Economics 326 Methods of Empirical Research in Economics. Lecture 14: Hypothesis testing in the multiple regression model, Part 2

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1 Economics 326 Methods of Empirical Research in Economics Lecture 14: Hypothesis testing in the multiple regression model, Part 2 Vadim Marmer University of British Columbia May 5, 2010

2 Multiple restrictions I Consider the model: ln (Wage i ) = β 0 + β 1 Experience i + β 2 Experience 2 i + + β 3 PrevExperience i + β 4 PrevExperience 2 i + β 5 Education i + U i, where Experience is the experience at current job, and PrevExperience is the previous experience. I Suppose that we want to test the null hypothesis that, after controlling for the experience at current job and education, the previous experience has no e ect on wage: H 0 : β 3 = 0, β 4 = 0. I We have two restrictions on the model parameters. I The alternative hypothesis is that at least one of the coe cients, β 3 or β 4, is di erent from zero: H 1 : β 3 6= 0 or β 4 6= 0. 1/23

3 t-statistics and multiple restrictions I Let T 3 and T 4 be the t-statistics associated with the coe cients of PrevExperience and PrevExperience 2 : T 3 = ˆβ 3 SE ˆβ 3 and T 4 = ˆβ 4 SE ˆβ 4. I We can use T 3 and T 4 to test signi cance of β 3 and β 4 separately by using two separate size α tests: I I Reject H 0,3 : β 3 = 0 in favor of H 1,3 : β 3 6= 0 when jt 3 j > t n k 1,1 α/2. Reject H 0,4 : β 4 = 0 in favor of H 1,4 : β 4 6= 0 when jt 4 j > t n k 1,1 α/2. 2/23

4 t-statistics and multiple restrictions I Rejecting H 0 : β 3 = 0, β 4 = 0 in favor of H 1 : β 3 6= 0 or β 4 6= 0 when at least one of the two coe cients is signi cant at level α, i.e. when jt 3 j > t n k 1,1 α/2 or jt 4 j > t n k 1,1 α/2, is not a size α test! I Recall that if A and B are two sets then (A \ B) A and therefore P (A \ B) P (A). I When β 3 = β 4 = 0 : P (Reject H 0,3 or H 0,4 ) = = P (jt 3 j > t n k 1,1 α/2 or jt 4 j > t n k 1,1 α/2 ) = P (jt 3 j > t n k 1,1 α/2 ) + P (jt 4 j > t n k 1,1 α/2 ) P (jt 3 j > t n k 1,1 α/2 and jt 4 j > t n k 1,1 α/2 ) = α + α P (jt 3 j > t n k 1,1 α/2 and jt 4 j > t n k 1,1 α/2 ) α. 3/23

5 Testing multiple exclusion restrictions I Consider the model Y i = β 0 + β 1 X 1,i β q X q,i + β q+1 X q+1,i β k X k,i + U i. Suppose that we want to test that the rst q regressors have no e ect on Y (after controlling for other regressors). I The null hypothesis has q exclusion restrictions: H 0 : β 1 = 0, β 2 = 0,..., β q = 0. I The alternative hypothesis is that at least one of the restrictions in H 0 is false: H 1 : β 1 6= 0 or β 2 6= 0 or... or β q 6= 0. 4/23

6 F-statistic I The idea of the test is to compare the t of the unrestricted model with that of the null-restricted model. I LetSSR ur denote the Residual Sum-of-Squares of the unrestricted model Y i = β 0 + β 1 X 1,i β q X q,i + β q+1 X q+1,i β k X k,i + U i. I The restricted model given H 0 : β 1 = 0,..., β q = 0 is Y i = β 0 + β q+1 X q+1,i β k X k,i + U i. I LetSSR r denote the Residual Sum-of-Squares of the restricted model. I Consider the following statistic: F = (SSR r SSR ur ) /q SSR ur / (n k 1). I Note that q = number of restrictions; I n k 1 = unrestricted residual df, where k is the number of regressors in the unrestricted model. 5/23

7 F-statistic F = (SSR r SSR ur ) /q SSR ur / (n k 1). I Since SSR can only increase when you drop some regressors, SSR r SSR ur 0, and therefore F 0. I If the null restrictions are true, the excluded variables do not contribute to explaining Y (in population), and therefore we should expect that SSR r SSR ur is small and F is close to zero. I If the null restriction are false, the imposed restriction should substantially worsen the t, and we should expect that SSR r SSR ur is large and F is far from zero. I Thus, we should reject H 0 when F > c where c is some positive constant. 6/23

8 I Similarly to the standard normal and t distributions, the F distribution has been tabulated and its critical values are available in statistical tables and statistical software such as Stata. 7/23 F test F = (SSR r SSR ur ) /q SSR ur / (n k 1). I We should reject H 0 when F > c. I There is a probability that F > c even when H 0 is true, thus we need to choose c so that P (F > cjh 0 is true) = α. I It turns out that when H 0 is true, the F -statistic has F distribution with two parameters: the numerator df (q) and the denominator df (n k 1): F F q,n k 1.

