4. Suppose that we have a binary outcome, and wish to show non-inferiority of treatment B relative to treatment A. In designing the trial, we assumed that the failure rate in each treatment group is PA = ps = 0.30. Given these rates, we consider that an increase in failure rate to at most pp = 0.36 to constitute non-inferiority and we enrolled 1300 subjects in each treatment group. We can parameterize the non-inferiority margin in (at least) two ways: . 6 = 0.36 - 0.30 = 0.06 . 6 = log OR = log(0.36/0.64) - log(0.30/0.70) = .272 corresponding to two null hypotheses: . Ho : PB - PA > 0.06 . H3 : logPB/(1 -PB) - logPA/(1 - PA) > .272 Suppose at the trial's end we observe the following: failures successes A 273 1027 B 299 1001 (a) (10 pts) Using the random allocation rule (re-allocate 1300 to each treatment group), conduct the randomization test of H,. Can we infer non-inferiority? (b) (10 pts) Using the random allocation rule (re-allocate 1300 to each treatment group), conduct the randomization test of Hi. Can we infer non-inferiority? (c) (5 pts) Comment on the differences and similarities among these analyses, and those from problem 4 of homework 5