STAT 557
Solutions to Assignment 2
Fall 2002
1. (a) Assuming a multinomial distribution for each country, the expected counts will be
Under Normal
Over
Obese
GB 77.42591 154.3160 29.73798 11.52012
CAN 123.37095 245.8881 47.38469 18.35623
USA 88.20314 175.7959 33.87733 13.12365
The test results for the null hypothesis of the homogeneous model:
X2
G2
statistic
34.3963
34.9478
d.f.
6
6
p-value
0.000+
0.000+
The null hypothesis is rejected; i.e., the distribution of women across the weight categories is not the same
for all three countries.
(b) The Pearson residuals (Yij m
^ ij )=m
^ ij are displayed in the following table and plot.
Under
Normal
Over
Obese
GB -1.639457 -0.1059374 2.6153232 0.4360123
CAN 2.217385 0.1984510 -2.5255033 -2.4171856
USA -1.086405 -0.1354480 0.5365021 2.4502312
Prob 1(b): Residuals
2
Country
0
-2
-1
residuals
1
US
Britain
Canada
Under
Normal
Over
Obese
Obesity
The residual plot reveals a similar pattern of residuals for Great Britain and United States women: the frequencies of heavy women are underestimated while those of underweight women are overestimated under
the null hypothesis of identical multinomial models for the three countries. On the other hand, in Canada,
the proportion of underweight women is larger than the proportion expected under the null hypothesis
while the proportions of overweight and obese women are smaller than expected under the null hypothesis.
2. (a) The use of a multinomial distribution would be appropriate if either
i. a simple random sample of twins is selected with replacement from all twins born at the hospital in
a specic time period, or
1
ii. if the number of twins born at the hospital is much larger than 98 and a simple random sample of
twins is selected without replacement from this large population of twins, or
iii. the 98 sets of twins represent all of the twins born at the hospital in the specied time period and there
is an underlying biological mechanism associated with the birth of twins that makes each occurance
of twins an independent event with a constant set of probabilities for the various types of twins.
It is unlikely that either (i) or (ii) is reasonable in this case.
(b) The log-likelihood function is given by
`( ; Y)
3
X
= log(n!)
To obtain the mle under the restriction that
g (; )
i=1
P3
i=1 i
log(Yi !) +
3
X
i=1
Yi log i :
= 1, maximize
3
X
= `( ; Y) + 1
i=1
!
i
:
Solving the likelihood equations,
the mle's turn out to be
or
0 =
@g
;
@i
0 =
@g
@
^i A
^A
pi
i = 1; 2; 3
3
X
=1
Y
= i;
i=1
i
i = 1; 2; 3:
n
= (^1 A ; ^2 A ; ^3 A ) = (0:2959; 0:3673; 0:3367)
(c) For model B, the probability of exactly j
j
2
1
1 boys is
j
1
(1
)3 j ;
Then, the probability for each category will be
Sex
Two Boys
Two Girls
One Boy / One Girl
for j = 1; 2; 3
prob under Model B
1B = 2
2B = (1 )2
3B = 2 (1 )
The log-likelihood for this model is
`( ; Y)
= log(n!)
3
X
i=1
log(Yi !) + 2Y1 log + 2Y2 log(1
Solving the likelihood equation:
0=
we have the mle of computed as
^ =
@`
@
=
2Y1 + Y3
) + Y3
flog 2 + log + log(1
2Y2 + Y3
1 2Y1 + Y3
total number of boys
=
2(Y1 + Y2 + Y3 )
total number of children
= 0:4643:
and the mle of is
^B
= (^1 B ; ^2 B ; ^3 B ) = (^ 2 ; (1
^)2 ;
2
2^ (1
^)) = (0:2156;
0:2870; 0:4974):
g
) :
(d) The expected counts under this model are ( 21.125, 28.125, 48.750) and the results of the tests are shown
in the following table. (Note that the estimated means satisfy the constraint on the observed counts
98 = n = Y1 + Y2 + Y3 = 29 + 36 + 33 = 21:125 + 28:125 + 48:750. This provides a check on your
computations.) Goodness of t statistics are shown below.
stat df
pval
G2 10.39752 1 0.00126
X2 10.22911 1 0.00138
We conclude that the model B does not t well, i.e., the sex of one twin is in some way related to the sex
of the other twin. In the next part we will see that this good be a result of the fact that some twins are
monozygotic.
