problema de diseño de experimentos
DESCRIPTION
Problema resuelto 7.17 del Cap 7 del libro de Douglas Montgomery de Diseño de ExperimentosTRANSCRIPT
Problema 7.17
Se cree que la adhesividad de un pegamento depende de la presión y de la temperatura, al ser aplicado se realiza un experimento con ambos factores fijos.Realice una prueba de NO aditividad.
PRESIÓN TEMPERATURA250 260 270 yi
120 9.60 11.28 9.00 27.88130 9.69 10.10 9.57 29.36140 8.43 11.01 9.03 28.47150 9.98 10.44 9.80 30.22
37.7 42.03 37.4 117.43
SS AT=∑
i=1
aYi2
b− y ..2
ab ¿
(29.88 )2+(29.36 )2+(28.47 )2+(30.22 )2
3=
(117.43)2
(4 ) (3 )
37786133
−13907.4812
=1159.53−1158.95=0.58
SSBT=∑
i=1
aYi2
b− y ..2
ab =
(37.7 )2+ (42.83 )2 +(37.4 )2
4−
(117.93)2
(4 ) (3 )
465.454
−13907.4812
=1163.61−1158.95=4.66
SST=∑i=1
a
∑j=1
b
y2ij−y ..
2
ab=1166.3493−1158.95=7.3493
(9.60 )2+(11.28)2+(9.00 )2+(9.69 )2…−(117.93 )2
12
SSResidual=S ST−SS A−SSB
= 7.394 – 0.58 – 4.66 = 2.159
Suma de cuadrados de la No aditividad
∑i=1
a
∑j=1
b
y2ij−y ..
2
ab=¿
(9.60 ) (29.88 ) (37.7 )+ (11.28 ) (29.88 ) (42.83 )+(9.00 ) (29.88 ) (37.4 )+(9.69 ) (29.36 ) (37.7 )+(10.10 ) (29.36 ) (42.83 )+(9.57 ) (29.36 ) (37.4 )+ (8.43 ) (28.47 ) (37.7 )+(11.01 ) (28.47 ) (42.83 )+(9.03 ) (28.47 ) (37.4 )+ (9.98 ) (30.22 ) (37.7 )+ (10.44 ) (30.22 ) ( 42.83 )+(9.80 ) (30.22 ) (37.4 )=137237.8068
SSN=[∑i=1
a
∑j=1
b
y2ij− y (S SA+S SB+ y ..
2
ab )]abS S AS SB
¿¿
¿¿¿
¿ 3038.203332.4336
=93.674
La suma de cuadrados del error
SS Error=S SResidual−S SN
SS E=2.159−93.676=−91.517
Los datos de la interacción de Fo ( ser aprobado o rechazada)
f o=SSN
ssError /(a−1)(b−1)−1
fo= 93.676−91.517/(4−1)(3−1)−1
fo= 93.676−91.517/(3)(2)−1
fo= 93.676−91.517/5
fo= 93.676−18.3034
=−5.1179
ANOVA
Fuente de variación Suma de cuadrados
Grados de libertad
Media de cuadrados
Fo
Temperatura 0.58 4-1 =3 0.2 -5.1179Presión 4.66 3-1= 2 2.33
No adivitidad 93.6744 1 93.6Error -91.515 (4-1)(3-1)=6 -15.2525Total 7.3922 12
Análisis de datos
Expected Mean Squares SectionSource Term Denominator ExpectedTerm DF Fixed? Term Mean SquareA: Temp 2 Yes S(AB) S+bsAB: Presion 3 Yes S(AB) S+asBS(AB) 6 No SNote: Expected Mean Squares are for the balanced cell-frequency case.
Analysis of Variance TableSource Sum of Mean Prob PowerTerm DF Squares Square F-Ratio Level
(Alpha=0.05)A: Temp 2 4.65765 2.328825 6.49 0.031618* 0.695279B: Presion 3 0.5806917 0.1935639 0.54 0.672704 0.109303S 6 2.153883 0.3589806Total (Adjusted) 11 7.392225Total 12* Term significant at alpha = 0.05