Ejercicio 1

H0: M0 = (9,5) Ha: M0 no es igual a (9,5)

# Información del problema
n        <- 3
p        <- 2

# Vector de medias específico a probar
mu0   <- as.vector(c(9,5))
# Vector de medias muestral
xbar  <- as.vector(c(8,6))
# Matriz de covarianza muestral
x<- matrix(c(6,9,10,6,8,3),nrow=3, byrow=T)
x
##      [,1] [,2]
## [1,]    6    9
## [2,]   10    6
## [3,]    8    3
S <- cov(x)
S
##      [,1] [,2]
## [1,]    4   -3
## [2,]   -3    9
T0   <- n*t((xbar-mu0)) %*% solve(S) %*% (xbar-mu0)
F0   <- ((n-p)/(p*(n-1)))*T0
round(F0,2)
##      [,1]
## [1,] 0.19
a     <- 0.01
Fc    <- qf(1-a,p,n-p)
round(Fc,2)
## [1] 4999.5

Como FC > F0 no rechazamos H0 es decir no tenemos evidencia significativa para decir que M0 no es = (9,5)

Ejercico 2

H0: M0 = (7,11) Ha: M0 no es igual a (7,11)

# Información del problema
n        <- 4
p        <- 2

# Vector de medias específico a probar
mu0   <- as.vector(c(7,11))
# Vector de medias muestral
xbar  <- as.vector(c(6,10))
# Matriz de covarianza muestral
x<- matrix(c(2,8,6,8,12,9,9,10),ncol=2)
x
##      [,1] [,2]
## [1,]    2   12
## [2,]    8    9
## [3,]    6    9
## [4,]    8   10
S <- cov(x)
S
##           [,1]      [,2]
## [1,]  8.000000 -3.333333
## [2,] -3.333333  2.000000
T0   <- n*t((xbar-mu0)) %*% solve(S) %*% (xbar-mu0)
F0   <- ((n-p)/(p*(n-1)))*T0
round(F0,2)
##      [,1]
## [1,] 4.55
a     <- 0.03
Fc    <- qf(1-a,p,n-p)
round(Fc,2)
## [1] 32.33

# Como FC > F0, es decir F0 cae dentro de la zona de no rechazo, por ende no rechazamos H0 ya que no tenemos evidencia significativa para decir que M0 no es = (7,11)

Ejercicio 3

H0: M0= (4,50,10) Ha: M0 no es igual a (4,50,10)

# Información
n      <- 20
p      <- 3
a      <- 0.03
# Vector de medias específico a probar
mu0   <- as.vector(c(4,50,10))
# Vector de medias muestral
data1 <- c(3.7,5.7,3.8,3.2,3.1,4.6,2.4,7.2,6.7,5.4,3.9,4.5,3.5,4.5,1.5,8.5,4.5,6.5,4.1,5.5)
data2 <- c(48.5,65.1,47.2,53.2,55.5,36.1,24.8,33.1,47.4,54.1,36.9,58.8,27.8,40.2,13.5,56.4,71.6,52.8,44.1,40.9)
data3 <- c(9.3,8,10.9,12,9.7,7.9,14,7.6,8.5,11.3,12.7,12.3,9.8,8.4,10.1,7.1,8.2,10.9,11.2,9.4)

xbar  <- as.vector(c(4.64,45.4,10))
# Matriz de covarianza muestral
x<- matrix(c(data1,data2,data3),ncol=3)
x
##       [,1] [,2] [,3]
##  [1,]  3.7 48.5  9.3
##  [2,]  5.7 65.1  8.0
##  [3,]  3.8 47.2 10.9
##  [4,]  3.2 53.2 12.0
##  [5,]  3.1 55.5  9.7
##  [6,]  4.6 36.1  7.9
##  [7,]  2.4 24.8 14.0
##  [8,]  7.2 33.1  7.6
##  [9,]  6.7 47.4  8.5
## [10,]  5.4 54.1 11.3
## [11,]  3.9 36.9 12.7
## [12,]  4.5 58.8 12.3
## [13,]  3.5 27.8  9.8
## [14,]  4.5 40.2  8.4
## [15,]  1.5 13.5 10.1
## [16,]  8.5 56.4  7.1
## [17,]  4.5 71.6  8.2
## [18,]  6.5 52.8 10.9
## [19,]  4.1 44.1 11.2
## [20,]  5.5 40.9  9.4
S <- cov(x)
S
##           [,1]     [,2]      [,3]
## [1,]  2.879368  10.0100 -1.809053
## [2,] 10.010000 199.7884 -5.640000
## [3,] -1.809053  -5.6400  3.627658
T0   <- n*t((xbar-mu0)) %*% solve(S) %*% (xbar-mu0)
F0   <- ((n-p)/(p*(n-1)))*T0
round(F0,2)
##      [,1]
## [1,] 2.97
a     <- 0.03
Fc    <- qf(1-a,p,n-p)
round(Fc,2)
## [1] 3.79

