Clase 9
Cesar Rincón
2022-11-27
Diseño Factorial simple
area = área de daño foliar
set.seed(22)
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)
head(dt)## area inoculo
## 1 0.119 D0
## 2 0.148 D0
## 3 0.166 D0
## 4 0.171 D0
## 5 0.200 D0
## 6 0.247 D0Comparación de medias
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 Ho')## Pr(>F)1
## "Rechazo Ho"## Pr(>F)1
## "Rechazo Ho"Prueba de comparaciones de Turkey
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 del diseño
\[y_{ij} = \mu + \tau_i + \epsilon_{ij} i=1,2,3,4~(inoculo) j=1,2,\cdots,6~(repeticiones)\]
Revisión de supuestos Normalidad \[[H_0:~los~residuales~del~modelo~tienen~distribución~normal\]
res = mod_1$residuals
n = shapiro.test(res)
n##
## Shapiro-Wilk normality test
##
## data: res
## W = 0.97101, p-value = 0.692#No rechazo Ho, los residuales se distribuyen normalmente pvalor>0.05Homoceasticidad de varianza \[var~_{D_0} = var~_{D_1} = \cdots = var~_{D_4}\]
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-05#Rechazo Ho: varianzas por grupo son distintas, No se cumple el supuesto de homocedasticidad. Pvalor < 0.05ANALISIS DE VARIANZA CON 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)
mod_3 = kruskal.test(dt$area~dt$inoculo)
mod_3##
## 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
mod_4 = RVAideMemoire::perm.anova(dt$area~dt$inoculo, nperm = 5000)##
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|======================================================================| 100%mod_4## 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$area[15] = NA
library(car)## Loading required package: carDatamod_5 = Anova(lm(dt_2$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