Christopher Dougherty EC220 - Introduction to econometrics (chapter 3) Slideshow: f tests in a...

Christopher Dougherty

EC220 - Introduction to econometrics (chapter 3)Slideshow: f tests in a multiple regression model

Original citation:

Dougherty, C. (2012) EC220 - Introduction to econometrics (chapter 3). [Teaching Resource]

© 2012 The Author

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Available in LSE Learning Resources Online: May 2012

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F TESTS OF GOODNESS OF FIT

1

This sequence describes two F tests of goodness of fit in a multiple regression model. The first relates to the goodness of fit of the equation as a whole.

uXXY kk ...221

0 :

0...:

1

20

H

H k

at least one


2

We will consider the general case where there are k – 1 explanatory variables. For the F test of goodness of fit of the equation as a whole, the null hypothesis, in words, is that the model has no explanatory power at all.

uXXY kk ...221

0 :

0...:

1

20

H

H k

at least one


3

Of course we hope to reject it and conclude that the model does have some explanatory power.

uXXY kk ...221

0 :

0...:

1

20

H

H k

at least one


4

The model will have no explanatory power if it turns out that Y is unrelated to any of the explanatory variables. Mathematically, therefore, the null hypothesis is that all the coefficients 2, ..., k are zero.

uXXY kk ...221

0 :

0...:

1

20

H

H k

at least one


uXXY kk ...221

5

The alternative hypothesis is that at least one of these coefficients is different from zero.

0 :

0...:

1

20

H

H k

at least one


uXXY kk ...221

6

In the multiple regression model there is a difference between the roles of the F and t tests. The F test tests the joint explanatory power of the variables, while the t tests test their explanatory power individually.

0 :

0...:

1

20

H

H k

at least one


uXXY kk ...221

7

In the simple regression model the F test was equivalent to the (two-sided) t test on the slope coefficient because the ‘group’ consisted of just one variable.

0 :

0...:

1

20

H

H k

at least one


)()1()1(

)(

)1(

)()1(

),1(

2

2

knRkR

knTSSRSS

kTSSESS

knRSSkESS

knkF

uXXY kk ...221

8

The F statistic for the test was defined in the last sequence in Chapter 2. ESS is the explained sum of squares and RSS is the residual sum of squares.

0 :

0...:

1

20

H

H k

at least one


)()1()1(

)(

)1(

)()1(

),1(

2

2

knRkR

knTSSRSS

kTSSESS

knRSSkESS

knkF

uXXY kk ...221

9

It can be expressed in terms of R2 by dividing the numerator and denominator by TSS, the total sum of squares.

0 :

0...:

1

20

H

H k

at least one


10

)()1()1(

)(

)1(

)()1(

),1(

2

2

knRkR

knTSSRSS

kTSSESS

knRSSkESS

knkF

uXXY kk ...221

ESS / TSS is the definition of R2. RSS / TSS is equal to (1 – R2). (See the last sequence in Chapter 2.)

0 :

0...:

1

20

H

H k

at least one


11

uSFSMASVABCS 4321

The educational attainment model will be used as an example. We will suppose that S depends on ASVABC, the ability score, and SM, and SF, the highest grade completed by the mother and father of the respondent, respectively.


12

0: 4320 H

The null hypothesis for the F test of goodness of fit is that all three slope coefficients are equal to zero. The alternative hypothesis is that at least one of them is non-zero.

uSFSMASVABCS 4321

. reg S ASVABC SM SF

Source | SS df MS Number of obs = 540-------------+------------------------------ F( 3, 536) = 104.30 Model | 1181.36981 3 393.789935 Prob > F = 0.0000 Residual | 2023.61353 536 3.77539837 R-squared = 0.3686-------------+------------------------------ Adj R-squared = 0.3651 Total | 3204.98333 539 5.94616574 Root MSE = 1.943

------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval]-------------+---------------------------------------------------------------- ASVABC | .1257087 .0098533 12.76 0.000 .1063528 .1450646 SM | .0492424 .0390901 1.26 0.208 -.027546 .1260309 SF | .1076825 .0309522 3.48 0.001 .04688 .1684851 _cons | 5.370631 .4882155 11.00 0.000 4.41158 6.329681------------------------------------------------------------------------------


13

Here is the regression output using Data Set 21.

uSFSMASVABCS 4321 0: 4320 H





14


In this example, k – 1, the number of explanatory variables, is equal to 3 and n – k, the number of degrees of freedom, is equal to 536.

