STAT 557
Solutions to Assignment 3
Fall 2002
1. (a) For this multinomial sample relative risk is dened as
RR =
11 =1+
21 =2+
and the natural logarithm of the relative risk is
g() = log( ) log( ) log( ) + log( )
First partial derivatives of g() with respect to the elements of = ( 1
(
) 1
G=
11
1+
21
2+
11
1+
11
11 1+
1+
2+
21 2+
21
12
21
22 )0
are
2+
Using the delta method, the large sample variance of log(RR) is
log RR = G( 0 )G0
= G G0 because G = 0
= + 2
(
)
12
11
Estimates are
^ = Y =Y = 1:932;
RR
Y =Y
11
1+
21
2+
1+
log(^RR) = 0:658
22
21
2+
var(
^ log(^RR)) = Y Y Y + Y Y Y = 0:01221
12
1+
22
11
2+
21
Then, an approximate 95% condence interval for log(RR) is
q
^
^ ) = (0:442; 0:875)
^ RR
log(RR) 1:96 var(log
(b)
and an approximate 95% condence interval for RR is
[exp(0:442); exp(0:875)] = (1:555; 2:399)
The risk of having a major CHD event when working under tension is about 55% to 140% higher than
that when working under no tension.
odds.ratio
stderr lower.limit upper.limit
2.225945 0.1388392
1.695634
2.922111
The odds ratio is about 25% larger than the direct estimate of relative risk, but both estimates indicate
the risk of coronary heart disease is approximately doubled by a stressful work environment.
2. (a) Test results are in the following table.
Female
Male
statistic d.f. p-value statistic d.f. p-value
G 47.8166 4 0.000+ 37.5094 4 0.000+
X 50.2535 4 0.000+ 35.9090 4 0.000+
For both males and females, there appear to be some associations between attitudes toward physical and
psychological demands of employment.
(b)
2
2
1
Gamma
95% condence interval
estimate standard error lower limit upper limit
Females 0.236
0.0385
0.160
0.311
Males
-0.125
0.0327
-0.189
-0.061
The is a signicant positive association between physical and psychological demands of work for females, but
a signicant negative association for males.
3. (a) The rst derivative of = log is
1
1
1 :
@ 1
=
+
=
@ 2 1 + 1 1 By the Æ-method,
@
var ^ @
var(^ ) = (1 1 ) var(^ ):
For females,
^f = 0:240; var ^f = 0:00166:
For males,
^m = 0:126; var ^m = 0:00110:
(b) For females, the approximate 95% condence interval for is
q
q
^ z : var(^); ^ + z : var(^) = (0160; 0:320):
1+
1
1
2
2
2
2 2
0 975
0 975
Since = 1 e2 , an approximate 95% condence interval of is (0:159; 0:310)
Similarly for males, an approximate 95% condence interval for is ( 0:189; 0:061). These closely
match the results
from
problem 2 because the sample sizes are large.
(c) Since = log is a strictly increasing function of , testing the hypothesis
H : f = m vs. Ha : f 6= m
is equivalent to testing the hypothesis
H : f = m vs. Ha : f 6= m :
Consider ^f ^m, the estimator of f m . Under the assumption that the counts for females and males
are independent,
var(^f ^m) = var(^f ) + var(^m ):
When H holds, the distribution of the test statistic Z = pvar ff mvar m is well approximated by a
standard normal distribution when the sample sizes are large enough.
This generally provides a more
m
accurate p-value than approximating the null distribution of pvar ff var
with a standard normal
m
distribution. For these data, Z = 6:96 and the standard normal approximation yields p-value < :0001 The
association between physical and psychological demands of work is not the same for males and females.
The association is positive for females, but weaker and negative for males.
4. (a) Odds that a mother has a college graduate divided by the odds that a father is a college graduate.
