Prueba De Normalidad Multivariante
PRUEBA DE ASIMETRIA MULTIVARIADA
1. Hipótesis nula de Asimetría
Ho: β1.p <=0
H1: β2.p > 0
b1.P= 1n2 injn Xi-XS-1Xi-X'3
FUNCION PIVOTAL:X02= n.b1.p6 ≈ X20.05, p(p+1)(p+2)6 gl
PARTIENDO DE LOS DATOS YA OBTENIDOS, HAREMOS LOS CALCULOS ANTERIORES PARA VERIFICAR SI NUESTROS DATOS PRESENTAN UNA DISTRIBUCION SIMETRICA PARA LUEGO CON LAPRUEBA DE KURTOSIS COMPROBAR LA HIPOTESIS DE NORMALIDAD CONJUNTA.
| 115.3 | 2.165 | 4.819 |
| 114.5 | 1.865 | 4.107 |
| 112.2 | 1.841 | 3.834 |
| 118.3 | 2.087 | 3.692 |
| 113.7 | 2.177 |3.593 |
| 115.7 | 2.091 | 4.767 |
| 116.5 | 2.127 | 3.944 |
| 116.1 | 1.806 | 3.922 |
| 119.1 | 2.189 | 4.086 |
X= | 117.4 | 1.656 | 4.596 |
| 115.9 | 1.581 | 4.331 |
| 114.3 |2.158 | 4.698 |
| 114.9 | 1.779 | 4.083 |
| 114.6 | 1.849 | 3.89 |
| 116.9 | 1.942 | 3.333 |
| 114.2 | 2.054 | 2.585 |
| 117.4 | 1.965 | 4.01 |
| 114.9 | 1.76 | 3.376 |
| 115.3 |1.494 | 3.526 |
| 113.5 | 2.027 | 3.284 |
TENEMOS 3 VARIABLES (RESISTENCIA, DIAMETRO, PESO), LOS CUALES OBTUVIMOS SU VECTOR DE MEDIAS Y MATRIZ DE VARIANZAS Y COVARIANZAS:
| 115.53 |
X= | 1.93|
| 3.92 |
| 9.60 | 0.11 | 0.13 |
S= | 0.11 | 0.33 | -0.08 |
| 0.13 | -0.08 | 1.43 |
| -0.2 | 0.2 | 0.9 |
| -1.1 | -0.1 | 0.2 |
| -3.3 | -0.1 | -0.1 |
| 2.7 | 0.2 | -0.2 || -1.8 | 0.2 | -0.3 |
| 0.1 | 0.2 | 0.8 |
| 0.9 | 0.2 | 0 |
| 0.6 | -0.1 | -0 |
X-X= | 3.6 | 0.3 | 0.2 |
| 1.9 | -0.3 | 0.7 |
| 0.4 | -0.3 | 0.4 |
| -1.2 | 0.2 | 0.8 |
| -0.7| -0.2 | 0.2 |
| -0.9 | -0.1 | -0 |
| 1.3 | 0 | -0.6 |
| -1.3 | 0.1 | -1.3 |
| 1.8 | 0 | 0.1 |
| -0.6 | -0.2 | -0.5 |
| -0.3 | -0.4 | -0.4 |
| -2 | 0.1 | -0.6 |
POR LO TANTO:b1.3= 5.459202=0.014
ENTONCES:
X02= n.b1.p6 = 20*0.0146=0.045
ADEMAS:
X20.05, p(p+1)(p+2)6 gl , gl=3*4*56=10→ X20.05,10 gl=18.31
| | | | |
| X2O <X2α
X2O <X2α
| |...
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