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1 ( ) (, ) (X, Y) Y Y = X =0601 Y = a + bx + cx X (nonlinea) ( ) X Y X Y b(016) ( ) log Y = log a + b log X = e Y = b ax 71 X (explanatoy va :independent ), Y (dependent : esponse) X, Y Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

2 711 : FITNESStxt SASUSER Fitness SAS/Insight Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

3 71 SAS PROC PLOT ( ) HPOS=50 VPOS=5 (Hoizontal) 50%, (Vetical) 5% Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

4 713 SAS PROC GPLOT PLOT I Intepolation( ), V value( ), C Colo Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

5 714 SAS SAS/Insight SAS (Solution) (scatte plot matix) Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

6 1 (18 ) 1 Oxygen Y, Runtime X Oxygen X, Runtime Y 7? Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

7 1 ( ) ( ) 1 ( ) ( ), 1( ) 1( 71 SAS PROC CORR VAR, PROC CORR NOSIMPLE Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

8 (Peason) ( : ρ = 0 ) p- 005( 5%) (un time) (-086) ( ) (AGE) (MAXPULSE) (p- =0003) (age) (RUNTIME) ( ) ( =019, p- =031) ( : 06 ) Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

9 7 HO: =ρ0 =0 HO: =ρ0 0 t = ~ T ( n ) (1 ) /( n ) 1+ z* = 05ln 1 1+ ~ N(05ln 1 ρ ρ 1, ) n 3 1+ h1 = 05 ln 196 / 1 n 3) 1+ h = 05ln / 1 n 3) 95% e ( e h1 h1 e + e h1 h1 e, e h h e + e h h ) ( ) 1+ z( x) = 05ln 1 z = 1/( n z( x) z( y) x x x 3) + 1/( n 1+, z( y) = 05ln 1 y ~ N(0,1) 3) y y 73 ( ) ( Y = a + bx ) 1) (X Y) ) b( ) : (1) X (unit) Y b () b 3) a, b X Y ( ) Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

10 731 (association),, ( ) ( ) ( ) ( ) ( ) Log-linea( ) ( ) ( ) Logistic egession, (coelation analysis) (egession analysis) (linea) (casual elationship) ( ),, IQ,? IQ IQ? (independent vaiable) (explanatoy vaiable) (, IQ) (esponse vaiable) (dependent vaiable) ( ) ( :??) F Galton, ( : egess) Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

11 73 y i = a + bxi + ei fo i = 1,,, n ( ) e i iid N (0, σ ) ~ =( a + bx i )+ ( e i ) (cf) iid: independently( ) and identically distibuted( ) 1) ) 3 3) (esidual analysis) ( aˆ, b ˆ ) OLS (Odinay Least Squae ) min e = ( ) i yi a bxi a, b Y ( ) unexplained by Model y ( ) Explained by Model R =SSR/SST ( yi y) = ( yi yˆ) + ( yˆ y) => SST = SSE + SSR ( = + ) X SSR / df F = : ( : H 0 : b = 0 t- ) SSE / df Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

12 733 Sy 1) b ˆ = Sx 1) ) 3)F- ) = R 734 SAS ( ) ( ) SSR SSE SST t = F Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

13 F- ( t- ) ( b = 0 ) : = * (t=-917, p<0001) 735 (esidual analysis) (,, : F- ) Y Residual, X (pedicted) ( ) ± STDENT=RES_S RES_S Y = MSE Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

14 F- ê Yˆ Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

15 Y Residual( ê ) Yˆ Y * = Y * Y = Log( Y ) Y * = 1/ Y 735 SAS Solution Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

16 ( 114) Sehyug Kwon, Dept of Statistics, HANNAM Univesity Fall,

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Microsoft PowerPoint - Info R(3) pptx Coelaton Analyss 개념 Bvaate analyss 측정형두변수간의관계분석 상관관계? 두측정형변수의산점도 : 상호직선적관련성을상관계수 (Coelaton Coeffcent 측정. 잠재설명 ( 원인 변수 (X s 상관관계, 잠재변인과결과변수 (Y 의상관관계 Peason 상관계수 측정형변수직선관계정도 cov( X, Y E( X E( X E( Y E( Y