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Marchés boursiers et croissance économique. Une analyse comparative entre l'Afrique subsaharienne et l'Asie du sud-est.

( Télécharger le fichier original )
par Larissa Nawo
Université de Dschang - Master of Sciences en Analyse et Politiques Economiques 0000
  

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Annexe 7 : Test d'hétéroscédascité

En ASS. Breusch-Pagan / Cook-Weisberg test for heteroskedasticity

Ho: Constant variance

Variables: fitted values of g

chi2(1) = 0.49

Prob > chi2 = 0.4831

Conclusion absence d'heterocedsticite

En ASE. Breusch-Pagan / Cook-Weisberg test for heteroskedasticity

Ho: Constant variance

Variables: fitted values of g

chi2(1) = 24.85

Prob > chi2 = 0.0000

Conclusion présence d'heterocedsticite

Annexe 8 : Correction de l'hétéroscédasticité de White

En ASE. Linear regression Number of obs = 92

F( 9, 82) = 10.67

Prob > F = 0.0000

R-squared = 0.4445

Root MSE = 3.209

------------------------------------------------------------------------------

| Robust

g | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+----------------------------------------------------------------

rto | .0249068 .0296569 0.84 0.403 -.0340903 .0839039

traval | .0521768 .0302668 1.72 0.088 -.0080336 .1123872

capbours | -11.24397 5.424776 -2.07 0.041 -22.03558 -.4523637

inv | .1322184 .0469673 2.82 0.006 .0387856 .2256513

open | -6.5728 1.53153 -4.29 0.000 -9.619502 -3.526099

dext | -4.907572 1.647457 -2.98 0.004 -8.184888 -1.630256

rto² | -.0000925 .000132 -0.70 0.485 -.0003551 .00017

traval² | -.0000872 .0000789 -1.11 0.272 -.0002443 .0000698

capbours² | 6.801071 4.602571 1.48 0.143 -2.35491 15.95705

_cons | 49.44272 9.902838 4.99 0.000 29.74282 69.14262

------------------------------------------------------------------------------

Annexes 9 : Résultat du modèle à effets fixe

En ASS. voir tableau 8.Chapitre IV

En ASE. FE (within) regression with AR(1) disturbances Number of obs = 88

Group variable: ident Number of groups = 4

R-sq: within = 0.4543 Obs per group: min = 22

between = 0.0769 avg = 22.0

overall = 0.2044 max = 22

F(9,75) = 6.94

corr(u_i, Xb) = -0.7491 Prob > F =0.0000

-------------------------------------------------------------------------------------

g | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+-----------------------------------------------------------------------

rto | .0809839 .0502882 1.61 0.112 -.0191952 .1811631

traval | -.0493666 .0942349 -0.52 0.602 -.237092 .1383589

capbours | 1.188945 12.23316 0.10 0.923 -23.18077 25.55866

inv | .2729787 .0650098 4.20 0.000 .1434724 .4024849

open | -.5247324 2.915851 -0.18 0.858 -6.333405 5.28394

dext | -4.746648 1.757022 -2.70 0.009 -8.246815 -1.24648

rto² | -.0003375 .0002534 -1.33 0.187 -.0008423 .0001673

traval²| .000094 .0002062 0.46 0.650 -.0003167 .0005048

capbours²| 4.005473 6.677849 0.60 0.550 -9.297485 17.30843

_cons | 17.49944 9.941724 1.76 0.082 -2.305494 37.30436

-------------+------------------------------------------------------------------------

rho_ar | .37107282

sigma_u | 4.1113565

sigma_e | 2.9639394

rho_fov | .6580165 (fraction of variance because of u_i)

--------------------------------------------------------------------------------------

F test that all u_i=0: F(3,75) = 3.21 Prob > F = 0.0276

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