Lecture Applied econometrics course - Chapter 3: Statistic inference and hypothesis testing

APPLIED ECONOMETRICS COURSE

Chapter III
Statistic Inference and
Hypothesis Testing

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I. THE DISTRIBUTION OF THE PARAMETERS

Assumpt 6. Ui ~ N(0, 
2)
Theorem 4.1: Normal Distribution of the 
parametors 

ˆ −β
β
0
βˆ0 ~ N ( β 0 , σ ) � Z = 0
~ N (0,1)
σ βˆ
2
βˆ0

0

ˆ −β
β
βˆ1 ~ N ( β1 , σ ) � Z = 1 1 ~ N (0,1)
σ βˆ

2
βˆ1

1

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I. THE DISTRIBUTION OF THE PARAMETERS
•

Under the assumptions from 1-6, we have the
following theorem:
Theorem 4.2: Student Distribution

βˆ j − β j
t=
~ t(n − k − 1)
se( βˆ )
j

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II. HYPOTHESES TESTING

2.1. Testing against One­Sided Alternatives: 
Right Hand Side
H 0 : β1 = 0
(1). Manifest the hypothesis:

H1 : β1 > 0

βˆ1 − β1
(2). Compute t­statistics: t = se( βˆ )
1
(3).  Search  to  find  t  (n­k­1),  for  example,    α=5%,  n­k­1=28,    t0.05 
(28)=1.701 ( excel: TINV(0.1,28)) 
(4). Decision Rule:
 Nếu t >  t  (n­k­1) thì có thể bác bo gia thiê
̉ ̉
́t H0
Nếu  t <  t  (n­k­1) thì có thể  chấp nhân gia thiê
̣
̉
́t H0 
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II. HYPOTHESES TESTING

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Example: expenditure.wf

(1). Let’s test the following hypothesis with α = 5%: 
H 0 : β1 = 0
H1 : β1 > 0
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II. HYPOTHESES TESTING

βˆ1 − β1 0.509
(2). Compute s­statistics: t = se( βˆ ) = 0.035 = 14.54
1
(3). Search to find: t0.05 (8) = 1.86

(4). Decision Rule: t­statistics = 14.54  >  t0.05 (8)=1.86, we therefore 
can reject the null hypothesis H0. In other word, we can accept the 
alternative hypothesis H1:  1 > 0

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II. HYPOTHESES TESTING 
2.2. Testing against One-Sided Alternatives: Left Hand
(1).
Side Suppose the hypothesis:
H 0 : β1 = 0
H1 : β1 < 0

(2).Compute t­statistics:

βˆ1 − β1
t=
se( βˆ1 )

(3).Search to find t  (n­k­1)
(4). Decision Rule :

 Nếu t < ­ t  (n­k­1) thì có thể bác bo gia thiê
̉ ̉
́t H0

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II. HYPOTHESES TESTING 

2.3.
(1). SetTwo-Sided
up the hypothesis:
H :β
Alternatives H : β
0

1

=0

1

1

0

βˆ1 − 0
(2). Compute t­statistics: t = se( βˆ )
1
(3).  Search  to  find  t /2  (n­k­1),  for  example  α  =  5%  và  n­k­1=8  thì 

t0.025 (8)=2.306 (TINV(0.05,8))
(4). Decision Rule:
if   t  >  t /2(n­k­1), we can reject the null hypothesis H0
if   t     t /2(n­k­1), we can accept the null hypothesis H0
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Example: Whether income affects to expenditure?
Solution:
(1)

Manifest the hypothesis:
H 0 : β1 = 0
H1 : β1

(2)

Compute s-statistics:
βˆ − β

0.5091 − 0
t=
=
= 14.243
0.035742
se( βˆ1 )
1

(3)

0

With

1

= 5%, we have: t0.025(n-k-1) = t0.025(8) =

2.306
(4)

Because t =14.243 > t0.025(8) = 2.36, we can reject
the null hypothesis H0: β1= 0. This means that
income significantly
impacts to expenditure.
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Example: expenditure.wf

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2.4. Testing Other Hypotheses
about that
Suppose
parameter
there is a research showing that β1= 0.3,
Whether this conclusion is true or not?
Solution:

H 0 : β1 = 0.3

(1)

(2)

H1 : β1
Set up the null hypothesis:

0.3

βˆ1 − β1
t=
Compute s-statistics:se( βˆ1 )

(3)

Search to find t /2(n-k-1)

(4)

