Lecture_12_Prof._Arkonac's_Slides_(Ch_8.3_-_end_of_9)

Lecture_12_Prof._Arkonac's_Slides_(Ch_8.3_-_end_of_9) -...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Nonlinear Regression III (cont’) Assessing Regression Studies I (Fall 2010) Lecture 12 Prof: Seyhan Erden Arkonac, PhD Solution to Problem Set 4 is posted. Problem Set 5 is due at the beginning of class on Tuesday October 19 th . 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 (b) Interactions between continuous and binary variables Y i = 0 + 1 D i + 2 X i + u i D i is binary, X is continuous As specified above, the effect on Y of X (holding constant D ) = 2 , which does not depend on D To allow the effect of X to depend on D , include the “interaction term” D i X i as a regressor: Y i = 0 + 1 D i + 2 X i + 3 ( D i X i ) + u i
Background image of page 2
3 Binary-continuous interactions: the two regression lines Y i = 0 + 1 D i + 2 X i + 3 ( D i X i ) + u i Observations with D i = 0 (the “ D = 0” group): Y i = 0 + 2 X i + u i The D=0 regression line Observations with D i = 1 (the “ D = 1” group): Y i = 0 + 1 + 2 X i + 3 X i + u i = ( 0 + 1 ) + ( 2 + 3 ) X i + u i The D=1 regression line
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4 Binary-continuous interactions, ctd.
Background image of page 4
5 Interpreting the coefficients Y i = 0 + 1 D i + 2 X i + 3 ( D i X i ) + u i General rule: compare the various cases Y = 0 + 1 D + 2 X + 3 ( D X ) (b) Now change X : Y + Y = 0 + 1 D + 2 ( X + X ) + 3 [ D ( X + X )] (a) subtract (a) – (b): Y = 2 X + 3 D X or Y X = 2 + 3 D The effect of X depends on D (what we wanted) 3 = increment to the effect of X , when D = 1
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
6 Example : TestScore, STR, HiEL (=1 if PctEL 10) TestScore = 682.2 – 0.97 STR + 5.6 HiEL – 1.28( STR HiEL ) (11.9) (0.59) (19.5) (0.97) When HiEL = 0: TestScore = 682.2 – 0.97 STR When HiEL = 1, TestScore = 682.2 – 0.97 STR + 5.6 – 1.28 STR = 687.8 – 2.25 STR Two regression lines: one for each HiSTR group. Class size reduction is estimated to have a larger effect when the percent of English learners is large.
Background image of page 6
7 Example, ctd: Testing hypotheses TestScore = 682.2 – 0.97 STR + 5.6 HiEL – 1.28( STR HiEL ) (11.9) (0.59) (19.5) (0.97) The two regression lines have the same slope the coefficient on STR HiEL is zero: t = –1.28/0.97 = –1.32 The two regression lines have the same intercept the coefficient on HiEL is zero: t = –5.6/19.5 = 0.29 The two regression lines are the same population coefficient on HiEL = 0 and population coefficient on STR HiEL = 0: F = 89.94 ( p -value < .001) !! We reject the joint hypothesis but neither individual hypothesis ( how can this be? )
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Why are they not significant individually, but significant jointly? Answer: Because of multicollinearity among
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 65

Lecture_12_Prof._Arkonac's_Slides_(Ch_8.3_-_end_of_9) -...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online