EJERCICIO DISTRIBUCIONES
Ana María Florián Pulido
11/4/2021
DISTRIBUCIÓN EXPONENCIAL
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~exponencial\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~exponencial \]
require(vcd)## Loading required package: vcd## Loading required package: gridrequire(MASS)## Loading required package: MASS# Generación de datos
exponencial <- rexp(1000, rate = 1.85) # generar datos con distribución exponencial
control <- abs(rnorm(1000)) # generar datos con cualquier otra distribución
# Estimación de parámetros
fit1 <- fitdistr(exponencial, "exponential")
fit2 <- fitdistr(control, "exponential")
# Bondad del ajuste del test
(pr_ex1=ks.test(exponencial, "pexp", fit1$estimate)) # p-value > 0.05 -> Ho no rechazada##
## One-sample Kolmogorov-Smirnov test
##
## data: exponencial
## D = 0.025309, p-value = 0.5436
## alternative hypothesis: two-sidedifelse(pr_ex1$p.value<0.05, "Rechazo Ho", "No rechazo Ho")## [1] "No rechazo Ho"(pr_ex2=ks.test(control, "pexp", fit2$estimate)) # significant p-value -> Ho rechazada##
## One-sample Kolmogorov-Smirnov test
##
## data: control
## D = 0.084311, p-value = 1.339e-06
## alternative hypothesis: two-sidedifelse(pr_ex2$p.value<0.05, "Rechazo Ho", "No rechazo Ho")## [1] "Rechazo Ho"# Graficar
hist(exponencial, freq = FALSE, breaks = 100, xlim = c(0, quantile(exponencial, 0.99)))
curve(dexp(x, rate = fit1$estimate), from = 0, col = "red", add = TRUE)
DISTRIBUCIÓN NORMAL
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~normal\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~normal \]
#Generación de datos
normal = rnorm(1000) #Distribución normal
control2 = rt(1000, df=3) #Distribución t
#Graficar densidad
plot(density(normal));plot(density(control2))
#Realizar el test
(pr_nor1=shapiro.test(normal))##
## Shapiro-Wilk normality test
##
## data: normal
## W = 0.99811, p-value = 0.3307ifelse(pr_nor1$p.value<0.05, "Rechazo Ho", "No rechazo Ho")## [1] "No rechazo Ho"(pr_nor2=shapiro.test(control2))##
## Shapiro-Wilk normality test
##
## data: control2
## W = 0.90091, p-value < 2.2e-16ifelse(pr_nor2$p.value<0.05, "Rechazo Ho", "No rechazo Ho")## [1] "Rechazo Ho"#Graficar usando qqplot
qqnorm(normal)
qqline(normal, col = 2)
qqnorm(control2)
qqline(control2, col = 2)
DISTRIBUCIÓN UNIFORME
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~uniforme\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~uniforme\]
#Generación de datos
library(swfscMisc)##
## Attaching package: 'swfscMisc'## The following object is masked from 'package:vcd':
##
## oddsuniforme <- runif(1000) #Uniforme
control3 <- rlnorm(1000) #Normal
#Realizar el test
(pr_uni1=uniform.test(hist(uniforme), B = 1000))
##
## Chi-squared test for given probabilities with simulated p-value (based
## on 1000 replicates)
##
## data: hist.output$counts
## X-squared = 6.72, df = NA, p-value = 0.6523ifelse(pr_uni1$p.value<0.05, "Rechazo Ho", "No rechazo Ho")## [1] "No rechazo Ho"(pr_uni2=uniform.test(hist(control3), B = 1000))
##
## Chi-squared test for given probabilities with simulated p-value (based
## on 1000 replicates)
##
## data: hist.output$counts
## X-squared = 7989.5, df = NA, p-value = 0.000999ifelse(pr_uni2$p.value<0.05, "Rechazo Ho", "No rechazo Ho")## [1] "Rechazo Ho"DISTRIBUCIÓN POISSON
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~poisson\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~poisson\]
#Crear función que se basa en el hecho de que la media de la distribución de Poisson es igual a su varianza, y la relación entre la varianza y la media en una muestra de n recuentos de una distribución de Poisson debe seguir una distribución de Chi-cuadrado, con n-1 grados de libertad.
