library(ggplot2)
library(dplyr)
library(readxl)
library(collapsibleTree)
library(lme4)

COMPLETAMENTE AL AZAR

Es muy dificil hacer la práctica en un escenario real

set.seed(123)
xy = expand.grid(x = seq(6), y = seq(4))
xy = sample_frac(xy)
xy$fa1 = gl (4, 6, 24, paste0('V', 1:4))
xy$fa2 = gl (3, 2, 24, paste0('R', 1:3))
xy$arep = gl (2, 1, 24, paste0('rep', 1:2))
xy$aname = paste0(xy$fa1, xy$fa2, xy$arep)
xy
ggplot(xy)+
  aes(x, y, label = aname, fill = fa2)+
  geom_tile(color= 'white')+
  geom_label(fill = 'white')+
  labs(title = 'COMPLETAMENTE AL AZAR')

NA

PARCELA DIVIDIDA

collapsibleTree::collapsibleTree(esqu, c('riego', 'variedad', 'rep', 'X'), collapsed = FALSE)

MODELO

\[Y_{ijk} = \mu + \alpha_i + \eta{k_{ij}} + \beta_j + (\alpha\beta)_{ij} + e_{k_{ij}}\]

set.seed(123)
xy = expand.grid(y = seq(4), x = seq(6))
f2 = gl(3, 8, 24, paste0('R',1:3))
lf1 = paste0('V',1:4)
f1 = c(sample(lf1),sample(lf1),
       sample(lf1),sample(lf1),
       sample(lf1),sample(lf1))
rep = rep(rep(paste0('r',1:2), each=4), 3)

data = data.frame(xy, f1, f2, rep)
data$name = with(data, paste0(f1, rep))
ggplot(data)+
  aes(x,y,label=name, fill=f1)+
  geom_tile(color='white')+
  geom_text()+
  facet_wrap(~f2, scales = 'free')+
  theme(axis.text = element_blank())

data$biom = rnorm(24, 8, 2)

ANALISIS DESCRIPTIVO

ggplot(data)+
  aes(f2, biom)+
  geom_boxplot()


ggplot(data)+
  aes(f1, biom)+
  geom_boxplot()


ggplot(data)+
  aes(f2, biom, fill = f1)+
  geom_boxplot()

ANOVA USANDO ENFOQUE VIEJO

mod1 = aov(biom ~ f2*f1 + Error(f2:rep), data)
Warning: Error() model is singular
summary(mod1)

Error: f2:rep
          Df Sum Sq Mean Sq F value Pr(>F)
f2         2  2.485   1.243   0.763   0.54
Residuals  3  4.886   1.629               

Error: Within
          Df Sum Sq Mean Sq F value Pr(>F)
f1         3  4.777   1.592   0.787  0.531
f2:f1      6 20.432   3.405   1.683  0.232
Residuals  9 18.209   2.023               

INTERPRETACIÓN

No hay intereacción f1:f2 p value es mayor a 5% no se rechaza la hipotesis nula que dice que no hay interaccion

No hay efecto del riego f2 sobre la respuesta porque el p value es mayor a 5%

No hay efecto de las variedades f1 sobre la respuesta porque el p value es mayor a 5%

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