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Unformatted text preview: Solution to Homework Assignment 2
1. (12 pts) In a small clinical trial, permuted block randomization was used to assign
one of two treatments (A or B) to six patients. Two blocks are chosen; the first block
size was four (two A's and two B's chosen at random) and the second block size was
two (one A and one B chosen at random). The end point of interest was change from
baseline in log HIV viral RNA counts after eight weeks on treatment. The results of
this trial are summarized as follows:
Treatment
Response Block 1
B
B
0.8
1.5
Block 2
B
A
1.0
0.2 A
0.6 Treatment
Response A
0.7 The test statistic used to evaluate whether treatment B has better response than
treatment A is based on
, where is the average response among patients
assigned to treatment B and is the average response for treatment A. We assume
the sharp null hypothesis and reject in favor of treatment B if the test statistic is
sufficiently large.
a. Find the distribution of the test statistic that is induced by the permuted block
randomization under the sharp null hypothesis.
Under the sharp null hypothesis, the distribution of the test statistic,
,
induced by the permuted block randomization is listed in the following table:
Patient
Response 1 2 3 4 5 6 .6 .8 1.5 .7 1.0 .2 A B B A B A 1 0.60 A A B B B A 2 0.53 A B A B B A 3 0.07 Test Statistics ( ) B A A B B A 4 0.07 B B A A B A 5 0.00 B A B A B A 6 0.47 A B B A A B 7 0.07 A A B B A B 8 0.00 A B A B A B 9 0.47 B A A B A B 10 0.60 B B A A A B 11 0.53 B A B A A B 12 0.07 b. What is the pvalue for the clinical trial described above?
Under the sharp null hypothesis, the test statistic can take on any of the 12
42
values , , , corresponding to the
12 combinations, each with
21
probability 1/12. Therefore, the pvalue for the clinical trial described in this
problem can be computed as
1
12 0.60sharp null hypothesis 2. (12 pts) Let the random variable ~
/.
1 a. Prove that , . The sample proportion is denoted by
1 is an unbiased estimator for Since ~
, and
1
/ . Hence, /, 0.083 2. , 2, we have that and 1 1 , or equivalently,
1
So 1 1 is an unbiased estimator for . 1 2. , Consider data from N studies using the same treatment regimen. We observe
,,
1, , , where denotes the number of patients in the ith study and denotes
the number of these patients that respond to treatment. It will be assumed that these
data are from a hierarchical model where
,, ,
1, , , are assumed to be
iid random vectors. The are assumed to be studyspecific response rates which
themselves represent an iid sample from some distribution. Conditional on and ,
is assumed to follow a binomial distribution. Specifically,
, ~ , .
1 b. Find an unbiased estimator for the expectation
unbiased.
1 From the result of part a. we have that and prove that it is
, 1 , which gives
1
,, ,
Since
easily see that ∑ 1
1, , , 1 , are assumed to be iid random vectors, we can
1
is an unbiased estimator for
1 . 3. (15 pts) Consider the following hierarchical model which may be appropriate in
conceptualizing results from different studies using the same regimen but where the
response of interest is a continuous measurement. The data from N such studies are
defined as
, 1, , , 1, , , where
denotes the response from the jth individual (among individuals) in the
ith study. The underlying hierarchical model assumes that there exists N iid random
vectors
, 1, , , , , ,
,
1, , are themselves assumed to
for
1, , . The random vectors
be iid unobserved random effects corresponding to the studyspecific mean and
variance of response which are allowed to vary according to some distribution. In
particular, we will denote the expected value of the studyspecific means by
and the variance of the studyspecific means by
Conditional on
, , , we will assume that the data from the ith study represent
iid observations with mean and variance . Specifically, we assume that for
the ith study
, , , , , , 1, , . The major reason for this conceptualization is to determine whether study to study
variation in the mean response is substantial. Specifically, we want to estimate
. In the table below, we summarize results from nine studies. For each
study
1, ,9, we give the sample size , the sample mean response
∑
and the sample variance
∑
1
Study ID
1
2
3
4
5
6
7
8
9 Sample Mean
59
23
41
28
40
31
34
20
30 Sample Variance
2475
1971
2009
1771
2484
2139
2275
1411
2379 Sample Size
47
48
37
183
13
150
37
141
62 a. In terms of
Since , , , , , 1, , , derive an unbiased estimator for , , , , , , 1, , . , we have that , and
, , , We know that the studyspecific sample variance
, i.e.
