Demand estimation

Department of Business Administration

FALL 2010-11
Demand Estimation

by
Assoc. Prof. Sami Fethi

Ch 4 : Demand Estimation

Demand Estimation


To use these important demand relationship in
decision analysis, we need empirically to
estimate the structural form and parameters of
the demand function-Demand Estimation.

 Qdx=

(P, I, Pc, Ps, T)

(-, + , - , +, +)



The demand for a commodity arises from the consumers’
willingness and ability to purchase the commodity.
Consumer demand theory postulates that the quantity
demanded of a commodity is a function of or depends on the
price of the commodity, the consumers’ income, the price of
2
related commodities, and the tastes of the consumer.

© 2004, Managerial Economics, Dominick Salvatore

© 2010/11, Sami Fethi, EMU, All Right Reserved.


Demand Estimation


In general, we will seek the answer for the
following qustions:



How much will the revenue of the firm change
after increasing the price of the commodity?
How much will the quantity demanded of the
commodity increase if consumers’ income
increase
What if the firms double its ads expenditure?
What if the competitors lower their prices?
Firms should know the answers the
abovementioned questions if they want to
achieve the objective of maximizing thier value.
3










The Identification Problem






The demand curve for a commodity is generally
estimated from market data on the quantity purchased
of the commodity at various price over time (i.e. Timeseries data) or various consuming units at one point in
time (i.e. Cross-sectional data).
Simply joinning priced-quantity observations on a graph
does not generate the demand curve for a commodity.
The reason is that each priced-quantity observation is
given by the intersection of a different and unobserved
demand and supply curve of commodity.
In other words, The difficulty of deriving the demand
curve for a commodity from observed priced-quantity
points that results from the intersection of different and
unobserved demand and supply curves for the
commodity is referred to as the identification
4
problem.



The Identification Problem



In the following demand
curve, Observed price-quantity
data points E1, E2, E3, and E4,
result respectively from the
intersection of unobserved
demand and supply curves D1
and S1, D2 and S2, D3 and S3,
and D4 and S4. Therefore, the
dashed
line
connecting
observed points E1, E2, E3, and
E4 is not the demanded curve
for the commodity. The
derived a demand curve for the
commodity, say, D2, we allow
the supply to shift or to be
different and correct, through
regression analysis, for the
forces that cause demand
curve D2 to shift or to be
different as can be seen at
5
points E2, E'2. This is done by
regression analysis.



Demand Estimation: Marketing Research Approaches








Consumer Surveys
Observational Research
Consumer Clinics
Market Experiments
These approaches are usually covered
extensively
in
marketing
courses,
however the most important of these are
consumer
surveys
and
market
experiments.
6





o



Consumer surveys: These surveys require the
questioning of a firm’s customers in an
attempt to estimate the relationship between
the demand for its products and a variety of
variables perceived to be for the marketing
and profit planning functions.
These surveys can be conducted by simply
stopping and questioning people at shopping
centre or by administering sophisticated
questionnaires to a carefully constructed
representative sample of consumers by
trained interviewers.
7









Major advantages: they may provide the
only information available; they can be made
as simple as possible; the researcher can
ask exactly the questions they want
Major disadvantages: consumers may be
unable or unwilling to provide reliable
answers; careful and extensive surveys can
be very expensive.

8






Market experiments: attempts by the
firm to estimate the demand for the
commodity by changing price and
other determinants of the demand for
the commodity in the actual market
place.

9








Major advantages: consumers are in a real
market situation; they do not know that they
being observed; they can be conducted on a
large scale to ensure the validity of results.
Major disadvantages: in order to keep cost
down, the experiment may be too limited so the
outcome can be questionable; competitors could
try to sabotage the experiment by changing
prices and other determinants of demand under
their control; competitors can monitor the
experiment to gain very useful information
about the firm would prefer not to disclose.
10




Purpose of Regression Analysis


Regression Analysis is Used Primarily
to Model Causality and Provide
Prediction
 Predict

the values of a dependent
(response) variable based on values of at
least one independent (explanatory)
variable
 Explain the effect of the independent
variables on the dependent variable
 The relationship between X and Y can be
shown on a scatter diagram
11



