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Lecture_9_Prof._Arkonac's_Slides_(Ch_7.3_-_8.3)

Lecture_9_Prof._Arkonac's_Slides_(Ch_7.3_-_8.3) - Multiple...

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Multiple Regression (cont’) Confidence Intervals in Multiple Regression. Nonlinear Regression I (Fall 2010) Lecture 9 Prof: Seyhan Arkonac, PhD PS #3 is due NOW! Solutions will be posted on Oct. 6 th after 9pm. PS#4 will be posted today after class. 1
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TA Information: Naihobe Gonzalez E-mail: [email protected] Office Hours: Thurs 12-1 (Uris Library), Recitation: Thurs 11-11:50 (PUP 424) TA Information: Ju Hyun Kim E-mail: [email protected] Office Hours: Recitation: Fri 2:10-3PM(PUP 424), Office Hours: Fri 3:10-4:10PM(Lehman) TA Information: WooRam Park E-mail: [email protected] Office Hours: Office hours : Thurs 2:00~3:00 IAB 1006A Recitation Thurs 3:10~4:00 IAB 403 TA Information: Ran Huo E-mail: [email protected] Office Hours: Recitation: Thursday 12:00-12:50 404IAB; Office Hour: Wednesday 1-2 1006A IAB TA Information: Shreya Agarwal E-mail: [email protected] Office Hours: Mon 12:30pm - 1:30 pm (Uris Library Common Area) Recitation: Fri 12:00pm - 12:50pm (Schermerhorn Extension 558) 2
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3 Testing Single Restrictions on Multiple Coefficients (SW Section 7.3) Y i = 0 + 1 X 1 i + 2 X 2 i + u i , i = 1,…, n Consider the null and alternative hypothesis, H 0 : 1 = 2 vs. H 1 : 1 2 This null imposes a single restriction ( q = 1) on multiple coefficients – it is not a joint hypothesis with multiple restrictions (compare with 1 = 0 and 2 = 0).
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4 Testing single restrictions on multiple coefficients, ctd. Here are two methods for testing single restrictions on multiple coefficients: 1. Rearrange (“transform”) the regression Rearrange the regressors so that the restriction becomes a restriction on a single coefficient in an equivalent regression; or, 2. Perform the test directly Some software, including STATA, lets you test restrictions using multiple coefficients directly
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5 Method 1: Rearrange (“transform”) the regression Y i = 0 + 1 X 1 i + 2 X 2 i + u i H 0 : 1 = 2 vs. H 1 : 1 2 Add and subtract 2 X 1 i : Y i = 0 + ( 1 2 ) X 1 i + 2 ( X 1 i + X 2 i ) + u i or Y i = 0 + 1 X 1 i + 2 W i + u i where 1 = 1 2 W i = X 1 i + X 2 i
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6 Rearrange the regression, ctd. (a) Original system : Y i = 0 + 1 X 1 i + 2 X 2 i + u i H 0 : 1 = 2 vs. H 1 : 1 2 (b) Rearranged (“transformed”) system : Y i = 0 + 1 X 1 i + 2 W i + u i where 1 = 1 2 and W i = X 1 i + X 2 i so H 0 : 1 = 0 vs. H 1 : 1 0 The testing problem is now a simple one: test whether 1 = 0 in specification (b).
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Does ACT and bgfriend have the same effect on colGPA ? . reg colGPA hsGPA ACT bgfriend skipped,r Linear regression Number of obs = 141 F( 4, 136) = 11.12 Prob > F = 0.0000 R-squared = 0.2446 Root MSE = .32832 ------------------------------------------------------------------------------ | Robust colGPA | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- hsGPA | .4186377 .0987409 4.24 0.000 .2233717 .6139038 ACT | .0140484 .0110336 1.27 0.205 -.0077711 .035868 bgfriend | .0780312 .0554459 1.41 0.162 -.0316163 .1876788 skipped | -.0825624 .0256374 -3.22 0.002 -.1332619 -.0318629 _cons | 1.344903 .3590842 3.75 0.000 .6347917 2.055014 ------------------------------------------------------------------------------ 7
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Are hsGPA and ACT jointly significant? . test hsGPA ACT ( 1) hsGPA = 0 ( 2) ACT = 0 F( 2, 136) = 12.95 Prob > F = 0.0000 Are hsGPA and bgfriend jointly significant? . test hsGPA bgfriend ( 1) hsGPA = 0 ( 2) bgfriend = 0 F( 2, 136) = 9.27 Prob > F = 0.0002 Then “transform” the regression to test H 0 : β ACT = β bgfriend 8
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Does ACT and bgfriend have the same effect on colGPA ?
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Lecture_9_Prof._Arkonac's_Slides_(Ch_7.3_-_8.3) - Multiple...

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