9 F test When H 0 is true, F = (SSR r SSR ur ) /q SSR ur / (n k 1) F q,n k 1. I Let F q,n k 1,τ be the τ-quantile of the F q,n k 1 distribution. I A size α test H 0 : β 1 = 0,..., β q = 0 against H 1 : β 1 6= 0 or... or β q 6= 0 is Reject H 0 when F > F q,n k 1,1 α. I One can nd the p-value by nding τ such that F = F q,n k 1,1 τ. The p-value is equal to τ. 8/23

10 F distribution in Stata I To compute F critical values use disp invftail(q, n k 1, α). I To compute p-values from F distribution use disp Ftail(q, n k 1, F ). 9/23

11 Example I Consider the model: ln (Wage i ) = β 0 + β 1 Experience i + β 2 Experiencei β 3 PrevExperience i + β 4 PrevExperiencei 2 + β 5 Education i + U i. I We test H 0 : β 3 = 0, β 4 = 0 against H 1 : β 3 6= 0 or β 4 6= 0. I q = 2. I α = /23

12 Example: the unrestricted model. regress lnwage Experience Experience2 PrevExperience PrevExperience2 Education Source SS df MS Number of obs = F( 5, 520) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = lnwage Coef. Std. Err. t P> t [95% Conf. Interval] Experience Experience PrevExperi~e PrevExperi~ Education _cons I SSR ur = I n k 1 = = /23

13 Example: the restricted model. regress lnwage Experience Experience2 Education Source SS df MS Number of obs = F( 3, 522) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = lnwage Coef. Std. Err. t P> t [95% Conf. Interval] Experience Experience Education _cons I SSR r = /23

14 Example: F statistic and test I To compute the statistic: F = (SSR r SSR ur ) /q ( ) /2 = SSR ur / (n k 1) / ( ) I The critical value:. disp invftail(2,520,0.05) I The test: 6.61 > and at 5% signi cance level we reject H 0 that previous experience has no e ect on wage. I The p-value:. disp Ftail(2,520,6.61) =) We reject H 0 for any α > /23

15 Example: Stata test command I Instead of running two models, restricted and unrestricted, one can use the Stata test command after estimation of the unrestricted model. I To test that previous experience has no e ect:. test (PrevExperience=0) (PrevExperience2=0) I The output of this command is: ( 1) PrevExperience = 0 ( 2) PrevExperience2 = 0 F( 2, 520) = 6.61 Prob > F = /23

16 Example: Stata test command I To test that the coe cient on previous experience equal to the coe cient on experience and the coe cient on previous experience squared is zero:. test (Experience==PrevExperience2) (PrevExperience2=0) I The output is: ( 1) Experience - PrevExperience2 = 0 ( 2) PrevExperience2 = 0 F( 2, 520) = Prob > F = /23

17 F and R 2 I Let R 2 ur denote the R 2 corresponding to the unrestricted model: Y i = β 0 + β 1 X 1,i β q X q,i + β q+1 X q+1,i β k X k,i + U i. I Let R 2 r denote the R 2 corresponding to the restricted model: Y i = β 0 + β q+1 X q+1,i β k X k,i + U i. I The two models have the same dependent variable and therefore the same Total Sum-of-Squares: SST = n (Y i Ȳ ) 2 = SST ur = SST r. i=1 16/23

18 F and R 2 I In this case, we can write then F = (SSR r SSR ur ) /q SSR ur / (n k 1) /q SSRr SST SSR ur SST = SSR ur SST / (n k 1) = 1 R2 r 1 Rur 2 /q (1 Rur 2 ) / (n k 1) Rur 2 R 2 r /q = (1 Rur 2 ) / (n k 1). 17/23

19 F test: more examples I Suppose that you want to test H 0 : β 1 = 1 against H 1 : β 1 6= 1 in Y i = β 0 + β 1 X 1,i + β 2 X 2,i β k X k,i + U i. I The restricted model is Y i = β 0 + X 1,i + β 2 X 2,i β k X k,i + U i, or Y i X 1,i = β 0 + β 2 X 2,i β k X k,i + U i. 1. Generate a new dependent variable Y i = Y i X 1,i. 2. Regress Y against a constant, X 2,..., X k to obtain SSR r. 3. Estimate the unrestricted model to obtain SSR ur. 4. Compute F = (SSR r SSR ur )/1 SSR ur /(n k 1). 18/23

20 F test: more examples I Suppose that you want to test H 0 : β 1 + β 2 = 1 against H 1 : β 1 + β 2 6= 1 in Y i = β 0 + β 1 X 1,i + β 2 X 2,i β k X k,i + U i. I The restricted model is or Y i = β 0 + (1 β 2 ) X 1,i + β 2 X 2,i β k X k,i + U i, Y i X 1,i = β 0 + β 2 (X 2,i X 1,i ) β k X k,i + U i. 1. Generate a new dependent variable Y i = Y i X 1,i. 2. Generate a new regressor X 2 = X 2,i X 1,i. 3. Regress Y against a constant, X 2, X 3,..., X k to obtain SSR r. 4. Estimate the unrestricted model to obtain SSR ur. 5. Compute F = (SSR r SSR ur )/1 SSR ur /(n k 1). 19/23

21 Relationship between F and t statistics I The F statistic can also be used for testing a single restriction. I In the case of a single restriction, the F test and t test lead to the same outcome because t 2 n k 1 = F 1,n k 1. 20/23

22 Test of model signi cance I Consider the model Y i = β 0 + β 1 X 1,i β k X k,i + U i. I Suppose that you want to test that none of the regressors explain Y : H 0 : β 1 = β 2 =... = β k = 0 (k restrictions) against H 1 : β j 6= 0 for some j = 1,..., k. I The restricted model is given by Y i = β 0 + U i, and since ˆβ 0 = Ȳ in this model, SSR r = n (Y i Ȳ ) 2 = SST and SSR ur = SSR. i=1 21/23

23 Test of model signi cance I The F statistic for model signi cance test is F = (SSR r SSR ur ) /k SSR ur / (n k 1) (SST SSR) /k = SSR/ (n k 1) SSE /k = SSR/ (n k 1) = R 2 /k (1 R 2 ) / (n k 1). 22/23

24 Test of model signi cance I The F statistic for the model signi cance test and its p-value is reported by Stata as in the top part of the regression output. Source SS df MS Number of obs = F( 5, 520) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = /23