title: “Clase 2” author: “Helian A. Cruz Lopez” date: “2023-02-10” output: html_document
pruebas de hipotesis
problemas 1
se desea comparar dos genotipos de papa con base al rendimiento (biomasa de tuberculos), un ensayo utilizo dos variades (criolla y pastusa) involucrado ciento ochenta plantas de la primera variedad y 200 de la segunda. Los datos de rendimientos de la cosecha se presentan en los siguientes vectores
options(digits = 2)
Criolla = rnorm(n=180, mean=2.8, sd=2.0)
pastusa= rnorm(n=200, mean=3.0, sd=2.1)
Criolla## [1] 1.529 3.696 3.412 3.500 -0.018 2.344 6.183 1.844 0.960 7.278
## [11] 5.728 -0.048 5.824 1.678 1.510 2.806 6.433 1.388 9.295 4.435
## [21] 2.408 2.180 4.403 6.766 3.389 8.102 0.173 -0.398 3.393 1.446
## [31] 2.762 5.677 4.061 0.912 0.532 6.720 2.000 4.004 2.277 -0.248
## [41] 2.409 -1.257 2.130 1.951 0.083 2.406 6.685 0.973 2.969 3.416
## [51] 2.196 0.280 0.296 2.471 3.381 3.424 1.473 -1.471 1.784 0.621
## [61] 2.014 1.567 0.433 2.519 1.554 3.804 4.832 1.050 2.344 2.520
## [71] 7.013 0.978 2.789 2.654 3.747 6.485 5.226 3.672 2.030 4.494
## [81] 5.520 1.401 2.506 2.664 4.774 4.951 3.108 3.335 1.361 2.087
## [91] 4.209 2.968 0.530 2.753 2.419 0.894 -0.173 0.665 3.757 0.933
## [101] 3.783 7.047 1.897 3.233 1.643 1.427 6.848 4.425 6.279 4.027
## [111] 7.303 4.905 4.027 2.908 1.422 3.471 3.835 0.092 3.393 5.771
## [121] 4.705 0.499 2.352 3.022 5.055 5.270 4.566 6.523 8.267 5.180
## [131] 3.765 1.398 1.800 -0.232 4.772 4.245 5.006 0.326 5.320 4.419
## [141] 3.401 3.573 3.558 0.962 1.736 4.087 5.641 1.084 2.735 2.277
## [151] 0.423 2.776 1.467 4.605 1.588 -1.264 4.163 -3.430 0.343 3.000
## [161] 1.923 3.697 3.260 3.891 2.084 2.443 -0.932 6.815 4.591 4.035
## [171] 1.015 1.730 2.770 1.553 3.664 3.837 6.958 2.326 2.031 2.173pastusa## [1] 5.348 -0.129 4.886 4.599 2.882 4.574 2.297 1.282 3.921 0.069
## [11] 2.688 2.693 2.372 5.012 1.785 2.475 2.476 2.478 5.974 1.901
## [21] 5.664 4.729 0.652 7.461 0.362 4.204 1.571 3.301 3.391 2.113
## [31] 3.474 0.850 -0.921 0.404 0.993 3.031 3.305 4.549 2.376 4.595
## [41] 1.757 3.918 2.437 4.134 2.624 7.041 3.635 1.059 4.106 4.199
## [51] 1.801 0.630 4.392 4.153 2.697 3.198 2.583 2.021 -1.091 0.931
## [61] 1.129 1.621 0.307 5.683 2.741 4.130 6.362 1.860 4.088 4.672
## [71] 3.702 2.269 -1.015 7.868 5.496 4.615 -0.360 3.579 3.646 6.840
## [81] 2.954 3.939 2.951 0.982 7.122 -0.772 -1.159 2.049 5.884 4.104
## [91] 5.646 2.994 5.968 2.873 1.599 1.880 4.725 1.224 0.486 3.362
## [101] 2.141 1.860 2.244 5.176 1.278 5.684 4.165 4.243 4.561 2.282
## [111] 1.909 2.108 1.720 5.001 6.283 3.335 2.300 1.101 -0.227 2.843
## [121] 0.757 3.084 1.125 5.357 3.163 1.919 0.141 3.359 1.682 5.250
## [131] 1.578 0.554 -0.395 3.043 