I have been making some calculations in preparatio
Post# of 153887
(Total Views: 541)
Posted On: 04/05/2020 1:41:16 PM
I have been making some calculations in preparation of what to expect in our trials results as they will come rapidly.
In this particular case, the trial is designed as:
Se we have two outcomes: our patients survive or not. I would assume that the placebo arm will serve as a reference in order to set the success criteria for our arm (patients treated with Leronlimab). As this is a true or false outcome, a one-tailed binomial distribution test is the best suited.
With n=114, trying different percentage of survivals (as we don't know yet) and using a binomial probability distribution (unfortunately unable to share pictures of the resulting binomial distribution probability chart, so below is a table with some values):
n Deaths n succ. p succ Min Probability
114 55% 51.3 0.45 61 0.4196
114 60% 45.6 0.40 55 0.0452
114 65% 39.9 0.35 49 0.047
114 70% 34.2 0.30 43 0.047
114 75% 28.5 0.25 37 0.044
114 80% 22.8 0.20 31 0.03931
114 90% 11.4 0.10 18 0.0344
For example, if in the control arm 75% of the patients survive (28.5 patients) we should have a minimum of 37 patients surviving for the area below the probability distribution to be less than 0.05 (with 37 is actually 0.044).
The data above fits the equation:
n= 6.21+ 1.072 * number of survivors placebo arm
So, roughly we should have the number of survivors in control arm times 1.072 plus 6.21 patients more. Since it is a 2:1 trial ( 228 patients will receive Leronlimab) this number should be multiplied by 2.
I am assuming a p<less than 0.05. the probabilities in the table are the maximum before it goes above the limit of 0.05, for example, in the 90% case we cannot have only 17 (34) patients surviving as this would bring the probability to 0.06157
This are only approximations and imo, don't make trading decisions based on the above
In this particular case, the trial is designed as:
Quote:
Patients enrolled in this trial are expected to be administered leronlimab for two weeks with the primary endpoint being the mortality rate at 14 days.
Se we have two outcomes: our patients survive or not. I would assume that the placebo arm will serve as a reference in order to set the success criteria for our arm (patients treated with Leronlimab). As this is a true or false outcome, a one-tailed binomial distribution test is the best suited.
With n=114, trying different percentage of survivals (as we don't know yet) and using a binomial probability distribution (unfortunately unable to share pictures of the resulting binomial distribution probability chart, so below is a table with some values):
n Deaths n succ. p succ Min Probability
114 55% 51.3 0.45 61 0.4196
114 60% 45.6 0.40 55 0.0452
114 65% 39.9 0.35 49 0.047
114 70% 34.2 0.30 43 0.047
114 75% 28.5 0.25 37 0.044
114 80% 22.8 0.20 31 0.03931
114 90% 11.4 0.10 18 0.0344
For example, if in the control arm 75% of the patients survive (28.5 patients) we should have a minimum of 37 patients surviving for the area below the probability distribution to be less than 0.05 (with 37 is actually 0.044).
The data above fits the equation:
n= 6.21+ 1.072 * number of survivors placebo arm
So, roughly we should have the number of survivors in control arm times 1.072 plus 6.21 patients more. Since it is a 2:1 trial ( 228 patients will receive Leronlimab) this number should be multiplied by 2.
I am assuming a p<less than 0.05. the probabilities in the table are the maximum before it goes above the limit of 0.05, for example, in the 90% case we cannot have only 17 (34) patients surviving as this would bring the probability to 0.06157
This are only approximations and imo, don't make trading decisions based on the above
(5)
(0)