Comparing Mean Vectors for Two Populations
• Independent random samples, one sample from each of two
populations
• Randomized experiment: n1 units are randomly allocated to
treatment 1 and n2 units are randomly allocated to treatment
2. Sample sizes need not be equal.
• Measure p outcomes (or variables or traits) on each unit
329
Comparing Mean Vectors for Two Populations
• Data vectors from population 1:


x1j = 


x1j1
x1j2
...
x1jn1





j = 1, 2, . . . , n1
• Data vectors from population 2:


x2j = 


x2j1
x2j2
...
x2jn2





j = 1, 2, . . . , n2
• We use x̄1 and x̄2 to denote the sample mean vectors, and
S1 and S2 to denote the estimated covariance matrices.
330
Comparing two mean vectors (cont’d)
• If n1, n2 are large, the following assumptions are all we need
to make inferences about the difference between treatments
µ1 − µ2 :
1. X11, X12, ..., X1n1 ∼ p-variate distribution(µ1, Σ1).
2. X21, X22, ..., X2n2 ∼ p-variate distribution(µ2, Σ2).
3. X11, X12, ..., X1n1 are independent
4. X21, X22, ..., X2n2 are independent.
5. X11, X12, ..., X1n1 are independent of X21, X22, ..., X2n2 .
331
Comparing two mean vectors (cont’d)
• When samples sizes are small, we also need that both
distributions are multivariate normal.
• Will will first examine the situation where the population
covariance matrices are the same: Σ1 = Σ2. This is a strong
assumption because it says that all p(p + 1)/2 variances and
covariances are the same in both populations.
332
Pooled estimate of the covariance matrix
• If Σ1 = Σ2 = Σ, then
n1
X
(x1j − x̄1)(x1j − x̄1)0,
j=1
n2
X
(x2j − x̄2)(x2j − x̄2)0
j=1
are estimates of (n1 − 1)Σ and of (n2 − 1)Σ, respectively.
Then, we can pool or average the information from the two
samples to obtain an estimate of the common covariance
matrix:
Spool =
=
Pn1
0 + Pn2 (x − x̄ )(x − x̄ )0
(x
−
x̄
)(x
−
x̄
)
1
1
2
2
1j
2j
j=1 1j
j=1 2j
n1 + n2 − 2
(n1 − 1)
(n2 − 1)
S1 +
S2 .
(n1 + n2 − 2)
(n1 + n2 − 2)
333
Testing hypotheses
• Consider testing H0 : µ1 − µ2 = δ0, where δ0 is some fixed
p × 1 vector. Often, δ0 = 0.
• An estimate of µ1 − µ2 is X̄1 − X̄2 and
1
1
Σ+
Σ,
n1
n2
Since units from treatment 1 are independent of units from
treatment 2, Cov(X̄1, X̄2) = 0.
Cov(X̄1 − X̄2) = Cov(X̄1) + Cov(X̄2) =
• An estimate of the covariance matrix of the difference
between the sample mean vectors is given by
1
1
Spool +
Spool =
n1
n2
!
1
1
+
Spool.
n1
n2
334
Testing hypotheses (cont’d)
• We reject H0 : µ1 − µ2 = δ0 at level α if
T 2 = (x̄1 − x̄2 − δ0)0
"
!
1
1
+
Spool
n1
n2
#−1
(x̄1 − x̄2 − δ0) > c2,
where
c2 =
(n1 + n2 − 2)p
Fp,n1+n2−p−1(α).
(n1 + n2 − p − 1)
• A 100(1 − α)% CR for µ1 − µ2 is given by all values of µ1 − µ2
that satisfy
(X̄1−X̄2−(µ1−µ2))0
"
!
1
1
+
Spool
n1
n2
#−1
(X̄1−X̄2−(µ1−µ2)) ≤ c2.
335
Example 6.3: Two Processes for Manufacturing
Soap
• Objective was to compare two processes for manufacturing
soap. Outcome measures were X1 = lather and X2 = mildness, and n1 = n2 = 50.
• Sample statistics for sample 1 were
"
x̄1 =
8.3
4.1
#
"
, S1 =
2 1
1 6
#
,
and for sample 2:
"
x̄2 =
10.2
3.9
#
"
, S2 =
2 1
1 4
#
.
336
Example 6.3: Two methods for manufacturing
soap
• The pooled estimate of the common covariance matrix and
the difference in sample mean vectors are
49
49
S1 +
S2 =
Spool =
98
98
"
2 1
1 5
#
"
,
x̄1 − x̄2 =
−1.9
0.2
#
.
• We reject H0 : µ1 − µ2 = 0 at level α = .05 because
T 2 = (x̄1 − x̄2 − 0)0
"
!
1
1
+
Spool
n1
n2
#−1
(x̄1 − x̄2 − 0) = 15.66
is larger than
(50 + 50 − 2)(2)
F2,97(.05) = 6.26.
(50 + 50 − 2 − 1)
337
Example 6.3: Two methods for manufacturing
soap
• Eigenvalues and eigenvectors of the pooled covariance matrix
are given by
"
λ=
5.303
1.697
#
"
,
E = [e1, e2] =
0.290
0.957
0.957 −0.290
#
.
338
Two methods for manufacturing soap (cont’d)
• A 95% confidence ellipse for the difference between the two
means is centered at x̄1 − x̄2.
• Since
!
1
1
+
c2 =
n1
n2
1
1
+
n1
n2
!
(n1 + n2 − 2)p
Fp,n1+n2−p−1(α)
(n1 + n2 − p − 1)
1
1 (98)(2)
=
+
F2,97(0.05) = 0.25,
50
50
(97)
√
√
we know that the ellipse extends 5.303 0.25 = 1.15 and
√
√
1.697 0.25 = 0.65 units in the e1 and e2 directions,
respectively.
339
Two methods for manufacturing soap (cont’d)
• Since µ1 − µ2 = 0 is not inside the ellipse, we conclude that
the two processes produce different results. There appears
to be no big difference in mildness (X2), but soaps made
with the second process produce more lather.
340
Confidence Intervals
• As before, we can obtain simultaneous confidence intervals
for any linear combination of the components of µ1 − µ2.
• In particular, in the case of p variables, we might be interested
in

