# 2010 ECON 7710 5.1 Hypothesis Testing 2: Joint Restrictions Testing joint hypotheses Chow test Objectives.

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2010 ECON 7710 5.1 Hypothesis Testing 2: Joint Restrictions Testing joint hypotheses Chow test Objectives

2010 ECON 7710 5.2 Y i =  0 +  1 X 1i +  2 X 1i 2 +  3 X 2i +  i Can the variation in X 1 explain the variation in Y significantly? Y i =  0 +  1 X 1i +  2 X 2i +  3 X 3i +  i Can X 2 and X 3 explain Y significantly simultaneously? 1. Adding or Dropping Variables

2010 ECON 7710 5.3 Regression model: Y i =  0 +  1 X 1i +  2 X 2i + … +  K X Ki +  i Ho:  1 =0,  2 = 0, ,  M = 0 H A :  i  0 for at least one i = 1, 2, , M. Unconstrained model: Y i =  0 +  1 X 1i +  2 X 2i + … +  K X Ki +  i  ESS U, RSS U. Constrained model: (delete M regressors) Y i =  0 +  M+1 X M+1,i + … +  K X Ki +  i  ESS C  ESS U, RSS C  RSS U.

2010 ECON 7710 5.4 Unconstrained model: Y i =  0 +  1 X 1i +  2 X 2i + … +  K X Ki +  i  ESS U, RSS U. Constrained model: (delete M regressors) Y i =  0 +  M+1 X M+1,i + … +  K X Ki +  i  ESS C  ESS U, RSS C  RSS U.

2010 ECON 7710 5.5 If  is normally distributed, then RSS C – RSS U has a chi-square distribution. Total variation in Y: Unexplained variation in the unconstrained model: RSS U =  e U 2 Increased unexplained variation in the constrained model: RSS C – RSS U =  e C 2 –  e U 2

2010 ECON 7710 5.6 F 0 f(F) F Distribution 1-   FcFc Ho:  1 =0,  2 = 0, ,  M = 0 H A :  i  0 for at least one i = 1, 2, , M.

2010 ECON 7710 5.7 Y i =  0 +  1 X 1i + … +  K X Ki +  i Unconstrained model  RSS U with N – K – 1 degrees of freedom Constrained model with M restrictions  RSS C with N – K – 1 + M degrees of freedom Critical value: F c = F M,N-K-1,  Reject H o if F > F c. Testing Procedures

2010 ECON 7710 5.8 Example 1: Consider the following regression model: Y i =  0 +  1 X 1i +  2 X 2i +  3 X 3i +  i. What are the unconstrained and constrained models that are used to test the following null hypotheses? a.  1 = 0 b.  2 = 0 and  3 = 0 c.  k = 0 for k = 1, 2, 3

2010 ECON 7710 5.9 Exclusion Restriction: 1 variable Unconstrained model: uniGPA =  0 +  1 hsGPA +  2 HKAL +  3 skipped +  Restriction:  3 =0 Constrained model: uniGPA =  0 +  1 hsGPA +  2 HKAL +  Example 2

2010 ECON 7710 5.10 Example 2 (Cont’d): Regression results Unconstrained model uniGPA’ = 1.390 +.412hsGPA +.015HKAL -.083skipped se (0.332) (0.094) (0.011) (0.026) R 2 = 0.2336, N = 141, RSS U = 14.87297 Constrained model uniGPA’ = 1.286 +.453hsGPA +.009HKAL se (0.341) (0.096) (0.011) R 2 = 0.1764, N = 141, RSS C = 15.98244

2010 ECON 7710 5.11 H o :  3 = 0; H A :  3  0 (RSS C  RSS U )/M RSS U /(N  K – 1 ) F = (15.98244  14.87297)/1 14.87297/(141 - 4) = = 10.2197 -0.083 0.026 t = = 3.1923 Note that when there is only one restriction, F = t 2. Example 2 (Cont’d): Hypothesis testing

2010 ECON 7710 5.12 Exclusion Restrictions: 2 variables TR i =  0 +  1 P i +  2 A i +  3 A 2 i +  i Example 3: A general functional form H 0 :   = 0,  3 = 0 H A : H 0 not true Testing the significance of advertising expenses on revenue. Constrained model: TR i =  0 +  1 P i +  i

2010 ECON 7710 5.13 Next run the constrained regression by dropping A i and A i 2 to get RSS C. First run unconstrained regression to get RSS U. (RSS C  RSS U )/M RSS U /(N  K – 1) F = = TR = 110.46 *** – 10.20 *** P + 3.36 *** A – 0.027 * A 2 se (3.74) (1.58) (0.42) (0.016) R 2 = 0.878, N = 78, RSS U = 2,592.30 ^ TR = 111.71*** + 5.06P se (8.85) (4.01) R 2 = 0.0205, N = 78, RSS C = 20,907.33 ^