Scatter Diagram






It is two dimensional graph of plotted points in
which the vertical axis represents values of the
dependent variable and the horizontal axis
represents values of the independent or
explanatory variable.
The patterns of the intersecting points of
variables can graphically show relationship
patterns.
Mostly, scatter diagram is used to prove or
disprove cause-and-effect relationship. In the
following example, it shows the relationship
between advertising expenditure and its sales
revenues.
12




Scatter Diagram-Example
Year

X

Y

1

10

44

2

9

40

3

11

42

4

12

46

5

11

48

6

12

52

7

13

54

8

13

58

9

14

56

10

15

60

Managerial Economics

Scatter Diagram

13



Scatter Diagram







Scatter diagram
shows a positive
relationship between
the relevant variables.
The relationship is
approximately linear.
This gives us a rough
estimates of the linear
relationship between
the variables in the
form of an equation
such as
Y= a+ b X

14



Regression Analysis




In the equation, a is the vertical
intercept
of
the
estimated
linear
relationship and gives the value of Y
when X=0, while b is the slope of the line
and gives an estimate of the increase in
Y resulting from each unit increase in X.
The difficulty with the scatter diagram is
that different researchers would probably
obtain different results, even if they use
same data points. Solution for this is to
use regression analysis.
15




Regression Analysis







Regression analysis: is a statistical
technique for obtaining the line that best
fits the data points so that all researchers
can reach the same results.
Regression Line: Line of Best Fit
Regression Line: Minimizes the sum of the
squared vertical deviations (et) of each
point from the regression line.
This is the method called Ordinary Least
Squares (OLS).
16




Regression Analysis
Year

X

Y

1

10

44

2

9

40

3

11

42

4

12

46

5

11

48

6

12

52

7

13

54

8

13

58

9

14

56

10

15

60







In the table, Y1 refers actual or observed
sales revenue of $44 mn associated with
the advertising expenditure of $10 mn in
the first year for which data collected.
In the following graph, Y^1 is the
corresponding sales revenue of the firm
estimated from the regression line for the
advertising expenditure of $10 mn in the
first year.
The symbol e1 is the corresponding
vertical deviation or error of the actual
sales revenue estimated from the
regression line in the first year. This can
be expressed as e1= Y1- Y^1.

17



Regression Analysis




In the graph, Y^1 is
the
corresponding
sales revenue of the
firm estimated from
the regression line
for the advertising
expenditure of $10
mn in the first year.
The symbol e1 is the
corresponding
vertical deviation or
error of the actual
sales
revenue
estimated from the
regression line in the
first year. This can be
expressed as e1= Y1Y^1.

18




Regression Analysis


Since there are 10
observation points, we
have
obviously
10
vertical deviations or
error (i.e., e1 to e10). The
regression line obtained
is the line that best fits
the data points in the
sense that the sum of
the squared (vertical)
deviations from the line
is minimum. This means
that each of the 10 e
values is first squared
and then summed.

19



Simple Regression Analysis








Now we are in a position to calculate
the value of a ( the vertical intercept)
and the value of b (the slope
coefficient) of the regression line.
Conduct tests of significance of
parameter estimates.
Construct confidence interval for the
true parameter.
Test for the overall explanatory power
of the regression.
20




Simple Linear Regression Model
Regression line is a straight line that describes
the dependence of the average value of one
variable on the other
Slope
Coefficient

Y Intercept

Random Error

Yi = β 0 + β1 X i + ε i
Dependent
(Response)
Variable

Regression
Line


Independent
(Explanatory)
Variable
21


Ordinary Least Squares (OLS)

Model:

Yt = a + bX t + et
ˆ ˆ ˆ
Yt = a + bX t
ˆ
et = Yt − Yt

22




Objective: Determine the slope and
intercept that minimize the sum of
the squared errors.
n

n

n

t =1

t =1

t =1

ˆ
ˆ ˆ
et2 = ∑ (Yt − Yt ) 2 = ∑ (Yt − a − bX t ) 2
∑
23




Estimation Procedure
n

ˆ
b=

∑(X
t =1

t

− X )(Yt − Y )

n

( X t − X )2
∑

ˆ
ˆ
a = Y − bX

t =1

24




Estimation Example
Time

Xt

1
2
3
4
5
6
7
8
9
10

10
9
11
12
11
12
13
13
14
15
120

Xt − X

Yt − Y

( X t − X )(Yt − Y )

