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Finding a sample size that would give a certain probability of sucess

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    a population of n people contain k individuals with a disease. a sample of m people without replacement are chosen. how big should m be so that the probability of having at least one person with the disease is approximately .9?

    find m when n= 1000 and k= 10

    hint. find the proabability of there being no one in the sample with the disease.


    2. Relevant equations

    professor states there is no closed form solution for this answer, but that we are supposed to find it via R. i, and my other ecn grad friends, can't figure out how to get an answer for either that, or excel or matlab, as we kept being told the answer doesn't exist.

    3. The attempt at a solution

    first, i tackled the hint and thought the probability of no one having the disease could be m choose 0 divided by n chose k, which gave me k!/n!. supposing this is right, or at least on the right path, i don't understand how knowing what the probability that no one get the disease gives me for finding the size of m needed.

    i then thought about the idea that to be absolutely sure that you get a sample size large enough to get at least one person with the disease, it would need to be n-k+1. however, i don't know how to extend this idea to a probability of .9 (except multiplying it by .9, which even i know is totally wrong).

    i don't really need help with the artihematic, but setting up the problem. thanks for any help you can give me.
     
  2. jcsd
  3. Oct 10, 2009 #2

    Dick

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    The number of ways of choosing m people without the disease is C(990,m), right? (Where the 'C's are binomial coefficients). The total number of ways of choosing m people is C(1000,m). What does that make the probability of choosing m people without the disease? You want that less than 0.1, correct? How large does m have to be?
     
    Last edited: Oct 10, 2009
  4. Oct 11, 2009 #3
    that definately helps, thank you.
     
  5. Oct 11, 2009 #4
    hmm, okay, so i looked at the problem again and saw that i can't use the binomial dist, since i an NOT replacing my elements. so i tried to use the hyper geometric, such that:

    {(k choose 1)(n-k choose m-1)} / (n choose m), and simplifying, got :

    km(n-k)! / n!(n-k-m-1)! = .90

    however, even by plugging in number for n and k, i can't get the function in terms soley of m.
     
  6. Oct 11, 2009 #5

    Dick

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    Yes, you CAN use the binomial distribution to find the probability of choosing m people who aren't sick. Of course you aren't replacing anything. But you are only choosing once.
     
  7. Oct 11, 2009 #6
    alright, so i finally did the problem by using the hyper geometric and finding how bug the sample size m should be by the very good old guess anad check standby. thanks y'all the help.
     
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