Diseño en bloques al azar

set.seed(2022)
xy<-expand.grid(x=seq(0,100,10),y=seq(0,20,10))
df_xy=data.frame(xy)
df_xy$trt=gl(3,11,33,c("v1","v2","v3"))
#df_xy$trt=sample(df_xy$trt,33)

Gráfico

plot(xy,asp=1,ann=F,axes=F,pch=15,cex=4,col=df_xy$trt)
grid(nx=11,col = "grey")
arrows(x0 = 0,y0 = -5,x1 = 100,y1 = -6)
text(x = 100,y=-9,"pH+")
text(x = 0,y=-9,"pH-")

Aleatorización

df_xy$trt=gl(3,1,33,c("v1","v2","v3"))
plot(xy,asp=1,ann=F,axes=F,pch=15,cex=4,col=df_xy$trt)
grid(nx=11,col = "grey")
arrows(x0 = 0,y0 = -5,x1 = 100,y1 = -6)
text(x = 100,y=-9,"pH+")
text(x = 0,y=-9,"pH-")

#mal porque tienen un patrón

Modelo de diseño \[y_{ij}=\mu+\tau_i+\beta_j+\epsilon_{ij}\\i:1\cdots3\\j:1\cdots11\]

\[H_0:\mu_{v1}=\mu_{v2}=\mu_{v3}\] Definición de variables y dataframe

set.seed(2022)
df_xy$biomasa = sort(rnorm(33,5,0.4))
df_xy$bloque = gl(11,1,33,paste0("bloque",1:11))

Comparación de medias

boxplot(df_xy$biomasa~df_xy$bloque,las=2,xlab=" ")

análisis de varianza para el diseño en bloques

mod2= aov(biomasa~bloque+trt,data = df_xy)
summary(mod2)
##             Df Sum Sq Mean Sq F value Pr(>F)
## bloque      10  0.663 0.06633   0.356  0.952
## trt          2  0.073 0.03675   0.197  0.823
## Residuals   20  3.726 0.18629
#No se rechaza la hipotesis nula, Pvalor>0.05

\[H_0:\text{Los residuales son normales}\] Prueba shapiro para comprobar normalidad de residuales

shapiro.test(mod2$residuals)
## 
##  Shapiro-Wilk normality test
## 
## data:  mod2$residuals
## W = 0.93226, p-value = 0.04061
# Si los p.valores son al menos 1% se puede asumir normalidad. 
#Pvalor =4%

Homoceasticidad de la varianza \[H_0:\sigma^2_{v1}=\sigma^2_{v2}=\sigma^2_{v3}\]

bartlett.test(mod2$residuals,df_xy$trt)
## 
##  Bartlett test of homogeneity of variances
## 
## data:  mod2$residuals and df_xy$trt
## Bartlett's K-squared = 0.21275, df = 2, p-value = 0.8991
#se