3.071 2.863 1.005 2.696 2.711 3.982
## [141] 3.669 1.413 4.861 0.346 3.154 1.825 2.632 2.257 6.670 2.486
## [151] 5.784 6.239 1.814 0.734 2.629 1.413 2.750 5.513 2.535 3.678
## [161] -1.121 -0.110 2.091 6.170 2.059 2.701 0.362 -0.293 1.854 2.131
## [171] 3.702 5.008 3.762 4.465 3.029 0.708 3.384 6.318 2.351 -0.102
## [181] 6.186 -1.874 2.386 1.681 4.196 3.377 2.125 2.433 2.111 2.571
## [191] 3.718 2.344 -0.396 3.588 0.069 2.970 2.946 2.494 3.702 1.048par( mfrow= c(1,2))
hist(Criolla)
abline(v=mean(Criolla), col="red", lwd=3)
hist(pastusa)
abline(v=mean(pastusa), col="red", lwd=3)
par(mfrow=c(1,2))
boxplot(Criolla, xlab="criolla", ylab="rto(kg/planta)")
boxplot(pastusa, xlab="pastusa", ylab="rto(kg/planta)")
summary(Criolla)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.4 1.5 2.8 3.0 4.2 9.3summary(pastusa)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -1.9 1.7 2.7 2.8 4.1 7.9library(psych)## Warning: package 'psych' was built under R version 4.2.2psych:: describe(Criolla)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 180 3 2.1 2.8 2.9 1.9 -3.4 9.3 13 0.27 0.06 0.16psych:: describe(pastusa)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 200 2.8 1.9 2.7 2.8 1.8 -1.9 7.9 9.7 0.15 -0.28 0.14disgresion
medA= 3.5; sdA=0.35
medB= 3.2; sdB=0.20
# ¿Cual seleccinar?
# Coeficiente de variacion cv=100 * sd/mean
cvA=100 * sdA/medA
cvB=100 * sdA/medB
cvA; cvB## [1] 10## [1] 11100* sd(Criolla)/mean(Criolla)## [1] 72100* sd(pastusa)/mean(pastusa)## [1] 68conclusiones desde el analisis descriptivo
Ambos coefficientes de variacion son altos (>20). selecciono la variedad con el menos cv, en este caso la pastusa.
Analisis inferencias a traves de pruebas de hipotesis
\[H_0: \mu_(pastusa) = \mu_{criolla}\\\] \[H_a: \mu_(pastusa) \neq\mu_{criolla}\\\]
prueba student para comparar dos muestras independientes
Modalidad 1: Varianza iguales Modalidad 2: Varianza desiguales
Entonces tengo que comparar las varianzas de las dos variedades
prueba para comparacion de dos varianzas
\[H_0: \sigma^2_{pastusa}=\sigma^2: {criolla}\\ H_0: \sigma^2_{pastusa}\neq\sigma^2_{criolla}\]
var(pastusa)## [1] 3.7var(Criolla)## [1] 4.5vt=var.test(pastusa, Criolla)
vt$p.value## [1] 0.15ifelse(vt$p.value<0.025, "rechazo Ho", "No rechazao Ho")## [1] "No rechazao Ho"## prueba t-student para comparar las dos medias con varianzas igualespt=t.test(pastusa, Criolla, alternative = "t", var.equal = TRUE)
ifelse(pt$p.value <0.025, "rechazo Ho", "No rechazao Ho")## [1] "No rechazao Ho"Conclusion final: Los datos proporcionan evidencia estadistica a favor de la hipotesis, es decir, que estadisticamene que ambas variedades son igual de rentables. Cualquiera de las variedades es igual de buena