Transcript Step 2 - City University

Multivariate Regression Model
y = b0 + b1x1 + b2x2 + b3x3 +… + e
y is the DEPENDENT variable
Each of the xj is an INDEPENDENT variable
The OLS estimates b0,b1 ,b2 , b3 .. ….
are sample statistics used to estimate
b0 , b1, b2 , b3 .... respectively
Conditions:
Each explanatory variable Xj is assumed
(1A) to be deterministic or non-random
(1B) : to come from a ‘fixed’ population
(1C) : to have a variance V(xj) which is
not ‘too large’
The above assumptions are best suited
to a situation of a controlled
experiment
Assumptions concerning the random
term ei :
(IIA) E(ei ) = 0 for all i
(IIB) Var(ei) = s2 = constant for all i
(IIC) Covariance (ei , ek) = 0 for any i
and k
(IID) Each of the ei has a normal
distribution
Properties of b0 , b1 , b2 , b3
1. Each of these statistics is a linear
functions of the Y values.
2. Therefore, they all have normal
distributions
3. Each is an unbiased estimator.
That is, E(bk) = bk;
4. Each bk is the most efficient
estimator of all unbiased
estimators.
Thus, each of b0 , b1 , b2 ….is
Best
Linear
Unbiased
Estimator of the respective parameter
Conclusion
Each estimator bi has a normal
distribution with mean = bi and
variance = sbi2 where sbi2 is
unknown.
Income (£ per week) of an individual is
regressed on a constant, education
(in years), age (in years) and wealth
inheritance (in £), using EViews.
Number of observations is 20 and the
regression output is given below:
Variable Coefficient Std.Error t-Stats Prob.
C
-1001.87
520.71 -1.92 0.0654
AGE
8.85
5.45
1.62 0.1168
EDUCATION 95.17
38.54
2.46 0.0252
0.46
3.26 0.0031
WEALTH
1.51
The Maximum
Type 1 Error
= Significance
Level
Significance
Level (a)
p-value
The smaller the p-value
the more significant is the test
The proposed regression model is:
Income = ß0 + ß1(Age) +ß2(Education)
+ ß3(Wealth Inheritance)
…..
(A)
We are proposing that Income is the
variable dependent on three independent
variables: Age, Education and Wealth.
b0 is a constant.
It measures the effect of other
deterministic factors on Income not
included in the model.
b1 , b2, b3 measure the effect of a marginal
change in Age, Education and Wealth,
respectively.
However, we recognise that there may be other
random factors affecting the dependent
variable Income.
So we add a random variable e to the model
which now becomes:
Income = ß0 + ß1(Age) +ß2(Education)
+ ß3(Wealth Inherited) + e
… . . (B)
We use the least squares technique
to estimate the model B.
Therefore, our estimation of the proposed
model B is
Ye = -1001.87 + 8.85*AGE
+ 95.17*EDUCATION
+ 1.51*WEALTH INHERITANCE
Here Ye is the estimated value of income
-1001.87 is the estimate of ß0, 8.85 is
the estimate of ß1,; 95.17 is the estimate
of ß2 and 1.51 is the estimate of ß3
The least-squares estimates of the ßvalues are denoted by b-values. Thus,
b1 is the estimate of ß1 and b2 is the
estimate of ß2 . In our case, b1 = 8.85
and b2 = 95.17.
We next make the following
assumptions on the specification of
model B so that the least-squares
method produces ‘good’ estimators.
i.
e is normally distributed with
mean 0 and an unknown
variance s2e .
In the context of the model B, e can be thought
of as a luck factor which can be good
(positive values) or bad (negative values),
If the positive and negative values cancel out
on average, we can say that mean value is 0.
The e values are uncorrelated across the
population
(Whether or not you are lucky does not
influence my being lucky/unlucky)
i. The e values have the same variance (s2e)
across it. (Every individual is exposed to the
same extent/chance of good or bad luck)
The e values are uncorrelated with
the independent variables Age,
Education and Wealth Inheritance.