(b) Consider a multinomial distribution
2
+1
1
2
1+
1
0
0
^
0
^
( ^ )+
^
(^ )
^
(^ )+
0
B
Y=B
@
Y11
Y12
Y21
Y22
1
C
C
A
0
0
B
B
B
Mult B
@n = 390; = @
2
11
12
21
22
(^
11
CC
CC
AA
)
Note that
log = log + log log log = log( + ) + log( + ) log( + ) log( + )
g()
1+
11
and
Let
+2
2+
12
+1
11
12
21
22
11
21
varfg = varfY=ng = n1 fdiag() 0g V():
h
G( )
= @@g11 @@g12 @@g21
= 11 12 11 21
Then, by the delta method,
1
+
1
+
and it is estimated by
@g
@22
1
i
11 +12
+ 12
1
+ 22
1
21 +22
1
11 +21
1
12 +22
1
21 +22
:
varflog(^)g G()V()G()0
var
^ flog(^)g = = G(^)V(^)G(^)0
= 0:01142:
(c) Note ^ = 1:202966 and log ^ = 0:18479. Then, a 95% condence interval for log() is is given by
p
log ^ 1:96 var(log
^ ^) = ( 0:0246; 0:3942):
(d) (0.9757, 1.4832)
(e) Applying the delta method to the asymptotic normal distribution for log(^) obtained in (b):
log(^) _ N (log(); var(log ^));
we have the large sample normal distribution for ^:
^ _ N (; fvar(log ^)g):
p
^ ^)g = (0:9510; 1:4549):
(f) ^ 1:96 fvar(log
(g) Yes. The distribution for log(^) is more symmetric and thus better approximated by Normal than that
for ^.
(h) As expected, obtained condence intervals are all closer to the condence interval in (d).
5. (a) Table 1: Kappa = -0.125 with 95% condence interval (-0.245, -0.004)
Table 2: Kappa = -0.074 with 95% condence interval (-0.180, 0.033)
There is some evidence, at least in white families, of less agreement between the mother's rating and the
child's rating than one would expect at random.
(b)
q
(K^ K^ ) (1:96) SK1 + SK2
)
( 0:212; 0:110)
Since the condence interval for the dierence in the Kappa values contains zero, the data do not
provide suÆcient evidence to conclude that the levels of agreement are dierent for white and black
mother/daughter pairs. This does not prove, however, the levels of agreement are exactly the same in
white and black families.
Incidentally, if the two tables in part (a) were the results of a single simple random sample, the counts
in Table 1 would have negative correlations with the counts in Table 2. The correlation between the two
Kappa values converges to zero as the sample size increases, however, and the formula shown above is also
an appropriate large sample formula in this second case.
You would not have a simple random sample, for example, if several classes were selected from a school
district and all of the children in the selected classes were used in the study. This would be a one-stage
cluster sample, and standard errors based on an assumption of simple random sampling (used in parts (a)
and (b)) would tend to be too small.
2
2
1
2
2
^
2
^
3
6. (a) Columns 2 and 3 of the following tables contain estimates of odds ratios and 95% condence intervals
where 0.5 was substituted for any observed zero count, but other counts were not changed. Columns 4
and 5 show corresponding values where 0.5 was added to each count in every table. Condence intervals
were constructed by constructing a 95% condence interval for the natural logarithm of each odds ratio
and applying the exponential function to each end of the interval.
Age Estimated 95% condence Estimated 95%condence
Group odds ratio
interval
odds ratio
interval
25-34
23.56
(0.74, 751.23)
33.63 (1.28, 883.73)
35-44
5.05
(1.27, 20.03)
5.08
(1.37, 18.84)
45-54
5.66
(2.80, 11.46)
5.56
(2.77, 11.19)
55-64
6.36
(3.45, 11.73)
6.25
(3.40, 11.47)
65-74
2.58
(1.22, 5.48)
2.56
(1.22, 5.38)
74+
38.75
(1.91, 785.55)
40.76 (2.04, 812.69)
The results in columns 4 and 5 are quite similar to those in columns 2 and 3. Adding 0.5 to every count in
each table tends to shrink the estimates of the odds ratios a little more toward one and results in slightly
shorter condence intervals. Consequently, variance is reduced at the expense of increased bias. The
results of some simulation studies suggest that adding .025 to each count is a better option for obtaining
an estimator with small mean squared error (MSE=variance+(bias) ). The changes in the estimated odds
ratios for any of these options are small relative to the size of the standard errors of the estimates, so all
of these options provide the same inference about the odds ratios.
