Solution to class problem set 6 - Class Assignment 6 Use...

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Class Assignment 6: Use Data 7-6 to test for structural break (Chow s Test) for the following model: Variable D90=0 for 1980 census and D90=1 for 1990 census. (1) Povrte= β 0 + β 1 urb+ β 2 famsize+ β 3 unemp+ β 4 highschl+ β 5 college+ β 6 medinc+u Method 1- H 0: There no structural break (model differences) H 1 : Null is not true Regress (1) using the entire sample to obtain SSR re =517.2600 in model 1. Regress (1) using obs. 1-58 (1980 census) to obtain SSR 1 =133.95, model 1. Regress (1) using obs. 59-116 (1990 census) to obtain SSR 2 =146.09. SSR ur =113.95+146.09=280.04 Test statistics ࠵? = 517.26−280.04 7 280.04 102 = 12.34 > ࠵? 7,102 (0.01) = 2.82 Hence, reject the null. There is structural break Model 1: OLS, using observations 1-116, Dependent variable: povrate Coefficient Std. Error t-ratio p-value const 46.7035 6.08573 7.6743 <0.0001 *** urb 0.00144673 0.00876398 0.1651 0.8692 . . . . . . . . . Sum squared resid 517.2600 S.E. of regression 2.178418 Model 2: OLS, using observations 1-58, Dependent variable: povrate Coefficient Std. Error t-ratio p-value const 27.8515 8.67518 3.2105 0.0023 *** urb 0.0238359 0.0108756 2.1917 0.0330 ** . . . . . . . . . Sum squared resid 133.9526 S.E. of regression 1.620655 Model 3: OLS, using observations 59-116 (n = 58), Dependent variable: povrate
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