Clase 11 Diseño experimental
Javier Stiven Bustos
2022-09-27
set.seed(22)
# area de daño foliar
area = c(rnorm(18, 3, 0.8),
rnorm(6, 0.2, 0.05))
inoculo = gl(4, 6, 24, paste0('D', 0:3))
dt = data.frame(area = round(sort(area), 3),
inoculo)
View(dt)boxplot(dt$area~dt$inoculo)
Hipótesis
\[H_0: \mu_{D0} = \mu_{D1}= \cdots = \mu_{D3}\]
mod_1 = aov (dt$area~dt$inoculo)
tb=summary(mod_1)
ifelse(unlist(tb)[9]<0.05,"Rechazo Ho", "No rechazo")## Pr(>F)1
## "Rechazo Ho"prueba de comparaciones de Tukey
TukeyHSD(x=mod_1)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = dt$area ~ dt$inoculo)
##
## $`dt$inoculo`
## diff lwr upr p adj
## D1-D0 2.3715000 1.84876038 2.894240 0.0000000
## D2-D0 2.8731667 2.35042705 3.395906 0.0000000
## D3-D0 4.0190000 3.49626038 4.541740 0.0000000
## D2-D1 0.5016667 -0.02107295 1.024406 0.0628165
## D3-D1 1.6475000 1.12476038 2.170240 0.0000001
## D3-D2 1.1458333 0.62309372 1.668573 0.0000299MODELO DE DISEÑO (Factorial simple)
\[y_{ij} = \mu + \tau_i + \epsilon_{ij}\] \[i=1,2,3,4~(inoculo)\] \[j=1,2,\cdots,6~(repeticiones)\]
Revision de supuestos
Normalidad
\[H_0:~los~residuales~del~modelo~tienen~distribucion~normal\]
res=mod_1$residuals
res## 1 2 3 4 5 6
## -0.056166667 -0.027166667 -0.009166667 -0.004166667 0.024833333 0.071833333
## 7 8 9 10 11 12
## -0.284666667 -0.180666667 -0.157666667 0.043333333 0.286333333 0.293333333
## 13 14 15 16 17 18
## -0.178333333 -0.115333333 -0.101333333 0.017666667 0.185666667 0.191666667
## 19 20 21 22 23 24
## -0.600166667 -0.505166667 -0.388166667 0.291833333 0.407833333 0.793833333n=shapiro.test(res)HOMOCEDASTICIDAD
\[var~D_0 = var~D1 = \cdots = var~D4\]
bartlett.test(dt$area~dt$inoculo)##
## Bartlett test of homogeneity of variances
##
## data: dt$area by dt$inoculo
## Bartlett's K-squared = 22.055, df = 3, p-value = 6.355e-05Rechazo Ho, varianzas por grupo son distintas … No se cumple el supuesto de homocedasticidad …
Anilisis de varianza para varianzas desiguales (ANOVA de WELCH)
mod_2 = oneway.test(dt$area~dt$inoculo)
mod_2##
## One-way analysis of means (not assuming equal variances)
##
## data: dt$area and dt$inoculo
## F = 727.7, num df = 3.0000, denom df = 8.8951, p-value = 5.925e-11Analisis de varianza no parametrico (KRUSKAL WALLIS) (ultima opcion)
kruskal.test(dt$area~dt$inoculo)##
## Kruskal-Wallis rank sum test
##
## data: dt$area by dt$inoculo
## Kruskal-Wallis chi-squared = 21.6, df = 3, p-value = 7.9e-05Anova permutacional
library(RVAideMemoire)## *** Package RVAideMemoire v 0.9-81-2 ***RVAideMemoire::perm.anova(dt$area~dt$inoculo, nperm =5000)##
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|======================================================================| 100%## Permutation Analysis of Variance Table
##
## Response: dt$area
## 5000 permutations
## Sum Sq Df Mean Sq F value Pr(>F)
## dt$inoculo 51.465 3 17.1552 163.94 2e-04 ***
## Residuals 2.093 20 0.1046
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova desbalanceado
dt_2 = dt
dt_2$area[15]=NAtipo II ajuste de varianzas
library(car)## Loading required package: carDatamod_5= Anova(lm(dt_2$area~dt_2$inoculo), type ="II")
mod_5## Anova Table (Type II tests)
##
## Response: dt_2$area
## Sum Sq Df F value Pr(>F)
## dt_2$inoculo 51.261 3 156.04 1.464e-13 ***
## Residuals 2.081 19
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1