All odds ratios appear to be signicantly bigger than one, indicating that the odds for cancer for the
high alcohol consumption group is signicantly higher than that for the low alcohol consumption group
for each age group. The condence intervals also show that the odds ratios are not well estimated in the
youngest and oldest age groups. Those estimates will receive less weight and have smaller inuence on
the MH estimator of a common odds ratio in the next part.
(b) The Mantel-Haenszel estimator of a common odds ratio does not require adding anything to the observed
counts in this case. The resulting estimator is ^MH = 5:16 with approximate 95% condence interval
(3.56, 7.47), using the formula in the notes. PROC FREQ in SAS inverts the CMH test to obtain (3.64,
7.31) for an approximate 95% condence interval. The information in the tables for the youngest and
oldest age groups does not have much inuence of the value of ^MH because of the relatively large variances
associated with the estimates of the odds ratios for those two tables. The value of ^MH is 5.08 when 0.5
is added to each count in every table and the condence interval (3.53, 7.30) is slightly shorter. PROC
FREQ in SAS gives (3.60, 7.15) as a 95% condence interval.
(c) The following results are for the original data:
stat d.f. p-value
Breslow-Day test 9.32 5 0.097
T test 0.956 1 0.169
The square of the T test can be compared to the percentiles of a central chi-squared distribution with 1
d.f. Neither test rejects the null hypothesis of homogeneous odds ratios.
(d) If you just had the data for the four middle age groups, the Breslow-Day test would be preferred because
you would have a few tables with relatively large counts in each table. Try applying the Breslow-Day test
to the tables for the four middle age groups. Would you expect the results to change? The T test would
be preferred if you had many tables with small counts in some tables.
(e) There appear to be no large dierences in the odds ratios for the individual age groups, and it is reasonable
to compute the value of an estimate for a common odds ratio. If either the Breslow-Day test or the T had
been signicant, you could have compared odds ratios by constructing condence intervals for dierences
in log-odds ratios for all pairs of age groups. Of courses, the variances of the estimated odds ratios for the
oldest and youngest age groups are so large that these estimates will not be found to dier signicantly
from odds ratios for other age groups.
(f) The value of the CMH test produced by PROC FREQ in SAS is X = 85:01 with 1 d.f. and p-value
< :0001. This agrees with the results in part (b). Within each age group the odds of oesophageal cancer
are about 5 times higher for people who consume at least 80g of alcohol per day.
2
4
4
4
4
2
4
7. (a) .
Birth Order Odds Ratio 95% condence interval
2
1.317
(0.572, 3.030)
3{4
1.189
(0.545, 2.596)
5+
2.016
(0.844, 4.816)
(b) The value of the Breslow-Day statistic is 0.85 with 2 degrees of freedom and p-value=0.653. The data are
consistent with the hypothesis of homogeneous odds ratios within age groups.
The use of the Breslow-Day test is appropriate in this case since the number of groups is small (three)
and each group has enough subjects.
(c) The estimate of the odds ratio for the marginal table of counts is 1.347. An approximate 95% condence
interval is (0.851, 2.132). This is consistent with the information in the tables for the three age groups,
and it is not an example of Simpson's paradox.
8. (a) From the software posted as negbin.ssc or negbin.sas, which was applied to the cavity data, we have
^ = :018888224 and k^ = :5883654 as the maximum likelihood estimates for the parameters in the negative
binomial model. Then the m.l.e. of P r(Y = 0) = k is ^k = 0:375.
(b) Dene g(; k) = k . The rst partial derivatives of this function are G = (k k ; k log()). A
consistent estimator of G is obtained by evaluating G at the m.l.e's of the parameters. Then, G^ =
(k^ ^k ; ^k log(^)) = (1:168553; 0:62574). The software from part (a) provided the estimated covariance
of (^; k^), the inverse of the estimated Fisher Information matrix,
2
3
:0009275 :0024042
5:
V^ = 4
:0024042 :0092353
Then, by the delta method, an estimate of the large sample variance of ^k is
G^ 0 V^ G^ = :001363
and the standard error of ^k is .03692.
