4.1 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Lecture #4 Studenmund (2006): Chapter 5...

4.1

All rights reserved by Dr.Bill Wan Sing Hung - HKBU

Lecture #4

Studenmund (2006): Chapter 5

• Review of hypothesis testing

• Confidence Interval and estimation

• t-test and Partial significance test

Objectives

4.2


Hypothesis Testing in regression:

1. Testing individual (partial) coefficient

2. Testing the overall significance of all coefficients

3. Testing restriction on variables (add or drop): Xk = 0 ?

4. Testing partial coefficient under some restrictions

Such as 1+ 2 = 1; or 1 = 2 (or 1+ 2 = 0); etc.

5. Testing the stability of the estimated regression model

-- over time

-- in different cross-sections

6. Testing the functional form of regression model.

4.3


Properties of OLS estimators–two variable case

0 2

1

1. UnbiasedUnbiased

ˆ ˆefficiency

3. ConsistencyConsistency: as n gets larger, estimator is more accurate

x

ˆ[Se(2

22ˆ1

2

2

2ˆ0

xnxSe2. Min. VarianceMin. Variance

4.4


x 2

)2n(2

2

~2n ˆ

Properties of OLS estimators (continue)

2ˆ0

00 ,N~ˆ

2ˆ

111,N~ˆ

6.

ˆˆ4 & 5. 0 and 1 are normal distribution

4.5


Hypothesis Testing and Confidence Interval

f1

Den

sity

11

1

1

Random interval (confidence interval)

true

Estimated falls in area

^

How reliable is the OLS estimation ?

How “close” is to (true or theoretical value)?

How “close” is to (true or theoretical value )?

^^

4.6


Hypothesis Testing and Confidence Interval

ˆ ˆ0.990.950.90

Pr( 1- < 1 < 1+ ) = (1-)

which (1-) is confidence coefficient: (0< <1)

is also called the level of significancelevel of significance.

is called lower confidencelower confidence bound bound

is called upper confidenceupper confidence bound bound

the interval between ( and ( is

called random intervalrandom interval (confidence intervalconfidence interval)

ˆ

ˆ ˆ

0.010.050.10

Select one level to construct the interval

4.7


Constructing Confidence IntervalConfidence Interval for i

xn

x2

i

2i

22ˆ

0

200 ˆ

0,N~ˆ

211 ˆ

1,N~ˆ

x2

i

22ˆ

1

2i ,0N~ E() = 0

Var() =

By assumptions:

4.8


Constructing Confidence IntervalConfidence Interval for (cont.)i

f1

1 E 1 1

Actual estimated 1 could be fallen intothese regions

^

True value

4.9


ˆSe

ˆZ

1

11

Zf

O

Constructing Confidence IntervalConfidence Interval for (cont.)i

Transform into Normal standard distributionNormal standard distribution

4.10



(cont.) Use the normal distribution to make

probabilistic statements about 2 provided the

true 1 is known

In practice this is unobserved

xˆ

1,0NˆSe

ˆZ

2

11

1

11 ~

4.11


For example:

0 96.196.1

%5.2%5.2

%95


(cont.)

Accept region

Critical Values

Rejected region

4.12


95.096.1Z96.1Pr

95% confidence interval:

96.1)ˆ(Se

ˆ96.1

1

11


(cont.)

ˆ 95.096.1

)(Se96.1Pr

1

11

4.13


ˆSe96.1ˆ 11

2In practice, is unknown, we have to use the unbiased estimator

2n

2 RSS i2ˆ

Instead of using normal standard distribution, t-distribution is used.


(cont.)

* ˆSe96.1ˆˆSe96.1ˆ11111

4.14


standard error of estimatorestimated - true parametert

)(Se

ˆt

1

11

ˆxˆ

t2

11

Use the tt to construct a confidence interval for 1


(cont.)

SEE

a specified value

4.15


Use the tc to construct a confidence interval for 1 as


(cont.)

where is critical t value at two-tailed

level of significance. is level of significance

and (n-2) is degree of freedom (in 2-variable case).

t c

2n,2

2

1*Pr

2,2

2,2

c

n

c

nttt

4.16



(cont.)

ˆˆˆˆ c*

90.0

SetSetPr11

c

1 11 2n,05.02n,05.0

Rearranging,

90.0tˆSe

ˆtPr

c

1

11c2n,05.02n,05.0

Therefore

Pr( -tc0.025, n-2 (1- 1)/Se(1) tc

0.025, n-2 ) = 0.95 ^ ^

4.17


Then 90% confidence intervalconfidence interval for 1 is:

ˆSetˆ1

c

1 2n,05.0

Se 1t

c2n,05.0 Check it from t-table

Check it from estimated result

1 &

The 95% confidence interval interval for becomes

ˆ*Set1

c1 0.025, n-2

4.18


tc

2N,2

CriticalCritical t-value:

α= 0.1two-tailed

= 1.860tc

8,05.0

= 2.306tc

8,025.0

α= 0.05two-tailed

4.19


ˆSe( )tˆ1

c

12n,

2

For exampleFor example:

Given = 0.5091, n = 10, Se( = 0.0357, ˆ ˆ

95% 95% confidence intervalconfidence interval is: is:

α= 0.05

)5914.0,4268.0(

0823.05091.0

)0357.0(t5091.0c

8,025.0

0.5091 ± 2.3062.306(0.0357)

