EJERCICIOS

1. MATRIZ DE COVARIANZA DESCONOCIDA

Información del problema

data <- matrix(c(6,9,10,6,8,3),nrow=3,byrow=T)
data
##      [,1] [,2]
## [1,]    6    9
## [2,]   10    6
## [3,]    8    3
n        <- 3
p        <- 2

Vector de medias específico a probar

mu0   <- as.vector(c(9,5))
mu0
## [1] 9 5

Vector de medias muestral

xbar  <- as.vector(apply(data,2,mean))
xbar
## [1] 8 6

Matriz de covarianza muestral

S <- cov(data)
S
##      [,1] [,2]
## [1,]    4   -3
## [2,]   -3    9

α=0.01, calculamos el valor crítico:

a     <- 0.01
Fc    <- qf(1-a,p,n-p)
round(Fc,2)
## [1] 4999.5

Calculamos el estadístico de prueba:

T0   <- n*t((xbar-mu0)) %*% solve(S) %*% (xbar-mu0)
F0   <- ((n-p)/(p*(n-1)))*T0
round(F0,2)
##      [,1]
## [1,] 0.19

Como F0=4999.5 ≥ Fc=0.64, rechazamos la hipótesis. # 2. MATRIZ DE COVARIANZA DESCONOCIDA

Información del problema

data2 <- matrix(c(2,12,8,9,6,9,8,10),nrow=4,byrow=T)
data2
##      [,1] [,2]
## [1,]    2   12
## [2,]    8    9
## [3,]    6    9
## [4,]    8   10
n2        <- 4
p2        <- 2

Vector de medias específico a probar

mu02   <- as.vector(c(7,11))
mu02
## [1]  7 11

###Vector de medias muestral

xbar2  <- as.vector(apply(data2,2,mean))
xbar2
## [1]  6 10

Matriz de covarianza muestral

S2 <- cov(data2)
S2
##           [,1]      [,2]
## [1,]  8.000000 -3.333333
## [2,] -3.333333  2.000000

α=0.03, calculamos el valor crítico:

a2     <- 0.03
Fc2    <- qf(1-a2,p2,n2-p2)
round(Fc2,2)
## [1] 32.33

Calculamos el estadístico de prueba:

T02   <- n2*t((xbar2-mu02)) %*% solve(S2) %*% (xbar2-mu02)
F02   <- ((n2-p2)/(p2*(n2-1)))*T02
round(F02,2)
##      [,1]
## [1,] 4.55

Como F0 = 4.55 < Fc = 32.33, no rechazamos la hipótesis H0, no tenemos evidencia significativa para rechazar que M0 no es (7,11).

3. MATRIZ DE COVARIANZA DESCONOCIDA

Informacion

n3 <- 20
p3 <- 3

Datos

data3 <- matrix(c(3.7,  48.5,   9.3, 5.7, 65.1, 8, 3.8, 47.2, 10.9, 3.2, 53.2, 12, 3.1, 55.5, 9.7,
4.6,    36.1,   7.9,2.4,    24.8,   14, 7.2,    33.1,   7.6,
6.7,    47.4,   8.5,5.4,    54.1,   11.3, 3.9,  36.9,   12.7, 4.5,  58.8,   12.3, 3.5,  27.8,   9.8,
4.5,40.2,   8.4, 1.5,   13.5,   10.1, 8.5,  56.4,   7.1, 4.5,   71.6,   8.2, 6.5,   52.8,   10.9, 4.1,  44.1,   11.2, 5.5,  40.9,   9.4),nrow=20,byrow=T)
data3
##       [,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

Vector de medias específico a probar

mu03   <- as.vector(c(4,50,10))
mu03
## [1]  4 50 10

Vector de medias

xbar3  <- as.vector(apply(data3,2,mean))
xbar3
## [1]  4.640 45.400  9.965

Matriz de covarianza

S3 <- cov(data3)
S3
##           [,1]     [,2]      [,3]
## [1,]  2.879368  10.0100 -1.809053
## [2,] 10.010000 199.7884 -5.640000
## [3,] -1.809053  -5.6400  3.627658

