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ECON2206 / ECON3290 Introductory EconometricsLecture 4: The Multiple Regression Model - Inference I

Lecturer: Garry Barrett

UNSW

Session 1, 2009

Lecturer: Garry Barrett (UNSW) Lecture 4: Inference I Session 1, 2009 1 / 40

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Outline

The Multiple Regression Model - Inference I1. Sampling Distribution of OLS Estimators

2. Testing hypotheses about 1 population parameter (The t test)

3. Con…dence Intervals

4. Summary

Inference II (Next Week)

5. Testing single linear combination of parameters

6. Testing multiple linear restrictions (The F test)7. SHAZAM

8. Summary


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Distribution of OLS Estimators

Assumption 6. Normality

The population error u is independent of the explanatory variables x and is normally distributed with mean 0 and variance σ 2 :

u s Normal (0, σ 2) (1)

This assumption includes Assumptions 4 (ZCM) and 5(Homoskedasticity)

Together assumptions 1-6 are known as the classical linear model

(CLM) assumptions) under these assumptions the OLS estimators are the minimum

variance unbiased estimators (stronger than the Gauss-Markovresults, since we are not limiting ourselves to estimators which arelinear in the y i )


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Distribution of OLS Estimators

A shorthand way to state the population assumptions of the CLM is:

y jx s Normal ( β0 + β1x 1 + .... + βk x k , σ 2) (2)

Whether normality of u is a good or bad assumption depends on theapplication at hand.

E.g. Is it reasonable to assume wages (conditional on educ , etc ) are

normally distributed when no wages < 0 and there are minimum wagelaws ? How about log (wages ) ?

Empirical evidence shows that the latter is reasonable


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Assumptions on the error term


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Normal Sampling Distribution

Normality of the error term translates into normal distributions for theOLS estimators

Theorem 7. Normal Sampling Distributions Under the assumptions 1-6 (CLM):

ˆ β j s Normal ( β j , Var ( ˆ β j )) (3)

where Var ( ˆ β j ) was given last week


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Normal Sampling Distribution

In turn, this implies:

( ˆ β j β j )

sd (ˆ β j )

s Normal (0, 1) (4)

A linear combinations of the ˆ β j is also normally distributed - this willbe useful later when we want to test things like β1 = β2

We’ll also see later on that the normality of the OLS is approximately

true in large samples even without normality of the error terms !


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Testing Hypotheses about a Single Parameter

2. Testing Hypotheses about a Single β:

The t TestTo construct tests regarding the value of β j we use the following result:Theorem 8. t Distribution for the Standardised Estimators

Under assumptions 1-6

(ˆ β j β j )se ( ˆ β j )

s t nk 1 (5)

Note that (5) is di¤erent to (4):

) we have replaced the σ 2 in (5) with σ̂ 2

) consequently the expression has a t -distribution

The t -distribution is very similar to a normal distribution but has‘fatter’ tails


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Testing Hypotheses about a Single Parameter

The result in (6) will be used to test hypotheses about β j .

We are often interested in testing the null hypothesis:

H 0 : β j = 0 (6)

) in words, this means we are interested in testing whether x j has no(partial) e¤ect on the expected value of y , after controlling for all theother explanatory variables.


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Example

Example: Wages Consider the model:

log(wage ) = β0 + β1educ + β2exper + β3tenure + u (7)The null hypothesis H 0 : β2 = 0 means that once education and tenurehave been accounted for, the number of years of labour market experiencehas no e¤ect on the wage. If β2 > 0 then prior work experience

contributes to higher wages


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The t-test

The statistic we used to test H 0 : β j = 0 against any alternative iscalled the “t -statistic” (or the “t -ratio”) which is given by:

t ̂ β j

ˆ β j

se ( ˆ β j )(8)

) this is simple to calculate once we have the parameter estimate andthe standard error which are reported by statistical software (the

software usually reports the t -statistics as well)


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The t-test

Even if β j = 0 is true, it is rare that the ‘point estimate’ ˆ β j = 0 in asample of data

) how far away from 0 must ˆ β j be before we reject the hypothesis?

