CLASE GRABADA EJEM 2 INCUMP DE SUPUESTOS
Geraldine Barbosa
2023-04-24
Diseño 2 incumplimiento de supuestos
# Factor
genotipo = gl(n = 6, k = 6, length = 36,
labels = paste0('gen', 1:6))
# Variable respuesta
set.seed(123)
PS = c(
rnorm(12, 1200, 120),
rnorm(12, 1500, 100),
rnorm(12, 1420, 250)
)
aleat = sample(36)
datos = data.frame(genotipo, PS)
head(datos)## genotipo PS
## 1 gen1 1132.743
## 2 gen1 1172.379
## 3 gen1 1387.045
## 4 gen1 1208.461
## 5 gen1 1215.515
## 6 gen1 1405.808library(ggplot2)## Warning: package 'ggplot2' was built under R version 4.2.3ggplot(datos)+
aes(genotipo, PS)+
geom_boxplot()
mod1b = aov(PS ~ genotipo, datos)
smod1b = summary(mod1b)
shapiro.test(mod1b$residuals)##
## Shapiro-Wilk normality test
##
## data: mod1b$residuals
## W = 0.98349, p-value = 0.8558bartlett.test(mod1b$residuals, datos$genotipo)##
## Bartlett test of homogeneity of variances
##
## data: mod1b$residuals and datos$genotipo
## Bartlett's K-squared = 12.401, df = 5, p-value = 0.02969*Como se rechaza la hipotesis nula de igualdad de variazan se incumple el supuesto lo cual complica la interpretación.
Analisis de varianza para un diseño factorial simple en arreglo completamente al azar en presencia de heterocedasticidad (en presencia de varianzas desiguales):
- Alternativa rápida, no permite extraer residuales:
mod1c =oneway.test(PS ~ genotipo, datos)
mod1c##
## One-way analysis of means (not assuming equal variances)
##
## data: PS and genotipo
## F = 8.6764, num df = 5.000, denom df = 13.702, p-value = 0.0006918Cuando se incumple normalidade igualdad de varianzas (Prueba de Kruskal para cuando no se cumplen los dos supuestos): permite comparar los tratamientos
*Analisis de varianza no parametrico para un diseño en arreglo factorial simple en arreglo completamente al azar
\[H_0: R1=R2=R3=R4=R5=R6=\]
mod1d = kruskal.test(PS, genotipo)
mod1d##
## Kruskal-Wallis rank sum test
##
## data: PS and genotipo
## Kruskal-Wallis chi-squared = 17.204, df = 5, p-value = 0.004128- Comparación de rangos promedios posterior a Kruskal wallis: Para comparar los rangos promedios en la prueba de Kuskal wallis, para conocer los rangos que se pueden considerar iguales o diferentes segun el enfoque
# library(PMCMR)
# posthoc.kruskal.nemenyi.test(PS, genotipo)
library(PMCMRplus)## Warning: package 'PMCMRplus' was built under R version 4.2.3kwAllPairsNemenyiTest(PS, genotipo)##
## Pairwise comparisons using Tukey-Kramer-Nemenyi all-pairs test with Tukey-Dist approximation## data: PS and genotipo## gen1 gen2 gen3 gen4 gen5
## gen2 0.998 - - - -
## gen3 0.172 0.058 - - -
## gen4 0.443 0.205 0.995 - -
## gen5 0.807 0.533 0.883 0.993 -
## gen6 0.046 0.012 0.995 0.904 0.587##
## P value adjustment method: single-step## alternative hypothesis: two.sidedlibrary(FSA)## Warning: package 'FSA' was built under R version 4.2.3## ## FSA v0.9.4. See citation('FSA') if used in publication.
## ## Run fishR() for related website and fishR('IFAR') for related book.dunnTest(PS, genotipo)## Dunn (1964) Kruskal-Wallis multiple comparison## p-values adjusted with the Holm method.## Comparison Z P.unadj P.adj
## 1 gen1 - gen2 0.4383973 0.6610983037 0.66109830
## 2 gen1 - gen3 -2.3563855 0.0184537565 0.22144508
## 3 gen2 - gen3 -2.7947828 0.0051934597 0.06751498
## 4 gen1 - gen4 -1.8357887 0.0663889146 0.66388915
## 5 gen2 - gen4 -2.2741860 0.0229548062 0.25250287
## 6 gen3 - gen4 0.5205968 0.6026476838 1.00000000
## 7 gen1 - gen5 -1.2603922 0.2075279011 1.00000000
## 8 gen2 - gen5 -1.6987895 0.0893588460 0.80422961
## 9 gen3 - gen5 1.0959932 0.2730817290 1.00000000
## 10 gen4 - gen5 0.5753965 0.5650232007 1.00000000
## 11 gen1 - gen6 -2.8769823 0.0040149814 0.05620974
## 12 gen2 - gen6 -3.3153796 0.0009151876 0.01372781
## 13 gen3 - gen6 -0.5205968 0.6026476838 1.00000000
## 14 gen4 - gen6 -1.0411936 0.2977857118 1.00000000
## 15 gen5 - gen6 -1.6165900 0.1059668029 0.84773442rangos = rank(PS, ties.method = 'average')
rangos## [1] 4 7 16 8 9 19 11 2 3 6 15 10 27 25 21 35 28 13 29 22 17 24 18 20 12
## [26] 1 32 23 5 36 26 14 34 33 31 30boxplot(rangos ~ genotipo)
Analisis de varianza permutacional
Se corre el analisis de varianza haciendo simulación,
library(RVAideMemoire)## Warning: package 'RVAideMemoire' was built under R version 4.2.3## *** Package RVAideMemoire v 0.9-81-2 ***##
## Attaching package: 'RVAideMemoire'## The following object is masked from 'package:FSA':
##
## seperm.anova(PS ~ genotipo, data = datos, nperm = 999)##
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|======================================================================| 100%## Permutation Analysis of Variance Table
##
## Response: PS
## 999 permutations
## Sum Sq Df Mean Sq F value Pr(>F)
## genotipo 627712 5 125542 5.3717 0.001 ***
## Residuals 701126 30 23371
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Rechaza la hipotesis nula, los genotipos son diferentes.