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# Ejercicios diseño de bloques completos al azar ejercicio 2

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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