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