( X t − X )2

-2
-3
-1
0
-1
0
1
1
2
3

Yt

-6
-10
-8
-4
-2
2
4
8
6
10

12
30
8
0
2
0
4
8
12
30
106

4
9
1
0
1
0
1
1
4
9
30

44
40
42
46
48
52
54
58
56
60
500

n = 10

n

∑X
t =1

n

X 120
X=∑ t=
= 12
10
t=1 n

t

= 120

n

n

∑(X

∑ Yt = 500

t =1

t =1

n

Y 500
Y=∑ t =
= 50
10
t =1 n


n

∑ (X
t =1

t

t

− X ) 2 = 30

− X )(Yt − Y ) = 106

ˆ 106 = 3.533
b=
30

ˆ
a = 50 − (3.533)(12) = 7.60

25




Estimation Example
n

X t 120
X =∑
=
= 12
10
t =1 n

n = 10
n

∑ X t = 120
t =1

t =1

t =1

t =1

t

= 500

Yt 500
Y =∑ =
= 50
10
t =1 n

t

− X ) = 30

t

− X )(Yt − Y ) = 106

ˆ
a = 50 − (3.533)(12) = 7.60
26

2

n

∑(X

∑Y

n

ˆ = 106 = 3.533
b
30

n

∑(X

n




The Equation of Regression Line




The equation of the regression line can be
constructed as follows:
Yt^=7.60 +3.53 Xt



When X=0 (zero advertising expenditures),
the expected sales revenue of the firm is
$7.60 mn. In the first year, when X=10mn,
Y1^= $42.90 mn.



Strictly speaking, the regression line should be
used only to estimate the sales revenues
resulting from advertising expenditure that are
within the range.
27




Crucial Assumptions







Error term is normally distributed.
Error term has zero expected value
or mean.
Error term has constant variance in
each time period and for all values
of X.
Error term’s value in one time
period is unrelated to its value in
any other period.
28




Tests of Significance: Standard Error




To test the hypothesis that b is
statistically
significant
(i.e.,
advertising positively affects sales),
we need first of all to calculate
standard error (deviation) of b^.
The
standard
error
can
be
calculated
in
the
following
expression:
29




Tests of Significance

Standard Error of the Slope Estimate

sbˆ =

ˆ )2
∑ (Yt − Y

(n − k )∑ ( X t − X )

2

=

∑e
( n − k )∑ ( X
2
t

t

− X )2

30




Example Calculation
Time

Xt

Yt

ˆ
Yt

ˆ
et = Yt − Yt

ˆ
et2 = (Yt − Yt )2

( X t − X )2

1

10

44

42.90

1.10

1.2100

4

2

9

40

39.37

0.63

0.3969

9

3

11

42

46.43

-4.43

19.6249

1

4

12

46

49.96

-3.96

15.6816

0

5

11

48

46.43

1.57

2.4649

1

6

12

52

49.96

2.04

4.1616

0

7

13

54

53.49

0.51

0.2601

1

8

13

58

53.49

4.51

20.3401

1

9

14

56

57.02

-1.02

1.0404

4

10

15

60

60.55

-0.55

0.3025

9

65.4830

30

∑ (Y − Yˆ )
( n − k )∑ ( X − X )
2

sbˆ =

t

t

2

65.4830
=
= 0.52
(10 − 2)(30)

n

n

∑ e = ∑ (Yt − Yˆt )2 = 65.4830
t =1

2
t

t =1

Yt^=7.60 +3.53 Xt =7.60+3.53(10)= 42.90

n

∑ (X
t =1

t

− X ) 2 = 30

31



Example Calculation
n

n

t =1

t =1

ˆ
et2 = ∑ (Yt − Yt ) 2 = 65.4830
∑
n

( X t − X ) 2 = 30
∑
t =1

ˆ )2
∑ (Yt − Y

65.4830
sbˆ =
=
= 0.52
2
(n − k )∑ ( X t − X )
(10 − 2)(30)
32





Calculation of the t Statistic
ˆ
b 3.53
t= =
= 6.79
sbˆ 0.52

Degrees of Freedom = (n-k) = (10-2) = 8
Critical Value (tabulated) at 5% level =2.306
33