(e) For model C,
1C
= P rf2 boysg = P rf2boysjmonozygotesgP rfmonozygotesg + P rf2boysjdizygotesgP rfdizygotesg
= (1=2) + (1=2)2 (1 ) = (1=4)(1 + )
Similarly, we can compute 2C and 3C as shown in the following table.
Sex
Two Boys
Two Girls
One Boy / One Girl
Probability under Model C
1C = (1 + )=4
2C = (1 + )=4
3C = (1 )=2
The log-likelihood under this model is written as:
`( ; Y)
= log(n!)
3
X
i=1
log(Yi !) + Y1 logf(1 + )=4g + Y2 logf(1 + )=4g + Y3 logf(1
g
)=2 :
Solving the likelihood equation:
0=
we have the mle of computed as
and the mle of is
@`
@
^ =
=
Y1
+1
+
Y2
+1
Y1 + Y2 Y3
Y1 + Y2 + Y3
1
Y3
= 0:3265
^ C = (^1 C ; ^2 C ; ^3 C ) = (0:3316; 0:3316; 0:3367):
(f) The expected counts under this model are (32.5, 32.5, 33.0) and the results of the tests are shown in the
following table.
stat df
pval
G2 0.7553101 1 0.3848002
X2 0.7538462 1 0.3852612
There is not enough evidence to reject the null hypothesis, so Model C is consistent with the data.
(g) No. Model B is a special case of Model A, and Model C is a special case of Model A, but neither Model
B nor Model C is nested in the other.
(h) For Model D,
1D
= P rf2 boysg = P rf2boysjmonozygotesgP rfmonozygotesg + P rf2boysjdizygotesgP rfdizygotesg
= + 2 (1 )
2D
= P rf2 girlsg = (1
) + (1
)2 (1
3
)
and
3D
= 2 (1
)(1
)
Since there are two functionally independent parameters (; ) and 1 + 2 + 3 = 1, the mle ^; ^ will be
such that
^1D = Y1 =n = ^1A ;
^2D = Y2 =n = ^2A ;
^3D = Y3 =n = ^3A :
Both Model A and Model D are saturated models, the estimated proportions are the observed proportions.
Then, there is no dierence between the estimates for Models A and D, and G2 = X 2 = 0 with 0 d.f.
Consequently, there is no information in the available data that can distinguish between Models A and D.
One alternative would be to classify twins into 5 categories,
Category
monozygotic twins, 2 boys
monozygotic twins, 2 girls
dizygotic twins, 2 boys
dizygotic twins, 2 girls
dizygotic twins, 1 boy and 1 girl
Count
Y1
Y2
Y3
Y4
Y5
prob under Model D
1 = 2 = (1 )
3 = (1 ) 2
4 = (1 )(1 )2
5 = 2(1 ) (1 )
The log-likelihood under this model is
`(; ; Y)
5
X
= log(n!)
i=1
log(Yi !) + Y1 logf g + Y2 logf(1
+Y4 logf(1
)(1
)2
g+Y
logf(1
) 2
g+Y
logf2(1
) (1
) :
)
5
3
g
g
Solving the likelihood equations:
0 =
0 =
@`
@
@`
@
=
=
Y1 + 2Y3 + Y5
Y2 + 2Y4 + Y5
1 Y1 + Y2
Y3 + Y4 + Y5
1 we have the mle's of and computed as
^
=
^
=
Y1 + Y2
number of monozygotic pairs
=
n
number of pairs of twins
Y1 + 2Y3 + Y5
Y1 + Y2 + 2(Y3 + Y4 + Y5 )
and X 2 and G2 would have 2 d.f. for testing the t of Model D against the general alternative.
3. (a) Without combining any birth weight categories, X 2 = 71:25 with d.f.=8 and the chi-square approximation
to the distribution of the Pearson statistic, when the null hypothesis is true, yields p value < :0001. We
conclude that SIDS rates are not the same for all birth weight categories.
Here the observed proportions of SIDS cases decrease across the last 7 weight categories, from over 13
cases per 1000 in weight category 3 to under 1 case per 1000 births for weight category 9.
(b) Since two of the estimated expected counts are smaller than one and four of the estimated expected
counts are smaller than 5, the large sample chi-square approximation to the distributions of the Pearson
and deviance statistics may not provide accurate p-values. One option would be to combine the results for
the three smallest birth weight categories. This yields a Pearson statistic that also rejects the hypothesis
of equal SIDS rates across all birth weight categories (X 2 = 60:28 with d.f.=5 and p value < :0001).
q
(c) The formula is p z:025 p(1n p) . Here, n = 2061 and p = 2=2061. This yields: lower limit=-0.00037,
upper limit=0.00231. The lower limit is negative because the mean count is too small for the large sample
normal approximation to the binomial distribution to give accurate results.
