77

6

Multivariate

Test 6.1 Multivariate—Single 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 dened 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−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−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.

Dening a multivariate noninferiority region presents some difculties.

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 satises the univariate denitions for noninfe-

riority, but the multivariate criterion is not satised.

Condence interval formulation:

In a multivariate situation, the condence region is a k-dimensional object

and its interior. In the case of mean vectors, it is an ellipsoid together with its

interior. Specically, it is the set of all vectors, μ, such that

Tn

kn

nk

F

ˆ

ˆ

ˆ

(1)

T

kn k

21

,,

()()

=µ−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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