)/()1/(

),1(knRSS

kESSknkF

3.104

536/20243/1181

)536,3( F





15


)/()1/(

),1(knRSS

kESSknkF

3.104

536/20243/1181

)536,3( F

The numerator of the F statistic is the explained sum of squares divided by k – 1. In the Stata output these numbers are given in the Model row.





16


)/()1/(

),1(knRSS

kESSknkF

3.104

536/20243/1181

)536,3( F

The denominator is the residual sum of squares divided by the number of degrees of freedom remaining.





17


)/()1/(

),1(knRSS

kESSknkF

3.104

536/20243/1181

)536,3( F

Hence the F statistic is 104.3. All serious regression packages compute it for you as part of the diagnostics in the regression output.





18


3.104536/20243/1181

)536,3( F

The critical value for F(3,536) is not given in the F tables, but we know it must be lower than F(3,500), which is given. At the 0.1% level, this is 5.51. Hence we easily reject H0 at the 0.1% level.

51.5)500,3(crit,0.1% F





19


3.104536/20243/1181

)536,3( F51.5)500,3(crit,0.1% F

This result could have been anticipated because both ASVABC and SF have highly significant t statistics. So we knew in advance that both 2 and 4 were non-zero.





20


3.104536/20243/1181

)536,3( F51.5)500,3(crit,0.1% F

It is unusual for the F statistic not to be significant if some of the t statistics are significant. In principle it could happen though. Suppose that you ran a regression with 40 explanatory variables, none being a true determinant of the dependent variable.





21


3.104536/20243/1181

)536,3( F51.5)500,3(crit,0.1% F

Then the F statistic should be low enough for H0 not to be rejected. However, if you are performing t tests on the slope coefficients at the 5% level, with a 5% chance of a Type I error, on average 2 of the 40 variables could be expected to have ‘significant’ coefficients.





22


3.104536/20243/1181

)536,3( F51.5)500,3(crit,0.1% F

The opposite can easily happen, though. Suppose you have a multiple regression model which is correctly specified and the R2 is high. You would expect to have a highly significant F statistic.





23


3.104536/20243/1181

)536,3( F51.5)500,3(crit,0.1% F

However, if the explanatory variables are highly correlated and the model is subject to severe multicollinearity, the standard errors of the slope coefficients could all be so large that none of the t statistics is significant.





24


3.104536/20243/1181

)536,3( F51.5)500,3(crit,0.1% F

In this situation you would know that your model is a good one, but you are not in a position to pinpoint the contributions made by the explanatory variables individually.


uXXXY 4433221

uXY 221 1RSS

2RSS

25

We now come to the other F test of goodness of fit. This is a test of the joint explanatory power of a group of variables when they are added to a regression model.


uXXXY 4433221

uXY 221 1RSS

2RSS

26

For example, in the original specification, Y may be written as a simple function of X2. In the second, we add X3 and X4.


uXXXY 4433221

uXY 221 1RSS

2RSS

27

The null hypothesis for the F test is that neither X3 nor X4 belongs in the model. The alternative hypothesis is that at least one of them does, perhaps both.

0 :

0:

31

430

H

H

04 04 3or or both and


uXXXY 4433221

uXY 221 1RSS

2RSS

28

For this F test, and for several others which we will encounter, it is useful to think of the F statistic as having the structure indicated above.

0 :

0:

31

430

H

H


F (cost in d.f., d.f. remaining) =reduction in RSS cost in d.f.

RSS remaining degrees of freedomremaining


uXXXY 4433221

uXY 221 1RSS

2RSS

29

The ‘reduction in RSS’ is the reduction when the change is made, in this case, when the group of new variables is added.

