CASO 10 ECONOMETRIA
Enviado por Adria Calonge • 24 de Febrero de 2020 • Prácticas o problemas • 1.269 Palabras (6 Páginas) • 128 Visitas
Modelo 1: MCO, usando las observaciones 1960-1999 (T = 40)
Variable dependiente: C
| Coeficiente | Desv. Típica | Estadístico t | valor p | |
const | 27.5939 | 1.58446 | 17.42 | <0.0001 | *** |
PB | 0.0921878 | 0.0398831 | 2.311 | 0.0266 | ** |
PP | −0.607160 | 0.157120 | −3.864 | 0.0004 | *** |
R | 0.244860 | 0.0110954 | 22.07 | <0.0001 | *** |
Media de la vble. dep. | 50.56725 | D.T. de la vble. dep. | 19.53879 | |
Suma de cuad. residuos | 143.0726 | D.T. de la regresión | 1.993549 | |
R-cuadrado | 0.990391 | R-cuadrado corregido | 0.989590 | |
F(3, 36) | 1236.776 | Valor p (de F) | 2.37e-36 | |
Log-verosimilitud | −82.24700 | Criterio de Akaike | 172.4940 | |
Criterio de Schwarz | 179.2495 | Crit. de Hannan-Quinn | 174.9366 | |
rho | 0.508409 | Durbin-Watson | 0.897776 |
H1 ; Linearity ; ei = 0
It is always positive as we are doing lineal least squares models.
H2 ; Homoskedasticity 🡪 White test 🡪 H0 = Homo , H1 = Hetero
P-Value = 0.085570 > 0.05
As the p-value is bigger than alpha we don’t reject the null hypothesis so the model is homoscedastic.
H3; Normality 🡪 H0; Ei = Normal , H1; =/ Normal
P-Value = 0,3840
As the p-value is bigger than alpha we don’t reject the null hypothesis, so the model is normal.
H4; Independence
Autocorrelation indexes:
[pic 2][pic 1]
AR(1) +
[pic 3]
AR(1) +
[pic 4]
3d)
Modelo 2: MCO, usando las observaciones 1961-1999 (T = 39)
Variable dependiente: uhat1
| Coeficiente | Desv. Típica | Estadístico t | valor p | |
uhat1_1 | 0.508409 | 0.132103 | 3.849 | 0.0004 | *** |
Media de la vble. dep. | 0.089700 | D.T. de la vble. dep. | 1.853310 | |
Suma de cuad. residuos | 94.14052 | D.T. de la regresión | 1.573970 | |
R-cuadrado no centrado | 0.280461 | R-cuadrado centrado | 0.278732 | |
F(1, 38) | 14.81163 | Valor p (de F) | 0.000441 | |
Log-verosimilitud | −72.52253 | Criterio de Akaike | 147.0451 | |
Criterio de Schwarz | 148.7086 | Crit. de Hannan-Quinn | 147.6419 | |
rho | −0.086183 | h de Durbin | −0.952323 |
As the coefficient value is 0.508409 > 0 and the model is significative (***) there is also AR(1)+
3.
Test de Durbin – Watson
et = p * et – 1 + i[pic 5]
H0; No autocorrelación : p = 0
H1; Autorcorrelación : p =/ 0
Estadístico de contraste: k = 3 m = 40
AR(1)+ NRHO AR(1)
0 dl du 2 4-du 4-d2 4[pic 6]
1,3384 1,6589 2,3425 2,6717
As the Durbin Watson coefficient is 0,89776 we can graphically see there is AR(1)+.
HOW TO AVOID AUTOCORRELATION
- Cambio de variable:
Variables Retardos New variables
y = Ct Ct – 1 CNt = - 0,508409 * Ct - 1
x1 = PBt PBt – 1 PBNt = - 0,508409* PBt - 1
x2 = PPt PPt – 1 PPNt = - 0,508409* PPt - 1
x3 = Rt Rt – 1 RNt = - 0,508409* Rt - 1
rho = 0,508409
k = 3 m = 39
dw = 1,490401
AR(1)+ NRHO AR(1)
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