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Econometrics Assignment Sample Questions and Answers:
Question 1: From the following data from the regression equations,
Ye = a + bX and Xe = a + bY
Use the normal equation method:
X :
Y :
1
15
3
18
5
21
7
23
9
22
Also, estimate the value of Y when X = 4, and the value of X when Y = 24.

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Solution:
Formulation of the regression equations
X Y X2
Y2
XY
1
3
5
7
9
15
18
21
23
22
1
9
25
49
81
225
324
441
529
484
15
54
105
161
198
Total X = 25 Y = 99 X
2
= 165 Y
2
=
2003
XY = 533
N = 5
(i) Regression equation of X on Y. This is given by
Xe =a + bY
To find the values of the constant a and b in the above formula, the following two normal
equations are to be simultaneously solved:
X = Na + b Y
XY = a Y + b Y2
Substituting the respective values in the above formula we get,
25 = 5a + 99b
533 = 99a + 2003b
Multiplying the equation (i) by 99 and eqn. (ii) by 5 and presenting them in the form of a
subtraction we get,
2475 = 495a + 9801b
(-)2665 = 495a + 10015b/-190 = -214b
or 214b = 190
∴ b = 190/214 = .888 approx.
Putting the above values of b in the eqn. (i) we get,
25 = 5a + 99(.888)
or 5a = 25 – 87.912 = -62.912
∴ a = -62.912/5 = -12.5824
Thus, a = -12.5824 and b = 01.888.
Substituting the above values of the constants a and b, we get the regression equation of X
on Y as,
Xe = -12.5824 + 0.888 Y

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Thus, when Y = 24, Xe = -12.5824 + 0.888 (24)
= -12.5824 + 21.312
= 8.7296.
(ii) Regression equation Y on X. This is given by
Ye = a + bX
To find the values of the constants a and b in the above formula, the following two normal
equations are to be simultaneously solved as under:
𝑌 = Na + b 𝑋 ….(i)
𝑋𝑌 = a 𝑋 + b 𝑋
2
….(ii)
Substituting the respective values in the above formula we get,
99 = 5a + 25b
….(i)
533 = 25a + 165b ….(ii)
Multiplying the equation (1) by 5 and getting the same subtracted from the equation (ii) we
get,
533 = 25a + 165b ….(ii)
(-)495 = 25a + 125b ….(iii)
Thus, 38 = 40b

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∴ b = 38/40 = 0.95
Putting the above values of b in the equation (i) we get,
99 = 5a + 25(0.95)
or 5a = 99 = 23.75 = 75.25
∴ a = 75.25/5 = 15.05
Thus, a = 15.05 and b = 0.95
Substituting the above values of the two constants a and b we get the regression equation
of Y on X as,
Ye = 15.05 + 0.95X
Thus, when X = 4, Ye = 15.05 + 0.95(4)
= 15.05 + 3.80
= 18.85.
Note. It may be noted that the above normal equation method of formulating the two
regression equations is very lengthy and tedious. In order to do away with such difficulties,
any of the following two methods of deviation may be used advantageously.
2. Method of deviation from the Means
Under this method, the two regression equation are developed in a modified form from the
deviation figures of the two variables from their respective actual Means rather than their
actual values. For this, the two regression equations are modified as under:
(i) Regression equation of X on Y. This is given by
X = X + bxy (Y - Y), or X - X = bxy (Y - Y)
(ii) Regression equation of Y on X. This is given by
Y = Y + byx (X - X), or Y - Y = byx (X - X)
In the above formulae,
X = given value of the X variable
Y = given value of the Y variable
X = arithmetic average of the X variable,
Y = arithmetic average of the Y variable,
bxy = regression coefficient of X on Y i.e., r
𝜎 𝑥
𝜎 𝑦
∴ bxy = r
𝜎 𝑥
𝜎 𝑦

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Where, 𝜎𝑥 = standard deviation of X variable
𝜎𝑌 = standard deviation of Y variable
and byx = regression coefficient of Y on X i.e., r
𝜎 𝑥
𝜎 𝑦
∴ byx = r
𝜎 𝑥
𝜎 𝑦
Question 2: Using the method of deviations from the actual Means from the data given
below find.
(i) the two regression equations
(ii) the correlation coefficient and
(iii) the most probable value of Y when X = 30
X :
Y :
25
43
28
46
35
49
32
41
31
36
36
32
29
31
38
30
34
33
32
39

