assign3key - Assignment #3 Clinical Trials (100 points) Due...

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Assignment #3 Clinical Trials (100 points) Due April 5, 2011 1. (20) The following data were obtained from a randomized, double-blind crossover clinical trial involving participants with mild to acute bronchial asthma. The treatments were single doses of two active drugs (A and B). The response is forced expiratory volume in one second (FEV1). Treatment Participant Sequence Baseline Period 1 Washout Period 2 1 AB 1.09 1.28 1.24 1.33 2 BA 1.74 3.06 1.54 1.38 3 AB 1.38 1.60 1.90 2.21 4 BA 2.41 2.68 2.13 2.10 5 AB 2.27 2.46 2.19 2.43 6 BA 3.05 2.60 2.18 2.32 7 AB 1.34 1.41 1.47 1.81 8 BA 1.20 1.48 1.41 1.30 9 AB 1.31 1.40 0.85 0.85 10 BA 1.70 2.08 2.21 2.34 11 AB 0.96 1.12 1.12 1.20 12 BA 1.89 2.72 2.05 2.48 13 AB 0.66 0.90 0.78 0.90 14 BA 0.89 1.94 0.72 1.11 15 AB 1.69 2.41 1.90 2.79 16 BA 2.41 3.35 2.83 3.23 17 BA 0.96 1.16 1.01 1.25 Group 1: AB (n=8) Baseline Period 1(A) Washout Period 2(B) A-B (diff) sum w-b Mean 1.3375 1.5725 1.4313 1.6900 -0.1175 3.2625 0.09375 Variance 0.2370 0.3269 0.2732 0.5345 0.1255 1.5975 0.2782 Group 1: BA (n=9) Baseline Period 1(B) Washout Period 2(A) A-B(diff) sum w-b Mean 1.8055 2.3411 1.7876 1.9456 -0.3956 4.2867 -0.01889 Variance 0.5277 0.5282 0.4442 0.5217 0.3373 1.7625 0.4198 a. (5) Using the period 1 and 2 measurements, determine the difference between treatments A and B in FEV1 and its 95% CI. 1
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Assuming there is no differential carry-over effects, then the difference between treatments A and B is: (d 1 +d 2 )/2 = (-0.1775 -0.3956) = -0.2566 The pooled variance estimate: 2385 . 0 2 9 8 ) 3373 . 0 ( 8 ) 1255 . 0 ( 7 2 ˆ ) 1 ( ˆ ) 1 ( ˆ 2 1 2 2 2 2 1 1 2 = - + + = - + - + - = n n n n d d dp σ The 95% CI for the treatment difference is: ) 1 1 ( 4 ˆ ] 2 / ) [( 2 1 2 ) 975 . 0 , 15 ( 2 1 n n t d d dp + ± + = ) 2361 . 0 )( 0596 . 0 ( 1315 . 2 2566 . 0 ± - = 2529 . 0 2566 . 0 ± - So the 95% CI is (-0.5095, -0.0037) b. (5) Use the period 1 and period 2 measurements to determine if there is evidence of differential carry over effects? (Construct the appropriate test statistic and test the hypothesis that the carry over effects are equal. Show your work.) Null hypothesis: the carry over effects are equal, Carry over effect = 3.2625 – 4.2867 = -1.0242 The pooled variance of the mean of the sum for both groups 1 and 2: 6855 . 1 2 9 8 ) 7625 . 1 ( 8 ) 5975 . 1 ( 7 2 ˆ ) 1 ( ˆ ) 1 ( ˆ 2 1 2 2 2 2 1 1 2 = - + + = - + - + - = n n n n s s sp The test statistic: df t n n m su m su T sp 15 2 1 2 2 1 * 6237 . 1 ) 9 / 1 8 / 1 ( 6855 . 1 2867 . 4 2625 . 3 ) / 1 / 1 ( ˆ - = + - = + - = Compare with t-tables with 15 df, p = 0.1253 > 0.05. So we cannot reject the null hypothesis. c. (5) Using the period 1 and 2 measurements, determine if there is period effect. Compute the test statistic and the corresponding p-value. Test null hypothesis of no period effect the test statistics for no treatment effect is
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This note was uploaded on 11/21/2011 for the course PUBH 7420 taught by Professor Ph7420 during the Spring '07 term at Minnesota.

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assign3key - Assignment #3 Clinical Trials (100 points) Due...

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