Decision:
If t > t /2(n-k-1) => Reject H0
If NGUYEN BA TRUNG ­ 2016
t
t /2(n-k-1) => Accept H0


H 0 : β1 = 0.3

(1). Null hypothesis:

•

H1 : β1

0.3

βˆ1 − β1 0.5091 − 0.3
t-statistics: t =
=
= 5.85
ˆ
0.035742
se( β1 )

•

Search to find: t0.025(n-k-1)=t0.025(8) = 2.306

•

Because t =5.85 > t /2(8) = 2.36, we can reject H0:
β1= 0.3 with

= 5%.
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2.5. Compute P-value for t-tests

•

•
•

•

With confident level α, P-value is the probability of
mistake if we reject the null hypothesis H0
P-value: Prob(|T|>|t|)
Thus, if small P-value provides the evidence against
H0, and if large P-value provide the evidence against
H1
Normally, we often choose confident level at 1%, 5%,
and 10%.

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Ví dụ: expenditure.wf

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III. Confident Interval
Confident Interval of the parameters

βˆ0
βˆ1

1; 2

tα / 2 se( βˆ0 )
t se( βˆ )
α /2

1

Confident Interval of error term Ui

σˆ (n − 2)
σˆ (n − 2)
2
<σ < 2
2
χα /2 (n − 2)
χ1−α /2 (n − 2)
2

2

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Example: expenditure (continue)
Let’s  compute  the  confident  intervals  for  the  parameters  at  confident 
level 95%?
Ø

 Because 1­   =0.95 ➨  = 0.05 ➨  /2 = 0.025

Ø

  t 0.025(8) =2.306 [TINV(0.05,8)]

ˆ

t

0

ˆ ) =  24.4545 ± 2.306 * 6.4138
se
(
/2
0
hay (9.6643 <  0 < 39.2448)

ˆ

1

t

ˆ)
se
(
/2
1

= 0.5091 ± 2.306 * 0.035742

hay (0.4268 <  1 < 0.5914)


Thí du: expenditure (continue)
̣


We can utilize confident interval in testing hypothesis,
for instance

•

H 0 : β1 = 0.3
H1 : β1

v

0.3

 The confident interval for  1: 
 (0.4268 <  2 < 0.5914)

v

 Because  1 = 0.3 is out of the confident interval, we can reject the H0 

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IV. F- test

(1). Suppose that X1…X2…Xk do not effect on Y, this
means that:
H 0 : β1 = β 2 = ... = β k = 0 � R 2 = 0
(2). Compute F­statistics as below:

ESS
(k )
2
ESS / (k )
R
/k
TSS
F (k , n − k − 1) =
=
=
RSS / (n − k − 1) RSS (n − k − 1) (1 − R 2 ) / (n − k − 1)
TSS

(3). Search to find: F(k,n­k­1,α) 
(4). Decision Rule: 
v
If F> F(k,n­k­1,α), we can reject H0.
v
 If F ≤ F(k,n­k­1,α), we can not reject H0.
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Example: continuted
•

Compute F-statistics:
F=

v

0.96206(10 − 8)
= 202.86
1 − 0.96206

 With α = 5%, search to find Fα (k, n­k­1), we have: F0.05 (1,8)= 5.32 

v

 Because F = 202.86 > F0.05 (1,8)= 5.32, we can reject H0

v

EVIEWs provide the p­value of F­test(next slide)

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Thí dụ: expenditure.wf

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V. Test the simple linear
combination
the
parameters
(1). Assume that we of
want
to
H test:
:β
=β
0

(2). Compute t­statistics as:
v

 Where:
Se( βˆ1 − βˆ2 ) =

1

2

H1 : β1

β2

βˆ1 − βˆ2
t=
se( βˆ1 − βˆ2 )

( se(βˆ )
1

2

(

− 2 cov( βˆ1 , βˆ2 ) + se( βˆ2

(3). With α = 5%, search to find: tα/2(n­k­1)
(4). Decision: If |t| > tα/2(n­k­1), We can reject H0
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)

2


Example:Twoyears.wf
•

We have the following model:
log(wage) = β 0 + β1 jc + β 2univ + β 3 exp er + u

Where:  Jc: trình độ cao đẳng (2 năm đại học) 
v

 Unive: trình độ đại học (4 năm đại học) 

v

 Exper: kinh nghiệm (tháng)

Question: whether wage differs between who those graduated from 
colleague and university? 
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v

 From the estimate result, the different wage: 
θ1 = βˆ1 − βˆ2 = −0.0102

v

 Is this difference statistically significant? 
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