, ,
, which gives
, , Considering that where 1, , , ∑
∑ . is an unbiased estimator for , . . Therefore, we have vectors, , , , for 1, , , are iid random can be an unbiased estimator for , and ∑ be an unbiased estimator for unbiased estimator for ,
. So an is
∑ ∑ . b. Derive this estimate for the data provided in the table above and comment on the
implication for using such data as historical controls for comparison against a
new treatment.
According to the data provided in the table, we have the estimate of
as
∑ ∑ 83.46, or the estimated standard deviation of the studyspecific mean is
√83.46
where the estimate of is ̂ 9.14,
∑ 3 4. We see that this is an enormous variation compared to the sample means of the
data, which, at this point, implies the study to study variation among these
groups. Hence, we may not use such data as historical controls for comparison
against a new treatment. 4. (15 pts) Suppose the goal of a clinical trial is to estimate the difference in mean
response between two treatments assumed to have the same variance. It costs 300
dollars to treat a patient on regimen A and 100 dollars to treat on regimen B. An
investigator has 400,000 dollars to spend on this trial.
a. What is the optimal allocation of patients to treatments A and B subject to the
budgetary constraints of the investigator; that is, what allocation of patients
would result in the smallest variance of the estimator of treatment difference?
be the number of patients allocated to treatments A and B,
Let ,
respectively, , be the mean responses for the two treatments A and B,
be the common variance for the two treatments.
respectively, and
Then the variance for the difference in mean response between treatments is
1 1 Suppose the investigator has 400,000 dollars to spend on this trial, then the
optimal allocation of patients to treatments A and B can be obtained from
min 1 1 , s.t. 300 100 From the restriction, we have 400,000,
4000 where equality can be reached at 3 4000 , or
3 . It is straightforward to get that the minimum of
845 (by minimizing the function
theory), and hence the minimum of
4000
is 0. can be reached at
using basic calculus is reached at 845 and 3
1465. So the optimal allocation of patients to treatments A and B
845 and
1465. b. If the investigator insisted on equal allocation to both treatments, then how much
more money would need to be spent to get the same degree of precision (i.e.
variance of estimator) as the allocation in part a.
If the investigator insisted on equal allocation to both treatments, say , we
would have the variance for the difference in mean response between treatments
as
. If we need to get the same degree of precision, we
should have
, i.e.
1072. So the money would need to be spent
for this allocation is 300
100
428,800, which is 28,800 more than the
money spent on the allocation in part a. 5. (30 pts) Suppose patient enter a clinical trial in the order given by the table below.
The variables and in the table refer to two binary variables that are considered
important prognostic factors
a. The patients are to be assigned to either treatment A or B. Using random
numbers, either from a table of random numbers, or generating them by
computer, fill in the treatment assignments in the previous table using
i. Simple randomization
This randomization is implemented by generating 20 random numbers
,,
uniformly from 0 to 1. For the ith individual entering the study,
we assign treatment as follows:
If 0.5 assign treatment A
.