Confidence interval








We can also construct confidence interval
for the true parameter from the estimated
coefficient.
Accepting the alternative hypothesis that
there is a relationship between X and Y.
Using tabular value of t=2.306 for 5% and
8 df in our example, the true value of b will
lies between 2.33 and 4.73
t=b^+/- 2.306 (sb^)=3.53+/- 2.036 (0.52)
34





Decomposition of Sum of Squares
Total Variation = Explained Variation + Unexplained Variation

ˆ − Y ) 2 + ∑ (Y − Y ) 2
ˆ
(Yt − Y ) = ∑ (Y
∑
t
t
2

35



Decomposition of Sum of Squares

36



Coefficient of Determination




Coefficient of Determination: is defined
as the proportion of the total variation or
dispersion in the dependent variable that
explained by the variation in the
explanatory variables in the regression.
In our example, COD measures how
much of the variation in the firm’s sales
is explained by the variation in its
advertising expenditures.
37




Coefficient of Determination

Explained Variation
R =
=
TotalVariation
2

ˆ − Y )2
∑ (Y

(Yt − Y ) 2
∑

373.84
R =
= 0.85
440.00
2

38



Coefficient of Correlation








Coefficient of Correlation (r): The
square root of the coefficient of
determination.
This is simply a measure of the degree
of association or co-variation that
exists between variables X and Y.
In our example, this mean that
variables X and Y vary together 92%
of the time.
The sign of coefficient r is always the
same as the sign of coefficient of b^.
39





Coefficient of Correlation

ˆ
r = R 2 with the sign of b

−1 ≤ r ≤ 1
r = 0.85 = 0.92
40



Multiple Regression Analysis
Model:

Y = a + b1 X 1 + b2 X 2 + L + bk ' X k '

41



Relationship between 1 dependent & 2 or more
independent variables is a linear function
Y-intercept
Slopes
Random error

Yi = β 0 + β 1 X 1i + β 2 X 2i + L + β k X ki + ε i

Dependent (Response) variable
Independent (Explanatory) variables
42




Y i = β0 + β1X

1i +
(O b s e rv e d Y )

Y
R esponse
P la n e

X

1

β0

β2X

2i

+ εi

εi

X
( X 1 i, X 2 i)
µ Y |X = β 0 + β 1 X

1i

+ β2X

2

2i

43



Too
complicated
by hand!

Ouch!

44



Multiple Regression Model: Example
Develop a model for estimating
heating oil used for a single family
home in the month of January,
based on average temperature and
amount of insulation in inches.

Oil (Gal) Temp
Insulation
275.30
40
3
363.80
27
3
164.30
40
10
40.80
73
6
94.30
64
6
230.90
34
6
366.70
9
6
300.60
8
10
237.80
23
10
121.40
63
3
31.40
65
10
203.50
41
6
441.10
21
3
323.00
38
3
52.50
58
10

45




Multiple Regression Model: Example

ˆ
Yi = b0 + b1 X 1i + b2 X 2i + L + bk X ki
Excel Output

Intercept
X Variable 1
X Variable 2

Coefficients
562.1510092
-5.436580588
-20.01232067

ˆ
Yi = 562.151 − 5.437 X 1i − 20.012 X 2i
For each degree increase in
temperature, the estimated
average amount of heating oil
used is decreased by 5.437
gallons, holding insulation
constant.

For each increase in one inch of
insulation, the estimated average use
of heating oil is decreased by 20.012
gallons,
holding
temperature
constant.