) the standard error of ˆ β j is a measure of the standard deviation of ˆ β j ,

so the t -statistic measures how many estimated standard deviationsˆ β j is away from 0

The precise rejection rule depends on the alternative hypothesis andthe chosen signi…cance level of the test

) the signi…cance level is the probability of rejecting the null hypothesiswhen it is true


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The t-test

Testing against One-Sided Alternatives

Consider:

H 0 : β j = 0 (9)

H 1 : β j > 0

The signi…cance level is the probability of rejecting the null hypothesiswhen it is true

) suppose we decide to use a 5% signi…cance level ) we are willing tomistakenly reject H 0 when it is true 5% of the time

) given our alternative, we are looking for a “su¢ciently large” positivevalue of t ̂ β j

in order to reject H 0 in favour of H 1

) “su¢ciently large” for a 5% signi…cance level is the 95th percentile of a t distribution with n-k-1 degrees of freedom (call this c for “critical

value”)


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The t-test

The rule is to reject H 0 in favour of H 1 at the 5% level of signi…canceif

t ̂ β j > c (10)

) this rejection rule is an example of a one-tailed test

(never reject H 0 in favour of H 1 if t ̂ β j is negative)


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Rejection Region


Th

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The t-test

To determine c we only need to know the level of signi…cance and df

Examples: From Table G.2If the signi…cance level is 10% and df =21 then c =1.323If the signi…cance level is 1% and df =21 then c =2.518

When the df of the t distribution is large, the t distributionapproaches the standard normal distribution, and for practical

purposes when df 120, we can use critical values from the standardnormal distribution


T bl f l

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Table of t-values


Th

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The t-test

Consider the test with the one-sided alternative:

H 0 : β j = 0

H 1 : β j < 0

) this rejection rule is the mirror image of the previous case

) given symmetry of the t -distribution, reject H 0 in favour of H 1 if

t ̂ β j < c (11)

where c is the critical value for the alternative H 1 : β j > 0


E l

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Example

Examples:

If the signi…cance level is 5% and df =18, c =1.734,and hence reject H 0: β j = 0 in favour of H 1: β j < 0 if t ̂ β j < 1.734


Th t t t

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The t-test


The t test

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The t-test

Two-Sided Alternatives

It is common to test the null hypothesis (6) against a 2-sidedalternative:

H 0 : β j = 0 (12)

H 1 : β j 6= 0

) when the alternative is 2-sided, we are interested in the absolute valueof the t statistic

) the rejection rule is reject H 0 when

jt ̂ β j j > c (13)


The t test

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The t-test

For a signi…cance level of 5%, c is chosen so that the area in each tailof the t distribution equals 2.5%

(e.g. c is the 97.5th percentile of the t distribution with n-k-1 df)

Terminology

) be explicit about the alternative hypothesis and the level of signi…cance

E.g. If we reject in favour of the alternative in (10) we say “x j is

statistically signi…cant, or statistically di¤erent from 0, at the 5%level”


Rejection Region

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Rejection Region


The t test

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The t-test

Testing Other Hypotheses about β j

Sometime we are interested in testing whether β j is equal to a speci…cvalue apart from 0. For instance, H 0 : β j = 1 or perhaps H 0 : β j = 1

For the null hypothesisH 0 : β j = a (14)

the t -statistic is given by

t =(estimate - hypothesised value)

standard error (15)

=( ˆ β

j

a)

se ( ˆ β j )

and follow exactly the same procedure as above


Example

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Example

Examples:

1. Wage EquationThe following model was estimated:

\ log(wage ) = 0.284 + 0.092 educ + 0.0041exper + 0.022 tenure

(0.104) (0.007) (0.0017) (0.003)

n = 526, R 2 = 0.316

where the standard errors are in (.)


Example

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Example

(i) Test whether the return to experience, after controlling for educationand tenure, in the population is 0 against the alternative it is positive.

Answer:Testing:

H 0 : βexper = 0

H 1

: βexper

> 0

Test statistic is:

t ̂ βexper

=0.0041

0.0017

= 2.41

The critical value at the 1% level of signi…cance is 2.326 and since t ̂ βexper

exceeds this value, we reject the null in favour of the alternativehypothesis (that the return to experience is positive).


Example

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Example

(ii) Test whether the return to education, after controlling for experienceand tenure, in population is 0 against the alternative it is not equal to 0.

Answer:Testing:

H 0 : βeduc = 0

H 1:

βeduc 6= 0

Test statistic is:

jt j =0.092

0.007

= 13.149

The critical value at the 1% level of signi…cance is 2.576 and since jt jexceeds this value, we reject the null in favour of the alternativehypothesis (that the return to education is not equal to 0).


