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Hypothesis Testing 2: Joint Restrictions

Hypothesis Testing 2: Joint Restrictions. Objectives. Testing joint hypotheses Chow test. 1 . Adding or Dropping Variables. Y i =  0 +  1 X 1 i +  2 X 1 i 2 +  3 X 2 i +  i Can the variation in X 1 explain the variation in Y significantly?.

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Hypothesis Testing 2: Joint Restrictions

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  1. Hypothesis Testing 2: Joint Restrictions Objectives • Testing jointhypotheses • Chow test 2010 ECON 7710

  2. 1. Adding or Dropping Variables Yi = 0 + 1X1i + 2X1i2 + 3X2i + i Can the variation in X1 explain the variation in Y significantly? Yi = 0 + 1X1i + 2X2i + 3X3i+ i Can X2 and X3 explain Y significantly simultaneously? 2010 ECON 7710

  3. Regression model: Yi = 0 + 1X1i + 2X2i + … + KXKi + i Ho: 1 =0, 2 = 0, , M = 0 HA: i  0 for at least one i = 1, 2, , M. Unconstrained model: Yi = 0 + 1X1i + 2X2i + … + KXKi + i  ESSU, RSSU. Constrained model: (delete M regressors) Yi = 0 + M+1XM+1,i + … + KXKi + i ESSC  ESSU,RSSC  RSSU. 2010 ECON 7710

  4. Unconstrained model: Yi = 0 + 1X1i + 2X2i + … + KXKi + i  ESSU, RSSU. Constrained model: (delete M regressors) Yi = 0 + M+1XM+1,i + … + KXKi + i ESSC  ESSU,RSSC  RSSU. 2010 ECON 7710

  5. Total variation in Y: Unexplained variation in the unconstrained model: RSSU = eU2 Increased unexplained variation in the constrained model: RSSC – RSSU = eC2–eU2 If  is normally distributed, then RSSC – RSSU has a chi-square distribution. 2010 ECON 7710

  6. f(F) F Distribution  1- F 0 Fc Ho: 1 =0, 2 = 0, , M = 0 HA: i  0 for at least one i = 1, 2, , M. 2010 ECON 7710

  7. Testing Procedures • Yi = 0 + 1X1i + … + KXKi + i • Unconstrained model  RSSU with N – K – 1 degrees of freedom • Constrained model with M restrictions  RSSC with N – K – 1 + M degrees of freedom • Critical value: Fc = FM,N-K-1, • Reject Ho if F > Fc. 2010 ECON 7710

  8. Example 1: Consider the following regression model: Yi = 0 + 1X1i + 2X2i + 3X3i + i. What are the unconstrained and constrained models that are used to test the following null hypotheses? • 1 = 0 • 2 = 0 and 3 = 0 • k = 0 for k = 1, 2, 3 2010 ECON 7710

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

  10. Example 2 (Cont’d): Regression results Unconstrained model uniGPA’ = 1.390 + .412hsGPA + .015HKAL - .083skipped se (0.332) (0.094) (0.011) (0.026) R2 = 0.2336, N = 141, RSSU = 14.87297 Constrained model uniGPA’ = 1.286 + .453hsGPA + .009HKAL se (0.341) (0.096) (0.011) R2 = 0.1764, N = 141, RSSC = 15.98244 2010 ECON 7710

  11. -0.083 (RSSC RSSU)/M F = t = 0.026 RSSU/(N K – 1) (15.98244  14.87297)/1 = 14.87297/(141- 4) Example 2 (Cont’d): Hypothesis testing Ho: 3 = 0; HA: 3  0 = 3.1923 = 10.2197 Note that when there is only one restriction, F = t2. 2010 ECON 7710

  12. H0: 2= 0, 3= 0 HA: H0 not true Exclusion Restrictions: 2 variables Example 3: A general functional form TRi = 0 + 1Pi + 2Ai+ 3A2i+ i Testing the significance of advertising expenses on revenue. Constrained model: TRi = 0 + 1Pi + i 2010 ECON 7710

  13. ^ TR = 110.46*** – 10.20***P + 3.36***A – 0.027*A2 se (3.74) (1.58) (0.42) (0.016) R2 = 0.878, N = 78, RSSU = 2,592.30 ^ TR = 111.71*** + 5.06P se (8.85) (4.01) R2 = 0.0205, N = 78, RSSC = 20,907.33 (RSSC RSSU)/M = F = RSSU/(N K – 1) First run unconstrained regression to get RSSU. Next run the constrained regression by dropping Aiand Ai2to get RSSC. 2010 ECON 7710

  14. Remark: Relation between F and R2 RSSC = TSS(1 - RC2) RSSU = TSS(1 - RU2) 2010 ECON 7710

  15. A Special Case: Testing the Overall Significance of the Regression Equation Yi = 0 + 1X1i + 2X2i + … + KXKi + i To test the overall significance of the regression equation, the null and alternative hypotheses are H0: 1 = 2 =  =K = 0 HA: H0 not true 2010 ECON 7710

  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 Ho if F > Fc = F1,2, 2010 ECON 7710

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

  18. ESS/K F = RSS/(N-K-1) Example 4: Picking Restaurant Locations (Table 3.1) Yhat = 102192 – 9075N + 0.3547P + 1.2879I se (12800) (2053) (0.0727) (0.5433) R2 = 0.6182, N = 33, RSS = 6133282062, TSS = 16062183882 2010 ECON 7710

  19. 2010 ECON 7710

  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.R2 = 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.R2 = 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

  21. 2. Are Two Equations Equal? 2010 ECON 7710

  22. Suppose there are 2 groups of data: Group A: (Yi, X2i,…,XKi), i = 1,…,N1. Group B: (Yi, X2i ,…,XKi), i = N1+1,…,N If the relation between X & Y is different, then (1) Yi = 0 + 1X1i +…+ KXKi + 1i, i = 1,…,N1 (2) Yi = 0 + 1X1i +…+  KXKi + 2i, i = N1+1,…,N If the relation between is identical for both groups, (3) Yi = 0 + 1X1i +…+ KXKi + i, i = 1,…,N 2010 ECON 7710

  23. Should the two groups be treated as one group? 1. Ho:  k= k, k= 0,1,…,K; H1:  k  k for at least one k 2. Estimate equations (1) and (2) to get RSS1 and RSS2. RSSU = RSS1 + RSS2. 3. Estimate equation (3) to get RSSC. 2010 ECON 7710

  24. Reject Ho if F > FK+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

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

  26. Example 7 (Cont’d): Empirical results of different periods 2 Dep. Constant Indep. V R SEE RSS N variable X Y 1.0161 0.0803 0.9021 0.1412 1785 12 (70-81) (11.6377) (0.00837) Y 153.4947 0.0148 0.2971 0.1660 10005 14 (82-95) (32.7123) (0.00839) Y 62.4226 0.0376 0.7672 0.1891 23248 26 (70-95) (12.7608) (0.0424) (RSSC - RSSU)/(K+1) (23248.30 – 1785.032 – 10005.22)/ 2 = F = (1785.032 +10005.22) / (26-2*2) RSSU / (N-2K-2) > Fc= F2,22, 0.01= 5.72 F = 10.69 2010 ECON 7710

  27. Exercises: • Find the following critical values •  = 5%, 1 = 3, 2 = 10; •  = 5%, 1 = 12, 2 = 5; •  = 1%, 1 = 2, 2 = 19; 2010 ECON 7710

  28. 2. Testing the joint significance of the estimated coefficients of X3, X4 and X5. ( = 5%) 2010 ECON 7710

  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. 2010 ECON 7710

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