Tarea07_JARR_MEMV_AnalysisFactores
Joel Alejandro Rodarte Rivera
2023-12-03
1.- Considera la siguiente matriz de correlaciones de 6 variables para 100 observaciones. Tratar de identificar factores no observables que estén relacionados con estas variables. \(\begin{array}{rrrrrr}1 & 0.83 & 0.81 & 0.8 & 0.71 & 0.54 \\ 0.83 & 1 & 0.87 & 0.62 & 0.59 & 0.58 \\ 0.81 & 0.87 & 1 & 0.63 & 0.37 & 0.3 \\ 0.8 & 0.62 & 0.63 & 1 & 0.49 & 0.3 \\ 0.71 & 0.59 & 0.37 & 0.49 & 1 & 0.34 \\ 0.54 & 0.58 & 0.3 & 0.3 & 0.34 & 1\end{array}\)
2.- Se tiene la tasa de retorno semanal de 5 acciones bursátiles del NYSE. Tratar de identificar factores no observables que estén relacionados con estas variables.
3.- Se colectaron datos sobre la contaminación del aire en cierta ciudad. Tratar de identificar factores no observables que estén relacionados con estas variables.
Ejercicio 1
options(warn=-1)
library(psych)
#Matriz correlaciónes
R <- matrix(c(1,.83,.81,.8,.71,.54,.83,1,.87,.62,.59,.58,.81,.87,1,.63,.37,.3,.8,.62,.63,1,.49,.3,.71,.59,.37,.49,1,.34,.54,.58,.3,.3,.34,1),ncol = 6,byrow = TRUE)
# Ho: La correlación entre cada par de variables es cero
# H1: La correlación entre cada par de variable diferente de cero
cortest.bartlett(R, n=100)## $chisq
## [1] 665.5447
##
## $p.value
## [1] 4.066112e-132
##
## $df
## [1] 15# Se infiere que si hay correlación entre las variables
#Analisis de componentes principales para determinar cual es el sweetspot de cantidad de componentes y por ende de factores que podríamos utilizar
fa.parallel(R,fm = "pa", n.obs=100, ylabel = "Eigenvalues")
## Parallel analysis suggests that the number of factors = 1 and the number of components = 1#Se identifican que para el segundo factor, cumplo con más del 80% de la variabilidad.
#Analisis de factores de matriz sin rotar
acp <- principal(R,nfactors = 2, rotate = "none")
acp## Principal Components Analysis
## Call: principal(r = R, nfactors = 2, rotate = "none")
## Standardized loadings (pattern matrix) based upon correlation matrix
## PC1 PC2 h2 u2 com
## 1 0.97 -0.05 0.94 0.058 1.0
## 2 0.93 0.04 0.86 0.135 1.0
## 3 0.84 -0.32 0.81 0.190 1.3
## 4 0.80 -0.31 0.74 0.263 1.3
## 5 0.71 0.10 0.51 0.485 1.0
## 6 0.60 0.76 0.94 0.061 1.9
##
## PC1 PC2
## SS loadings 4.01 0.79
## Proportion Var 0.67 0.13
## Cumulative Var 0.67 0.80
## Proportion Explained 0.83 0.17
## Cumulative Proportion 0.83 1.00
##
## Mean item complexity = 1.3
## Test of the hypothesis that 2 components are sufficient.
##
## The root mean square of the residuals (RMSR) is 0.09
##
## Fit based upon off diagonal values = 0.98#Analisis de factores de matriz matriz rotarda (Varimax)
acp2 <- principal(R,nfactors = 2, rotate = "varimax")
