Econometrics Assignment Help

Econometrics Assignment Help

Economics Help Desk Mark Austin

Contact Us:

Web: http://economicshelpdesk.com/

Email: [email protected]

Twitter: https://twitter.com/econ_helpdesk Facebook: https://www.facebook.com/economicshelpdesk Tel: +44-793-744-3379

Copyright © 2012-2015 Economicshelpdesk.com, All rights reserved

http://economicshelpdesk.com/

mailto:[email protected]

https://twitter.com/econ_helpdesk

https://www.facebook.com/economicshelpdesk


About Economics Help Desk:

At Economicshelpdesk.com we offer all sorts of

assistance needed in econometrics subject. As we

all know how tough is econometrics. With the use

of statistical concepts and applications of statistical

software’s, it becomes very difficult to grasp this

subject. We have a dedicated team of economics

and statistics experts who can help you to solve

various tasks involved in econometrics. We offer

econometrics assignment help in all topics

including probability approach, conditional

expectation, regression, least squares regression,

analysis of variance (ANOVA), restricted estimation,

hypothesis testing, regression extension, bootstrap, endogeneity, matrix algebra, numerical

optimization etc.

Why Choose Us?

Accuracy: We are a company employed with highly qualified Economics experts to ensure

fast and accurate homework solutions aimed at any difficult homework – both Micro

economics and Macro economics.

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.

http://economicshelpdesk.com/econometrics-assignment-homework-help.php


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 X2 = 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


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 𝑋2 ….(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


∴ 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 𝜎𝑥

𝜎𝑦

∴ bxy = r 𝜎𝑥

𝜎𝑦


Where, 𝜎𝑥 = standard deviation of X variable

𝜎𝑌 = standard deviation of Y variable

and byx = regression coefficient of Y on X i.e., r 𝜎𝑥

𝜎𝑦

∴ byx = r 𝜎𝑥

𝜎𝑦

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


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 𝑥2 = 140 𝑌

2 = 398 𝑥𝑦 = -93

(a)(i) Regression equation of X on Y

This is given by

X = X + r 𝜎𝑥

𝜎𝑦 (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

𝑁𝜎𝑥𝜎𝑦 = -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


X = 40.8806 – 0.2337Y

Aliter

Substituting the regression coefficient of X on Y

i.e., r 𝜎𝑥

𝜎𝑦 by

𝑥𝑦

𝑦2 we get,

X = 𝑋 + 𝑥𝑦

𝑦2 (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 𝜎𝑦

𝜎𝑥 (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 𝜎𝑦

𝜎𝑥 by

𝑥𝑦

𝑥2 we get,

y = 𝑦 + 𝑥𝑦

𝑥2 (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


(b) Coefficient of Correlation

The coefficient of correlation between the two variables, X and Y is given by

rxy = 𝑥𝑦

𝑁𝜎𝑥𝜎𝑦 = -93/10 × 3.74 × 6.31

= -93/235.994 = -0.394

Alternatively

By the method of regression coefficients we have,

rxy = 𝑏𝑥𝑦 × 𝑏𝑦𝑥

= 𝑥𝑦

𝑦2 ×

𝑥𝑦

𝑥2

= −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.

Documents

Econometrics Assignment Help