Stat 557
Fall 2000
Assignment 4 Solutions
Problem 1
Consider the vector of sample proportions
2
3
p
11
6
7
6
7
6
p
12 7
6
p = 66 777 =
4 p21 5
p22
2
3
Y
11
6
7
6
6
1 6 Y12 777
n 664 Y21 775
:
Y22
From the Multivariate Central Limit Theorem, we have
pn(p ; ) _ N (0; ; 0 )
for large samples. Here
log() = g() = log(11 + 12 ) ; log(11 + 21 ) ; log(21 + 22 ) + log(12 + 22 )
and the vector of rst partial derivatives of g() with respect to the elements of is
"
; +1
G = 1+
1++1
#
;2+ ; +1
1+ + +2
1+ +2
;2+ + +2 :
2+ +1
2+ +2
Then, from the delta method, the large sample normal distribution of log(^) can be expressed
as
log(^) _ N log(); n;1 G ( ; 0)G0 :
Problem 2
(a) From the Multivariate Central Limit Theorem, we have
p _ N ; n;1 ; 0
for large samples. Here g() = A, and by the delta method
Ap _ N A; n;1A( ; 0)A0 ;
and
AY _ N (nA; nA( ; 0)A0) :
1
Note that
A(
2
1+
6
6
; 0 )A0 = 64 0
0
(b) The required matrix is
B
2
6
= 664
0
0
0
2+
0 3+
3 2
3
1+ 7
7
6
7
6
7 ; 6 2+ 7
7 [1+
5 4
5
2+
3+
3+ ] :
3
1 1 1 0 0 0 0 0 07
0 0 0 1 1 1 0 0 0 775
0 0 0 0 0 0 1 1 1
(c) For large samples
(A ; B )p _ N (A ; B); n;1(A ; B)( ; 0)(A ; B)0 :
and
(A ; B )Y _ N (n(A ; B); n(A ; B)( ; 0)(A ; B)0) :
Note that when the null hypothesis of marginal homogeneity is true, then (A;B ) = 0,
and
(A ; B )p _ N 0; n;1(A ; B)( )(A ; B)0 :
and
(A ; B )Y _ N (0; n(A ; B)( )(A ; B)0) :
(d) Test the null hypothesis of marginal homogeneity
Ho : (A ; B )
2
3 2
3
1+ 7 6 +1 7
6
6
= 64 2+ 775 ; 664 +2 775 = 0
Ha : (A ; B )
2
3 2
3
1+
+1
6
7 6
7
= 664 2+ 775 ; 664 +2 775 6= 0
against
3+
3+
+3
+3
:
Using a Type I error level of , reject the null hypothesis if
Xa2 = np0 (A ; B)0 ((A ; B)(p ; pp0 )(A ; B)0) (A ; B)p
= n;1 Y0(A ; B)0 ((A ; B)(p ; pp0)(A ; B)0) (A ; B)Y
2
exceeds the upper percentile of a central chi-square distribution with 2 degrees of
freedom.
Since we assume the null hypothesis is true in computing a p-value for a test, we
could use a consistent estimator of the covariance matrix in the asymptotic normal
distribution for (A ; B )p or (A ; B )Y when the null hypothesis is true to construct
a Wald statistic. Then, using a Type I error level of , reject the null hypothesis if
Xo2 = np0 (A ; B)0 ((A ; B)p(A ; B)0) (A ; B)p
= n;1Y0(A ; B)0 ((A ; B)p (A ; B)0) (A ; B)Y
exceeds the upper percentile of a central chi-square distribution with 2 degrees of
freedom. This approach maintains a true Type I error level closer to the nominal level, because the simplication of the covariance matrix obtained by assuming the null
hypothesis is true generally reduces variability in the covariance estimator. Although
both Xo2 and Xa2 tend to have inated true Type I error levels, the true Type I error
level will tend to be slightly more inated for Xa2. Of course, the true Type I error
level for both Xo2 and Xa2 tends to be closer to the nominal level for larger sample
sizes.
(e) For the data on physical and psychological demands of work reported by females in
Denmark, Xo2 = 72:161 and Xa2 = 75:725, both with 2 degrees of freedom and pvalues that are essentially zero.. The null hypothesis of marginal homogeneity can be
rejected. There is a greater tendency for females to nd work physically demanding
than psychologically demanding.
Problem 3
(a) Use
C
2
6
= 664
0 1 0 ;1 0 0 0 0 0
0 0 1 0 0 0 ;1 0 0
0 0 0 0 0 1 0 ;1 0
3
7
7
7
5
Using a Type I error level of , reject the null hypothesis of symmetry if
Xa2 = np0 C0 (C(p ; pp0 )C0);1 Cp
3
A Wald test with a Type I error level closer to the nominal level is obtained by
realizing that C = 0 when the null hypotheis is true. The the null hypothesis of
symmetry is rejected if
Xo2 = np0C0 (Cp C0);1 Cp
exceeds the upper percentile of a central chi-square distribution with 3 degrees of
freedom. Here p = n;1Y is the vector of sample proportions. Applying this test to
the Danish female reports on physical and psychological demands of work, we have
Xo2 = 75:126, on 3 degrees of freedom, with a p-value that is essentially zero. The
symmetry hypothesis is rejected. Although p23 = p32 , both p12 > p21 and p13 > p31 .
There is a tendency for Danish females to encounter the situation where work is usually
physically demanding and not often psychologically demanding more often than the
opposite situation where work is usually found to be psychologically demanding but
not often physically demanding.
(b) Yes. The null hypothesis of symmetry implies the null hypotheis of marginal homogeneity. When the null hypothesis of marginal homogeneity was rejected, it followed
that the symmetry hypothesis must also be incorrect.
4