46



Adjusted Coefficient of Determination
(n − 1)
R = 1 − (1 − R )
(n − k )
2

2

Regression Statistics
Multiple R
0.982654757
R Square
0.965610371
Adjusted R Square
0.959878766
Standard Error
26.01378323
Observations
15

rY2,12

SSR
=
SST

47



Interpretation of Coefficient of Multiple Determination

96.56% of the total variation
in heating oil can be explained
by temperature and amount of
insulation


2
Y •12

r

SSR
=
= .9656
SST

95.99% of the total fluctuation
in heating oil can be explained
by temperature and amount of
insulation after adjusting for the
number of explanatory
variables and sample size


2
radj =
.9599

48



Testing for Overall Significance





Shows if Y Depends Linearly on All of the X
Variables Together as a Group
Use F Test Statistic
Hypotheses:






H0: β1 = β2 = … = βk = 0 (No linear relationship)
H1: At least one βi ≠ 0 ( At least one independent
variable affects Y )

The Null Hypothesis is a Very Strong Statement
The Null Hypothesis is Almost Always Rejected
49




Analysis of Variance and F Statistic

Explained Variation /(k − 1)
F=
Unexplained Variation /(n − k )

R /(k − 1)
F=
2
(1 − R ) /(n − k )
2

50



k = 3, no of
parameters

Test for Overall Significance
Excel Output: Example

ANOVA
df
Regression
Residual
Total

k -1= 2, the number

SS
MS
F
Significance F
2 228014.6 114007.3 168.4712
1.65411E-09
12 8120.603 676.7169
14 236135.2

n-1

of explanatory
variables and
dependent variable

p-value

51


Test for Overall Significance:
Example Solution
H0: β1 = β2 = … = βk = 0
H1: At least one βj ≠ 0
α = .05

Test Statistic:

F

= 168.47

df = 2 and 12

Critical Value :
α = 0.05
0

3.89


Decision:
Reject at α = 0.05.
Conclusion:

There is evidence that at
least one independent
variable affects Y.
52


t Test Statistic
Excel Output: Example

t Test Statistic for X1
(Temperature)
Coefficients Standard Error
t Stat
Intercept
562.1510092
21.09310433 26.65094
Temp
-5.436580588
0.336216167 -16.1699
Insulation -20.01232067
2.342505227 -8.543127

bi
t=
Sbi

P-value
4.77868E-12
1.64178E-09
1.90731E-06

t Test Statistic for X2
(Insulation)
53



t Test : Example Solution


Does temperature have a significant effect on monthly
consumption of heating oil? Test at α = 0.05.
Test Statistic:

H0: β1 = 0

t Test Statistic = -16.1699

H1: β1 ≠ 0

Decision:

df = 12

Critical Values:

Conclusion:

Reject H 0 There is evidence of a significant

Reject H 0

.025
-2.1788

Reject H0 at α = 0.05.

.025
0

2.1788


effect of temperature on oil
consumption holding constant the
effect of insulation.
54


Problems in Regression Analysis










Multicollinearity: Two or more explanatory
variables are highly correlated.
Heteroskedasticity: Variance of error term is not
independent of the Y variable.
Autocorrelation: Consecutive error terms are
correlated.
Functional form: Misspecified by the omission of
a variable
Normality: Residuals are normally distributed or
not
55




Practical Consequences of Multicollinearity










Large variance or standard error
Wider confidence intervals
Insignificant t-ratios
A high R2 value but few significant tratios
OLS estimators and their Std. Errors
tend to be unstable
Wrong signs for regression coefficients
56




Multicollinearity









How can Multicollinearity be overcome?
Increasing number of observation
Acquiring additional data
A new sample
Using an experience from a previous study
Transformation of the variables
Dropping a variable from the model
This is the simplest solution, but the worse
one referring an economic model (i.e., model
specification error)
57




Heteroskedasticity




Heteroskedasticity: Variance of error term is
not independent of the Y variable or
unequal/non-constant variance. This means
that when both response and explanatory
variables increase, the variance of response
variables does not remain same at all levels of
explanatory variables (cross-sectional data).
Homoscedasticity: when both response and
explanatory variables increase, the variance of
response variable
around its mean value
remains same at all levels of explanatory
variables (equal variance).
58