(d) The 95 % exact condence interval is ( 0.00012, 0.00350). This would give a more reasonable condence
interval. The interval in part (c) would have a coverage probability smaller than 0.95.
4
(e) If you condition on the total counts in the birth weight categories, the SIDS case counts, Yi for the i-the
birth weight category, have independent binomial distributions with probability i of a SIDS case and
sample size ni . From the large sample normal approximation,
p5
p7
_
N
5
p5 (1 p5 )
7 ;
n5
Then, an approximate 95 % condence interval for 5
0
@p
s
5
p7
1:96
p5 (1 p5 )
n5
+
p7 (1 p7 )
; p5
n7
7
+
p7 (1 p7 )
n7
is
s
p7 + 1:96
p5 (1 p5 )
n5
1
+
p7 (1 p7 ) A
n7
= (0:00050; 0:00281)
Since the condence interval does not include zero, there is an evidence that the incidence of SIDS is
signicantly higher in Group 5 than in Group 7.
4. (a) The m.l.e. for the mean number of corn borers per location in the Poisson model is
m
^=
16
1X
(i
n
i=1
1)Yi = 3:1566:
This estimate was computed without combining any categories.
(b) Combining some categories to avoid small expected counts, as suggested in the statement of the problem,
The Pearson statistic for testing the t of the i.i.d. Poisson model is X 2 = 26:578 with 7 d.f. and p-value
= .0004. The Poisson model does not appear to be appropriate.
(c) In this case, Y = 3:1566; and S 2 = 5:5899: The deviance statistic is nS 2 =Y = 293:9618 with 165 d.f. and
p-value < :0001: This test shows that the variance of the numbers of accidents is larger than the mean.
Hence, the i.i.d. Poisson model is not appropriate.
(d) Maximum likelihood estimates of the expected counts for the Poisson and negative binomial models are
shown in the following table. Maximum likelihood estimates of the parameters in the negative binomial
^ = 4:2716.
probability function are ^ = 0:5751 and K
Number of
Accidents
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15 or more
Number of
Drivers
15
32
26
29
22
19
9
8
3
1
0
0
1
0
0
1
Poisson
Model
7.066
22.306
35.206
37.044
29.234
18.456
9.710
4.379
1.728
0.606
0.191
0.055
0.014
0.004
0.001
0.000
Neg. Binomial
Model
15.619
28.353
31.757
28.212
21.795
15.322
10.061
6.274
3.756
2.176
1.227
0.677
0.366
0.195
0.102
0.107
(e) Combining some categories to avoid small expected counts, as suggested in the statement of the problem,
the value of the Pearson goodness-of-t statistic is X 2 = 3:990 with 8 d.f. and p-value = 0.858. The
negative binomial model is consistent with the observed data.
5
(f) Do not use the i.i.d. Poisson model to construct the condence interval because it was shown that the
Poisson model is inappropriate. There are several ways, however, to obtain an approximate 95% condence
interval for the mean number of accidents per bus driver in a two year period.
i. You could use the central limit to show that Y has a limiting normal distribution. You must also show
that S 2 is a consistent estimator for 2 = V ar(Yi ). Then an approximate 95% condence interval 1s
Y
(1:96)
p
)
S 2 =n
(2:80; 3:52)
This method does not assume any particular distribution for the counts, but it does assume that
counts of the n drivers are independent and identically distributed random variables.
ii. Assuming the negative binomial model is appropriate, V ar(Yi ) = K (1 )= 2 and V ar(Y ) = K (1
)=n 2 . Then an approximate 95% condence interval is
s
Y
(1:96)
^ (1 ^ )
K
n
^2
)
(2:80; 3:51)
iii. You could use the Delta method to nd the limiting normal distribution for the m.l.e. of the mean
m
^ nb
^) =
= g (^ ; K
^ (1 ^ )
K
^
(4:2716)(1 0:5751)
= 3:1566 = Y :
0:5751
=
The computer output inverts the estimated Fisher information matrix to estimate the covariance
^ )0 as
matrix of (^ ; K
0:004623843 0:07736258
^
V =
0:077362578 1:35233094
Compute rst partial derivatives of g (; K ),
G
Then
^
G
=
=
@g @g
@ @K
^
K
2
^
1
=
^
^
K
2
1
!