4.20


90% 90% confidence intervalconfidence interval is is:

)5755.0,4427.0(

0664.05091.0

)0357.0(860.15091.0

)0357.0(t5091.0 c8,05.0

α= 0.1

4.21


Procedures for Hypothesis Testing

1. Determine null (H0) and alternative (H1) hypotheses

2. Specify the test statistic and its distribution as if the null hypothesis were true.

3. Select and determine the rejection region.

4. Calculate the sample value of test statistic (t*).

5. State your conclusion.

4.22


Test-Significance Approach: Two-Tailed T-test

ˆ

ˆSet*

1

112. Compute

111

110

ˆ:H

ˆ:H1. State the hypothesis

tc

2n,2

3. Check t-table for critical t value:

4.23


Two-Tailed t-test decision rule

tc4. Compare t* and

1

Accept region

reject H0 region

ˆSetˆ

1

c

12n,2

reject H0 region

ˆSetˆ

1c

12n,2

5. If t* > tc or –t* < - tc , then reject Ho

or | t* | > | tc |

Decision RuleDecision Rule:

4.24


Two-Tailed t-test

(1) From Confidence-intervalConfidence-interval approach: 95% confidence-interval is (0.4268, 0.5914)which does not contain the true 1.

The estimated 1 is not equal to 0.30.3

3.0:H

3.0:H

11

10

11

Suppose we postulatepostulate that

Is the observed compatible with true ?

ˆ

ˆ

4.25


(2) From Significance testSignificance test approach:

Compare “tt*-value”*-value” and the “critical critical tt-value”-value”:

5.8570357.02091.0

0357.03.05091.0 1ˆSe

ˆt* 11

t c0.025, 8 = 2.306 check from the t-table

,

306.2857.5 8,025.0* ctt

c

==> reject H0

It means the estimated 1 is not equal 0.3

4.26


The t-statistic in computer outputcomputer output

SEE= RSS^

H0: 1=0

H1: 10

^

^Var(βi)^

1ˆSe

ˆt* 11

=0.5091 - 0

0.0357t*

4.27


The p-valuesp-values Reported by Regression Software

When it (p-valuep-value) is the lowest level of significance at which we could reject HoHo.

2. Testing

The p-valuep-value of a coefficient reported by the computer is the probability, probability, or marginal significor marginal significant level,ant level, of obtaining that estimated coefficient (βi ) if the null hypothesis (HoHo: ββii =0 =0) were true.

^^

4.28


One-tailed T-test

tc

2n, Step 3: check t-table for

look for critical t value

Computed valueStep 2:

ˆSe

ˆt* 11

1

111111

110110

ˆ:Hˆ:H

ˆ:Hˆ:HStep 1: State thehypothesis

Step 4: compare tc and t*

4.29


One-tailed t-test decision rule

Left-tail

(If t < - tc ==> reject H0 )

(If t > - tc ==> not reject H0 )

Decision RuleStep 5: If t > tc ==> reject H0

If t < tc ==> not reject H0

Right-tail

0 tc < t

Right-tail

0-tct <

left-tail

4.30


Example: One-Tailed t-test

5.8570357.02091.0

0357.03.05091.0

t*

ˆSe

ˆt*

1

11

1. Compute:3.0:H

3.0:H

11

10

We also could postulate that: ˆ

ˆ

4.31


One-Tailed t-test (cont.)

tc8,05.0

tc8,05.0

= 0.052. Check t-table for where =1.860

Hreject

860.1t857.5t

0

c8,05.0

3. Compare t and the critical t

4.32


One-Tailed t-test (cont.)

*111 :H

*110 :H ˆ

ˆ“ Decision rule for left-tail test”

If t < - tc df => reject H0

^ *

*- tc• Se()

left-tail

^ ^

4.33


“Accepting” or “Rejecting”

"Accept "the null hypothesis (HH00): All we are saying is that on the basis of the sample sample evidence we have no reason to reject it; We are not saying that the null hypothesis is true beyond any doubt.

Therefore, in “accepting” a HHoo , we should always be aware that another null hypothesis may be equally compatible with the data. So, the conclusion of a statistical test is “do not reject” rather than “accept”.

4.34


1. Individual partial coefficient test

t* =

1 - 0^

Se (1)

^ =0.726

0.048= 14.906

Compare with the critical value tc0.025, 12 = 2.179

Since t* > tc reject Ho

Answer : Yes, 1 is statistically significant and is

significantly different from zero.

^

Y = 0 + 1X1 + 2X2 ^ ^ ^^

H0 : 1 = 0

H1 : 1 0

Holding X2 constant: Whether XX11 has the effecteffect on YY ?1

Y

X1

= 1 = 0?^^

From the printout

4.35


1. Individual partial coefficient test (cont.)

t =2 - 0^

Se (2)^

=2.736-0

0.848= 3.226

Critical value: tc0.025, 12 = 2.179

Since | t | > | tc | ==> reject Ho

Answer: Yes, 2 is statistically significant and is

significantly different from zero.

^

holding X1 constant: Whether XX22 has the effecteffect on YY?2

H0 : 2 = 0

H1 : 2 0

Y

X2

= 2 = 0?^

^

From the printout

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4.1 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Lecture #4 Studenmund (2006): Chapter 5...

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