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An investigational drug study is being conducted. The initial screening phase needs to include enough people so that the final...Asked by Craig
An investigational drug study is being conducted. The initial screening phase needs to include enough people so that the final (approved) group has at least 10 patients. The probability of a person being approved for the study is .80. what initial sample size is needed to make sure with a .939 probability that the study will have at least 10 patients?
i don't know how to go about working this problem. i know its a discrete random variable problem and i know that i somehow have to work backwards, but i have NO idea how to set it up or work it. any and all guidance would be appreciated. thank you!!!
i don't expect you to work it for me, but i do need some help.
i don't know how to go about working this problem. i know its a discrete random variable problem and i know that i somehow have to work backwards, but i have NO idea how to set it up or work it. any and all guidance would be appreciated. thank you!!!
i don't expect you to work it for me, but i do need some help.
Answers
Answered by
Craig
would this be a binomial-like should i consider it
applicable for study
not applicable for study
applicable for study
not applicable for study
Answered by
MathMate
Yes, the binomial distribution is a discrete random variable problem.
Basically, if you screen N people, each having a success rate of 0.8. You would like to calculate the minimum candidates to have a minimum of 10 success with a probability of 0.939.
It is a very similar process to flipping coins, except the probability of success is 0.8 instead of 0.5.
The probability can be obtained by tabulation, but the a formula exists to calculate the probability of success of n times out of N candidates, with a probability of success of p.
P(n,N,p)=<sub>N</sub>C<sub>n</sub>*p<sup>N</sup>*(1-p)<sup>(N-n)</sup>
Let say, since the proability of success is 0.8, and 10 people are required, a good starting point is to screen 10/0.8=13 candidates.
So N=13, p=0.8,
P(10,13,0.8) = 0.246
P(11,13,0.8) = 0.268
P(12,13,0.8) = 0.179
P(13,13,0.8) = 0.055
Probability of retaining 10 or more candidates
= 0.246+0.268+0.179+0.055
= 0.747 < 0.939 required.
So you could screen 14 candidates and calculate the probability. Keep increasing the number of candidates until you get the probability of 0.939 or higher.
Total =
Basically, if you screen N people, each having a success rate of 0.8. You would like to calculate the minimum candidates to have a minimum of 10 success with a probability of 0.939.
It is a very similar process to flipping coins, except the probability of success is 0.8 instead of 0.5.
The probability can be obtained by tabulation, but the a formula exists to calculate the probability of success of n times out of N candidates, with a probability of success of p.
P(n,N,p)=<sub>N</sub>C<sub>n</sub>*p<sup>N</sup>*(1-p)<sup>(N-n)</sup>
Let say, since the proability of success is 0.8, and 10 people are required, a good starting point is to screen 10/0.8=13 candidates.
So N=13, p=0.8,
P(10,13,0.8) = 0.246
P(11,13,0.8) = 0.268
P(12,13,0.8) = 0.179
P(13,13,0.8) = 0.055
Probability of retaining 10 or more candidates
= 0.246+0.268+0.179+0.055
= 0.747 < 0.939 required.
So you could screen 14 candidates and calculate the probability. Keep increasing the number of candidates until you get the probability of 0.939 or higher.
Total =
Answered by
MathMate
Oh yes, here's a link for more explanation:
http://stattrek.com/Lesson2/Binomial.aspx
http://stattrek.com/Lesson2/Binomial.aspx
Answered by
MathMate
Also, in case you are not familiar with the notation,
<sub>n</sub>C<sub>r</sub> is the combination function defined as:
<sub>n</sub>C<sub>r</sub>
= n!/((n-r)!r!)
=n(n-1)(n-2)...(n-r+1)/r!
For example,
<sub>13</sub>C<sub>10</sub>
=13*12*11/(1*2*3)
=286
<sub>n</sub>C<sub>r</sub> is the combination function defined as:
<sub>n</sub>C<sub>r</sub>
= n!/((n-r)!r!)
=n(n-1)(n-2)...(n-r+1)/r!
For example,
<sub>13</sub>C<sub>10</sub>
=13*12*11/(1*2*3)
=286
Answered by
Craig
thank you so much! i think i got it!
Answered by
Jason
what was the answer you finally got
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