Suponiendo que α=0.03, calculamos el valor crítico:

a3    <- 0.03
Fc3    <- qf(1-a3,p3,n3-p3)
round(Fc3,2)
## [1] 3.79

Calculamos el estadístico de prueba:

T03   <- n3*t((xbar3-mu03)) %*% solve(S3) %*% (xbar3-mu03)
F03   <- ((n3-p3)/(p3*(n3-1)))*T03
round(F03,2)
##      [,1]
## [1,]  2.9

Como F0 = 2.9 < Fc = 3.79, No rechazamos la hipótesis nula. Es decir, que en efecto el vector de medias de las variables de la transpiracion de mujeres saludables es igual a Mu0 (4,50,10). .

4. MATRIZ DE COVARIANZA CONOCIDA

Información

n4a    <- 15
n4b    <- 12
p4     <- 7

Vectores de medias muestrales

xbar4a <- as.vector(c(168.78,   63.89, 38.98,   73.46,  45.85,  57.24,  43.09))
xbar4b <- as.vector(c(177.58,   74.25,  41.67,  77.75,  49.00,  58.00,  45.62))

Matrices de covarianza muestrales

S4a     <- matrix(c(
37.64,  22.1,   6.38,   15.65,  9.49,   2.75,   9.02,     22.1,   80.4, 7.36,   12.94, 14.39,    7.2,   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,      7.2, 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)
S4a
##       [,1]  [,2] [,3]  [,4]   [,5]  [,6] [,7]
## [1,] 37.64 22.10 6.38 15.65  9.490 2.750 9.02
## [2,] 22.10 80.40 7.36 12.94 14.390 7.200 9.31
## [3,]  6.38  7.36 1.92  3.06  1.490 0.760 1.98
## [4,] 15.65 12.94 3.06  7.41  3.990 1.170 4.53
## [5,]  9.49 14.39 1.49  3.99  9.420 2.559 1.12
## [6,]  2.75  7.20 0.76  1.17  2.559 2.940 0.95
## [7,]  9.02  9.31 1.98  4.53  1.120 0.950 3.78
S4b     <- matrix(c(45.53,  48.84,  9.48,   14.34,  14.86,  9.45,   8.92, 48.84,    74.2,   9.63,   19.34,  19.77,  9.9,    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, 9.9,    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)
S4b
##       [,1]  [,2] [,3]  [,4]  [,5] [,6] [,7]
## [1,] 45.53 48.84 9.48 14.34 14.86 9.45 8.92
## [2,] 48.84 74.20 9.63 19.34 19.77 9.90 5.23
## [3,]  9.48  9.63 2.79  2.09  3.23 1.86 2.31
## [4,] 14.34 19.34 2.09 12.57  6.18 2.36 1.21
## [5,] 14.86 19.77 3.23  6.18  6.77 3.02 1.84
## [6,]  9.45  9.90 1.86  2.36  3.02 3.13 2.63
## [7,]  8.92  5.23 2.31  1.21  1.84 2.63 6.14

Calculamos la matriz de covarianza conjunta.

Sp4     <- ((n4a-1)*S4a + (n4b-1)*S4b)/(n4a+n4b-2)
Sp4
##         [,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     <- xbar4a - xbar4b
T04     <- ((n4a*n4b)/(n4a+n4b))*t(dif)%*%solve(Sp4)%*%dif
T04
##          [,1]
## [1,] 26.08934

Entonces, el valor de prueba es:

F04     <- ((n4a+n4b-2-p4)/(p4*(n4a+n4b-2-1)))*T04
round(F04,2)
##      [,1]
## [1,]  2.8

Suponiendo que α=0.05, calculamos el valor crítico:

a4     <- 0.06
Fc4    <- qf(1-a4,p4,(n4a+n4b-2-p4))
round(Fc4,2)
## [1] 2.44

Como F0=2.80 ≥ Fc=2.58,cae en la zona de rechazo por lo que rechazamos la hipótesis nula. Es decir, no existe diferencias detectables entre las variables/caracteristicas de mujeres y hombres.