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Hypothesis Testing: Binomial Experiment

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A drug company markets a medication that cures about 60% of cases with depression. A CB program is thought to be more effective. It was delivered to 15 depressed people. Determine the minimum number of cured people required to support the claim that the CB program is more effective than the drug. Use alpha=.10.

    2. Relevant equations

    nCx p^(x)q^(n-x)

    3. The attempt at a solution

    This is a binomial distribution problem.

    - Upper tail test (H1: p>.6000)
    n = number of trials = 15
    p = probability of a success on a given trial = .6000
    x = ?

    I am trying to solve for x. However, I have no idea as to how to go about this. If I plug in the known values into the binomial distribution equation (written under "relevant equations") it becomes too difficult for me to solve, beyond the scope of the course I'm taking. I cannot use the normal approximation to solve the problem, because nxp does not equal 10.

    Could someone please give me some detailed guidance? It would be greatly, greatly appreciated.
     
  2. jcsd
  3. Nov 16, 2009 #2
    I just had a thought. I could use the probability of failure instead of success (i.e. 1 - .6000 = .4000).

    Then, I could look up n=15, p=.40 on the binomial distribution table. Starting at the lowest value of x (x=0, probability = .0005), it is evident that if one person was not cured, we could reject the null hypothesis because the p-value would be lower than alpha (.10).

    I could work my way down the list, adding on each probability for the next highest value of x. When the probability exceeded the alpha level (this occurs at x=3), I would know I'd gone too high, because I could not reject the null hypothesis. The x value one down would be the key to my answer (x=2).

    Thus, a maximum of 2 people must not be cured. To rephrase this in terms of the question, a minimum of 13 people must be cured.

    Am I on the right track here?
     
    Last edited: Nov 16, 2009
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