= ( 12:9176; 0:7390)
and V ar(m^nb ) is estimated as G^ V^ G^ 0 = 0:03307: Using the large sample normal approximation to the
distribution of the m.l.e., m
^ nb ; an approximate 95% condence interval for the mean number of corn
borers in a location is
p
)
3:156626 (1:96) 0:03307
(2:8002; 3:5130):
This method is also based on the belief that the negative binomial model is appropriate. Note that
the condence interval based on the incorrect Poisson model
m
^
p
(1:96)
m=n
^
)
(2:8863; 3:4269)
is too short because the Poisson model does not allow for enough dispersion in the counts.
iv. You could use the delta method and large sample normality of m.l.e.'s to rst construct a condence
interval for
log(mnb ) = log(K ) + log(1 ) log( )
Then apply the exponential function to the endpoints of that interval to obtain an approximate 95%
condence interval for mnb . When would this be better than approach (iii)?
5. See problem 8 on assignment 3.
6. There are only seven tables with the same row and column totals as the observed table. These tables can be
distinguished by the Y11 value. The exact probabilities are presented in the following table.
6
Table
Number
1
2
3
4
5
6
7
Y11
0
1
2
3
4
5
6
Exact
Probability
0.010098
0.080784
0.239828
0.338580
0.239828
0.080784
0.010098
observed data
Looking at the proportions of high blood pressure cases for the aspirin treatment group and the control
group, only table 1 is less consistent with the null hypothesis than the observed table, in the direction of the
alternative that aspirin is better at controlling blood pressure than the placebo. Consequently, the p-value,
0.239827+0.080784=0.0909, is the sum of the probabilities for the tables 1 and 2. These data are inconclusive,
they do not allow us to reject the null hypothesis that the high blood pressure rates are the same for the aspirin
group and the control group at the 0:05 level of signicance.
7. Since the objective is to show that the IFN-B treatment is better, we should test the null hypothesis that
the IFN-B treatment has the same eect as the control treatment against a directional alternative where the
IFN-B treatment is "better". Dierent denitions of what it means for the IFN-B treatment to be "better"
result in dierent answers. The row totals in this table are xed by the randomization procedure that places
10 subjects in each treatment group. Under the null hypothesis that the IFN-B and control treatments are
equally eective, the column totals are also xed for these particular subjects. There are 43 possible tables
with these row and column totals. Each table is distinguished by the values of Y11 and Y12 , the rst two counts
in the rst row of the table. The values of these two counts and the corresponding probability that each table
occurs by random assignment of subjects to treatment groups are shown below.
Table
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Y11
6
6
6
6
6
5
5
5
5
5
5
4
4
4
4
4
4
Y12
0
4
1
3
2
5
0
1
4
2
3
6
0
1
2
3
4
Pearson
X2
14.67
12.00
10.50
9.17
8.67
9.17
13.33
7.83
5.33
4.67
3.83
8.67
14.67
7.83
3.33
1.17
1.33
7
Exact
Probability
15/184756
70/184756
160/184756
336/184756
420/184756
336/184756
25/184756
600/184756
2520/184756 (observed table)
2800/184756
4200/184756
420/184756
15/184756
600/184756
6300/184756
16800/184756
15750/184756
Table
Number
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
Y11
4
3
3
3
3
3
3
3
2
2
2
2
2
2
2
1
1
1
1
1
1
0
0
0
0
0
Pearson
X2
Y12
5
7
6
5
4
3
2
1
3
4
5
6
7
8
2
5
4
8
7
6
3
6
5
7
4
8
3.83
10.50
4.67
1.17
0.00
1.17
4.67
10.50
3.83
1.33
1.67
3.33
7.83
14.67
8.67
4.67
5.33
13.33
7.83
4.67
9.17
8.67
9.17
10.50
12.00
14.67
Exact
Probability
4200/184756
160/184756
2800/184756
16800/184756
28000/184756
16800/184756
2800/184756
160/184756
4200/184756
15750/184756
16800/184756
6300/184756
600/184756
15/184756
420/184756
2800/184756
2520/184756
25/184756
600/184756
2800/184756
336/184756
420/184756
336/184756
160/184756
70/184756
15/184756
Note that in this case the exact probabilities provide the same ordering of the tables the Pearson X 2 values.
Table 9 is the observed table and the EXACT option in PROC FREQ in SAS computes a multi-sided p-value
of 0.0642 by adding the probabilities for tables 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 19, 25, 30, 31, 32, 34,
35, 36, 38, 39, 40, 41, 42, 43. The sher.test( ) function in SPLUS yields the same p-value by summing the
probabilities for all tables that occur with probability no larger than the probability of the observed table. This
is not necessarily a good way to dene a critical region or dene a p-value. This set of tables includes many
tables that have fewer treated patients that either improve or stay the same than in the observed table, so it
may not provide an appropriate p-value with respect to the objective of this study.
