Introductory Econometrics. Lecture 13: Hypothesis testing in the multiple regression model, Part 1

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1 Introductory Econometrics Lecture 13: Hypothesis testing in the multiple regression model, Part 1 Jun Ma School of Economics Renmin University of China October 19, 2016

2 The model I We consider the classical normal linear regression model: 1. Y i = b 0 X 1,i b k X k,i + U i. 2. Conditional on X ís, E (U i ) = 0 for all iís. 3. Conditional on X ís, E! Ui 2 " = s 2 for all iís. 4. Conditional on X ís, E! " U i U j = 0 for all i 6= j. 5. Conditional on X ís, U i ís are jointly normally distributed. I We also continue to assume no perfect multicolinearity: The k regressors and constant do not form a perfect linear combination, i.e. we cannot Önd constants c 1,..., c k, c k+1 (not all equal to zero) such that for all iís: c 1 X 1,i c k X k,i + c k+1 = 0. 1/15

3 Testing a hypothesis about a single coe cient I Take the j-th coe cient b j, j 2 f0, 1,..., kg. I Under our assumptions, its OLS estimator ˆb j satisöes that # # $$ conditional on X ís: ˆb j % N b j, Var ˆb j, where # $ Var ˆb j = s 2 / Â n X i=1 j,i 2 (see Lecture 12). # $ r # $ I Therefore, ˆb j & b j / Var ˆb j % N (0, 1). # $ I The conditional variance Var ˆb j is unknown because s 2 is # $ unknown.the estimator for Var ˆb j is # $ dvar ˆb j = s 2 Â n i=1 X 2 j,i, where s 2 = Â n i=1 Û2 i / (n & k & 1) (see Lecture 11). 2/15

4 Testing a hypothesis about a single coe cient I We have that conditional on X ís, # $ I Standard error: SE ˆb j = ˆb j & b r j # $ % t n&k&1. dvar ˆb j r # $ q dvar ˆb j = s 2 / Â n X i=1 j,i 2. 3/15

5 Testing a hypothesis about a single coe cient: Two-sided alternatives I Consider testing H 0 : b j = b j,0 against H 1 : b j 6= b j,0. I Under H 0, we have that T = ˆb j & b j,0 r # $ % t n&k&1. dvar ˆb j I Let t df,t be the t-th quantile of the t df distribution. I Test: Reject H 0 when jt j > t n&k&1,1&a/2. I P-value: Find t n&k&1,1&t such that jt j = t n&k&1,1&t. The p-value=t ( 2. 4/15

6 Testing a hypothesis about a single coe cient: One-sided alternatives I Consider testing H 0 : b j ) b j,0 against H 1 : b j > b j,0. I When b j = b j,0 we have that T = ˆb j & b r j,0 # $ % t n&k&1. dvar ˆb j I Let t df,t be the t-th quantile of the t df distribution. I Test: Reject H 0 when T > t n&k&1,1&a. I P-value: Find t n&k&1,1&t such that T = t n&k&1,1&t. The p-value=t. 5/15

7 Testing a hypothesis about a single linear combination of the coe cients I Let c 0, c 1,..., c k, r be some constants. Consider testing H 0 : c 0 b 0 = r against H 1 : c 0 b 0 6= r. I Example 1: Consider the model ln Y i = b 0 ln L i + b 2 ln K i + U i. I We want to test for constant returns to scale H 0 : b 1 + b 2 = 1 I In this case: c 0 = 0, c 1 = 1, c 2 = 1, r = 1. 6/15

8 Testing a hypothesis about a single linear combination of the coe cients I Let r, c 0, c 1,..., c k are some constants. Consider testing H 0 : c 0 b 0 = r against H 1 : c 0 b 0 6= r. I Example 2: Consider the model ln (Wage i ) = b 0 Experience i + b 2 PrevExperience i + b 3 X 3,i +...b k X k,i + U i. I I We want to test that Experience and PrevExperience have the same e ect on wage: H 0 : b 1 = b 2 or H 0 : b 1 & b 2 = 0. In this case: c 0 = 0, c 1 = 1, c 2 = &1, c 3 =...= c k = 0, r = 0. 7/15