(For example, an old person is as likely to
be lucky as a young one;
or a university graduate is as likely to be
unlucky as someone with no A-levels).
We now test (at 10% significance) the
following hypothesis:
Education has a positive effect on income
Step 1: Set up the hypotheses
H0 : ß2 = 0 (Education has no effect)
H1 : ß2 > 0(Education has a positive
effect)
one-tailed test
Step 2: Select statistic
The estimator b2 is the test-statistic
Step3 : Identify the distribution
of b2
Assumptions i-iii above imply
that b2 is
Best
Linear in the dependent variable income
Unbiased
Estimator of b2
Since b2 is unbiased, E(b2) = b2
b2 has a normal distribution because it
is linear in Income
Thus, b2~ N(b2, s22) where s22 is unknown.
Step 4: Construct test statistic
We use the standard error of b2
because we do not know what s22 is
Therefore, the test statistic is
t  (b2- b2) / (standard error of b2)
has a Student’s t-distribution with
20-4 = 16 d.o.f.
As b2 = 0 under the null hypothesis (H0)
t = b2 / (standard error of b2)
EViews therefore gives us a t-statistic
regarding education of 2.46907
The corresponding probability value
is 0.0252.
Select fx /TDIST. For X, enter 2.469607,
the t-Statistic value. The degree of freedom
is 16. EViews calculates two-tail probability
So number of tails is 2. You now get the
2-tail probability of 0.025165 from Excel.
Since we are performing a one-tail test,
take half the probability value, or 0.0126 .
Step 5: Compare with critical value tC
tC = 1.336757
for a one-tailed test with
significance level (a) = 0.1
and d.o.f. = 16
tC = 1.336757 < 2.469607
Step 6 : Draw conclusion
The test is significant. Reject H0 at 10%
and at 5% (1.745884 < 2.469607) but
not at 1% (2.583492 > 2.469607)
Step 7: Interpret result
The data supports (with at least 98%
accuracy) the hypothesis that
EDUCATION is an important
explanatory variable affecting income.
In rejecting H0, we are prone to make a
Type 1 Error.
The probability of a type 1 error is
nothing but the area to the right of tstatistic, or 0.0126.
Example 2: Use output 2 to test the
hypothesis (at 5% significance) that
weightgain is proportional to foodvalue.
The Model :: y = a + bx + e
and add the assumptions (Lec17)
Step 1:
H0 : a = 0 (proportionality)
H1 : a  0 (non-proportionality)
Step 2:
The estimator a is the test-statistic
Conditions:
The explanatory variable X is assumed
(1A) to be deterministic or non-random
(1B) : to come from a ‘fixed’ population
(1C) : to have a variance V(x) which is
not ‘too large’
The above assumptions are best suited
to a situation of a controlled
experiment
Assumptions concerning the random
term ei :
(IIA) E(ei ) = 0 for all i
(IIB) Var(ei) = s2 = constant for all i
(IIC) Covariance (ei , ej) = 0 for any i
and j
(IID) Each of the ei has a normal
distribution
Step 3:
Thus, a~ N(a, s2 ) where s2 is unknown.
Step 4:
Therefore, the test statistic
t  (a- a) / (standard error of a)
has a Student’s t-distribution with
10-2 = 8 d.o.f.
Step 5: Compare with critical value tC
tC = -2.31 for a two-tailed test with
significance level (a) = 0.05 and d.o.f.= 8
Step 6: Draw conclusion
The test is significant. Reject H0 at 5%
tC = -2.31 > -3.005262
The p-value is 0.0169 < 0.05
Step 7: Interpret
Foodvalue is not the only variable that
affects weightgain
Example 3:
Use output 3 to test (at 5% significance)
the following hypothesis:
Exercise has a negative effect on weight
gain
The proposed regression model is:
Weightgain
= ß0 + ß1(Foodvalue) +ß2(Exercise) + e
Step 1: Set up the hypotheses
H0 : ß2 = 0 (Exercise has no effect)
H1 : ß2 < 0(Exercise has a negative effect)