77
6
Multivariate
Test 6.1 MultivariateSingle Mean Vector (Two-Sided)
Parameters:
k = number of univariate variables being considered simultaneously
µ
= mean vector (k × 1)
Σ = covariance matrix (k × k)
µ
0
= target mean vector (k × 1)
Δ = tolerable distance between μ and μ
0
, that is, noninferiority is dened as:
µ−µ≤
0
where:
dd d
k1
2
2
22
δ= +++
where:
d
d
d
k
1
2
δ=
is a k × 1 column vector. 1 – β = power to reject the null if
0
µ−
µ>
.
Hypotheses:
µ−µ>H :
00
µ−µ≤H :
10
78 Equivalence and Noninferiority Tests
Data:
X = (n × k) matrix of observations, x
ij
= the ith observation of the jth variable.
µ=
=
x
x
x
ˆ
thesamplemeanvecto
r
k
1
2
.
ˆ
thesamplecovariancematri
x
Σ=
.
Critical value(s):
Let
Tn
ˆ
ˆ
ˆ
T
2
0
1
0
()()
−µ Σµ−µ
.
Then reject H
0
if:
F
nk
nk
TFkn kn
(1)
(, ,
ˆ
)
T2
1
1
=
−δ
Σδ
−β
where
Fknkn(, ,
ˆ
)
T
1
1
−δ
Σδ
−β
is the 100(1 − β) percentile of a noncentral F-distribution with degrees of
freedom k (numerator) and n – k (denominator) and noncentrality parameter
n
ˆ
T 1
δΣ δ
.
Discussion:
The statistic:
Tn
ˆ
ˆ
ˆ
T
2
0
1
0
()()
−µ Σµ−µ
has a distribution known as Hotelling’s T
2
. It can be shown (Anderson, 1958)
that
F
nk
nk
T
(1)
2
=
79Multivariate
has a noncentral F-distribution with degrees of freedom k (numerator) and
n – k (denominator); noncentrality parameter
n
T
0
1
δΣ δ
. ∑ 
0
is the population
covariance matrix, which generally is unknown. Thus, the critical value for
the hypothesis test is based on the sample covariance matrix. Strictly speak-
ing, the critical value should be based in part on the population covariance
matrix. However, since it is generally unknown, substituting the sample
covariance matrix is a reasonable approximation.
Dening a multivariate noninferiority region presents some difculties.
For one, it is possible that every dimension could satisfy the univariate
criteria:
μ
i0
d
i
μ
i
μ
i0
+ d
i
for i = 1, k
and not satisfy the multivariate criterion:
0
µ−µ≤
.
Example:
Suppose the population mean vector is
µ=
µ
µ
=
0.3
0.5
1
2
and
µ=
µ
µ
=
0.4
0.4
0
1,0
2,0
so that:
δ=
=
d
d
0.1
0.1
1
2
.
Furthermore, suppose that for each univariate mean, as well as in a
multivariate sense, the maximum tolerable difference was
Δ = 0.125.
Then
μ
1,0
– d
1
= 0.4 – 0.1 = 0.3 ≤ (μ
1
= 0.3) ≤ μ
1,0
+ d
1
= 0.4 + 0.1 = 0.5
μ
2,0
– d
2
= 0.4 – 0.1 = 0.3 ≤ (μ
2
= 0.5) ≤ μ
2,0
+ d
2
= 0.4 + 0.1 = 0.5.
80 Equivalence and Noninferiority Tests
However,
()()0.14142 0.125
011,0
2
22,0
2
µ−µ= µ−µ+µ−µ≈ >∆=
Thus, each univariate mean satises the univariate denitions for noninfe-
riority, but the multivariate criterion is not satised.
Condence interval formulation:
In a multivariate situation, the condence region is a k-dimensional object
and its interior. In the case of mean vectors, it is an ellipsoid together with its
interior. Specically, it is the set of all vectors, μ, such that
Tn
kn
nk
F
ˆ
ˆ
ˆ
(1)
T
kn k
21
,,
()()
−µ Σµ−µ
α−
where F
α, k, n − k
= the 100(1 − α) percentile of a (central) F-distribution with k
numerator degrees of freedom and n − k denominator degrees of freedom, and
µ=
=
x
x
x
ˆ
thesamplemeanvector
k
1
2
.
Computational considerations:
While it is possible to use JMP scripting language (JSL) or SAS Proc IML to
compute and invert covariance matrices, it is easier to do so in R. Use the R
function cov() to compute the covariance matrix, and the solve() function to
invert the covariance matrix. Recall that in R, a statement of the form:
>x <- c(1, 2, 3)
creates a column vector, not a row vector, called x.
R:
> df1 <- read.table("H:\\Personal Data\\Equivalence &
Noninferiority\\Programs & Output\\d20121109_test_6_1_example.
csv",header = TRUE,sep = ",")
> attach(df1)
> xmat <- as.matrix(df1)
> xmat
X1 X2 X3
[1,] 100.21 33.37 102.22
[2,] 101.22 33.79 100.33
[3,] 97.16 32.32 95.55
[4,] 98.72 33.05 97.60
81Multivariate
[5,] 97.26 32.41 97.01
[6,] 100.71 33.34 101.12
[7,] 101.30 33.75 101.62
[8,] 98.88 32.99 97.98
[9,] 100.25 33.55 99.88
[10,] 97.19 32.19 96.81
[11,] 105.16 35.07 105.63
[12,] 98.30 32.70 96.40
[13,] 100.74 33.38 101.39
[14,] 97.69 32.62 98.26
[15,] 100.02 33.12 101.32
[16,] 101.43 33.94 102.32
[17,] 95.40 31.76 96.44
[18,] 99.85 33.08 99.45
[19,] 97.43 32.43 98.82
[20,] 100.63 33.59 99.84
[21,] 102.00 33.90 101.51
[22,] 97.45 32.30 97.14
[23,] 102.03 34.03 100.69
[24,] 99.79 33.43 99.40
[25,] 98.61 33.00 97.65
[26,] 95.31 31.77 93.97
[27,] 98.34 32.57 99.32
[28,] 97.27 32.49 96.80
[29,] 98.46 32.82 99.50
[30,] 100.42 33.50 98.54
[31,] 99.90 33.39 99.34
[32,] 101.10 33.59 99.98
[33,] 98.40 32.72 96.44
[34,] 100.80 33.73 101.81
[35,] 95.56 31.62 94.11
[36,] 98.63 32.72 100.50
[37,] 99.13 32.80 98.35
[38,] 103.18 34.52 102.33
[39,] 101.08 33.64 100.17
[40,] 102.08 33.93 103.59
> cmat <- cov(xmat)
> cmat
X1 X2 X3
X1 4.578826 1.5819854 4.853564
X2 1.581985 0.5612728 1.654349
X3 4.853564 1.6543490 6.324297
> cmatinv <- solve(cmat)
> cmatinv
X1 X2 X3
X1 10.549891 -25.641264 -1.3890829
X2 -25.641264 70.101515 1.3407270
X3 -1.389083 1.340727 0.8734525
> mu1_est <- mean(X1)
> mu2_est <- mean(X2)

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