Date post: | 18-Jul-2015 |
Category: |
Health & Medicine |
Upload: | instituto-tecnologico-de-pachuca |
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EstadisticaII DiseñoDeBloquesCompletosAlAzar
InstitutoTecnologicoDePachuca
1 2 31 14.823 25.151 32.605
2 14.676 25.401 32.46 b ó j= 53 14.72 25.131 32.256 t ó i= 34 14.514 25.031 32.6695 15.065 25.277 32.111
Sumatorias:
73.798 125.991 162.101 361.8905,446.145 15,873.732 26,276.734 47,596.61114.760 25.198 32.420
219.721 632.573 1,063.086215.385 645.211 1,053.652216.678 631.567 1,040.450210.656 626.551 1,067.264226.954 638.927 1,031.1161,089.395 3,174.828 5,255.567 9,519.791
72.579 5,267.711 24.193 1,915.38072.537 5,261.616 24.179 1,914.24772.107 5,199.419 24.036 1,888.69572.214 5,214.862 24.071 1,904.47172.453 5,249.437 24.151 1,896.997
361.890 26,193.046 9,519.791
NivelesDeNitrogeno
Ejercicio2
Losdatosquesepresentanacontinuacionsonrendimientos(entoneladaspor hectarea) de un pasto con 3 niveles de fertilizacion nitrogenado. eldiseñofuealeatorizado,con5repeticionesportratamiento.Contrastarlahipotesisiconα=0.01ytomardecicion.
Yi =Yi2 =yi =
∑ =
yj yj2 yj
(yij )2∑
(yij )2∑
∑ =
EstadisticaII DiseñoDeBloquesCompletosAlAzar
InstitutoTecnologicoDePachuca
8,730.958
788.364
0.057
788.832
0.411
394.182 0.0514
7,669.095
S.C.MedioSumaDeCuadrados GradosDeLibertadϒ
SCTα=788.364 t‐13‐1=2
SCBβ=0.057 b‐15‐1=4
SCT=788.832 (b)(t)‐1(5)(3)‐115‐1=14
"LosRendimientosFueronDiferentesEnLosTratamientos"
SCR=0.411 (b‐1)(t‐1)(5‐1)(3‐1)=(4)(2)=8
FactorTratamiento
y2 ⇒ de ∑ yi
2 ó yj
2
(i)( j)=
(361.89)2
(5)(3)=
130,964.37215
=
SCTα =yi2
bi=1
t
∑ −y2
bt=47,596.611
5−(361.89)2
(5)(3)=
SCBβ =yj2
tj=1
b
∑ −y2
bt=26,193.046
3−(361.89)2
(5)(3)=
SCT =i=1
t
∑j=1
b
∑ (yij )2 −
yi2
bt= (9,519.791) − (361.89)
2
(5)(3)=
SCR = SCT − SCTα − SCBβ = 788.832 − 788.364 − 0.057 =
SCMTα =SCTαt −1
=788.3643−1
= 394.182
SCMBβ =SCBβ
b −1=0.0575 −1
= 0.01425
SCT(bt) −1
=788.832(5 * 3) −1
= 56.345
SCMR =SCR
(b −1)(t −1)=
0.411(5 −1)(3−1)
= 0.051375
SCMTα = SCMR =
FCal =SCMTαSCMR
=394.1820.0514
=
H0;α1 = α2 = α 3
H1; Al Menos Un Tratamiento Diferente
FTab = (t −1);(b −1)(t −1)FTab = (3−1);(5 −1)(3−1)FTab = (2);(4)(2)FTab = (2);(8) = 8.65 ν1 ; ν2 con α=0.01
7,669.095 > 8.65FCal > FTabl ∴ Rechazo H0