Example

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Example

Example 2. Housing Prices and Air Pollution.Using a sample of 506 communities in a large US city, a model was

estimated which related median house price in the community (price ) tovarious community characteristics: nox is the amount of nitrous oxide inthe air (in part per million), dist is the average distance of the communityfrom 5 employment centres, rooms is the average number of rooms inhouses in the community, and stratio is the average student-teacher ratio

in the community. The population model is:

log(price ) = β0 + β1 log(nox ) + β2 log(dist ) + β3rooms + β4stratio + u

Hence β1 is the elasticity of price with respect to nox . We wish to test

H 0 : β1 = 1

H 1 : β1 6= 1

The t -statistic for doing this test is t =( ˆ β1 +1)

se ( ˆ β1

)


Example

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Example

Using the data, the estimated model is:

\ log(price ) = 11.08 0.954 log(nox ) 0.134 log(dist )

(0.32) (0.117) (0.043)

+0.255rooms 0.052stratio

(0.019) (0.006)

n = 506, R 2 = 0.581

Each slope coe¢cient is signi…cantly di¤erent from 0 even at verysmall signi…cance levels

) however we want to test H 0 : β1 = 1 against H 1 : β1 6= 1) for this test, the t -statistics is (0.954 + 1)/0.117 = 0.393) this test statistic is very small and the null is not rejected: after

controlling for the factors included in the model, there is little evidencethat the elasticity is di¤erent from 1.


P-values

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P values

Computing p-values for t-tests

There is a degree of arbitrariness in choosing the signi…cance level

) di¤erent researchers prefer di¤erent signi…cance levels (e.g. 10%, 5%,

1%), and there is no “correct” signi…cance level (though researcherstend to adopt higher signi…cance levels when they use smaller samplesof data)

) rather than testing at di¤erent signi…cance levels, it is moreinformative to calculate the p -value for a test:


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P-values

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Economic vs Statistical Signi…cance

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g

Economic vs Statistical Signi…cance

We have been focussing on statistical signi…cance

) the statistical signi…cance of a variable, x j , is determined by the sizeof the t -statistic t ̂ β

j

) however, the economic (or practical) signi…cance also depends onthe size of ˆ β j

) too much focus on statistical signi…cance can lead to the falseconclusion that a variable is important for explaining y even thoughits estimated e¤ect is small


Con…dence Intervals

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3. Con…dence IntervalsIt is straightforward to construct con…dence intervals (CI) for β j

) con…dence intervals are also known as “interval estimates”

) under the CLM assumptions

( ˆ β j β j )

se ( ˆ β j )s

t nk 1 (16)

manipulating this expression gives a CI for the unknown β j :

) the 95% CI is given byˆ β j c

.

se (ˆ β j ) (17)

where c is the critical values for the t -distribution withdf = n k 1.

(e.g. c = 97.5th percentile of a t nk 1 distribution)



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More precisely, the lower and upper lower bounds are given by:

β j ˆ β j c

.

se (ˆ β j ) (18)

β j ˆ β j + c .se ( ˆ β j )



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Interpretation: if random samples were obtained over and over again,with β

j and β j computed each time, the population value β j would lie

in the interval ( β j , β j ) for 95% of the samples

) constructing a CI is very simple, just needˆ

β j ,

se (ˆ

β j ),

c (for c need toknow df and the signi…cance level)

) when df > 120, the t nk 1 distribution is close to the normaldistribution, and so you can use the standard normal distribution (e.g.for a 95% CI, c =1.96....as a rule of thumb, for a 95% CI take ˆ β j

2 se ( ˆ β j ))



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Note:

- lower con…dence levels (e.g. 90%) lead to narrower CIs- higher con…dence levels (e.g. 99%) lead to wider CIs

) Once a CI is constructed it is easy to carry out 2-tailed hypothesistests: if the null is H 0: β j =a, then H0 is rejected against H 1: β j 6=a atsay the 5% signi…cance level if (and only if) a is not in the 95% CI.


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Looking Ahead

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Next Week:

Continue with inference and consider:

- Testing single linear combination of parameters- Testing multiple linear restrictions (The F test)

Running SHAZAM


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E2206L0409b