acp2## Principal Components Analysis
## Call: principal(r = R, nfactors = 2, rotate = "varimax")
## Standardized loadings (pattern matrix) based upon correlation matrix
## RC1 RC2 h2 u2 com
## 1 0.88 0.42 0.94 0.058 1.4
## 2 0.80 0.48 0.86 0.135 1.6
## 3 0.89 0.12 0.81 0.190 1.0
## 4 0.85 0.11 0.74 0.263 1.0
## 5 0.58 0.42 0.51 0.485 1.8
## 6 0.16 0.96 0.94 0.061 1.1
##
## RC1 RC2
## SS loadings 3.29 1.52
## Proportion Var 0.55 0.25
## Cumulative Var 0.55 0.80
## Proportion Explained 0.68 0.32
## Cumulative Proportion 0.68 1.00
##
## Mean item complexity = 1.3
## Test of the hypothesis that 2 components are sufficient.
##
## The root mean square of the residuals (RMSR) is 0.09
##
## Fit based upon off diagonal values = 0.98#Se determina que se puede trabajar con dos factores
#La variable 1,2,3 y 4 se pueden
#La variable 6 iría en otro factor
#Me queda la duda que cuando se utiliza varimax pareciera que la 6 va ir sola. ¿Es esto posible? Esperar resultados de la maestra. Ejercicio 2
P2 <- read.csv(file = 'C:/Users/Joel Rodarte/Desktop/T7_2.csv')
corP2 <- cor(P2)
# Ho: La correlación entre cada par de variables es cero
# H1: La correlación entre cada par de variable diferente de cero
cortest.bartlett(corP2, n=100)## $chisq
## [1] 158.632
##
## $p.value
## [1] 6.207325e-29
##
## $df
## [1] 10# Se infiere que si hay correlación entre las variables
library(MVN)
mvn(P2) # los datos si son normales multivariados## $multivariateNormality
## Test HZ p value MVN
## 1 Henze-Zirkler 0.9423938 0.1490478 YES
##
## $univariateNormality
## Test Variable Statistic p value Normality
## 1 Anderson-Darling Allied.Chemical 0.3032 0.5673 YES
## 2 Anderson-Darling Du.Pont 0.7152 0.0601 YES
## 3 Anderson-Darling Union.Carbide 0.4967 0.2081 YES
## 4 Anderson-Darling Exxon 0.8761 0.0239 NO
## 5 Anderson-Darling Texaco 0.4896 0.2167 YES
##
## $Descriptives
## n Mean Std.Dev Median Min Max
## Allied.Chemical 100 0.00543372 0.04037235 0.0000000 -0.096654 0.122807
## Du.Pont 100 0.00482707 0.03506246 -0.0024895 -0.075758 0.118812
## Union.Carbide 100 0.00565418 0.03944713 0.0000000 -0.091463 0.102616
## Exxon 100 0.00629143 0.02832547 0.0032590 -0.053133 0.088549
## Texaco 100 0.00370852 0.02754518 0.0045765 -0.050505 0.082474
## 25th 75th Skew Kurtosis
## Allied.Chemical -0.01988250 0.03146875 0.2001254 -0.09307684
## Du.Pont -0.01932550 0.02443875 0.5468161 0.34697628
## Union.Carbide -0.01788425 0.02890450 0.1518975 -0.01779615
## Exxon -0.01389625 0.02088050 0.6120059 0.36646575
## Texaco -0.01525350 0.01896225 0.5035896 0.19129955# Analisis de componentes principales para determinar cual es el sweetspot de cantidad de componentes y por ende de factores que podríamos utilizar
# Se determina por analisis de componentes principales que se captura el 80% de la variabilidad con 3 factores
acp_corr <- prcomp(P2)
summary(acp_corr)## Importance of components:
## PC1 PC2 PC3 PC4 PC5
## Standard deviation 0.05996 0.02815 0.02714 0.02255 0.01854
## Proportion of Variance 0.60159 0.13255 0.12322 0.08511 0.05752
## Cumulative Proportion 0.60159 0.73414 0.85737 0.94248 1.00000#Analisis de factores de matriz sin rotar
acp <- principal(corP2,nfactors = 3, rotate = "none")
acp## Principal Components Analysis
## Call: principal(r = corP2, nfactors = 3, rotate = "none")
## Standardized loadings (pattern matrix) based upon correlation matrix
## PC1 PC2 PC3 h2 u2 com
## Allied.Chemical 0.78 -0.22 -0.45 0.86 0.14 1.8
## Du.Pont 0.77 -0.46 0.13 0.82 0.18 1.7
## Union.Carbide 0.79 -0.23 0.25 0.75 0.25 1.4
## Exxon 0.71 0.47 0.40 0.89 0.11 2.4
## Texaco 0.71 0.52 -0.32 0.88 0.12 2.3
##
## PC1 PC2 PC3
## SS loadings 2.86 0.81 0.54
## Proportion Var 0.57 0.16 0.11
## Cumulative Var 0.57 0.73 0.84
## Proportion Explained 0.68 0.19 0.13
## Cumulative Proportion 0.68 0.87 1.00
##
## Mean item complexity = 1.9
## Test of the hypothesis that 3 components are sufficient.
##
## The root mean square of the residuals (RMSR) is 0.1
##
## Fit based upon off diagonal values = 0.96#Analisis de factores de matriz matriz rotarda (Varimax)
acp2 <- principal(corP2,nfactors = 3, rotate = "varimax")