# Como F0 cae dentro de la zona de No rechazo de Fc, podemos concluir que no existe evidencia suficiente para decir que el vector de medias no es igual a (4,50,10)

Ejercicio 4

H0 = Existen diferencias detectables entre n1 y n2 Ha = NO existen diferencias detectables entre n1 y n2

# Información
n1     <- 15
n2     <- 12
p      <- 7
# Vectores de medias muestrales
xbar1    <-as.vector(c(168.78,63.89,38.98,73.46,45.85,57.24,43.09))
xbar2    <-as.vector(c(177.58,74.25,41.67,77.75,49.00,58.00,45.62))

# Matrices de covarianza muestrales
S1     <- matrix(c(37.64,22.10,6.38,15.65,9.49,2.75,9.02,22.10,80.4,7.36,12.94,14.39,7.20,9.31,6.38,7.36,1.92,3.06,1.49,0.76,1.98,15.65,12.94,3.06,7.41,3.99,1.17,4.53,9.49,14.39,1.49,3.99,9.42,2.559,1.12,2.75,7.20,0.76,1.17,2.559,2.94,0.95,9.02,9.31,1.98,4.53,1.12,0.95,3.78), nrow = 7, byrow = TRUE)

S2     <- matrix(c(45.53,48.84,9.48,14.34,14.86,9.45,8.92,48.84,74.20,9.63,19.34,19.77,9.90,5.23,9.48,9.63,2.79,2.09,3.23,1.86,2.31,14.34,19.34,2.09,12.57,6.18,2.36,1.21,14.86,19.77,3.23,6.18,6.77,3.02,1.84,9.45,9.90,1.86,2.36,3.02,3.13,2.63,8.92,5.23,2.31,1.21,1.84,2.63,6.14),nrow = 7, byrow = TRUE)

#Calculamos la matriz de covarianza conjunta.
Sp     <- ((n1-1)*S1+(n2-1)*S2)/(n1+n2-2)
Sp
##         [,1]    [,2]   [,3]    [,4]     [,5]    [,6]   [,7]
## [1,] 41.1116 33.8656 7.7440 15.0736 11.85280 5.69800 8.9760
## [2,] 33.8656 77.6720 8.3588 15.7560 16.75720 8.38800 7.5148
## [3,]  7.7440  8.3588 2.3028  2.6332  2.25560 1.24400 2.1252
## [4,] 15.0736 15.7560 2.6332  9.6804  4.95360 1.69360 3.0692
## [5,] 11.8528 16.7572 2.2556  4.9536  8.25400 2.76184 1.4368
## [6,]  5.6980  8.3880 1.2440  1.6936  2.76184 3.02360 1.6892
## [7,]  8.9760  7.5148 2.1252  3.0692  1.43680 1.68920 4.8184
#El estadístico es:
dif    <- xbar1 - xbar2
T0     <-  ((n1*n2)/(n1+n2))*t(dif)%*%solve(Sp)%*%dif
T0
##          [,1]
## [1,] 26.08934
#Entonces, el valor de prueba es:
F0     <- ((n1+n2-2-p)/(p*(n1+n2-2-1)))*T0
round(F0,2)
##      [,1]
## [1,]  2.8
#Suponiendo que 𝛼=0.06, calculamos el valor crítico:
a     <- 0.06
Fc    <- qf(1-a,p,(n1+n2-2-p))
round(Fc,2)
## [1] 2.44

Como F0 > FC rechazamos H0, por ende concluimos que no existen diferencias detectables entre las dos muestras.