Koefisien Determinasi dan Korelasi Berganda

Koefisien Determinasi dan Korelasi Berganda Y = 1 + 2 X2 + 3 X3 +…+ k Xk + u PowerPoint® Slides

byYana Rohmana Education University of Indonesian

© 2007 Laboratorium Ekonomi & Koperasi Publishing

Jl. Dr. Setiabudi 229 Bandung, Telp. 022 2013163 - 2523

Koefisien Determiniasi dan Korelasi Berganda Ingin diketahui berapa proporsi (presentase) sumbangan X2 dan X3 terhadap variasi (naik turunnya) Y secara bersama-sama.  Besarnya proporsi/persentase sumbangan ini disebut koefisien determinasi berganda,dengan symbol R2.  Rumus R2 diperoleh dengan menggunakan definisi : 

ESS R   TSS 2

R  2

Chapter

 yˆ y

2 i 2 i

b12.3  x2i yi  b1 3. 2  x3i yi

y

2 i

Analisis Regresi Linier Berganda : Persoalan Estimasi dan Pengujian Hipotesis

2

Penerapan pada Kasus 2 R  2

b12.3  x2i yi  b1 3. 2  x3i yi

y

2

i

1.203,533  20,0028  1.260,889  0,9387

Chapter


3

Persamaan garis regresi linier berganda (kasus 2)

Ŷ = b1.23 + b12.3 X2 + b13.2 X3 Ŷ = -17,8685 + 0,9277 X2 + 0,2532 X3 Standar error: (0,0972) (0,1464) R2 = 0,9387 Se = 3,5907

Chapter


4

The adjusted R2 (R2) as one of indicators of the overall fitness ESS R2 =

RSS =1-

=1-

TSS _ R2 = 1 -

TSS

e^i2 yi2

ei2 / (n-k) yi2 /

_ R2 = 1 -

k: (n-1)

Se2

# of independent variables plus the constant term.

Sy2

_ R2 = 1 -

e2

(n-1)

y2

(n-k)

_ R2 = 1 - (1 - R2)

n : # of obs.

n-1 n-k

_

R2  R2 0 < R2 < 1 Adjusted R2 can be negative: R2  0 5

Y = 1 + 2 X2 + 3 X3 + u

Y

u ^

Y

TSS n-1

6

Suppose X4 is not an explanatory Variable but is included in regression

C X2 X3 X4

7

Koefisien Korelasi Parsial Dan Hubungan Berbagai Koefisien Korelasi dan Regresi



Y = b1.23 + b12.3 X2 + b13.2 X3 + ei r12 = koefisien korelasi antara Y dan X2 (antara X2 dan Y) r13 = koefisien korelasi antara Y dan X3 (antara X3 dan Y) r23 = koefisien korelasi antara X2 dan X3 (antara X3 dan X2)



Antara X dan Y :



Antara X2 dan Y :

  

Chapter

r

r12 

 xi yi  xi2

 yi2

 x2i yi  x22i

 yi2


8

Koefisien Korelasi Parsial Dan Hubungan Berbagai Koefisien Korelasi dan Regresi





Antara X3 dan Y :

Antara X2 dan X3 :

Chapter

r13 

r23 

 x3i yi x

2 3i

y

2 i

 x2i x3i x

2 2i

x

2 3i


9

Partial Correlation Coefficient r12.3 = koefisien korelasi antara Y dan X2, kalau X3 konstan  r13.3 = koefisien korelasi antara Y dan X3, kalau X2 konstan  r23.1 = koefisien korelasi antara X2 dan X3, kalau Y konstan 

r12.3 

r13.2 

r23.1 

Chapter

r12  r13r23 (1  r132 ) (1  r232 ) r13  r12 r23 (1  r122 ) (1  r232 )

r23  r12 r13 (1  r122 ) (1  r132 )


10

1. Individual partial coefficient test 1

holding X3 constant: Whether X2 has the effect on Y ? Y

H0 : 2 = 0

X2

H1 : 2  0

t=

^ 2 - 0

= 2 = 0?

0.9277

=

Se (^2)

= 9.544 0.0972

Compare with the critical value tc0.025, 6 = 2.447 Since

t > tc

==> reject Ho

^ Answer : Yes, 2 is statistically significant and is significantly different from zero.

11

1. Individual partial coefficient test (cont.) 2

holding X2 constant: Whether X3 has the effect on Y? Y

H0 : 3 = 0

X3

H1 : 3  0

t=

^ 3 - 0

= 3 = 0?

0.2532 - 0 =

Se (^3) Critical value:

= 1.730 0.1464

tc0.025, 6 = 2.447

Since | t | < | tc |

==> not reject Ho

^ Answer: Yes, 3 is statistically not significant and is not significantly different from zero. 12

2. Testing overall significance of the multiple regression

3-variable case:

Y = 1 + 2X2 + 3X3 + u H0 : 2 = 0, 3 = 0, (all variable are zero effect) H1 : 2  0 or 3  0 (At least one variable is not zero)

1. Compute and obtain F-statistics 2. Check for the critical Fc value (F c , k-1, n-k)

3. Compare F and Fc , and if F > Fc ==> reject H0

13

Analysis of Variance:

y=^ y+u

Since

^2 ^2 + u ==> y2 = y TSS = ESS + RSS

ANOVA TABLE (SS) Sum of Square

df

Due to regression(ESS)

