9 de febrero de 2016
Distribuciones
#Test hipotesis una poblacion y supuestos de normalidad #Utilice los siguientes datos round(option=3)
## [1] 3
set.seed(2016) dfr<-rnorm(100, 25, 3.7) round(dfr,0)
## [1] 22 29 25 26 15 24 22 22 26 26 23 24 30 22 31 26 22 29 22 22 23 23 24 ## [24] 27 26 21 23 30 29 21 21 25 30 19 25 30 22 22 23 19 25 27 24 23 28 24 ## [47] 20 33 28 19 24 22 16 21 26 19 31 20 24 24 28 27 23 23 27 30 24 21 24 ## [70] 28 29 26 18 25 25 28 26 28 25 24 27 27 27 22 19 32 23 27 24 26 30 27 ## [93] 21 28 24 23 26 24 17 30
d <- density(dfr) plot(d) polygon(d, col = "wheat")
Distribucion y Tamano de Muestras
hist(dfr, col=144, main="Histograma de Frecuencias")
##Revisamos supuestos de Pruebas shapiro.test(dfr)
## ## Shapiro-Wilk normality test ## ## data: dfr ## W = 0.99281, p-value = 0.8764
Valoracion Descriptiva
## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 14.67 22.35 24.02 24.58 27.19 33.12
Prueba de Hipotesis
t.test(dfr, mu=26)
## ## One Sample t-test ## ## data: dfr ## t = -3.9195, df = 99, p-value = 0.0001635 ## alternative hypothesis: true mean is not equal to 26 ## 95 percent confidence interval: ## 23.86352 25.29964 ## sample estimates: ## mean of x ## 24.58158
wilcox.test(dfr, mu=26)
## ## Wilcoxon signed rank test with continuity correction ## ## data: dfr ## V = 1468, p-value = 0.0002806 ## alternative hypothesis: true location is not equal to 26
par(mfrow=c(1,2)) boxplot(dfr); hist(dfr)
#Generamos un segundo grupo de muestras rnorm(100, 27, 7)-> dfr2 #Elabore un cuadro de estadisticas de resumenes. #Incluya #Promedio, sd, max, min, mediana, cuantiles, IQR, #normalidad, Kolmogorov-test.
#Dos poblaciones #test pareado t.test(dfr, dfr2, paired = T)
## ## Paired t-test ## ## data: dfr and dfr2 ## t = -3.4477, df = 99, p-value = 0.0008317 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -4.073105 -1.097364 ## sample estimates: ## mean of the differences ## -2.585234
Muestras independientes
t.test(dfr, dfr2, paired = F)
## ## Welch Two Sample t-test ## ## data: dfr and dfr2 ## t = -3.41, df = 152.75, p-value = 0.000831 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -4.082997 -1.087471 ## sample estimates: ## mean of x mean of y ## 24.58158 27.16681
Homogeneidad de Var
var.test(dfr, dfr2)
## ## F test to compare two variances ## ## data: dfr and dfr2 ## F = 0.2951, num df = 99, denom df = 99, p-value = 4.025e-09 ## alternative hypothesis: true ratio of variances is not equal to 1 ## 95 percent confidence interval: ## 0.1985531 0.4385821 ## sample estimates: ## ratio of variances ## 0.2950963
Suponiendo var desiguales
t.test(dfr, dfr2, var.equal=F)
## ## Welch Two Sample t-test ## ## data: dfr and dfr2 ## t = -3.41, df = 152.75, p-value = 0.000831 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -4.082997 -1.087471 ## sample estimates: ## mean of x mean of y ## 24.58158 27.16681
Suponiendo var iguales
t.test(dfr, dfr2, var.equal=T)
## ## Two Sample t-test ## ## data: dfr and dfr2 ## t = -3.41, df = 198, p-value = 0.0007871 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -4.080269 -1.090199 ## sample estimates: ## mean of x mean of y ## 24.58158 27.16681
Kolmogorov
ks.test(dfr, dfr2)
## ## Two-sample Kolmogorov-Smirnov test ## ## data: dfr and dfr2 ## D = 0.3, p-value = 0.0002468 ## alternative hypothesis: two-sided
Chi-cuadrado
H0: Independiente /No asociacion /Son Homogenea / No interaccion
H1:Dependiente / Asociacion / No son Homogenea / Interaccion.