(c) An approximate 95% condence interval for k is
:375 (1:96)(:03692)
)
(:303; :447):
Alternatively, you could have used the delta method to rst construct an approximate 95% condence
interval for log(k = k log() and applied the exponential function to the end points of the resulting
interval to get an approximate 95% condence interval for k . Which method would provide a coverage
probability closer to 0.95?
(d) You could do a simulation study of the coverage probability of the procedure for constructing condence
intervals used in part (c). Select values of and k. You could try several sets of values, using some
that are close or equal to the m.l.e.'s of the parameters evaluated in the previous parts of this problem.
For each set of parameter values, simulate a large number of samples (say 10000 samples) from the
corresponding negative binomial distribution with n independent observations in each sample. Then,
construct a condence interval from the data for each sample, using the method from part (c). Record the
proportion of the 10,000 condence intervals that contain the true value for k . Repeat this for several
choices of n, including the number of children in the original study.
The upper and lower condence limits are random. To simulation the coverage probability of a method
of constructing condence intervals, you must simulate values of the random upper and lower limits and
monitor how often those random limits enclose the true value of the quantity you are trying to estimate.
Some students proposed simulating estimates of the probability of observing a child with no cavities
and then monitoring how often those simulated estimates fell between the particular condence limits
computed in part (c). This incorrectly considers the upper and lower limits of a condence interval as
xed (non-random) quantities.
^
1
^
1
^
^
^
5
9. (a) ^ = 2:933 with approximate 95% condence interval (1:67; 5:16). This suggests that the odds of contracting Hodgkin's Disease are between 67% to 500% greater among people who have had their tonsils
removed.
This condence interval was constructed by rst constructing an approximate 95% condence interval for
log() and then applying the exponential function to the endpoints of the interval to obtain an approximate
95% condence interval for the odds ratio. Directly using the large sample normal distribution for the
odds ratio, we have
XX
p
^ (1:96) (^
Yij )
)
(1:27; 4:59):
2
2
1
2
i=1 j =i
This does not adequately account for the right skewness in the distribution of the odds ratio in this case.
Notice how much this interval is shifted to the left of the previous interval.
(b) Some people questioned the use of the odds ratio as a good approximation to relative risk in the Vianna
study. Since this is a retrospective study, you cannot directly estimate relative risk, but the odds ratio does
provide a good approximation because Hodgkin's disease is a rare disease. Some people worried about the
restrictions put on the control group in the Vianna study, and others noted that the sibling controls used
by Johnson and Johnson may dier substantially from the controls used in the Vianna study.
(b) While there are advantages in using siblings as controls, Johnson and Johnson did not do an appropriate
analysis, because they did not allow for the eects of matched sibling pairs. The "controls" are not an
independent sample of persons without Hodgkin's disease. Each sibling pair, not each individual, should
be thought of as an independent experimental unit. Siblings are likely to provide correlated responses.
Johnson and Johnson should have reported the data in the following table with one count for each sibling
pair.
C ontrol Sibling
Had T. No T.
Sibling with Had T. Y
Y
Hodgkin's
disease
No T.
Y
Y
There are n = Y + Y + Y + Y = 85 sibling pairs. Unfortunately, this table cannot be obtained from
the table reported by Johnson and Johnson (1972).
The Pearson chi-square test for independence performed by Johnson and Johnson is incorrect. They
should have performed a test of marginal homogeneity using the counts in the table shown above. This is
McNemar's test (or the sign test). Reject the null hypothesis of marginal homogeneity if
(Y Y ) > X =
;
Y + Y 21
(this is a 2-sided test).
Using the above table Johnson and Johnson could have estimated an odds ratio as
(Y + Y )(Y + Y )
^ =
(Y + Y )(Y + Y ) :
Then you can construct an approximate condence interval using the methods you derived in problem 4
on this assignment. There was a reason for doing problem 4.
Finally, note that Vianna, et al. sampled from a dierent population than Johnson and Johnson. What
are the consequences of this?
11
12
21
11
12
21
22
22
12
2
21
2
12
2
1
11
12
12
22
21
22
11
12
6