a0i(µ1 − µ2) =
h
i

0 0 ··· 1 ··· 0 

µ11 − µ21
µ12 − µ22
...
µ1p − µ2p





= µ1i − µ2i,
where the vector ai has zeros everywhere except for the one
in the ith position.
• Typically, we would be interested in p such comparisons.
341
Confidence Intervals (cont’d)
• In general,
s
a0(X̄1−X̄2)±
(n1 + n2 − 2)p
Fp,n1+n2−p−1
(n1 + n2 − p − 1)
v
u
u
(α)ta0
!
1
1
+
Spoola
n1
n2
will simultaneously cover the true values of a0(µ1 − µ2) with
probability of at least (1 − α)%.
• One-at-a-time t intervals would be computed as
a0(X̄1 − X̄2) ± tn1+n2−2
v
u
u
(α/2)ta0
!
1
1
+
Spoola
n1
n2
and have less than (1 − α)% simultaneous probability of
coverage unless only one comparison is made. To apply the
Bonferroni method, divide α by the number m of comparisons
of interest.
342
Heterogeneous covariance matrices
• Life gets difficult when we cannot assume that Σ1 = Σ2.
• We must modify the standardized distance measure, and the
modification will not exacty be a multiple of an F-distibution
when the null hypothesis of equal mean vectors is true.
• How to decide whether the assumption of equal covariance
matrices is reasonable? Tests such as Bartlett’s test are
sensitive to departures from normality.
• A crude rule of thumb is the following: if σ1,ii ≥ 4σ2,ii or
σ2,ii ≥ 4σ1,ii, then it is likely that Σ1 6= Σ2.
343
Bartlett’s test for equality of k covariance
matrices
• Tests for testing homogeneity of variances are touchy in that
they are sensitive to the assumption of multivariate normality.
They tend to reject homogeneity of covariance matrices too
often when the samples are not selected from multivariate
normal distributions.
• Bartlett’s test is a likelihod ratio test (independent random
samples from multivariate normal distributions) for testing
H0 : Σ1 = Σ2 = ... = Σk = Σ
versus the alternative where at least two Σi are different.
344
Bartlett’s test (cont’d)
• The test-statistic is a generalization of a likelihood ratio
statistic:
M =
k
X
i=1
(ni − 1) ln |Spool| −
k
X
(ni − 1) ln |Si|,
i=1
where |Spool| is the determinant of the pooled estimate of
the covariance matrix, |Si| is the determinant of the sample
covariance matrix for the ith treatment group, and k is the
number of treatments (or populations).
• In this part of the course we focus on the case k = 2 but this
test is more general.
345
Bartlett’s test (cont’d)
• When all k samples come from multivariate normal populations and when ni − p is relatively large for all i = 1, ..., k, it
has been shown that
1
−1
2
MC
∼ χdf , df = (k − 1)(p + 1)p,
2
where the scale factor C −1 is given by
C −1 = 1 −
2p2 + 3p − 1
6(p + 1)(k − 1)

