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  • 1. Estadistica
II Diseño
De
Bloques
Completos
Al
Azar Ejercicio
2
 Los
datos
que
se
presentan
a
continuacion
son
rendimientos
(en
toneladas
 por 
 hectarea) 
 de 
 un 
 pasto 
 con 
 3 
 niveles 
 de 
 fertilizacion 
 nitrogenado. 
 el
 diseño
fue
aleatorizado,
con
5
repeticiones
por
tratamiento.
 Contrastar
la
hipotesisi
con
α=0.01
y
tomar
decicion.
 Niveles
De
Nitrogeno 1 2 3 1 14.823 25.151 32.605 2 14.676 25.401 32.46 b ó j= 5 3 14.72 25.131 32.256 t ó i= 3 4 14.514 25.031 32.669 5 15.065 25.277 32.111 Sumatorias
: Yi = 73.798 125.991 162.101 361.890 Yi 2 = 5,446.145 15,873.732 26,276.734 47,596.611 yi = 14.760 25.198 32.420 219.721 632.573 1,063.086 215.385 645.211 1,053.652 216.678 631.567 1,040.450 210.656 626.551 1,067.264 226.954 638.927 1,031.116 ∑ (y ) ij 2 ∑ = 1,089.395 3,174.828 5,255.567 9,519.791 yj y2 j yj ∑ (y )ij 2 72.579 5,267.711 24.193 1,915.380 72.537 5,261.616 24.179 1,914.247 72.107 5,199.419 24.036 1,888.695 72.214 5,214.862 24.071 1,904.471 72.453 5,249.437 24.151 1,896.997 ∑ = 361.890 26,193.046 9,519.791 Instituto
Tecnologico
De
Pachuca
  • 2. Estadistica
II Diseño
De
Bloques
Completos
Al
Azar y 2 ⇒ ∑ de y 2i ó y 2j (361.89)2 130,964.372 = = = 8,730.958 (i)( j) (5)(3) 15 yi2 y 2 47, 596.611 (361.89)2 t SCTα = ∑ − = − = 788.364 i =1 b bt 5 (5)(3) b y2 y 2 26,193.046 (361.89)2 SCBβ = ∑ − = − = 0.057 j j =1 t bt 3 (5)(3) t b yi2 (361.89)2 SCT = ∑ ∑ (yij ) − = (9,519.791) − 2 = 788.832 i=1 j =1 bt (5)(3) SCR = SCT − SCTα − SCBβ = 788.832 − 788.364 − 0.057 = 0.411 Suma
De
Cuadrados Grados
De
Libertad
ϒ S.C.
Medio SCTα 788.364 SCTα=788.364 t‐1





3‐1=2 SCMTα = = = 394.182 t −1 3−1 SCBβ 0.057 SCBβ=0.057 b‐1





5‐1=4 SCMBβ = = = 0.01425 b −1 5 −1 SCT=788.832 (b)(t)‐1


(5)(3)‐1

15‐1=14 SCT 788.832 = = 56.345 (bt ) − 1 (5 * 3) − 1 SCR 0.411 SCR=0.411 (b‐1)(t‐1)


(5‐1)(3‐1)=(4)(2)=8 SCMR = = = 0.051375 (b − 1)(t − 1) (5 − 1)(3 − 1) Factor
Tratamiento SCMTα = 394.182 SCMR = 0.0514 SCMTα 394.182 FTab = (t − 1);(b − 1)(t − 1) FCal = = = SCMR 0.0514 7,669.095 FTab = (3 − 1);(5 − 1)(3 − 1) FTab = (2);(4)(2) H 0 ;α1 = α 2 = α 3 FTab = (2);(8) = 8.65 H 1; Al Menos Un Tratamiento Diferente ν1 ; ν 2 con α =0.01 7, 669.095 > 8.65 FCal > FTabl ∴ Rechazo H 0 "Los
Rendimientos
Fueron
Diferentes
En
Los
Tratamientos" Instituto
Tecnologico
De
Pachuca