Residual Analysis for Homoscedasticity
Y

Y

X

SR

X

SR

X

Heteroscedasticity

X

Homoscedasticity
59


Autocorrelation or serial correlation




Autocorrelation: Correlation between
members of observation ordered in
time as in time series data (i.e.,
residuals
are
correlated
where
consecutive errors have the same
sign).
Detecting Autocorrelation: This can be
detected by many ways. The most
common used is DW statistics.
60




Durbin-Watson Statistic

Test for Autocorrelation
n

d=

(et − et −1 ) 2
∑
t =2

n

et2
∑
t =1

If d=2, autocorrelation is absent.
61



Residual Analysis for Independence
 The

Durbin-Watson Statistic
– Used when data is collected over time to detect
autocorrelation (residuals in one time period are
related to residuals in another period)
– Measures violation of independence assumption
n

D=

(ei − ei −1 ) 2
∑
i =2

n

∑e
i =1

2
i


Should be close to 2.

If not, examine the model
for autocorrelation.
62


Residual Analysis for Independence
Graphical Approach
Not Independent



e

Independent

e
Time

Cyclical Pattern

Time

No Particular Pattern

Residual is Plotted Against Time to Detect Any Autocorrelation
63



Using the Durbin-Watson Statistic
H0

: No autocorrelation (error terms are independent)

H1

: There is autocorrelation (error terms are not)

Reject H0
(positive
autocorrelation)

0

dL

Inconclusive
Accept H0
(no autocorrelation)

dU


2

4-dU

Reject H0
(negative
autocorrelation)

4-d
64 L

4



Steps in Demand Estimation







Model Specification: Identify Variables
Collect Data
Specify Functional Form
Estimate Function
Test the Results

65



Functional Form Specifications

Linear Function:
QX = a0 + a1PX + a2 I + a3 N + a4 PY + L + e

Power Function:
b
QX = a ( PX1 )( PYb2 )

Estimation Format:
ln QX = ln a + b1 ln PX + b2 ln PY
66




Dummy-Variable Models


When the explanatory variables are
qualitative in nature, these are known
as dummy variables. These can also
defined as indicators variables, binary
variables, categorical variables, and
dichotomous
variables
such
as
variable D in the following equation:

Q x = c 0 + c1 Px + c 2 I + c3 D + ...... + e
67





Categorical Explanatory Variable with 2 or More Levels



Yes or No, On or Off, Male or Female,



Use Dummy-Variables (Coded as 0 or 1)



Only Intercepts are Different



Assumes Equal Slopes Across Categories



Regression Model Has Same Form



Can the dependent variable be dummy?
68




Given:

ˆ
Yi = b0 + b1 X 1i + b2 X 2i

Y = Assessed Value of House
X1 = Square Footage of House
X2 = Desirability of Neighbourhood =

0 if undesirable
1 if desirable

Desirable (X2 = 1)

ˆ
Yi = b0 + b1 X 1i + b2 (1) = (b0 + b2 ) + b1 X 1i
Undesirable (X2 = 0)

ˆ
Yi = b0 + b1 X 1i + b2 (0) = b0 + b1 X 1i

Same
slopes
69




Simple and Multiple Regression Compared: Example

70



71



72



73



Regression Analysis in Practice











Suppose we have an
Employment (Labor
Demand) Function as follows:
N=Constant+K+W+AD+P+WT
N: employees in employment
K: capital accumulation
W: value of real wages
AD: aggregate deficit
P: effect of world manufacturing exports on
employment
WT: the deviation of world trade from trend.
74




Output by Microfit v4.0w
Ordinary Least Squares Estimation
*******************************************************************************
Dependent variable is LOGN
39 observations used for estimation from 1956 to 1994
*******************************************************************************
Regressor
Coefficient
Standard Error
T-Ratio[Prob]
CON
4.9921
.98407
5.0729[.000]
LOGK
.040394
.012998
3.1078[.004]
LOGW
.024737
.010982
2.2526[.032]
AD
-.9174E-7
.1587E-6
.57798[.567]
LOGP
.026977
.0099796
2.7032[.011]
LOGWT
-.053944
.024279
2.2219[.034]
*******************************************************************************
R-Squared
.82476
F-statistic F( 6, 33)
20.8432[.000]
R-Bar-Squared
.78519
S.E. of Regression
.012467
Residual Sum of Squares
.0048181
Mean of Dependent Variable
10.0098
S.D. of Dependent Variable
.026899
Maximum of Log-likelihood
120.1407
DW-statistic
1.8538
*******************************************************************************

