y<-c(1.5,0.8,1.2,1.4,0.2,0.8,0.6,1.3,0.4,0.6)
x1<-c(8.4,3.3,5.8,10,4.7,7.7,4.5,8.6,5.9,6.3)
x2<-c(7.7,4.5,8.4,7.8,2.4,4.8,2.5,3.4,2,4.1)
uno<-seq(1,1,length.out =10)
uno
## [1] 1 1 1 1 1 1 1 1 1 1
X<-matrix(c(uno,x1,x2),nrow=10)
X
## [,1] [,2] [,3]
## [1,] 1 8.4 7.7
## [2,] 1 3.3 4.5
## [3,] 1 5.8 8.4
## [4,] 1 10.0 7.8
## [5,] 1 4.7 2.4
## [6,] 1 7.7 4.8
## [7,] 1 4.5 2.5
## [8,] 1 8.6 3.4
## [9,] 1 5.9 2.0
## [10,] 1 6.3 4.1
Y<-matrix(c(y),nrow=10)
Y
## [,1]
## [1,] 1.5
## [2,] 0.8
## [3,] 1.2
## [4,] 1.4
## [5,] 0.2
## [6,] 0.8
## [7,] 0.6
## [8,] 1.3
## [9,] 0.4
## [10,] 0.6
La ecuacion se obtiene mediante \(\hat \beta=(X'X)^{-1}X'y\)
\[ \hat \beta=(X'X)^{-1}X'y \]
\[ (X'X) \]
XtX<-t(X)%*%X
XtX
## [,1] [,2] [,3]
## [1,] 10.0 65.20 47.60
## [2,] 65.2 465.18 332.61
## [3,] 47.6 332.61 278.36
\[ (X'X)^{-1} \]
XtX.inv<-solve(XtX)
XtX.inv
## [,1] [,2] [,3]
## [1,] 1.19364308 -0.14664660 -0.02888807
## [2,] -0.14664660 0.03277723 -0.01408844
## [3,] -0.02888807 -0.01408844 0.02536653
\[ X'y \]
XY<-t(X)%*%Y
XY
## [,1]
## [1,] 8.80
## [2,] 63.32
## [3,] 49.65
\[ \beta=(X'X)^{-1}X'y \]
B<-solve(t(X)%*%X)%*%(t(X)%*%Y)
B
## [,1]
## [1,] -0.21589662
## [2,] 0.08547343
## [3,] 0.11315334
\[ Y_{i} = -0.2159 + 0.0855X_{1}+0.1132X_{2} \]
\[ Syy = y'y - n \bar y^2 \]
Syy<-t(Y)%*%Y-nrow(Y)*mean(Y)^2
Syy
## [,1]
## [1,] 1.796
\[ SSE=y'y - \hat \beta'X'y \]
SSE=t(Y)%*%Y - t(B)%*%XY
SSE
## [,1]
## [1,] 0.4096497
\[ SSR= \hat \beta'X'y \]
SSR<-t(B)%*%XY-nrow(Y)*mean(Y)^2
SSR
## [,1]
## [1,] 1.38635
gl1=(ncol(X)-1)
gl2=(nrow(Y)-ncol(X))
MSE<-SSE/gl2
MSE
## [,1]
## [1,] 0.05852139
MSR<-SSR/gl1
MSR
## [,1]
## [1,] 0.6931751
\[ H_o: \beta_1=0 \\ \] \[ H_1: \beta_1 \neq 0 \]
Función de Prueba
\[ t=\frac{\hat \beta_1}{D.S(\hat \beta_1)} \]
V.B1<-MSE*XtX.inv[2,2]
t<-B[2]/sqrt(V.B1)
t
## [,1]
## [1,] 1.951586
v.p.B1<-2*pt(t,gl2,lower.tail=FALSE)
v.p.B1
## [,1]
## [1,] 0.09195252
El valor p es mayor que el 5%. Entonces la hipótesis nula no se rechaza. La variable \(X_1\) no está asociada con la variable \(Y\).
\[ H_o: \beta_2=0 \] \[ H_1: \beta_2 \neq 0 \]
Función de Prueba
\[ t=\frac{\hat \beta_2}{D.S(\hat \beta_2)} \]
V.B2<-MSE*XtX.inv[3,3]
t<-B[3]/sqrt(V.B2)
t
## [,1]
## [1,] 2.936835
v.p.B2<-2*pt(t,gl2,lower.tail=FALSE)
v.p.B2
## [,1]
## [1,] 0.0218112
El valor p es mayor que el 5%. Entonces la hipótesis nula no se rechaza. La variable \(X_2\) no está asociada con la variable \(Y\).