9 Testing a hypothesis about a single linear combination of the coe cients I We have that under H 0 : c 0 b 0 = r c 0 ˆb 0 + c 1 ˆb c k ˆb k & r qvar! c 0 ˆb 0 + c 1 ˆb c k ˆb k " = = c 0 ˆb 0 + c 1 ˆb c k ˆb k & (c 0 b 0 ) qvar! c 0 ˆb 0 + c 1 ˆb c k ˆb k " % N (0, 1). I Note that Var! " c 0 ˆb 0 + c 1 ˆb c k ˆb k = k # $ = cj 2 Var ˆb j +  j=1 k   j=1 l6=j c j c l Cov # ˆb j, ˆb l $. 8/15

10 Testing a hypothesis about a single linear combination of the coe cients I Consider T = c 0 ˆb 0 + c 1 ˆb c k ˆb k & r q d Var! c 0 ˆb 0 + c 1 ˆb c k ˆb k ". I Under H 0 : c 0 b 0 = r, T % t n&k&1. I Two-sided Test: Reject H 0 when jt j > t n&k&1,1&a/2. I One-sided: When testing H 0 : c 0 b 0 ) r against H 1 : c 0 b 0 > r, reject H 0 when T > t n&k&1,1&a. 9/15

11 Testing a hypothesis about a single linear combination of the coe cients I Consider the model ln Y i = b 0 ln L i + b 2 ln K i + U i. I We want to test for constant returns to scale: H 0 : b 1 + b 2 = 1. I The test statistic: T = ˆb 1 + ˆb 2 &1 q d Var(ˆb 1 + ˆb 2 ). I dvar! ˆb 1 + ˆb 2 " = dvar! ˆb 1 " + dvar! ˆb 2 " + 2 dcov! ˆb 1, ˆb 2 ". I I dvar! " ˆb 1 and dvar! " ˆb 2 can be computed from the corresponding standard errors reported by Stata. In Stata, dcov! " ˆb 1, ˆb 2 can be obtained (together with the variances) by using the command "matrix list e(v)" after running a regression. I Reject H 0 : b 1 + b 2 = 1 if jt j > t n&3,1&a/2. 10/15

12 Example I 1000 observations were generated using the following model: L i = e l i K i = e k i ( where l i, k i are iid N (0, 1), Cov (l i, k l ) = 0.5, U i % iid N (0, 1) is independent of l i, k i, Y i = L 0.35 i Ki 0.52 e U i. I The following equation was estimated: ln Y i = b 0 ln L i + b 2 ln K i + U i. I We test H 0 : b 1 + b 2 = 1 against H 1 : b 1 + b 2 6= 1 at 5% signiöcance level. 11/15

13 Example. regress lny lnl lnk Source SS df MS Number of obs = F( 2, 997) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = lny Coef. Std. Err. t P> t [95% Conf. Interval] lnl lnk _cons matrix list e(v) symmetric e(v)[3,3] lnl lnk _cons lnl lnk _cons 5.066e display invttail(997,0.025) /15

14 Example I We obtained: I ˆb 1 = , I ˆb 2 = I dvar! ˆb 1 " = = I dvar! " ˆb 2 = = I dcov! " ˆb 1, ˆb 2 = & I t 997,0.975 = q I Var! " d ˆb 1 + ˆb 2 = p & 2 ( = I T = ( & 1) / &2.35, I jt j = 2.35 > = t 997,0.975 =) We reject H 0. I Note that ignoring the covariance leads to an incorrect result: ( & 1) / p & /15

15 An alternative approach I We want to test b 1 + b 2 = 1 in ln Y i = b 0 ln L i + b 2 ln K i + U i. I DeÖne d = b 1 + b 2 or b 2 = d & b 1 so that ln Y i = b 0 ln L i + b 2 ln K i + U i = b 0 ln L i + (d & b 1 ) ln K i + U i = b 0 (ln L i & ln K i ) + d ln K i + U i. I Generate a new variable D i = ln L i & ln K i. I Estimate ln Y i = b 0 D i + d ln K i + U i. I Test H 0 : d = 1 against H 1 : d 6= 1. 14/15

16 Example. gen D=lnL-lnK. regress lny D lnk Source SS df MS Number of obs = F( 2, 997) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = lny Coef. Std. Err. t P> t [95% Conf. Interval] D lnk _cons I The 95% CI for the coe cient on ln K in the transformed mode does not include 1 =) We reject H 0. I Note that in the original equation ˆb 1 + ˆb 2 = and q d Var! ˆb 1 + ˆb 2 " = /15