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Estadistica II DiseñoDeCuadradosLatinos
Instituto Tecnologico De Pachuca
t= 4
1 A 1.4 B 1.38 C 1.4 D 1.6 5.78 33.4084
2 B 1.35 A 1.28 D 1.45 C 1.62 5.7 32.49
3 C 1.38 D 1.4 B 1.42 A 1.63 5.83 33.9889
4 D 1.39 C 1.39 A 1.4 B 1.6 5.78 33.4084
5.52 5.45 5.67 6.45
30.4704 29.7025 32.1489 41.6025
AlCuadrado
A= 1.4 + 1.28 + 1.4 + 1.63 = 5.71 32.6041
B= 1.35 + 1.38 + 1.42 + 1.6 = 5.75 33.0625
C= 1.38 + 1.39 + 1.4 + 1.62 = 5.79 33.5241
D= 1.39 + 1.4 + 1.45 + 1.6 = 5.84 34.1056Sumas 23.09 133.2963
Pisos(Filas)Casilleros(Columnas)
1 2 3 4
23.09133.2957
133.9243
33.32176
Ejercicio3:
Seprobaron4racionesalimenticiasparapollos,criadosenjaula(TipoBateria)de4pisos(Filas)y4casilleros(Columnas).Lavariableanalizadafue:pesodelpolloenKgalas8semanasdeedad.TablaRealizelapruebadehipotesiscorrespondiente.Useα=0.05.
yi yi2
yjy
j
2
yi2∑yij∑
yj2∑
TC =yi2
t 2=(23.09)2
42=
Estadistica II DiseñoDeCuadradosLatinos
Instituto Tecnologico De Pachuca
D 1.96 A 1.9044 C 1.96 B 2.56A 1.8225 B 1.6384 D 2.1025 C 2.6244C 1.9044 D 1.96 B 2.0164 A 2.6569B 1.9321 C 1.9321 A 1.96 D 2.56
Sumas 7.619 7.4349 8.0389 10.4013
0.17234
0.0023188
0.00216875
0.15931875
0.0085375
8.38448.18788.53778.3842
33.4941
1 2 3 4DatosAlCuadrado
Tot = yijk2∑∑ − TC Tot = 33.4941− 33.3217
SCT =yi2∑
t− TC SCT =
133.29634
− 33.3217
SCF =yj∑t
− TC SCF =133.2957
4− 33.3217
SCC =yK∑t
− TC SCC =133.9243
4− 33.3217
SCE = Tot − SCT − SCF − SCCSCE = 0.17234 − 0.0023188 − 0.00216875 −15931875
Estadistica II DiseñoDeCuadradosLatinos
Instituto Tecnologico De Pachuca
GradosDeLibertadϒ
t‐14‐1=3
t‐14‐1=3
t‐14‐1=3
(t‐1)(t‐2)(4‐1)(4‐2)(3)(2)=6
15
0.5432032
Pisos(Filas) SCF=0.00216875
FuenteDeVariacion
SumaDeCuadrados CuadradosMedios
"SeConcluyeQueNoHayDiferenciasEntreLosTratamientosDeLasRacionesAlimenticias"
Totales 0.17234
FCalculada FDeTablas
HipotesisiAProbar
Casilleros(Columnas)
SCC=0.15931875
Error SCE=0.0085375
Tratamientos SCT=0.0023188
0.543203 < 4.76FCal < FTabl ∴ Acepto H0
CMT =0.0023188
3= 0.000772933
CMF =0.00216875
3= 0.000722917
CMC =0.15931875
3= 0.05310625
CME =0.0085375
6= 0.001422917
FTab = (t −1);(t −1)(t − 2)FTab = (4 −1);(4 −1)(4 − 2)FTab = (3);(3)(2)FTab = (3);(6) = 4.76 ν1 ; ν2 con α=0.05
FCal =CMTCME
=0.0007729330.001422917
=
H0 ;µA = µB = µC = µD
H1; µi ≠ µ j