Estadistica II DiseñoDeBloquesCompletosAlAzar

Instituto Tecnologico De Pachuca

T1 T2 T3 T4Bloque1 4,268 4,482 4,699 4,549

Bloque2 4,500 4,500 4,880 4,750 b ó j= 5Bloque3 4,300 4,250 4,500 4,480 t ó i= 4Bloque4 4,250 4,290 4,470 4,420Bloque5 4,220 4,200 4,450 4,300

Sumatorias:

21,538 21,722 22,999 22,499 88,758

463,885,444 471,845,284 528,954,001 506,205,001 1,970,889,730

4,308 4,344 4,600 4,500

18,215,824 20,088,324 22,080,601 20,693,401

20,250,000 20,250,000 23,814,400 22,562,500

18,490,000 18,062,500 20,250,000 20,070,400

18,062,500 18,404,100 19,980,900 19,536,400

17,808,400 17,640,000 19,802,500 18,490,000

92,826,724 94,444,924 105,928,401 101,352,701 394,552,750

17,998 323,928,004 4,499.5 81,078,15018,630 347,076,900 4,657.5 86,876,90017,530 307,300,900 4,382.5 76,872,90017,430 303,804,900 4,357.5 75,983,90017,170 294,808,900 56.0 73,740,900

88,758 1,576,919,604 394,552,750

Tratamientos

Ejercicio1

Se Realizo un experimento para probar el efecto de suministrar diferentes niveles deenergeticossuministradosadiferentesdietasabasedesorgo,enelconsumodealimentoparapollosabarcadesdelos5diasdeedadhastalas8semanas.Paraelloseutilizaron160pollosde5lineascomercialesmismasqueconstituyeronlosbloques;dentrodeloscuales,fuerondistribuidoslostratamientos;entoncescadagrupode8pollosconstituyounaunidadexperimental,obteniendoselasiguienteinformacion.

Yi =

Yi2 =yi =

(yij )2∑

yj yj2 yj

∑ =

∑ =

(yij )2∑

Estadistica II DiseñoDeBloquesCompletosAlAzar

Instituto Tecnologico De Pachuca

393,899,128.2

278,817.8

330,772.8

653,621.8

44,031.2

92,939.27 3,669.27

25.329

SumaDeCuadrados GradosDeLibertadϒ S.C.Medio

SCTα=278,817.8 t‐14‐1=3

FactorTratamiento

"Ladietadesorgoesadecuadaparapollosdeunaedadde5diashasta8semanas"

SCBβ=330,772.8

SCT=653,621.8

SCR=44,031.2

b‐15‐1=4

(b)(t)‐1(5)(4)‐120‐1=19

(b‐1)(t‐1)(5‐1)(4‐1)=(4)(3)=12

y2 ⇒ de ∑ yi

2 ó yj

2

(i)( j)=

(88,758)2

(5)(4)=

7,877,982,56420

=

SCTα =yi2

bi=1

t

∑ −y2

bt=1,970,889,730

5−(88,758)2

(5)(4)=

SCBβ =yj2

tj=1

b

∑ −y2

bt=1,576,919,604

4−(88,758)2

(5)(4)=

SCT =i=1

t

∑j=1

b

∑ (yij )2 −

yi2

bt= (394,552,750) − (88,758)

2

(5)(4)=

SCR = SCT − SCTα − SCBβ = 653,621.8 − 278,817.8 − 330,772.8

SCMTα =SCTαt −1

=278,817.84 −1

= 92,939.26

SCMBβ =SCBβ

b −1=330,772.85 −1

= 82,693.2

SCT(bt) −1

=653,621.8(5 * 4) −1

= 34,401.14

SCMR =SCR

(b −1)(t −1)=

44,031(5 −1)(4 −1)

= 3,669.25

SCMTα = SCMR =

FCal =SCMTαSCMR

=92,939.273,669.27

=FTab = (t −1);(b −1)(t −1)FTab = (4 −1);(5 −1)(4 −1)FTab = (3);(4)(3)FTab = (3);(12) = 3.49 ν1 ; ν2 con α=0.05

H0;α1 = α2 = α 3 = α 4

H1; Al Menos Un Tratamiento Diferente

25.329 > 3.49FCal > FTabl ∴ Rechazo H0