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Solution_HW2

# Solution_HW2 - Solution to Homework Assignment 2 1(12 pts...

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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 p-value 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 p-value for the clinical trial described in this problem can be computed as 1 12 0.60|sharp 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 i-th 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 study-specific 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 j-th individual (among individuals) in the i-th 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 study-specific mean and variance of response which are allowed to vary according to some distribution. In particular, we will denote the expected value of the study-specific means by and the variance of the study-specific means by Conditional on , , , we will assume that the data from the i-th study represent iid observations with mean and variance . Specifically, we assume that for the i-th 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 study-specific 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 study-specific 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 i-th 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. Pocock-Simon minimization where all weights are equal By Pocock-Simon 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 i-th level of prognostic factor and j-th 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 j-th level of prognostic factor . ...
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