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Math Help - Shed some light on this? (Due tomorrow)

  1. #1
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    Shed some light on this? (Due tomorrow)

    I'm confused as to how to answer this question.

    Water witching, the practice of using the movements of a forked twig to locate underground water or minerals dates back over 400 years. Reliable evidence supporting or refuting water witching is hard to find. Personal accounts of isolated successes or failures tend to be strongly biased by the attitude of the observer. However, consider the following data of outcomes of wells dug in Fence Lake, New Mexico:

    Of 29 "witched" wells, 5 were unsuccessful;
    Of 32 "non-witched" wells, also 5 were unsuccessful;

    What would you conclude? Give your reasoning.

    I have no idea how to approach this problem! We are currently learning hypothesis testing for multiple samples (2). Please help me, this is due in 12 hours!
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is a hypothesis test of a two proportions.
    You obtain the two sample proportions by taking the percent of successes in your samples.
    NOW it's not that clear, but I would do a two sided test, to see if there is a difference.
    Look this up in your book and obtain the z-score via the central limit theorem.
    Post your work and I'll look it over.
    Last edited by matheagle; April 23rd 2009 at 09:41 PM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    this is a hypothesis test of a two proportions.
    You get the two smaple proportions by taking the percent of success in your samples.
    NOW it's not that clear, but I woul do a two sided test, to see if there is a difference.
    Look this up in your book and obtain the z-score via the central limit theorem.
    Post your work and I'll look it over.
    Thank you here's what I have.

    \mbox{Witched:} \ m = 29, x = 5, p_{1} = \frac{5}{29} = .1724
    \mbox{Non witched:} \ n= 32, y - 5, p_{2} = \frac{5}{32} = .1563
    p = \frac{5+5}{29+32}=.1639, q = 1 - p = .8361

    z = \frac{p_{1}-p_{2}}{\sqrt{pq(\frac{1}{m}+\frac{1}{n}})}=\frac{.  1724 - .1563}{\sqrt{(.1639)(.8361)(\frac{1}{29}+\frac{1}{  32}})} = .1696

    Is this reasonably all I need? What else do I have to explain? Thank you.
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  4. #4
    MHF Contributor matheagle's Avatar
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    These are all sample estimates.
    We usually put 'hats' on them to make that clear.
    The p's are the population values, the true probabilities.
    And it's good that you pooled the estimate under the null hypothesis.
    (It's y=5, not y-5. I thought you were looking at the complement.)

    Quote Originally Posted by VENI View Post
    Thank you here's what I have.

    \mbox{Witched:} \ m = 29, x = 5, \hat p_{1} = \frac{5}{29} = .1724
    \mbox{Non witched:} \ n= 32, y =5, \hat p_{2} = \frac{5}{32} = .1563
    \hat p = \frac{5+5}{29+32}=.1639, \hat q = 1 - \hat p = .8361

    z = \frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat p\hat q(\frac{1}{m}+\frac{1}{n}})}=\frac{.1724 - .1563}{\sqrt{(.1639)(.8361)(\frac{1}{29}+\frac{1}{  32}})} = .1696

    Is this reasonably all I need? What else do I have to explain? Thank you.
    If you have an alpha you can obtain the rejection region and make a decision.
    Otherwise you can state that the p-value is 2P(Z>.17) which is near one, if your calculations are correct.
    Clearly for any reasonable alpha we cannot conclude that there is a difference in the probabilities of detecting water.
    Last edited by matheagle; April 23rd 2009 at 09:42 PM.
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