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Jan,

One minor point, correcting an earlier post of mine, is that the
standard deviation is equal to the square root of (the sum of squared
deviations divided by the degrees of freedom, not by the number of
observations).  In this case, the numerator would be divided by 33, not
34, including all observations in the data set.

 From your posting, I infer that the standard deviation was constructed
from the deviation of the mean of the individual observations from the
"actual" value as determined by chronometer.  So that for Bolt's data
there were 34 sets of observations with 34 deviations (ignoring for the
present the subdivision into sun lunars, star lunars and two-sided
lunars).  The error for these observations should be the same as what
would be obtained if each of the 34*6 individual observations was
calculated, unless Bolt's sextant had a significant change in accuracy
over the course of the voyage or unless his technique were improving or
getting worse over the course of the voyage.  You could get a feel for
this by looking for a change in the deviations over the course of the
voyage, perhaps by plotting them against time and perhaps looking for a
significant difference from zero of the slope of a straight line fit to
the data.  I am saying that the errors within sets of observations
would equal errors between sets, for this particular model.

Yes, I was altering the standard deviation to the standard error of the
mean.  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).  For large numbers of observations, t is equal
to 2 at 95% confidence, 2.8 at 99% and 3.6 at 99.9% confidence.  I will
take your word for it that a value of 3 gives 99.7% confidence.

Given the above, it is my belief it would be appropriate to divide by
the square root of the number of observations making up a single
observation.  So perhaps it's not quite as bleak as your analysis
indicated.

Fred

On Dec 31, 2003, at 6:20 AM, Jan Kalivoda wrote:

> No, Fred,
>
> Errors reported in my posting were given for averaged measurements
> from six shots in every case. Each such average was taken as the
> *single* observation and the error was compared with the corrected
> chronometer time only for this average, not for each of six original
> shots. Therefore, no error referred to in my posting can be divided by
> the square root of six, as you propose (as if you search for the
> standard error of the mean from the set of six measurements).
>
>
> Yours, Jan Kalivoda
>

   

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