0.5 assign treatment B Therefore, using simple randomization, we have the assignment for the 20
patients as follows:
Random Num ( ) Patient ID
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20 1
1
0
1
1
0
1
1
0
0
0
1
1
0
1
0
1
0
1
0 0
1
0
0
0
0
0
1
1
1
1
0
1
0
1
1
0
0
0
1 Simple Randomization 0.119
0.413
0.797
0.801
0.234
0.552
0.051
0.169
0.177
0.637
0.697
0.235
0.833
0.054
0.455
0.465
0.098
0.008
0.721
0.381 A
A
B
B
A
B
A
A
A
B
B
A
B
A
A
A
A
A
B
A ii. permuted block randomization within strata using blocks of size 2 or 4 chosen
at random with equal probability
These 20 patients are divided into 4 strata according to the values of
prognostic factors and . Within each stratum, we first generate the
random numbers uniformly from 0 to 1 to decide which block sized need to
0.5 use block size 2
be used, specifically, if
, and then generate the
0.5 use block size 4
random numbers to associate each of the letters “AB” if block size 2 or “AABB” if block size 4. Finally, we will rank the letters by the random
numbers and assign the corresponding treatment to the patients in order.
Block Size
(Random Number) Patient ID
1
4
5
7
12
17
19
2
8
13
15
3
6
14
18
9
10
11
16 1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1 20 0 1 Size 4 ( 0.591) Size 4 ( 0.784) Size 2 ( 0.465) Size 4 ( 0.774) Size 4 ( 0.654) Size 2 ( 0.280) Size 2 ( 0.400) Size 4 ( 0.871) Random
Num
()
0.175
0.424
0.485
0.644
0.464
0.529
0.761
0.848
0.934
0.221
0.386
0.071
0.095
0.212
0.695
0.848
0.934
0.077
0.475 Stratified
Randomization 0.226 B
A
B
A
A
B
A
B
A
B
B
B
A
A
B
B
A
B
A
A * When the # of patients, k, is less than the block size, b, I assigned the treatments,
corresponding to the first k letters (after ranking), to the k patients in order. iii. PocockSimon minimization where all weights are equal
By PocockSimon minimization, where all weights are equal, the next patient
that enters the study is assigned either treatment A or treatment B according
to whichever makes the subsequent measure of marginal discrepancy
smallest. In case of a tie, the next patient is associated with a random numbers
0.5 assign treatment A
uniformly from 0 to 1, specifically if
.
0.5 assign treatment B Hence, we have the assignments for the 20 patients as follows:
Random Num ( ) Patient ID
1 1 0 2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20 1
0
1
1
0
1
1
0
0
0
1
1
0
1
0
1
0
1
0 1
0
0
0
0
0
1
1
1
1
0
1
0
1
1
0
0
0
1 MD after this assign Pocock Simon 0.165 3 A 2
3
4
3
2
3
0
3
0
3
4
5
4
5
2
3
0
3
4 B
B
A
B
A
B
A
A
B
B
A
A
B
B
A
B
A
A
B 0.372
0.95
0.146 0.008 The following table show the treatment assignments for the 20 patients by the above three
methods:
Patient ID
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20 1
1
0
1
1
0
1
1
0
0
0
1
1
0
1
0
1
0
1
0 0
1
0
0
0
0
0
1
1
1
1
0
1
0
1
1
0
0
0
1 Simple Randomization
A
A
B
B
A
B
A
A
A
B
B
A
B
A
A
A
A
A
B
A Stratified Randomization
B
B
B
A
B
A
A
A
B
A
B
A
B
A
B
A
B
B
A
A Pocock Simon
A
B
B
A
B
A
B
A
A
B
B
A
A
B
B
A
B
A
A
B b. For each of the three allocation schemes summarize the balance you achieved
i. Overall
For the allocation of simple randomization, the overall balance achieved is
1 3, 7 6 ii. by strata (in combination)
For the allocation of stratified randomization, the balance achieved is
2, 2 3, 2 4, 3 1, 3 ∑ 4 Here,
,,
0,1, is the number of patients that are on treatment A for the
ith level of prognostic factor and jth level of prognostic factor .
iii. by prognostic factor (marginally)
For the allocation of Pocock Simon, the balance achieved is
4, 5 6, 5 6, 5 4, 5 1 0,   ∑ 4 10 ,
1,2,
0,1, is the number of patients that are on treatment A
Here,
for the jth level of prognostic factor . ...
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This note was uploaded on 01/15/2012 for the course PHC 6020 taught by Professor Staff during the Fall '11 term at University of Florida.
 Fall '11
 Staff

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