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

30

The ‘cost in d.f.’ is the reduction in the number of degrees of freedom remaining after making the change. In the present case it is equal to the number of new variables added, because that number of new parameters are estimated.

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

31

(Remember that the number of degrees of freedom in a regression equation is the number of observations, less the number of parameters estimated. In this example, it would fall from n – 2 to n – 4 when X3 and X4 are added.)

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

32

The ‘RSS remaining’ is the residual sum of squares after making the change.

0 :

0:

31

430

H

H





33

uXXXY 4433221

uXY 221 1RSS

2RSS

The ‘degrees of freedom remaining’ is the number of degrees of freedom remaining after making the change.

0 :

0:

31

430

H

H




. reg S ASVABC


------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval]-------------+---------------------------------------------------------------- ASVABC | .148084 .0089431 16.56 0.000 .1305165 .1656516 _cons | 6.066225 .4672261 12.98 0.000 5.148413 6.984036------------------------------------------------------------------------------


34

We will illustrate the test with an educational attainment example. Here is S regressed on ASVABC using Data Set 21. We make a note of the residual sum of squares.





35

Now we have added the highest grade completed by each parent. Does parental education have a significant impact? Well, we can see that a t test would show that SF has a highly significant coefficient, but we will perform the F test anyway. We make a note of RSS.


uXXXY 4433221

uXY 221 1RSS

2RSS

36

The improvement in the fit on adding the parental variables is the reduction in the residual sum of squares.

16.13536/6.2023

2/)6.20230.2123()4540(2)(

)4540,2(2

21

RSS

RSSRSSF

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

37

The cost is 2 degrees of freedom because 2 additional parameters have been estimated.

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)4540,2(2

21

RSS

RSSRSSF

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

38

The remaining unexplained is the residual sum of squares after adding SM and SF.

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)4540,2(2

21

RSS

RSSRSSF

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

39

The number of degrees of freedom remaining is n – k, that is, 540 – 4 = 536.

16.13536/6.2023

2/)6.20230.2123()4540(2)(

)4540,2(2

21

RSS

RSSRSSF

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

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)4540,2(2

21

RSS

RSSRSSF

40

The F statistic is 13.16.

0 :

0:

31

430

H

H





uXXXY 4433221

uXY 221 1RSS

2RSS

41

The critical value of F(2,500) at the 0.1% level is 7.00. The critical value of F(2,536) must be lower, so we reject H0 and conclude that the parental education variables do have significant joint explanatory power.

00.7)500,2(crit,0.1% F

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2/)6.20230.2123()4540(2)(

)4540,2(2

21

RSS

RSSRSSF

0 :

0:

31

430

H

H





1RSS

2RSS

uXXY 33221

uXXXY 4433221

42

This sequence will conclude by showing that t tests are equivalent to marginal F tests when the additional group of variables consists of just one variable.


1RSS

2RSS

uXXY 33221

uXXXY 4433221

43

Suppose that in the original model Y is a function of X2 and X3, and that in the revised model X4 is added.


1RSS

2RSS

0 :

0:

41

40

H

H

uXXY 33221

uXXXY 4433221

44

The null hypothesis for the F test of the explanatory power of the additional ‘group’ is that all the new slope coefficients are equal to zero. There is of course only one new slope coefficient, 4.


45

1RSS

2RSS

The F test has the usual structure. We will illustrate it with an educational attainment model where S depends on ASVABC and SM in the original model and on SF as well in the revised model.

0 :

0:

41

40

H

H

uXXY 33221

uXXXY 4433221



. reg S ASVABC SM


------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval]-------------+---------------------------------------------------------------- ASVABC | .1328069 .0097389 13.64 0.000 .1136758 .151938 SM | .1235071 .0330837 3.73 0.000 .0585178 .1884963 _cons | 5.420733 .4930224 10.99 0.000 4.452244 6.389222------------------------------------------------------------------------------


46

Here is the regression of S on ASVABC and SM. We make a note of the residual sum of squares.


47

Now we add SF and again make a note of the residual sum of squares.





0 :

0:

41

40

H

H

uXXXY 4433221

uXXY 33221 1RSS

2RSS

48

The reduction in the residual sum of squares is the reduction on adding SF.