frec<-c(15,19, 22) chisq.test(frec)
## ## Chi-squared test for given probabilities ## ## data: frec ## X-squared = 1.3214, df = 2, p-value = 0.5165
qchisq(0.95,2)
## [1] 5.991465
chisq.test(frec)
## ## Chi-squared test for given probabilities ## ## data: frec ## X-squared = 1.3214, df = 2, p-value = 0.5165
chisq.test(frec)$expected
## [1] 18.66667 18.66667 18.66667
habitat1<-c(3,6,8) habitat2<-c(3,12,5) habt<-data.frame(habitat1,habitat2) habt
## habitat1 habitat2 ## 1 3 3 ## 2 6 12 ## 3 8 5
rownames(habt)<-c("machos","hembras", "no_sexados")
habt
## habitat1 habitat2 ## machos 3 3 ## hembras 6 12 ## no_sexados 8 5
chisq.test(habt)
## Warning in chisq.test(habt): Chi-squared approximation may be incorrect
## ## Pearson's Chi-squared test ## ## data: habt ## X-squared = 2.4653, df = 2, p-value = 0.2915
prop.table(habt)
## habitat1 habitat2 ## machos 0.08108108 0.08108108 ## hembras 0.16216216 0.32432432 ## no_sexados 0.21621622 0.13513514
fisher.test(habt,simulate.p.value=TRUE)
## ## Fisher's Exact Test for Count Data with simulated p-value (based ## on 2000 replicates) ## ## data: habt ## p-value = 0.3018 ## alternative hypothesis: two.sided
mosaicplot(habt, color=TRUE, main="Plot de mosaico")
chisq.test(c(28,49,27), p=c(1/4,2/4,1/4))
## ## Chi-squared test for given probabilities ## ## data: c(28, 49, 27) ## X-squared = 0.36538, df = 2, p-value = 0.833
pro<-chisq.test(c(28,49,27), p=c(1/4,2/4,1/4)); pro
## ## Chi-squared test for given probabilities ## ## data: c(28, 49, 27) ## X-squared = 0.36538, df = 2, p-value = 0.833
pro$expected
## [1] 26 52 26
Correlacion
head(Orange)
## Tree age circumference ## 1 1 118 30 ## 2 1 484 58 ## 3 1 664 87 ## 4 1 1004 115 ## 5 1 1231 120 ## 6 1 1372 142
cor.test(Orange$age,Orange$circumference,alternative="two.sided", method="pearson")
## ## Pearson's product-moment correlation ## ## data: Orange$age and Orange$circumference ## t = 12.9, df = 33, p-value = 1.932e-14 ## alternative hypothesis: true correlation is not equal to 0 ## 95 percent confidence interval: ## 0.8342364 0.9557955 ## sample estimates: ## cor ## 0.9135189
cor.test(Orange$age,Orange$circumference,alternative="two.sided", method="spearman")
## Warning in cor.test.default(Orange$age, Orange$circumference, alternative = ## "two.sided", : Cannot compute exact p-value with ties
## ## Spearman's rank correlation rho ## ## data: Orange$age and Orange$circumference ## S = 668.09, p-value = 6.712e-14 ## alternative hypothesis: true rho is not equal to 0 ## sample estimates: ## rho ## 0.9064294
cor(Orange[,c("age","circumference")], use="complete.obs")
## age circumference ## age 1.0000000 0.9135189 ## circumference 0.9135189 1.0000000
plot(Orange$age, Orange$circumference)
library(corrplot) M <- cor(mtcars)
corrplot(M, method = "circle")
corrplot(M, method = "ellipse")
