#PRUEBAS DE HIPOTESIS

##PROBLEMA1: SE DESEA COMPARAR 2 GENOTIPOS DE PAPA CON BASE EN ELREDNIMIENTO (BIOMASA DE TUBERCULOS). UN ESNAYO UTILIZO 2 VARIEDADES (CRIOLLA Y PASTUSA) INVOLUCRANDO 180 PLANTAS DE LA PRIMERA VARIEDAD Y 200 DE LA SEGUNDA. LOS DATOS DE RENDIMIENTO EN LA COSECHA SE PRESENTAN EN LOS SIGUIENTES VECTORES

options(digits = 3)
criolla = rnorm(n =180, mean = 2.8, sd = 0.2); criolla
##   [1] 2.79 2.75 2.83 2.58 2.81 2.92 3.23 2.47 2.60 2.83 2.68 3.22 2.33 2.55 2.64
##  [16] 3.07 2.54 2.49 2.72 2.71 2.61 2.93 3.04 3.02 2.92 2.96 2.45 2.78 2.89 2.80
##  [31] 2.63 2.98 2.78 2.56 2.67 2.95 2.85 2.84 2.86 2.93 2.62 2.72 2.60 2.85 2.77
##  [46] 2.78 2.95 3.13 2.78 2.72 2.69 2.55 2.82 2.59 2.76 2.86 2.99 3.05 2.75 2.56
##  [61] 2.84 2.73 2.94 2.73 2.76 2.70 3.02 2.54 2.52 3.05 2.92 2.34 2.56 3.14 2.63
##  [76] 3.08 2.96 3.24 2.62 2.83 2.84 2.99 2.68 2.75 2.94 3.02 2.40 2.72 2.63 2.89
##  [91] 3.23 2.83 2.91 2.89 2.32 2.80 2.60 3.18 2.77 2.68 2.95 2.39 3.21 2.55 2.94
## [106] 2.77 2.71 2.65 2.60 2.67 2.68 2.38 2.97 2.51 2.91 3.02 2.70 2.73 2.99 2.88
## [121] 2.73 2.66 2.68 2.74 2.30 2.74 2.74 2.80 2.69 2.88 2.57 2.80 2.81 2.50 3.22
## [136] 2.78 2.96 2.51 3.01 2.83 2.80 2.87 2.40 2.84 2.60 3.05 2.79 2.74 2.80 2.80
## [151] 2.84 3.12 2.65 3.08 3.08 3.01 2.82 3.10 3.21 2.79 2.80 2.74 2.77 2.71 2.64
## [166] 2.71 2.71 3.08 2.89 2.85 2.62 2.69 2.93 3.16 2.86 2.87 2.72 2.76 2.89 2.86
pastusa = rnorm(n = 200, mean = 2.8, sd =0.2); pastusa
##   [1] 2.77 2.92 2.81 2.91 2.79 2.63 3.09 3.16 2.90 2.81 2.85 2.97 2.92 2.82 2.72
##  [16] 2.89 2.76 2.48 3.01 2.60 2.71 2.76 2.74 2.86 2.72 2.90 2.73 2.84 3.11 2.89
##  [31] 3.13 2.84 2.84 2.82 3.04 2.58 2.44 2.62 2.62 2.62 2.79 2.38 2.77 2.96 3.19
##  [46] 2.94 2.59 2.62 2.92 2.48 3.04 2.88 2.77 2.90 3.07 2.74 2.61 2.75 2.70 3.02
##  [61] 2.71 2.86 2.55 2.72 2.54 2.53 3.08 2.52 2.88 2.96 2.76 2.84 3.02 3.09 3.01
##  [76] 2.73 2.67 2.97 2.85 2.79 2.83 2.74 2.83 3.15 2.62 2.72 3.06 2.64 2.78 2.71
##  [91] 2.66 2.57 2.67 2.83 3.05 2.88 2.96 2.81 2.92 2.57 3.32 2.56 3.08 3.05 3.11
## [106] 3.04 2.98 2.96 2.61 2.92 2.88 2.62 3.18 2.82 2.70 2.74 2.93 3.28 2.78 2.68
## [121] 2.60 2.45 2.79 2.74 2.86 2.54 2.75 2.60 2.85 2.92 2.66 2.85 2.80 2.77 2.93
## [136] 2.61 2.76 3.13 2.97 3.09 3.01 2.78 2.68 2.85 3.13 2.67 2.66 2.48 3.23 2.90
## [151] 2.74 2.75 2.85 2.81 2.65 2.60 3.11 2.50 2.63 2.84 2.83 3.03 2.89 2.73 2.60
## [166] 2.67 2.83 2.65 2.96 2.69 2.41 2.89 2.72 2.88 2.86 2.74 2.81 2.64 2.50 2.43
## [181] 3.03 2.88 2.93 2.90 2.48 2.46 3.08 2.76 2.99 2.79 3.12 3.04 2.88 2.80 3.13
## [196] 3.19 3.04 2.54 2.93 2.65
par(mfrow=c(1,2))
hist(criolla, col="darkblue")
abline(v=mean(criolla), col="red", lwd=3)
hist(pastusa, col="yellow")
abline(v=mean(pastusa), col="red", lwd=3)

