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    Re: Lunars using Bennett
    From: Frank Reed
    Date: 2008 Apr 06, 02:19 -0400

    Hi Peter.
    You wrote:
    "Secondly; the amounts by which the values are rounded up or down, when
    added together (as they are in Bennett) will NOT tend to accumulate,
    but will tend to cancel each other out, and trend towards accuracy,
    rather than inaccuracy.
    In Jan 05 I tested this and presented the results here:" etc.
    I looked at those earlier posts and you were on the right track but I think
    you missed the final point. As Alex has pointed out, the errors DO
    accumulate, but much more slowly than many people expect (sqrt(N) rather
    than in proportion to N). You should try your spreadsheet example again.
    Consider a simple case where you have ten random numbers in the range from 0
    to 100 (or 0 to 10 or 0 to 50, any integral value >=1 will work as the upper
    limit) with two digits beyond the decimal in column A, and those same
    numbers rounded to the nearest whole number in column B. You can think of
    these as prices on a shopping list, for example. Now add up the numbers in
    column A. This is the "correct" or "exact" sum. Then add up the numbers in
    column B. This is an estimated sum. We've all done this in the real world...
    you add up a list of prices rounding the cents to the closest dollar [insert
    local currency here as appropriate]. It's usually close enough for a quick
    double-check because, in fact, the individual rounding errors tend to cancel
    each other out --to some extent. If you set this up and re-calculate your
    spreadsheet a few times, you should find that the difference between the
    correct sum and the estimated sum bounces around with a standard deviation
    of about 0.92 units for a list of ten numbers. Now try it again with 20
    numbers in columns A and B. Now the standard deviation is about 1.28 units.
    Do it again with 100 random numbers in the columns. The s.d. is now about
    2.89 units. So the net error does increase, but rather slowly. For simple
    rounding like this, where we're never more than +/-0.5 from the actual
    value, the error, expressed as a standard deviation, goes as approximately
    0.289*sqrt(N) where N is the number of steps in the sum (the 0.289 arises
    because the actual errors at each step are uniformly distributed between
     -0.5 and +0.5). You're RIGHT that the maximum error, 5.0 for a set of ten,
    is quite rare, but you're WRONG in suggesting that the net error diminishes
    with increasing N. It increases in proportion to sqrt(N). On the other hand
     --there's always that other hand up to no good-- you might also note that
    the *percentage* error in a summing situation actually decreases with
    sqrt(N) since the expected sum is proportional to N. But that's not relevant
    for a navigation problem. Here's a nice Wikipedia article on this topic:
    Finally, note that observational accuracy DOES improve as sqrt(N) where N is
    the number of observations under the assumption of zero systematic error.
    This issue is distinct from the question of tabular accuracy.
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