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Math Help - Prediction interval for normal distribution

  1. #1
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    Prediction interval for normal distribution

    My textbook only has formulas for prediction intervals of t-distributed variables. If I have a binomial distribution, with an estimated p, how can I create a prediction interval that with 95% probability will contain the number of successes of the next 10000 trials? Thanks for any help with this!
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  2. #2
    MHF Contributor matheagle's Avatar
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    with large n you need the CLT.
    you need to tell me the TWO sample sizes.
    You will take the difference between the two p hat's.
    The mean is zero, since p-p=0
    and you will just sum the two variances.
    I've never seen this before, but it's just like the usual prediction situtaion.
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    All right, so n = 10000. I've estimated the p to be between 0,0015 and 0,0035. And what I want to do is to find a 95% confidence interval for the number of successes on the next sample of 10000 trials. Thanks for helping out.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Ok, let's get this straight.
    A confidence interval is for estimating a parameter.
    A prediction interval is for predicting a future random variable.
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    That is correct and I meant a prediction interval!
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  6. #6
    MHF Contributor matheagle's Avatar
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    well I know I'm right
    but you need to be clear on what you want.
    The idea is the same as a prediction of a new observation.
    You will pivot on the difference between two P hats and then apply the CLT.

    \hat P_n - \hat P_m \approx Normal

    where E\bigl(\hat P_n - \hat P_m\bigr)=p-p=0
    since I assume we are sampling from the same underlying binomial distribution.
    Otherwise this doesn't make sense.

    And  V\bigl(\hat P_n- \hat P_m\bigr)=V\bigl(\hat P_n)+V\bigl(\hat P_m\bigr)

     = {p(1-p)\over n}+{p(1-p)\over m}

    But here we have some options, since we don't know p, we need to use the p hat's but one
    has been oberved while the other hasn't.
    Using Slutsky's theorem I'm sure we can use the known one say, \hat P_n

    So {\hat P_n- \hat P_m\over\sqrt{\hat P_n(1- \hat P_n)\biggl({1\over n}+{1\over m}\biggr)}}\approx N(0,1)

    Now pivot and solve for \hat P_m and then multiply by m to obtain the estimate of number of successes in this second future sample.
    Last edited by matheagle; May 3rd 2010 at 10:34 PM.
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    All right, so I use p=0,0025 as my p. But I'm not sure which formula to put it in. I know this will be a standard normal distribution so anything within 95% would mean a value between two standard deviations. If I knew the standard deviation I could just write

    25 - 2 standard deviations < 25 < 25 + 2 standard deviations

    I might be way off, but if I'm on to something, how do I find 1 standard deviation here?
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  8. #8
    MHF Contributor matheagle's Avatar
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    you pivot and solve for the second p hat
    you want 95 percent, so you place that mess I created between -1.96 and 1.96 and solve for p hat m.
    Look at your book where they pivoted and solved for the parameter or future observation as I am doing.
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