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Solution:
Determination of the regression equations
by the method of deviation from the Means
X Y (X – 32)
x
(Y – 38)
Y
X2
Y2
Xy
25
28
35
32
31
36
29
38
34
32
43
46
49
41
36
32
31
30
33
39
-7
-4
3
0
-1
4
-3
6
2
0
5
8
11
3
-2
-6
-7
-8
-5
1
49
16
9
0
1
16
9
36
4
0
25
64
121
9
4
36
49
64
25
1
-35
-32
33
0
2
-24
21
-48
-10
0
𝑋 = 320 𝑌 = 380 𝑋 = 0 𝑌 = 0 𝑥
2
= 140 𝑌
2
= 398 𝑥𝑦 = -93
(a)(i) Regression equation of X on Y
This is given by
X = X + r
𝜎 𝑥
𝜎 𝑦
(Y - Y)
where X = X/N = 320/10 = 32
𝑌 =
𝑌
𝑁
= 380/10 = 38
𝜎𝑥 =
𝑥2
𝑁
=
140
10
= 3.74 approx.
𝜎𝑦 =
𝑦2
𝑁
=
398
10
= 6.31 approx.
and r =
𝑥𝑦
𝑁𝜎 𝑥 𝜎 𝑦
= -93/10 × 3.74 × 6.31
= -93/10 × 23.5994 = -93/235.99 = -0.394
Putting the respective values in the above equation we get,
X = 32 + -0.394 × -93/235.99 = -0.394
Putting the respective values in the above equation we get,
X = 32 + -0.394 × 3.74/6.31 (Y – 38)
= 32 – 0.2337 (Y – 38)
= 32 + 8.8806 – 0.2337 Y

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X = 40.8806 – 0.2337Y
Aliter
Substituting the regression coefficient of X on Y
i.e., r
𝜎 𝑥
𝜎 𝑦
by
𝑥𝑦
𝑦
2 we get,
X = 𝑋 +
𝑥𝑦
𝑦
2 (Y - 𝑌)
= 32 + -93/398 (Y – 38)
= 32 – 0.2337 (Y – 38) = 32 + 8.8806 – 0.2337Y
= 40.8806 – 0.2337Y
(ii) Regression equation of Y on X
This is given by Y = 𝑌 + r
𝜎 𝑦
𝜎 𝑥
(X - X)
Substituting the respective values in the above we get,
Y = 38 + -0.394 × 6.31/3.74 (X – 32)
= 38 – 0.6643 (X – 32)
= 38 + 21.2576 – 0.6643X
= 59.25876 – 0.6643X
∴ Y = 59.2576 – 0.643 X
Aliter
Replacing the formula of regression coefficient.
r
𝜎 𝑦
𝜎 𝑥
by
𝑥𝑦
𝑥
2 we get,
y = 𝑦 +
𝑥𝑦
𝑥
2 (X - X)
= 38 + -93/140 (X – 32)
= 380 – 0.6643 (X – 32) = 38 + 21.2576 – 0.6643X
∴ Y = 59.2576 – 0.6643X
Thus, the two regression equations are:
X on Y : X = 40.8806 – 0.2237 Y
and Y on X : Y = 59.2576 – 0.6643 X

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(b) Coefficient of Correlation
The coefficient of correlation between the two variables, X and Y is given by
rxy =
𝑥𝑦
𝑁𝜎 𝑥 𝜎 𝑦
= -93/10 × 3.74 × 6.31
= -93/235.994 = -0.394
Alternatively
By the method of regression coefficients we have,
rxy = 𝑏𝑥𝑦 × 𝑏 𝑦𝑥
=
𝑥𝑦
𝑦
2 ×
𝑥𝑦
𝑥
2
= −0.2337 × −0.6643
= 0.1552 = -0.394
Note. Since the regression coefficients are negative, the correlation coefficient has been
negative.
(c) Probable value of Y when X = 30
This will be determined by the regression equation of Y on X as follows :
We have, Y = 59.2576 – 0.6643X
Thus, when X = 30, Y = 59.257 – 0.6643(30)
= 59.2576 – 19.929 = 39.3286.