75


Diagnostic Tests
******************************************************************************
*
*
Test Statistics
*
LM Version
*
F Version
*
******************************************************************************
*
*
*
*
*
* A:Serial Correlation *CHI-SQ( 1)= .051656[.820]*F(1,30)=.039788[.843]*
*
*
*
*
* B:Functional Form
*CHI-SQ( 1)= .056872[.812]*F(1,30)=.043812[.836]*
*
*
*
*
* C:Normality
*CHI-SQ( 2)=
1.2819[.527]*
Not applicable
*
*
*
*
*
* D:Heteroscedasticity *CHI-SQ( 1)=
1.0065[.316]*F( 1,37)=.98022[.329]*
******************************************************************************
*
A:Lagrange
B:Ramsey's
C:Based on
D:Based on

multiplier test of residual serial correlation
RESET test using the square of the fitted values
a test of skewness and kurtosis of residuals
the regression of squared residuals on squared fitted values

76



Dependent Variable: LOGN
Explanatory Variables
CON
LOGK
LOGW
AD
LOGP
LOGWT
R2 bar
R2

4.9921
(5.07)
0.40394
(3.10)
0.0247
(2.25)
-0.9174
(-0.577)
0.0269
(2.70)
-0.0539
(-2.22)
0.87
0.83

DW

2.16

SER

0.021

X

2
SC

X
X

2

2

.05165[.820]

FF

05687[.812]

NORM

1.2819[.527]

X2HET


77

1.0065[.316]



Interpretation
t-test (individual significance)
Let’s first see the significance of each
variable;
n=39
k=6 and hence d.f.=39-6=33
α=0.05 (our confidence level is 95%).
With α=0.05 and d.f.=33, ttab=2.045
Our Hypothesis are:
Ho:βs=0 (not significant)

H1: βs≠0 (significant)
This is t- distribution and using this
distribution, you can decide whether
individual t-values (calculated or
estimated) of the existing variables are
significant or not according to the
tabulated t-values as appears in the fig
above.

78


F-test (overall significance)

Our result is F(6,33)=20.8432
k-1=5 and n-k=33
α= 0.05 (our confidence level is
95%).
With α= 0.05 and F(6,33), the
Ftab=2.34
Our hypothesis are

Ho:R2s=0 (not significant)
H1: R2≠0 (significant)
79



Diagnostic Tests:
Serial Correlation:
Ho:ρ=0(existence of autocorrelation )
H1:ρ≠0 (no autocorrelation)

Since
CHI-SQ(1)=0.051656<
X2=3.841, we reject Ho that
estimate regression does not have
first order serial correlation or
autocorrelation.
80



Functional Form:
Ho:γ=0 (existence misspecification)
H1: γ≠0 (no of misspecification)
The estimated LM version of CHISQ is 0.0568721 and with α= 0.05
the tabular value is X2=3.841.
Because CHI-SQ (1)=0.0568721<
X2=3.841, then we reject the null
hypothesis that there is functional
misspecification.
81













Normality:
Ho:ut=0
(residuals
normally distributed)
H1:ut≠0(residuals are
distributed)

are

not

normally

Our estimated result of LM version
for
normality
is
CHISQ(2)=1.28191, and the tabular
value with 2 restrictions with α=
0.05 is X2=5.991.
Since
CHI-SQ(2)=1.28191<
X2=5.991, the test result shows that
the null hypothesis of normality of
the residuals is accepted.


82









Heteroscedasticity:
Ho:σ yt2=σ 2 (heteroscedasticity)
H1: σ yt2≠σ 2(homoscedasticity)

LM version of our result for
the heteroscedasticity is CHISQ(1)=1.00651 and table
critical
value
with
1
restriction with α= 0.05 is
X2=3.841.
Since
CHISQ(1)=1.00651< X2=3.841,
we accept the null hypothesis
that error term is constant
for
all
the
independent
variables.


83


The End

Thanks

84


Demand estimation

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