^  y2

k-1

Due to residuals(RSS)

^  u2

Source of variation

k-1 ^2 u n-k

n-k

 y2

Total variation(TSS)

(MSS) Mean sum of Sq. ^ y2

= ^u2

n-1

Note: k is the total number of parameters including the intercept term. MSS of ESS F=

= MSS of RSS

H0 : 2 = … = k = 0 H1 : 2  …  k  0

^ y2/(k-1)

ESS / k-1 = RSS / n-k if F > Fck-1,n-k

^ u 2 /(n-k)

==> reject Ho 14

Tabel Anavar, untuk Regresi Tiga Variabel

Sumber Variasi

Dari regresi

Jumlah Kuadrat (SS)

b12.3 Σ x2iyi + b13.2 Σ x3iyi

(ESS)

Derajat

Rata-Rata

Kebebasan

Jumlah Kuadrat

(df)

(MSS)*

2

b12.3 Σ x2iyi + b13.2 Σ x3iyi 2

(k-1)

Kesalahan

Σ ei2

pengganggu

n-3

Σ ei2 / n - 3 = Se2

(n-k)

(RSS) TSS *Mean

Σ yi2

n-1

Sum of Squares.

Chapter


15

^ ^ y = 2x2 + 3x3 + u^

Threevariable case

^ ^ ^  y2 = 2  x2 y + 3  x3 y +  u2 TSS =

ESS

+ RSS

ANOVA TABLE Source of variation ESS

SS ^ ^ x y 2 x2 y +  3 3

df(k=3)

MSS

3-1

ESS/3-1 RSS/n-3

RSS

^2 u

n-3 (n-k)

TSS

y2

n-1

ESS / k-1

F-Statistic = RSS / n-k

=

(^2 x2y + ^ 3 x3y) / 3-1 ^2 / n-3 u 16

An important relationship between R2 and F ESS / k-1

ESS (n-k)

F=

= RSS / n-k

RSS (k-1)

ESS

n-k

TSS-ESS

k-1

ESS/TSS

n-k

=

=

ESS TSS

1-

For the three-variables case : R2 / 2 F=

k-1

R2

n-k

1 - R2

k-1

(1-R2) / n-3

=

F

(k-1) F

R2 / (k-1)

=

Reverse : (1-R2) / n-k

R2 = (k-1)F + (n-k) 17

Overall significance test: H0 : 2 = 3 = 4 = 0 H1 : at least one coefficient is not zero. 2  0 , or 3  0 , or 4  0 R2 / k-1

F* =

(1-R2)

=

/ n- k

0.9710 / 3

=

(1-0.9710) /16

= 179.13 Fc(0.05, 4-1, 20-4) = 3.24 

k-1

n-k Since F* > Fc ==> reject H0. 5.18

18

Construct the ANOVA Table (8.4) .(Information from EViews)

Source of variation Due to regression (SSE)

SS

Df

MSS

2

k-1

R (y )/(k-1)

=3

=5164.3903

2

R (y ) =(0.971088)(28.97771)2x1 =15493.171 9

2 2 2 Due to n-k (1- R )(y ) or (   ) Residuals =(0.0289112)(28.97771) )2x19 (RSS) =16 =461.2621

Total (TSS)

2

( y ) =(28.97771) 2x19 =15954.446

2

2

2

2

(1- R )(y )/(n-k) =28.8288

n-1 =19

Since (y)2 = Var(Y) = y2/(n-1) => (n-1)(y)2 = y2 MSS of regression F* =

5164.3903 =

MSS of residual

= 179.1339 28.8288 19

Example:Gujarati(2003)-Table6.4, pp.185)

H0 : 1 = 2 = 3 = 0 2

F* =

R / k-1 ESS / k= 1 (1-R2) / n- k RSS/(nk)

0.707665 / 2

=

(1-0.707665)/ 61 F* = 73.832

Fc(0.05, 3-1, 64-3) = 3.15 

k-1

n-k

Since F* > Fc ==> reject H0.

20

Construct the ANOVA Table (8.4) .(Information from EVIEWS)

Source of variation Due to regression (SSE)

SS

Df

MSS

2

k-1

R (y )/(k-1)

=2

=130723.67

n-k

(1- R )(y )/(n-k)

2

R (y ) =(0.707665)(75.97807)2x6 =261447.33 4

2 2 2 Due to (1- R )(y ) or (   ) Residuals =(0.292335)(75397807)2x64 (RSS) =108003.37

Total (TSS)

2

( y ) =(75.97807)2x64 =369450.7

=61

2

2

2

2

=1770.547

n-1 =63

Since (y)2 = Var(Y) = y2/(n-1) => (n-1)(y)2 = y2 MSS of regression F* =

130723.67 =

MSS of residual

= 73.832 1770.547 21

Y = 1 + 2 X2 + 3 X3 + u H0 : 2 = 0, 3= 0, H1 : 2  0 ; 3  0 Compare F* and Fc, checks the F-table:

Decision Rule: Since F*= .73.832 > Fc = 4.98 (3.15)

Fc0.01, 2, 61 = 4.98 Fc0.05, 2, 61 = 3.15

==> reject Ho

Answer : The overall estimators are statistically significant different from zero.

22

QUIZ    

1 2 3 4

Chapter


23

TERIMA KASIH

Chapter


24

Koefisien Determinasi dan Korelasi Berganda

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