2010 ECON 7710 5.14 Remark: Relation between F and R 2 RSS C = TSS(1 - R C 2 ) RSS U = TSS(1 - R U 2 )

2010 ECON 7710 5.15 To test the overall significance of the regression equation, the null and alternative hypotheses are A Special Case: Testing the Overall Significance of the Regression Equation Y i =  0 +  1 X 1i +  2 X 2i + … +  K X Ki +  i H 0 :  1 =  2 =  =  K = 0 H A : H 0 not true

2010 ECON 7710 5.16 The test statistic is ESS / K = average explained sum of squares RSS / (N – K – 1) = average unexplained sum of squares Larger F means higher explanatory power. Degrees of freedom: 1 = K, 2 = N – K – 1 Reject H o if F > F c = F 1, 2, 

2010 ECON 7710 5.17 Example 4: Picking restaurant locations pp. 75 – 78) Y i =  0 +  1 N i +  2 P i +  3 I i +  i N: Competition P: Population I: Income If the model cannot explain the variation of Y:  1 =  2 =  3 = 0.

2010 ECON 7710 5.18 Example 4: Picking Restaurant Locations (Table 3.1) Yhat = 102192 – 9075N + 0.3547P + 1.2879I se (12800) (2053) (0.0727) (0.5433) R 2 = 0.6182, N = 33, RSS = 6133282062, TSS = 16062183882 F = ESS/K RSS/(N-K-1)

2010 ECON 7710 5.19

2010 ECON 7710 5.20 Example 6 : Consider the following estimated saving function: Shat = -42.75 + 0.015Y + 0.007W + 7.67r se (-3.30) (2.09) (1.75) (3.81) Adj.R 2 = 0.962, F = 251.5, RSS = 2470.8, N = 31 Another regression has been run with the same data set, Shat = 9.4 + 0.06Y se (2.1) (17.2) Adj.R 2 = 0.908, F = 296.4, RSS = 6372.8, N = 31. Are the coefficients for wealth and interest rate jointly significant at 1% level?

2010 ECON 7710 5.21 2. Are Two Equations Equal?

2010 ECON 7710 5.22 Suppose there are 2 groups of data: Group A: (Y i, X 2i,…,X Ki ), i = 1,…,N 1. Group B: (Y i, X 2i,…,X Ki ), i = N 1 +1,…,N If the relation between X & Y is different, then (1) Y i = 0 + 1 X 1i +…+ K X Ki +  1i, i = 1,…,N 1 (2) Y i =  0 +  1 X 1i +…+  K X Ki +  2i, i = N 1 +1,…,N If the relation between is identical for both groups, (3) Y i =  0 +  1 X 1i +…+  K X Ki +  i, i = 1,…,N

2010 ECON 7710 5.23 Should the two groups be treated as one group? 1. H o : k =  k, k= 0,1,…,K; H 1 : k   k for at least one k 2. Estimate equations (1) and (2) to get RSS 1 and RSS 2. RSS U = RSS 1 + RSS 2. 3. Estimate equation (3) to get RSS C.

2010 ECON 7710 5.24 Reject H o if F > F K+1,(N-2K-2), . Remarks: a. This method is called the Chow test. b.It is assumed that the variances of the two groups are equal. c. One can use dummy variables to test for this change of equation structure.

2010 ECON 7710 5.25 Example 7 : Structural change in the US saving function (BE4_Tab0809) savings =  0 +  1 Income + 

2010 ECON 7710 5.26 Example 7 (Cont’d) : Empirical results of different periods Dep. variable ConstantIndep. V X R 2 SEERSS N Y (70-81) 1.0161 (11.6377) 0.0803 (0.00837) 0.90210.141217851212 Y (82-95) 153.4947 (32.7123) 0.0148 (0.00839) 0.29710.1660100051414 Y (70-95) 62.4226 (12.7608) 0.0376 (0.0424) 0.76720.1891232482626 F = (RSS C - RSS U )/(K+1) RSS U / (N-2K-2) = (23248.30 – 1785.032 – 10005.22)/ 2 (1785.032 + 10005.22) / (26-2*2) F = 10.69 F c = F 2,22, 0.01 = 5.72 >

2010 ECON 7710 5.27 Exercises: 1.Find the following critical values a.  = 5%, 1 = 3, 2 = 10; b.  = 5%, 1 = 12, 2 = 5; c.  = 1%, 1 = 2, 2 = 19;

2010 ECON 7710 5.28 2. Testing the joint significance of the estimated coefficients of X 3, X 4 and X 5. (  = 5%)

2010 ECON 7710 5.29 3. Consider the following regression results using the weight-height data of some students in 2004. Test whether the weight-height relation is different for male and female. AllFemaleMale Intercept-66.63 *** (18.7198) -39.00* (21.0006) -43.99 (51.6617) Height0.7287 *** (0.1110) 0.5491 *** (0.1291) 0.6097* (0.2946) R2R2 0.62380.58180.2802 N281513 RSS960.8552274.6812559.7543

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