dispersion_test <- function(x)
{
res <- 1-2 * abs((1 - pchisq((sum((x - mean(x))^2)/mean(x)), length(x) - 1))-0.5)
cat("Dispersion test of count data:\n",
length(x), " data points.\n",
"Mean: ",mean(x),"\n",
"Variance: ",var(x),"\n",
"Probability of being drawn from Poisson distribution: ",
round(res, 3),"\n", sep = "")
invisible(res)
}
#Crear datos
poisson=rpois(1000,1)
control4=rnorm(1000)
#Probar la función
dispersion_test(poisson)## Dispersion test of count data:
## 1000 data points.
## Mean: 0.983
## Variance: 0.9096206
## Probability of being drawn from Poisson distribution: 0.09dispersion_test(control4)## Dispersion test of count data:
## 1000 data points.
## Mean: 0.02478714
## Variance: 1.039307
## Probability of being drawn from Poisson distribution: 0DISTRIBUCIÓN BINOMIAL
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~binomial\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~binomial\]
require("fitdistrplus")## Loading required package: fitdistrplus## Loading required package: survival#Crear datos
set.seed(100)
binomial = rbinom(1000,50,0.4)
#Probar el test
fit1 = fitdist(binomial, dist="binom", #método "mle"
fix.arg=list(size = 50),
start=list(prob = 0.1))
summary(fit1)## Fitting of the distribution ' binom ' by maximum likelihood
## Parameters :
## estimate Std. Error
## prob 0.4044602 0.002194857
## Fixed parameters:
## value
## size 50
## Loglikelihood: -2678.442 AIC: 5358.884 BIC: 5363.792plot(fit1)
DISTRIBUCIÓN LOGNORMAL
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~lognormal\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~lognormal \]
require("fitdistrplus")
#Crear datos
set.seed(100)
lognormal = rlnorm(1000,50,0.3)
#Probar el test
fit2 = fitdist(lognormal, dist="lnorm") #método "mle"
summary(fit2)## Fitting of the distribution ' lnorm ' by maximum likelihood
## Parameters :
## estimate Std. Error
## meanlog 50.0050415 0.009772126
## sdlog 0.3090218 0.006909611
## Loglikelihood: -50249.64 AIC: 100503.3 BIC: 100513.1
## Correlation matrix:
## meanlog sdlog
## meanlog 1 0
## sdlog 0 1plot(fit2)
DISTRIBUCIÓN BETA
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~lognormal\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~lognormal \]
require("fitdistrplus")
#Crear datos
set.seed(100)
beta <- rbeta(1000,0.9,0.5)
#Probar el test
fit3 = fitdist(beta, dist="beta")
summary(fit3)## Fitting of the distribution ' beta ' by maximum likelihood
## Parameters :
## estimate Std. Error
## shape1 0.9449265 0.04250818
## shape2 0.4809618 0.01824865
## Loglikelihood: 320.3926 AIC: -636.7851 BIC: -626.9696
## Correlation matrix:
## shape1 shape2
## shape1 1.0000000 0.5452701
## shape2 0.5452701 1.0000000plot(fit3)
DISTRIBUCIÓN BETA
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~lognormal\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~lognormal \]
require("fitdistrplus")
#Crear datos
set.seed(100)
gamma <- rgamma(1000,0.9,0.6)
#Probar el test
fit4 = fitdist(gamma, dist="gamma")
summary(fit4)## Fitting of the distribution ' gamma ' by maximum likelihood
## Parameters :
## estimate Std. Error
## shape 0.8553526 0.03320887
## rate 0.5492155 0.02841333
## Loglikelihood: -1434.735 AIC: 2873.471 BIC: 2883.286
## Correlation matrix:
## shape rate
## shape 1.0000000 0.7504593
## rate 0.7504593 1.0000000plot(fit4)
DISTRIBUCIÓN CHI cuadrado
\[ Ho:~Los~datos~se~ajustan~a~una~distribución~chicuadrado\\ Ha:~Los~datos~no~se~ajustan~a~una~distribución~chicuadrado \]
require("fitdistrplus")
#Crear datos
set.seed(100)
chi <- rchisq(n = 1000,df = 20,ncp = 2)
hist(chi)
#Probar el test
print(chisq.test(chi))##
## Chi-squared test for given probabilities
##
## data: chi
## X-squared = 2047.4, df = 999, p-value < 2.2e-16```