##Diseño Cuadrado Latino ###factorial simple en bloques al azar (FSBA)
Un solo un factor Dos razones de bloqueo
lote <- c(rep("Lote1",1), rep("Lote2",1), rep("Lote3",1), rep("Lote4",1), rep("Lote5",1))
genotipo <- c(rep("genA",5), rep("genB",5), rep("genC",5), rep("genD",5), rep("genE",5))
prov <- c("A","E","C","B","D", "C","B","A","D","E", "B","C","D","E","A", "D","A","E","C","B", "E","D","B","A","C")
biom <- c(42,45,41,56,47, 47,54,46,52,49, 55,52,57,49,45, 51,44,47,50,54, 44,50,48,43,46)
data <- data.frame(genotipo, lote, prov, biom)
head (data)
## genotipo lote prov biom
## 1 genA Lote1 A 42
## 2 genA Lote2 E 45
## 3 genA Lote3 C 41
## 4 genA Lote4 B 56
## 5 genA Lote5 D 47
## 6 genB Lote1 C 47
###Graficos descriptivos
library(lattice)
bwplot(biom ~ genotipo | prov + lote,
data)
##Modelo
\[y= \mu + \tau_i + \beta_j + \delta_k + \epsilon_{ijk}\] \[i=1, \dots,p\] \[j=1, \dots,p\] \[k=1, \dots,p\]
tbl= matrix(data$prov, 5)
colnames(tbl) = unique(data$genotipo)
rownames(tbl) = unique(data$lote)
tbl
## genA genB genC genD genE
## Lote1 "A" "C" "B" "D" "E"
## Lote2 "E" "B" "C" "A" "D"
## Lote3 "C" "A" "D" "E" "B"
## Lote4 "B" "D" "E" "C" "A"
## Lote5 "D" "E" "A" "B" "C"
\[H_0: \mu_ {B_{G_1}} = \mu_ {B_{G_2}} = \mu_ {B_{G_3}} = \mu_ {B_{G_4}} = \mu_ {B_{G_5}}\]
mod<- lm (biom ~ lote + genotipo + prov, data)
anova(mod)
## Analysis of Variance Table
##
## Response: biom
## Df Sum Sq Mean Sq F value Pr(>F)
## lote 4 17.76 4.440 0.7967 0.549839
## genotipo 4 109.36 27.340 4.9055 0.014105 *
## prov 4 286.16 71.540 12.8361 0.000271 ***
## Residuals 12 66.88 5.573
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
bwplot(biom ~ genotipo | prov,
data)
interaction.plot(genotipo, prov, biom, lwd=2)
library(ggplot2)
ggplot(data)+aes (x= prov, y= biom, fill=genotipo)+ geom_col( position = 'dodge')
#Revisión de supuesto
res_mod = mod$residuals
#1. Normalidad
shapiro.test(res_mod)
##
## Shapiro-Wilk normality test
##
## data: res_mod
## W = 0.97691, p-value = 0.8178
#Se cumple el supuesto de normalidad
#2.Igualdad de varianzas
bartlett.test(res_mod, genotipo)
##
## Bartlett test of homogeneity of variances
##
## data: res_mod and genotipo
## Bartlett's K-squared = 5.9223, df = 4, p-value = 0.205
#Se cumple el supuesto (Varianzas iguales)
#install.packages("TuckeyC)
library(TukeyC)
tt = TukeyC(mod, 'genotipo')
plot(tt)