boxplot(criolla, main="criolla", ylab="Rto(kg/planta)")
boxplot(pastusa, main="pastusa", ylab="Rto(kg/planta)")

summary(criolla)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    2.30    2.68    2.79    2.79    2.92    3.24
summary(pastusa)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    2.38    2.67    2.81    2.81    2.93    3.32
library(psych)
## Warning: package 'psych' was built under R version 4.2.2
psych::describe(criolla)
##    vars   n mean  sd median trimmed  mad min  max range skew kurtosis   se
## X1    1 180 2.79 0.2   2.79    2.79 0.19 2.3 3.24  0.94 0.02    -0.15 0.01
psych::describe(pastusa)
##    vars   n mean   sd median trimmed mad  min  max range skew kurtosis   se
## X1    1 200 2.81 0.19   2.81    2.81 0.2 2.38 3.32  0.94 0.11    -0.46 0.01

#disgresion

medA = 3.5; sdA = 0.35
medB =3.2; sdB = 0.20

#cual seleccionar?
## coeficiente de variacion cv = 100 * sd/mean
cvA = 100 * sdA/medA
cvB = 100 * sdB/medB

cvA; cvB
## [1] 10
## [1] 6.25
100 * sd(criolla) / mean(criolla)
## [1] 7.15
100 * sd(pastusa) / mean(pastusa)
## [1] 6.79

#conclusion desde el analisis descriptivo

Ambos coeficientes de bariacion son najos Se puede omitir el problema de diferencia variablididad *Se selecciona la variedad de mayor rendimiento promedio

#ANALISIS INFERENCIAL A TRAVES DE PRUEBAS DE HIPOTESIS \[H_0: \mu_{pastusa} =\mu_{criolla} \\ H_a: \{pastusa} \neq \mu_{criolla}\]

prueba t-student para comparar dos muestras independientese

modalidad 1: varianzas iguales modalidad 2: varianzas desiguales

prueba de compraracion de dos varianzas

\[H_0: \sigma {pastusa = \sigma^2_{criolla} H_a: \sigma {pastusa} \neq \sigma^2_{criolla}\]

var(pastusa)
## [1] 0.0365
var (criolla)
## [1] 0.0399
var.test(pastusa, criolla)
## 
##  F test to compare two variances
## 
## data:  pastusa and criolla
## F = 0.9, num df = 199, denom df = 179, p-value = 0.5
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.687 1.218
## sample estimates:
## ratio of variances 
##              0.916

vt = var.test(pastusa, criolla) vt\(p.value ifelse(vt\)p.value < 0.025, “rechazo H0” , “no rechazo H0)”No rechazo H0

pt =t.test(pastusa, criolla, 
       alternative ="t", 
       var.equal = TRUE)
ifelse(pt$p.value < 0.025, "rechazo Ho", "No rechazo")
## [1] "No rechazo"

CONCLUSION FINAL: LOS DATOS PROPORCIONAN EVIDENCIA ESTADISTICA A FAVOR DE LA HIPOTESIS NULA, ES DECIR, QUE ESTADISTICAMENTE SE CONSIDERAN AMBAS VARIEDADES COMO DE IGUAL RENDIMIENTO, CUALQUIERA DE LAS VARIEDADES ES IGUAL DE BUENA