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    Re: Real accuracy of the method of lunar distances
    From: Fred Hebard
    Date: 2004 Jan 1, 10:43 -0500

    Jan,
    
    If you denote the observed mean (mean of lunar distances during one
    observing session) by Y and the known distance (computed from
    chronometer time, what was being used to check these lunars) by u, the
    standard deviation by s and the number of observations used to compute
    Y by n, then
    
    t = (Y - u) / (s / square root of n).
    
    This is the standard form for the test statistic for Student's t
    distribution, given in any elementary statistics test.  u is usually
    denoted by the Greek letter, mu,  s by the Greek letter, sigma, and a
    bar is usually placed over Y and it called "Y bar" in English, but
    those characters are difficult to transmit by email.  I apologize for
    not defining u more carefully in my earlier post.
    
    For a 95% confidence interval, t(p<.05) = 2 = (Y - u) / (s / square
    root of n).  One can then solve for (Y - u).  u is zero in this case,
    by the way.
    
    If two means are being compared, then the calculation of the average
    number of observations can be more complicated if they are not equal,
    having the form 1 / (1/n1 + 1/n2).  The standard deviations either can
    be calculated from the observations themselves, or, in this case, from
    a sort of analysis of variance of a larger dataset.  I say "sort of"
    because I don't believe analysis of variance had been invented when the
    paper you are describing was written.
    
    It can be complicated to calculate the standard deviation from a larger
    dataset, but this particular case is the simplest possible analysis of
    variance.  The key is that there is no systematic change in the
    standard deviation over the course of the voyage, as I pointed out in
    my previous post.
    
    People might quibble about correlated errors and not following all the
    assumptions of analysis of variance, but the t statistic is very robust
    to failures to follow all assumptions.  I also expect these data
    conform to the assumptions.
    
    It should be reasonably easy to check whether your confidence intervals
    or mine are the more appropriate by plotting a histogram of the
    distribution of of (Y-u).  Your distribution would be much broader than
    mine.  You would need to combine both the english and german datasets
    to get enough observations, I expect.
    
    Fred
    
    On Dec 31, 2003, at 8:33 PM, Jan Kalivoda wrote:
    
    > Dear Fred,
    >
    > I agree with all in your last message, with the exception of the last
    > but one paragraph, which was the most important, of course. I repeat
    > only this passage:
    >
    >> You appeared to be constructing confidence intervals based upon
    > the t statistic.  The test statistic in this case would be t =
    > (deviation from chronometer time) / (standard deviation/square root of
    > number of observations).
    >
    > The "(standard deviation/square root of number of observations)" seems
    > to be the standard error of the (any) mean. But what is meant by
    > "(deviation from chronometer time)"? Is it an error of one observation
    > from six in each subset or an average error of a subset (more
    > probably)? And what sense has the whole fraction, giving the value
    > "t"? Sorry, I didn't find it in my manuals of mathematics during very
    > short searches at the beginning of the New Year in Europe.
    >
    > Thank you for an explanation.
    >
    >
    > Jan Kalivoda
    >
    
    
    ------------------------------------------------------------------------
    Frederick V. Hebard, PhD                      Email: mailto:Fred@acf.org
    Staff Pathologist, Meadowview Research Farms  Web: http://www.acf.org
    American Chestnut Foundation                  Phone: (276) 944-4631
    14005 Glenbrook Ave.                          Fax: (276) 944-0934
    Meadowview, VA 24361
    
    
    

       
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