X


1
1
.
−P
i (ni − 1)
i (ni − 1)
• The null hypothesis H0 : Σ1 = Σ2 = ... = Σk = Σ is rejected
at level α if M C −1 ≥ χ2
df (α), with degrees of freedom as
defined above.
346
Example: soap manufacturing
• We test the hypothesis H0 : Σ1 = Σ2 = Σ. Here, k = 2 and
p = 2.
• From earlier results, we have:
|Spool| = 9,
|S1| = 11,
|S2| = 7.
• Then
M = 98 × ln(9) − 49 × ln(11) − 49 × ln(7) = 2.4794
1
1
2(22) + 3(2) − 1 1
−1
+
−
= 1 − 0.0221 = 0.9779
C
= 1−
6(3)(1)
49
49 98
df =
1
(1)(3)(2) = 3.
2
347
Example: soap manufacturing (cont’d)
• We reject the null hypothesis if
M C −1 = 2.4794 × 0.9779 = 2.4246 ≥ χ2
3 (0.05) = 7.82.
• In this case, we fail to reject the null hypothesis and
conclude that Σ1 is similar to Σ2. Thus, pooling the
samples to obtain a single estimate of the common
population variance is reasonable.
348
What to do when Σ1 6= Σ2
• Suppose that we reject H0, so that Σ1 6= Σ2. For n1 − p and
n2 − p large, an approximate 100(1 − α)% CR for µ1 − µ2 is
given by all µ1 − µ2 satisfying
[x̄1−x̄2−(µ1−µ2)]0
"
1
1
S1 +
S2
n1
n2
#−1
[x̄1−x̄2−(µ1−µ2)] ≤ χ2
p (α).
• For large samples, an approximate test of H0 : µ1 = µ2 is
obtained by rejecting H0 at level α if
T 2 = [x̄1 − x̄2]0
"
1
1
S1 +
S2
n1
n2
#−1
[x̄1 − x̄2] ≥ χ2
p (α).
349
Heterogeneous covariance matrices (cont’d)
• Note that if n1 = n2 = n, then
1
1
1
S1 +
S2 =
(S1 + S2)
n1
n2
n
(n − 1)S1 + (n − 1)S2 1
1
=
+
n + n −2
n
n
1
1
.
= Spool
+
n
n
• With equal sample sizes, the test statistic is the same as the
statistic for homogeneous covariance matrices. Thus, the
effect of heterogeneous matrices is less pronounced when
sample sizes are equal.
350
Steel Dataset
library(ICSNP)
HotellingsT2(steel[steel$temperature == 1, -1],
steel[steel$temperature == 2, -1])
# Hotelling’s two sample T2-test
#
#data: steel[steel$temperature == 1, -1] and steel[steel$temperature
#T.2 = 10.7603, df1 = 2, df2 = 9, p-value = 0.004106
#alternative hypothesis: true location difference is not equal to
#c(0,0)
#
# So there is evidence that temperature has an effect on yield or
# strength.Actually, both of them individually have are not
# significantly affected.
351