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Thread: SD/SE of Percentage Question

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    SD/SE of Percentage Question




    Have a question that's probably painfully easy to address. Little slow.

    I am sampling from a hypothetical infinite population with an unknown mean and SD. I performing 10 separate trials, and each trial calculates a "hit rate" out of 100 attempts.

    Trial 1: 65 hits out of 100, 65%
    Trial 2: 66 hits out of 100, 66%
    Trial 3: 55 hits out of 100, 55%
    Trial 4: 61 hits out of 100, 61%
    Trial 5: 69 hits out of 100, 69%
    Trial 6: 62 hits out of 100, 62%
    Trial 7: 63 hits out of 100, 63%
    Trial 8: 63 hits out of 100, 63%
    Trial 9: 64 hits out of 100, 64%
    Trial 10: 59 hits out of 100, 59%

    Overall, there are 627 hits out of 1000, or 62.7%

    Using binomial theory, the SD of the overall hit rate is sqrt(p*q)=0.48, and the SE of the overall hit rate is sqrt(p*q/1000) = 0.01529.

    The standard error of the individual trials ranges from 0.0462 to 0.0492 [sqrt(p*q/100)].

    The standard deviation, calculated the "regular way", of the 10 percentages from the 10 different trials, though, is 0.0386.

    I am a little confused as to what the SD of the 10 different trials is really telling me versus the SE of the individual trials--why aren't they the same, or at least a little closer? Wouldn't the SE calculated from a single trial indicate the SD (given the same N of 100) of infinite additional trials? But when I calculate the SD of the 10 trials, it's only 0.0386.

    Is this some sort of sample bias, or a misunderstanding on my part? Thanks!

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    Re: SD/SE of Percentage Question


    hi,
    I think you are comparing apples to oranges.The SD refers to the standard deviation of the binomial distribution of your infinite population that is calculated using the estimate of p and q.

    The SE refers to the precision of your estimate of p and q, so they are definitely not the same.

    regards
    rogojel

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