x_bar = c(4.64, 45.4, 9.96)
S = matrix(c(2.88, 10.01, -1.81,
10.01, 199.79, -5.64,
-1.81, -5.64, 3.63), ncol=3)
miu_0 = c(4, 50, 10)
\[H_{0}:\mu_{0}=\begin{bmatrix} 4\\ 50\\ 10 \end{bmatrix}\]
\[H_{1}:\mu_{0}\neq \begin{bmatrix} 4\\ 50\\ 10 \end{bmatrix}\]
S.inv = solve(S); S.inv
## [,1] [,2] [,3]
## [1,] 0.58606819 -0.022082652 0.257916605
## [2,] -0.02208265 0.006066921 -0.001584619
## [3,] 0.25791660 -0.001584619 0.401623087
n = 20
p = 3
# Misalkan d = (x_bar - miu_0)
d = x_bar - miu_0
# Mencari T.square
T.square = n*t(d)%*%S.inv%*%d; T.square
## [,1]
## [1,] 9.706127
F.tab = 2.44 #diketahui dari soal
c.square = (n-1)*p/(n-p)*F.tab; c.square
## [1] 8.181176
Kesimpulan : Tolak \(H_0\) pada taraf nyata 10% karena \(T^2\) > \(c^2\)
Artinya : Dengan tingkat kepercayaan 90%, dapat
disimpulkan bahwa minimal ada salah satu wainta dengan rata-rata kadar
gula, garam, dan potasium yang tidak sama dengan 4,
50, atau 10.
X = matrix(c(6,10,8,9,6,3), nrow=3, ncol=2)
X
## [,1] [,2]
## [1,] 6 9
## [2,] 10 6
## [3,] 8 3
xbar = apply(X[,1:2], 2, mean)
xbar
## [1] 8 6
cov_m = cov(X[,1:2])
cov_m
## [,1] [,2]
## [1,] 4 -3
## [2,] -3 9
\[H_{0}:\mu_{0}=\begin{bmatrix} 9\\ 5 \end{bmatrix}\]
\[H_{1}:\mu_{0}\neq \begin{bmatrix} 9\\ 5 \end{bmatrix}\]
library(MVTests)
##
## Attaching package: 'MVTests'
## The following object is masked from 'package:datasets':
##
## iris
mean0 = c(9,5)
result = OneSampleHT2(X[,1:2],mu0=mean0,alpha=0.1)
summary(result)
## One Sample Hotelling T Square Test
##
## Hotelling T Sqaure Statistic = 0.7777778
## F value = 0.194 , df1 = 2 , df2 = 1 , p-value: 0.849
##
## Descriptive Statistics
##
## [,1] [,2]
## N 3 3
## Means 8 6
## Sd 2 3
##
##
## Detection important variable(s)
##
## Lower Upper Mu0 Important Variables?
## 1 -8.248077 24.24808 9 FALSE
## 2 -18.372115 30.37212 5 FALSE
library(ICSNP)
## Loading required package: mvtnorm
## Loading required package: ICS
test = HotellingsT2(X[,1:2],mu=mean0,test="f")
print(test)
##
## Hotelling's one sample T2-test
##
## data: X[, 1:2]
## T.2 = 0.19444, df1 = 2, df2 = 1, p-value = 0.8485
## alternative hypothesis: true location is not equal to c(9,5)
Kesimpulan : Tidak tolak \(H_0\)
pada taraf nyata 10% karena \(p-value = 0.8845
> alpha = 0.1\)
Artinya : Dengan tingkat kepercayaan 90%, dapat disimpulkan bahwa
belum cukup bukti untuk mengatakan minimal ada salah
satu dari data sampel acak berukuran \(n=3\) yang memiliki rata-rata tidak sama
dengan 9 atau 5.