##Comparacion de la media de dos tratamientos independientes##

set.seed(7234)
rto_tomate= c(sort(rnorm(120,2.0, 0.3), decreasing = T))
aporq<-gl(2,60,120, labels = c('aporque','sin aporque'))
df2=data.frame(rto_tomate,aporq)

Diagrama de arbol

library(collapsibleTree)
collapsibleTree(df2, hierarchy = c("rto_tomate","aporq"), root = 'rto_tomate')

##Prueba T student para comparar do tratamientos independientes

\[H_0: \mu_{aporq}= \mu_{sin~aporq}\] \[H_a: \mu_{aporq} \neq \mu_{sin~aporq} \]

medias3= tapply(df2$rto_tomate,df2$aporq,mean);medias3
##     aporque sin aporque 
##    2.282802    1.737791
dv3= tapply(df2$rto_tomate,df2$aporq, sd);dv3
##     aporque sin aporque 
##   0.2130656   0.2104965
cv3= 100*(dv3/medias3); dv3
##     aporque sin aporque 
##   0.2130656   0.2104965

##Cambio Relativo## ##Cambio absoluto

Porcentaje en que el tramiento aporq supera al no aporq

camb_a= medias3[1]-medias3[2]; camb_a
##  aporque 
## 0.545011

##Cambio relativo

camb_r= medias3[1]-medias3[2]/medias3[2]; camb_r
##  aporque 
## 1.282802

#Prueba T de una muestra

\[t= \frac{\bar x- \mu}{S/\sqrt{n}}\] #Prueba T para dos muestras independientes

\[ \require{cancel} t=\frac{(\bar x_1-\bar x_2)-(\cancelto{0}{\mu_1-\mu_2)}}{\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n1}+\frac{1}{n2})}} \]

dt_sep= split(df2$rto_tomate,df2$aporq)
curve(dt(x,30), from = -5, to = 5, col = c('blue'), xlab = "t~tabulado", ylab = "densidad", lwd = 2)

#df2= n1+n2-2; 60+60-2
t_tab1= qt(p = 0.025, df=118, lower.tail = T);t_tab1
## [1] -1.980272
curve(dt(x,30), from = -5, to = 5, col = c('blue'), xlab = "t~tabulado", ylab = "densidad", lwd = 2, main="Dos colas")
abline(v=c(t_tab1,-t_tab1), col= "red", lty= 2)
text(0,0.2,"Zona de no \n rechazo H_0" )
text(0,0.1,"Nivel de confianza 0.95")
abline(v= 14.15, col= "green")

pb1= t.test(dt_sep$aporque,dt_sep$`sin aporque`, alternative = "t", mu= 0, var.equal = T, conf.level = 0.95)
pb1
## 
##  Two Sample t-test
## 
## data:  dt_sep$aporque and dt_sep$`sin aporque`
## t = 14.095, df = 118, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.4684409 0.6215812
## sample estimates:
## mean of x mean of y 
##  2.282802  1.737791
ifelse(pb1$p.value<0.05,"rechao H0","no rechazo H0")
## [1] "rechao H0"

\[H_0:\sigma_1^2=\sigma_2^2\]

\[H_a:\sigma_1^2 \neq \sigma_2^2\]

Prueba para homogeneidad de varianzas

pb2=var.test(dt_sep$aporque,dt_sep$`sin aporque`,ratio = 1,alternative = "t",conf.level = 0.95)
ifelse(pb2$p.value<0.05,"Rechazo Ho","No rechazo Ho")
## [1] "No rechazo Ho"

Varianza desiguales

#prueba1=t.test(datos_sep\(aporque,datos_sep\)sin_aporque,alternative = “t”,mu = 0,var.equal = F,conf.level = 0.95)

###Prueba T student para muestras pareadas

set.seed(7234)
crec=c(rnorm(60,15,2),rnorm(60,20,2.2))
dias_ev=gl(2,60,120,c("15d","30d"))
df_crec4=data.frame(crec,dias_ev)
medias4= tapply(df_crec4$crec,df_crec4$dias_ev, mean);medias4
##      15d      30d 
## 15.28730 19.83499
dv4= tapply(df_crec4$crec,df_crec4$dias_ev, sd); dv4
##      15d      30d 
## 2.395067 2.426501
cv4= 100*(dv4/medias4);cv4
##      15d      30d 
## 15.66704 12.23344
cambio_a4= medias4[1]-medias4[2];cambio_a4
##       15d 
## -4.547691
cambio_r4= medias4[1]-medias4[2]/medias4[2]; cambio_r4
##     15d 
## 14.2873

Hipótesis

\[H_0: \mu_{15d} = \mu_{30d} \\ H_a: \mu_{15d} \neq \mu_{30d} \]

df_sep4= split(df_crec4$crec,df_crec4$dias_ev)
pb3=t.test(df_sep4$`15d`,df_sep4$`30d`,alternative = "t",mu = 0,paired = T,conf.level = 0.95)
ifelse(pb3$p.value<0.05,"Rechazo Ho","No rechazo")
## [1] "Rechazo Ho"

Comparación de dos proporciones (porcentajes)

\[Prevalencia=\frac{No.~casos~positivos}{Total ~de ~ casos} 100\]

library(colorspace)
set.seed(7234)
lt1=round(runif(144,0,0.55),0)
xy=expand.grid(x=seq(1,12,1),y=seq(1,12,1))
color=ifelse(lt1==0,"blue","red")
plot(xy,col=color,pch=8,main="distribución palmas lote 1")

set.seed(7234)
lt2=round(runif(144,0,0.6),0)
xy=expand.grid(x=seq(1,12,1),y=seq(1,12,1))
color2=ifelse(lt2==0,"blue","red")
color2[14]= 'white'
color2[24]= 'white'
color2[11]= 'white'
color2[90]= 'white'
plot(xy,col=color,pch=8,main="distribución palmas lote 2")

Prevalencia d ela enfermedad

table(lt1)
## lt1
##   0   1 
## 128  16
pv1= (16/144)*100; pv1
## [1] 11.11111
table(color2)
## color2
##  blue   red white 
##   116    24     4
pv2= (16/140)*100;pv2
## [1] 11.42857

Hipótesis

\[H_0:prev_1=prev_2\]

\[H_0:prev_1 \neq prev_2\] #Prueba Z

pb4= prop.test(x=c(16,16), n=c(144,141));pb4
## 
##  2-sample test for equality of proportions with continuity correction
## 
## data:  c(16, 16) out of c(144, 141)
## X-squared = 8.085e-31, df = 1, p-value = 1
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.07804588  0.07331775
## sample estimates:
##    prop 1    prop 2 
## 0.1111111 0.1134752
ifelse(pb4$p.value<0.05, "Rechazo H0","No rechazo H0")
## [1] "No rechazo H0"