ( p0 + p1 ) × 100
given by q0 ( p0 + p1 )

∑ ∑p1q0 + p0q1 12. Walsh’s Price Index Number is given by
∑ ∑ (a) p0q1 p1q0 × 100
2 ∑ p1 q0q1
∑ (a) p0 q0q1 × 100

90

∑∑ (b) p0 q0q1 × 100 4. Price Index Number by WeightedAggregate
p1 q0q1 Method is given by _______________.

∑∑ (c) q1 p0 p1 × 100 5. Quantity Index Number by Weighted
q0 p0 p1 Aggregate Method is given by
_______________.
∑∑ (d) q0 p0 p1 × 100
q1 p0 p1 6. ValueIndexNumberbyWeightedAggregate
Method is given by __________________.
13. The Cost of Living Index Number using
Aggregate Expenditure Method is given by 7. Laspeyre’s Price Index Number is given by
_____________.
∑∑ (a) p1q0 × 100
p0q0 8. Paasche’s Price Index Number is given by
________________.
∑ (b) p1q1 × 100
p0q1 9. Dorbish-Bowley’s Price Index Number is
given by ______________.
∑ (c) p1q1 × 100
∑ p0q1 10. Fisher’s Price Index Number is given by
______________.
∑ p1q0
11. Marshall-Edgeworth’s Price Index Number
(d) p0q0 × 100 is given by _________________.

12. Walsh’s Price Index Number is given by
_________________.

14. The Cost of Living Index Number using III) State whether each of the following is
Weighted Relative Method is given by True or False.

∑ IW ∑∑1. p1 × 100 is the Price Index Number by
(a) ∑W p0

Simple Aggregate Method. .

(b) ∑ W ∑ q0
IW 2. ∑ q1 × 100 is the Quantity Index Number

∑W by Simple Aggregate Method.
(c) ∑ IW
∑3. p0q0
IW p1q1 × 100 is Value Index Number by
W
(d) ∑ Simple Aggregate Method.

II) Fill in the blanks. ∑ p1q0

1. Price Index Number by Simple Aggregate 4. p0q0 × 100 is Paasche’s Price Index
Method is given by _______________. Number.

2. Quantity Index Number by Simple ∑∑5. p1q1 × 100 is Laspeyre’s Price Index
Aggregate Method is given by _________. p0q1

3. Value Index Number by Simple Aggregate Number.
Method is given by ________________.

91