set.seed(5)
##### Simuler
##### "n" valeurs de deux distributions normales: "M" et "N"
n <- 1000
M <- rnorm(n, mean = 230, sd = 1)
N <- rnorm(n, mean = 230, sd = 8)
##### Correlation et Covariance
cor.test(M, N, method = "pearson")
##
## Pearson's product-moment correlation
##
## data: M and N
## t = 0.57364, df = 998, p-value = 0.5663
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.04388724 0.08005821
## sample estimates:
## cor
## 0.01815523
cor.test(M, N, method = "kendall")
##
## Kendall's rank correlation tau
##
## data: M and N
## z = 0.93451, p-value = 0.35
## alternative hypothesis: true tau is not equal to 0
## sample estimates:
## tau
## 0.01973574
cor.test(M, N, method = "spearman")
##
## Spearman's rank correlation rho
##
## data: M and N
## S = 161922418, p-value = 0.3685
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
## rho
## 0.02846452
cov(M, N)
## [1] 0.1459067
#### Définir la variable aléatoire "M-N" et
### Calculer la "Moyenne" et "Ecart-type"
################## Incertitude 1 ##############################################
SoustractionMN <- (M-N)
Moyenne_SoustractionMN <- mean(SoustractionMN)
EcartType_SoustractionMN <- sd(SoustractionMN)
#################### Incertitude de "M-N" en utilisant la Définition
I_SoustractionMN_1 <- ((1.96*EcartType_SoustractionMN/sqrt(n))
/abs(Moyenne_SoustractionMN))*100
########## Incertitude de "M-N" en utilisant l'Eq. 3.2 , Vol. 1, Chap. 3 (GIEC 2006, GIEC 2019)
################## Incertitude 2 ######################################################
MoyenneM <- mean(M)
MoyenneN <- mean(N)
EcartType_M <- sd(M)
EcartType_N <- sd(N)
I_M <- (1.96 * EcartType_M /sqrt(n)/MoyenneM) * 100
I_N <- (1.96 * EcartType_N /sqrt(n)/MoyenneN) * 100
I_SoustractionMN_2 <- sqrt((I_M * MoyenneM)^2 + (I_N * MoyenneN)^2)/abs(MoyenneM - MoyenneN)
########## Incertitude de "M-N"en utilisant l'Eq. 3.2 , Vol. 1, Chap. 3 (GIEC 2006, GIEC 2019)
################## Incertitude 3 #####################################################
MoyenneM <- mean(M)
MoyenneN <- mean(N)
EcartType_M <- sd(M)
EcartType_N <- sd(N)
I_SoustractionMN_3 <- sqrt((I_M * MoyenneM)^2 + (I_N * MoyenneN)^2)/abs(MoyenneM + MoyenneN)
### Imprimer les résultats ###
print(I_SoustractionMN_1)
## [1] 87.28667
print(I_SoustractionMN_2)
## [1] 87.48608
print(I_SoustractionMN_3)
## [1] 0.1077246
############ Afficer quelques graphiques ################################
par(mfrow=c(2,2))
par(mar = c(2, 2, 2, 2)) # Set the margin on all sides to 2
hist(M, main = "Histogramme de M", ylab = "Fréquence", col = "blue")
hist(N, main = "Histogramme de N", ylab = "Fréquence", col = "blue")
hist(M-N, main = "Histogramme de M-N", ylab = "Fréquence", col = "blue")
hist(M+N, main = "Histogramme de M+N", ylab = "Fréquence", col = "blue")
######## Fin ############################################################