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    Re: "Vernier acuity" of horizon IC tests
    From: Nicol�s de Hilster
    Date: 2009 Jul 10, 10:53 +0200

    In NavList 9005 engineer@clear.net.nz wrote:
    > If we write down the 95% standard errors for the means (1.96 x SD/root n)...
    The variance is known to be the summed squared residuals (difference 
    between mean and observation) divided by "the number of observations 
    minus 1" (minus one is for small populations, can be neglected at large 
    The SD (standard deviation) of the normal distribution is the square 
    root of the variance and represents 68.3% of the observations (so these 
    are within plus and minus SD from the mean).
    In order to get 95.4% probability (again normal distribution) you simply 
    multiply the SD by 2 which will give a wider range from the mean to fit 
    all the observations in (there should be ample sources on the internet 
    to verify this).
    A 99.7% probability is obtained by multiplying SD by 3.
    Just an example:
    30 observations: 
    2,4,3,5,2,3,5,3,1,3,2,4,3,5,2,3,5,3,1,3,2,4,3,5,2,3,5,3,1,3 (yes, three 
    times the same first 10 obs.)
    Average = 3.1
    Variance = 1.54
    SD (68.3%) = square root 1.54 = 1.24
    2SD (95.4%) = 2 x 1.24 = 2.48
    3SD (99.7%) = 3 x 1.24 = 3.72
    According to your formula we would get (1.96 x 1.24 / sqrt(30)) = 0.44, 
    which differs quite a bit from the 96% probability (1.24) as given above.
    If you re-calculate the sextant observations of you and Greg, you might 
    find that the results with and without telescope will overlap and that 
    there thus is no difference (statistically) between the two observations.
    If you and Greg send me the original observations I will be more than 
    happy to do the math for you guys.
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