Campo de R -> “Ctrl+Alt+i”
## [1] 4.933228 4.670382 4.325635 4.792240 4.060783 4.241770 4.852331 5.235716
## [9] 4.526417 4.952648 5.724638 4.297757 4.360860 5.998547 5.683482 4.285298
## [17] 4.689697 5.779251 4.639697 5.418032 5.256303 4.876345 4.679792 5.866185
Campo de ecuación \(\LaTeX\)
\[t=\frac{\bar x - \mu}{s/\sqrt{n}}\]
La ecuación anterior representa un estadistico de prueba, t-student
set.seed(2023)
# Redondear a 2 decimales
diam=round(diam,2)
orient = gl(n=2, k =12, length=24, labels=c("Ecuatorial","Longitudinal"))
df=data.frame(orient, diam)
head(df)## orient diam
## 1 Ecuatorial 4.93
## 2 Ecuatorial 4.67
## 3 Ecuatorial 4.33
## 4 Ecuatorial 4.79
## 5 Ecuatorial 4.06
## 6 Ecuatorial 4.24
## orient diam
## 19 Longitudinal 4.64
## 20 Longitudinal 5.42
## 21 Longitudinal 5.26
## 22 Longitudinal 4.88
## 23 Longitudinal 4.68
## 24 Longitudinal 5.87
# Distribución exponencial
diam2 = rexp(n = 24,rate = 1/4)
# boxplot
boxplot(diam2~orient,horizontal = T,col=c("lightgreen","lightblue"))Resumen estadístico descriptivo
Agregarle promedios a las cajas
## Ecuatorial Longitudinal
## 4.717500 5.129167
## Ecuatorial Longitudinal
## 3.798262 3.315129
# boxplot + medias
boxplot(diam~orient,horizontal = T,col=c("lightgreen","lightblue"),xlab = "Diametro (cm)",ylab="Orientación")
points(y=1:2,x=m1,pch=16,col="red",cex=1.5)
rug(diam[which(orient=="Ecuatorial")],lwd = 3,side = 3,col="darkgreen")
rug(diam[which(orient=="Longitudinal")],lwd = 3,side = 1,col="blue")# Densidad
set.seed(1234)
# Generación de datos
diam = runif(n = 240,min = 4,max = 6)
orient = gl(n = 2,k = 120,length = 240,labels = c("ecuatorial","longitudinal"))## Installing package into '/cloud/lib/x86_64-pc-linux-gnu-library/4.3'
## (as 'lib' is unspecified)
# Usando ggplot2
library(ggplot2)
# Grafico de densidades
ggplot(df,aes(x=diam,fill=orient))+
geom_density(alpha=0.4)## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
df2=split(diam,orient)
df2 = data.frame(ecuatorial=df2$ecuatorial,
longitudinal = df2$longitudinal)
ggplot(df2,aes(x=ecuatorial,
y=longitudinal))+ geom_point(size=2)
Investigar: Bihistograma(quiz) codigo y como interpretarlo.
##
## Attaching package: 'psych'
## The following objects are masked from 'package:ggplot2':
##
## %+%, alpha
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 240 4.98 0.57 4.99 4.97 0.7 4.02 6 1.98 0.11 -1.14 0.04
##
## Descriptive statistics by group
## group: ecuatorial
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 120 4.85 0.57 4.65 4.83 0.6 4.02 5.98 1.97 0.34 -1.18 0.05
## ------------------------------------------------------------
## group: longitudinal
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 120 5.1 0.53 5.13 5.11 0.65 4.04 6 1.95 -0.05 -0.96 0.05
\[\%~CV = \frac{s}{\bar{x}}\times 100\]
# Mi primera función
fun_cv=function(datos)
{
media=mean(datos)
desv=sd(datos)
if(media==0 & desv == 0){print("Indeterminación")}
else{
cv=100*desv/media
return(cv)}
}
# Evaluando la función
cat("%CV ecuatorial",fun_cv(df2$ecuatorial))## %CV ecuatorial 11.7698
## %CV longitudinal 10.44618
## [1] "Indeterminación"
# Validando inicialmente la funcion
fun_cv= function(datos){
media=mean(datos)
desv = sd(datos)
if(media==0 & desv==0){
print("Indeterminación")
}else{
cv = 100*desv/media
return(cv)
}
}## [1] "Indeterminación"