Econometrics homework help

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Econometrics Assignment Sample Questions and Answers:
Question 1: From the data given below, obtain the two regression coefficients and the two
regression equations using the value based method:
X :
Y :
2
5
4
7
6
9
8
8
10
11
Solution:
Computation of the Regression Coefficients and the Regression Equations by the
Value Based Method
X Y X2
Y2
XY
2
4
6
8
10
5
7
9
8
11
4
16
36
64
100
25
49
81
64
121
10
28
54
64
110
X = 30 Y = 40 X
2
= 220 Y
2
=
340
XY =
266
N = 5
(a) Regression Coefficients
(i) bxy =
𝑁 XY − X. Y
𝑁 𝑌2− ( 𝑌)
2

=
5 ×266 − (30 ×40)
5 ×340 − (40)2
=
1330 −1200
1700 −1600
=
130
100
= 1.3
(i) byx =
𝑁 𝑋𝑌− 𝑋. 𝑌
𝑁 𝑋
2
− ( 𝑋)
2
=
5 ×266 − (30 ×40)
5 ×220 − (30)2
=
130
(1100 −900)
= 130/200 = 0.65
(b) Regression Equations
(i) Regression Equation of X on Y
This is given by X = X + bxy (Y - Y)
Where, X =
𝑋
𝑁
=
30
5
= 6
And Y =
𝑌
N
=
40
5
= 8
Thus, X = 6 + 1.3 (Y – 8)
= 6 – 10.4 + 1.3Y
= -4.4 + 1.3Y
∴ X = -4.4 + 1.3Y
(ii) Regression Equation of Y on X
This is given by Y = Y + byx (X - X)
= 8 + 0.65 (X – 6)
= 8 – 3.9 + 0.65 X
∴ Y = 4.10 + 0.65X
3. Method of Deviation from Assumed Mean
This method is also otherwise known as short-cut method. Under this method, the
regression between any two related variables is studied on the basis of the deviations of the
items from their respective assumed Means rather than their deviations of the items from
their respective assumed Means rather than their actual values or deviations from their
respective actual Means.
The formula for the two regression equations remain the same as cited above under the
method of deviation from the actual Means except that the two regression coefficients are
determined by the following methods :

bxy =
𝑁 𝑑 𝑥 .𝑑 𝑦 − 𝑑 𝑥 . 𝑑 𝑦
𝑁 𝑑 𝑥2− ( 𝑑 𝑥2 )
byx =
𝑁 𝑑 𝑥 .𝑑 𝑦 − 𝑑 𝑥 . 𝑑 𝑦
𝑁 𝑑 𝑥2− ( 𝑑 𝑥2 )
In the above formulae,
N = number of pairs of the observations,
dx = deviation of items of X series from its assumed Mean,
dy = deviation of items of Y series from its assumed Mean, and all other
factors carry the same meanings as explained before.
Notes. 1. While taking deviations, if each of them has been divided by a common factor to
reduce its magnitude, then each of the above formula will be modified slightly as under:
The formula of bxy will be multiplied by
𝑖 𝑥
𝑖 𝑦
, and the formula of byx will be multiplied by
𝑖 𝑦
𝑖 𝑥
.
Where ix = common factor of X deviations
And iy = common factor of Y deviations.

Regression of Frequency Distribution
2. In case of grouped data given in two-way frequency tables, the above formulae of the
regression coefficients will be slightly modified as under:
bxy =
𝑁 F𝑑 𝑥 𝑑 𝑦 − Fd 𝑥 . Fd 𝑦
Fd 𝑦2− ( Fd 𝑦2)
×
𝑖 𝑥
𝑖 𝑦
byx =
𝑁 F𝑑 𝑥 𝑑 𝑦 − Fd 𝑥 . Fd 𝑦
Fd 𝑥2− ( Fd 𝑥2)
×
𝑖 𝑦
𝑖 𝑥
In the above formula, the various factors involved carry the same meanings as explained in
the previous formulae except that each factor is multiplied by its respective frequency.
The above formulae are multiplied respectively by
𝑖 𝑥
𝑖 𝑦
, and
𝑖 𝑦
𝑖 𝑥
as shown above interval of the
respective series. This adjustment in the formula is done because; the regression
coefficients are affected by the change of scale thought not by the change of their origin.
It may be noted that the short-cut method explained above is used only when the actual
Means of the two variables are in fraction. However, this method also, can be used when
the Means of the variables are not in fraction.
Question 2:
Obtain the regression equations of X on Y, and of Y on X from the following data by the
short-cut method.
X :
Y :
1
6
5
1
3
0
2
0
1
1
1
2
7
1
3
5
Also, determine r, and predict Y, when X = 10 and X, when Y = 2.5
Solution:
Regression Analysis by the short-cut method
X Y (X – 3)
dx
(Y – 2)
dy
dx
2
dy
2
dxdy
1
5
3
2
1
1
7
3
6
1
0
0
1
2
1
5
-2
2
0
-1
-2
-2
4
0
4
-1
-2
-2
-1
0
-1
3
4
4
0
1
4
4
16
0
16
1
4
4
1
0
1
9
-8
-2
0
2
2
0
-4
0
𝑋 = 23 𝑌 =
16
𝑑 𝑥 = -
1
𝑑 𝑦 = 0 𝑑 𝑥
2
=
33
𝑑 𝑦
2
=
36
𝑑 𝑥 𝑑 𝑦 = -
10
N = 8

Under the short-cut method we have
(i) bxy =
N dx dy− dx. dy
N dy2 − ( dy2 )
Putting the respective values in the above we get,
bxy =
8 −10 − (−1 ×0)
8 36 − (0)2
= -80/288 = -5/18 = -0.28 approx.
(ii) byx =
N 𝑑 𝑥 𝑑 𝑦 − 𝑑 𝑥 𝑑 𝑦
𝑑
𝑥
2−( 𝑑
𝑥2)
Putting the respective values in the above we get,
byx =
8 −10 − −1 (0)
8 33 − (−1)2
= -80/264 – 1 = -80/263 = -0.3 app.

(iii) Regression Equation of X on Y
This is given by
X = X + bxy (Y - Y)
Where, X = Ax +
𝑑 𝑥
𝑁
= 3 + -1/8 = 3 – 0.125 = 2.875
and Y = Ay +
𝑑 𝑦
N
= 2 + 0/8 = 2
Note. Ax = assumed Mean of X variable i.e. 3
Ay = assumed Mean of Y variable i.e. 2
X = actual Mean of X variable
Y = actual Mean of Y variable.
Thus, X = 2.875 + -0.28 (Y – 2)
= 2.875 + 0.56 – 0.28Y
= 3.435 – 0.28Y
∴ X = 3.435 – 0.28Y
(iv) Regression Equation of Y on X
This is given by
Y = Y + byx (X - X)
Putting the respective values in the above formula we get,
Y = 2 + (-0.3) (X – 2.875)
= 2 + 0.8625 – 0.3X = 2.8625 – 0.3X
∴ Y = 2.8625 – 0.3X
(v) Coefficient of Correlation
This is given by
r = 𝑏 𝑥𝑦 . 𝑏 𝑦𝑥 = −0.28 (−0.3)
= ± 0.084 = -0.29
Note. Since, the regression coefficients are negative, the correlation coefficient will be
negative.

(vi) Value of Y, when X = 10
Using the regression equation of Y on X we get,
Y = 2.8625 – 0.3 (10)
= 2.8625 – 3 = -0.1375
(vii) Value of X, when Y = 2.5
We have, X = 3.435 – 0.28 (2.5)
= 3.435 – 0.700 = 2.735

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