Solution_HW4

Solution_HW4 - Solution to Homework Assignment 4 1 Question...

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Solution to Homework Assignment 4 1 Question 1 a. Let U i = min( T i , C i ) be the observed time, Δ i = I ( T i C i ) be the failure indicator and n ( U i ) be the number of patients at risk at time U i for the ith individual, where T i and C i are the underlying survival and censoring times, respectively, i = 1 , . . . , 12. If combing the two treatments together, we get the ordered time U i listed in the left part of the following table. The Kaplan-Meier estimator KM ( t ) and its standard error seKM ( t ), for any time t 0, are described in the right part of the following table, where KM ( t ) = Y i : U i t 1 - 1 n ( U i ) Δ i and se { KM ( t ) } = KM ( t ) " X i : U i t Δ i n ( U i ) { n ( U i ) - 1 } # 1 / 2 . Table 1: Kaplan-Meier estimator of the survival curve and its standard error U i Δ i n ( U i ) t KM ( t ) se { KM ( t ) } 0.73 1 12 [0.00, 0.73) 1.000 0.000 0.95 0 11 [0.73, 1.43) 0.917 0.080 1.43 1 10 [1.43, 1.58) 0.825 0.113 1.58 1 9 [1.58, 1.98) 0.733 0.132 1.98 1 8 [1.98, 2.25) 0.642 0.144 2.25 1 7 [2.25, 2.30) 0.550 0.150 2.30 1 6 [2.30, 3.24) 0.458 0.150 3.24 1 5 [3.24, 3.30) 0.367 0.146 3.30 1 4 [3.30,+ ) 0.275 0.135 3.89 0 3 4.03 0 2 4.20 0 1 b. Let S 1 ( t ) = P ( T t | trt = A ) and S 2 ( t ) = P ( T t | trt = B ). To test the null hypothesis H 0 : S 1 ( t ) = S 2 ( t ), t 0, we consider the log-rank test T = i i =1 n d 1 ( u i ) - d ( u i ) n ( u i ) n 1 ( u i ) o q i i =1 n 1 ( u i ) n 2 ( u i ) d ( u i ) { n ( u i ) - d ( u i ) } n 2 ( u i ) { n ( u i ) - 1 } , where d 1 ( u i ) is the number of deaths at time u i from treatment A, d ( u i ) is the total number of deaths from both treatments, n 1 ( u i ), n 2 ( u i ) and n ( u i ) are the number of patients at risk at time u i for treatment A, B and the both treatments, respectively. 1

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Figure 1: Kaplan-Meier estimator of the survival curve 0 1 2 3 4 0.4 0.6 0.8 1.0 Time in years S(x) From the provided survival data, we see that a total of 8 deaths were observed, but no more than 1 death occurred at the same time. Table 2: Death times u i : d ( u i ) = 1 0.73 1.43 1.58 1.98 2.25 2.30 3.24 3.30 d 1 ( u i ) 1 1 0 0 1 0 1 0 n 1 ( u i ) 6 4 3 3 3 2 2 1 n ( u i ) 12 10 9 8 7 6 5 4 According to the summary information of the 8 death times as described in the above table 2, we have the estimated log-rank test T = i i =1 n d 1 ( u i ) - n 1 ( u i ) n ( u i ) o q i i =1 n 1 ( u i ) n 2 ( u i ) n 2 ( u i ) = 0 . 9798 1 . 8412 = 0 . 722 , and the corresponding p-value for a two-sided test is 0.4703. 2
2 Question 2 Let λ 0 and λ 1 be the hazards of the standard and new treatments, respectively. Since the survival distribution is approximated by an exponential distribution, we have that m 0 m 1 = λ 1 λ 0 = exp ( γ A ) , where m 0 and m 1 are the median survival times for these two treatments, respectively, which gives γ A = log(4 / 5) = - 0 . 2231. With 90% power to detect such significant difference using a one-sided logrank test at 0.05 level of significance, we have the required number of deaths to be d = 4( z α + z β ) 2 γ 2 A = 4(1 . 645 + 1 . 28) 2 log 2 (0 . 8) = 685 . a. Let a be the constant accrual rate, A be the accrual period and L be the total length of study.

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