One reasonable criterion is that tables provide more evidence than the observed table that the IFN-B treatment
is better if at least 5 of the treated patients show improvement and at least 9 of the treated patients either
improve or stay the same (or no more than one of the treated patients becomes worse). This includes tables 2,
4, 6, 9 and the p-value is 0.01766.
Another criterion is that dierence between the number of treated patients that improve and the number of
treated patients that get worse should be at least 5 1 = 4. This results in a p-value of 0.0222. Many students
failed to clearly describe the criterion used to identify the possible tables that were included in the evaluation
of the p-value.
q
^ )
8. (a) Here it means that a 95 % condence interval is no wider than ^ :03. Hence, (1:96) ^ (1
< :03, for
1100
any value of ^ . (The maximum occurs at ^ = :50).
(b) Use = :50 and nd n such that
r
^ (1 ^ )
< :01:
(1:96)
n
The solution is n 9604.
8
9. (a) 442 subjects in each group.
(b) Using the worst case scenario, we need
r
:03
(1:96)
(:5)(:5)
n
+
(:5)(:5)
n
:
The solution is n 2135 subjects in each group.
Using the researchers' best guess at what the results will be, we need
r
:03
(1:96)
(:15)(:85)
n
The solution is n 1277 subjects in each group.
+
(:22)(:78)
n
:
10. (a) Consider a multinomial distribution
0
@
Y11
Y12
Y22
1
A
0
0
M ult @n = 156; @
11
12
22
11
AA
Then, the null hypothesis that the conditional probability of a secondary infection, given a primary
infection, is equal to the probability of a primary infection is written as
H0 : 11 + 12
or
H0 : 11
=
12
11 + 12
= 2 ; 12 = (1
); 22
=1
:
(b) Under the assumption in (a), the log-likelihood is given as
`( ; Y)
= log(n!)
(log(Y11 !) + log(Y12 !) + log(Y22 !)) + Y11 log( 2 ) + Y12 log( (1
)) + Y22 log(1
):
(c) Solving the likelihood equation,
@`
@
0=
=
2Y11 + Y12
Y12 + Y22
1 the mle of is computed by
2Y11 + Y12
= 0:4940:
2Y11 + 2Y12 + Y22
Evaluating the negative second derivative using the observed data gives the observed information
^ =
I (^
)
=
@ 2 ` @ 2 =^
= 996:1423
Then,
var(^
^ ) = I (^ ) 1 = 0:0010:
It is better to use the expected information, the negative of the expectation of the second derivative of
the log-likelihood, instead of the observed information. This is computed as
E
2Y11 + Y12
2
+
Y12 + Y22
(1 )2
=
n(1
(1
)
The inverse of this quantity gives a formula for the large sample variance of ^, var(^ ) (1 )=(n(1+ )).
Evaluating at ^ = 0:494 and n = 156, we have
var(^
^ ) 0:00107
Then an approximate 95
^
p
1:96
var(^
^ )
)
p
0:494 (1:96) :00107
9
)
(0:430; 0:558):
(d) The test results are as follows. The model in part (a) is not an appropriate model for the data.
stat df
G2 17.73787 1
X2 19.70606 1
p-value
.0000+
.0000+
(e) This question was poorly stated. It should have asked about the probability of primary infection instead
of . Under the general alternative, the probability of primary infection is estimated as ^11 + ^12 =
(Y11 + Y12 )=n = 0:596. From the multinomial distribution, the standard error for this estimate is estimated
as
s(^
11
+ ^12 ) =
p
r
var(^
^ 11 + ^12 ) =
^11 (1
= 0:0393
^11 )
n
+
^12 (1
^12 )
n
2
^11 ^12
n
Alternatively, using the fact Y11 + Y12 Bin(n, 11 + 12 ) gives the same standard error.
r
s(^
11 + ^12 ) =
(^11 + ^12 )(1
n
^11
^12 )
= 0:0393
(f) Using the standard error in (e), a large sample approximate condence interval is
0:596 1:96 s(^11 + ^12 ) = (0:519; 0:673):
This condence interval is expected to perform better than the one in (c), with respect to more nearly
providing a coverage probability of 0.95, because the model in (a) does not appear to be appropriate for
these data.
10