acp2## Principal Components Analysis
## Call: principal(r = corP2, nfactors = 3, rotate = "varimax")
## Standardized loadings (pattern matrix) based upon correlation matrix
## RC1 RC2 RC3 h2 u2 com
## Allied.Chemical 0.59 -0.01 0.72 0.86 0.14 1.9
## Du.Pont 0.88 0.13 0.18 0.82 0.18 1.1
## Union.Carbide 0.77 0.36 0.16 0.75 0.25 1.5
## Exxon 0.27 0.89 0.18 0.89 0.11 1.3
## Texaco 0.07 0.54 0.77 0.88 0.12 1.8
##
## RC1 RC2 RC3
## SS loadings 1.79 1.22 1.20
## Proportion Var 0.36 0.24 0.24
## Cumulative Var 0.36 0.60 0.84
## Proportion Explained 0.43 0.29 0.28
## Cumulative Proportion 0.43 0.72 1.00
##
## Mean item complexity = 1.5
## Test of the hypothesis that 3 components are sufficient.
##
## The root mean square of the residuals (RMSR) is 0.1
##
## Fit based upon off diagonal values = 0.96# Se determina que Du Pont y Union Carbide son el primer factro
# Exxon es el segundo factor
# Allied Chemiocal y Texaco son el tercer factor Ejercicio 3
P3 <- read.csv(file = 'C:/Users/Joel Rodarte/Desktop/T7_3.csv')
corP3 <- cor(P3)
# Ho: La correlación entre cada par de variables es cero
# H1: La correlación entre cada par de variable diferente de cero
cortest.bartlett(corP3, n=42)## $chisq
## [1] 70.52667
##
## $p.value
## [1] 2.892456e-07
##
## $df
## [1] 21# Se infiere que si hay correlación entre las variables
library(MVN)
mvn(P3)## $multivariateNormality
## Test HZ p value MVN
## 1 Henze-Zirkler 0.9828227 0.03504611 NO
##
## $univariateNormality
## Test Variable Statistic p value Normality
## 1 Anderson-Darling wind 0.9891 0.0118 NO
## 2 Anderson-Darling solar.radiation 0.9623 0.0137 NO
## 3 Anderson-Darling CO 2.3631 <0.001 NO
## 4 Anderson-Darling NO 2.1283 <0.001 NO
## 5 Anderson-Darling NO2 0.6975 0.0636 YES
## 6 Anderson-Darling O3 1.2030 0.0034 NO
## 7 Anderson-Darling HC 3.9150 <0.001 NO
##
## $Descriptives
## n Mean Std.Dev Median Min Max 25th 75th Skew
## wind 42 7.500000 1.5811388 8.0 5 10 6.00 8.75 0.03614032
## solar.radiation 42 73.857143 17.3353881 76.5 30 107 68.25 84.75 -0.73323765
## CO 42 4.547619 1.2337209 4.0 2 7 4.00 5.00 0.57793497
## NO 42 2.190476 1.0873574 2.0 1 5 1.00 3.00 0.62983752
## NO2 42 10.047619 3.3709837 9.5 5 21 8.00 12.00 0.98543943
## O3 42 9.404762 5.5658345 8.5 2 25 6.00 11.00 1.13370267
## HC 42 3.095238 0.6917466 3.0 2 5 3.00 3.00 0.31349052
## Kurtosis
## wind -1.10809524
## solar.radiation 0.30901245
## CO -0.25497268
## NO -0.52220171
## NO2 1.22197318
## O3 1.02388767
## HC 0.04343283#Los datos no son normales multivariados
# Analisis de componentes principales para determinar cual es el sweetspot de cantidad de componentes y por ende de factores que podríamos utilizar
# Se determina por analisis de componentes principales que se captura el 80% con solo un componente. Sin emargo para poder hacer diferenciación de variables utilizaré dos dado que con dos se cubren .95%
summary(acp_corr)## Importance of components:
## PC1 PC2 PC3 PC4 PC5
## Standard deviation 0.05996 0.02815 0.02714 0.02255 0.01854
## Proportion of Variance 0.60159 0.13255 0.12322 0.08511 0.05752
## Cumulative Proportion 0.60159 0.73414 0.85737 0.94248 1.00000acp3 <- principal(corP3,nfactors = 2, rotate = "varimax")
acp3## Principal Components Analysis
## Call: principal(r = corP3, nfactors = 2, rotate = "varimax")
## Standardized loadings (pattern matrix) based upon correlation matrix
## RC1 RC2 h2 u2 com
## wind -0.12 -0.47 0.24 0.76 1.1
## solar.radiation -0.08 0.69 0.48 0.52 1.0
## CO 0.70 0.47 0.71 0.29 1.7
## NO 0.76 -0.11 0.60 0.40 1.0
## NO2 0.77 0.22 0.63 0.37 1.2
## O3 0.05 0.83 0.69 0.31 1.0
## HC 0.61 -0.04 0.37 0.63 1.0
##
## RC1 RC2
## SS loadings 2.05 1.67
## Proportion Var 0.29 0.24
## Cumulative Var 0.29 0.53
## Proportion Explained 0.55 0.45
## Cumulative Proportion 0.55 1.00
##
## Mean item complexity = 1.2
## Test of the hypothesis that 2 components are sufficient.
##
## The root mean square of the residuals (RMSR) is 0.15
##
## Fit based upon off diagonal values = 0.68# Se determina que CO,NO,NO2 y HC pertenecen al primer factor.
# Wind, Solar radiation y 03 pertenecen a el segundo factor.