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)4540,1(2

21

RSS

RSSRSSF




0 :

0:

41

40

H

H

uXXXY 4433221 1RSS

2RSS

49

The cost is just the single degree of freedom lost when estimating 4.

uXXY 33221

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1/)6.20233.2069()4540(1)(

)4540,1(2

21

RSS

RSSRSSF




0 :

0:

41

40

H

H

uXXXY 4433221 1RSS

2RSS

50

The RSS remaining is the residual sum of squares after adding SF.

uXXY 33221

10.12536/6.2023

1/)6.20233.2069()4540(1)(

)4540,1(2

21

RSS

RSSRSSF




0 :

0:

41

40

H

H

uXXXY 4433221 1RSS

2RSS

51

The number of degrees of freedom remaining after adding SF is 540 – 4 = 536.

uXXY 33221

10.12536/6.2023

1/)6.20233.2069()4540(1)(

)4540,1(2

21

RSS

RSSRSSF




0 :

0:

41

40

H

H

uXXXY 4433221 1RSS

2RSS

10.12536/6.2023

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)4540,1(2

21

RSS

RSSRSSF

uXXY 33221

52

Hence the F statistic is 12.10.




0 :

0:

41

40

H

H

uXXXY 4433221 1RSS

2RSS

10.12536/6.2023

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)4540,1(2

21

RSS

RSSRSSF

uXXY 33221

53

96.10)500,1( crit,0.1% F

The critical value of F at the 0.1% significance level with 500 degrees of freedom is 10.96. The critical value with 536 degrees of freedom must be lower, so we reject H0 at the 0.1% level.




0 :

0:

41

40

H

H

uXXXY 4433221 1RSS

2RSS

10.12536/6.2023

1/)6.20233.2069()4540(1)(

)4540,1(2

21

RSS

RSSRSSF

uXXY 33221

54

The null hypothesis we are testing is exactly the same as for a two-sided t test on the coefficient of SF.

96.10)500,1( crit,0.1% F




55

We will perform the t test. The t statistic is 3.48.

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F


56

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

The critical value of t at the 0.1% level with 500 degrees of freedom is 3.31. The critical value with 536 degrees of freedom must be lower. So we reject H0 again.

31.3crit,0.1% t


57

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% tIt can be shown that the F statistic for the F test of the explanatory power of a ‘group’ of one variable must be equal to the square of the t statistic for that variable. (The difference in the last digit is due to rounding error.)

11.1248.3 2


58

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% t11.1248.3 2 96.1031.3 2 It can also be shown that the critical value of F must be equal to the square of the critical value of t. (The critical values shown are for 500 degrees of freedom, but this must also be true for 536 degrees of freedom.)


59

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% t11.1248.3 2 Hence the conclusions of the two tests must coincide.

96.1031.3 2


60

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% t11.1248.3 2 96.1031.3 2 This result means that the t test of the coefficient of a variable is a test of its marginal explanatory power, after all the other variables have been included in the equation.


61

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% t11.1248.3 2 96.1031.3 2 If the variable is correlated with one or more of the other variables, its marginal explanatory power may be quite low, even if it genuinely belongs in the model.


62

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% t11.1248.3 2 96.1031.3 2 If all the variables are correlated, it is possible for all of them to have low marginal explanatory power and for none of the t tests to be significant, even though the F test for their joint explanatory power is highly significant.


63

96.10crit,0.1% F




10.12536/6.2023

1/)6.20233.2069()536,1(

F

31.3crit,0.1% t11.1248.3 2 96.1031.3 2 If this is the case, the model is said to be suffering from the problem of multicollinearity discussed in the previous sequence.

Copyright Christopher Dougherty 2011.

These slideshows may be downloaded by anyone, anywhere for personal use.

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used as a resource for teaching an econometrics course. There is no need to

refer to the author.

The content of this slideshow comes from Section 3.5 of C. Dougherty,

Introduction to Econometrics, fourth edition 2011, Oxford University Press.

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Individuals studying econometrics on their own and who feel that they might

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or the University of London International Programmes distance learning course

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11.07.25

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