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Fred,

please consider following:

You wrote:

>Jan was using the test statistic for the normal distribution, usually
denoted z, with y equal to the value of a _single_ observation, u the
actual or parametric mean and s the standard deviation is:

>z = (y - u) / s6,

>whereas I was using the test statistic for the t distribution, denoted
t, with Y equal to the observed mean of a set of observations, u their
parametric mean, s the standard deviation and n the number of
observations is:

>t = (Y - u) / (s1 / the square root of 6).


I reply: I added indexes "6" and "1" into those formulas to the symbol "s" and 
changed your symbol "n" for "6", the used number of individual measurements 
during each set = lunar observation.

In the first ("my") formula, "s6" denotes the standard deviation found for 
averages of sets of six measurements, that created each Bolte's lunar 
observation. This "s6" can be deduced from the summarized "probable error", 
given by Bolte, by multiplying it by the factor 1/0.6745 (see my first 
posting in the thread) or it can be directly calculated from the errors also 
given by Bolte for each his set (= lunar observation) evaluated as the 
average of six measurements. I have sent those 34 errors to you in my last 
posting. Both results agree.

In the second ("your") formula, "s1" denotes the standard deviation of 
individual measurements (six in each observation), which is not directly 
known, as they were not evaluated separately. But "s1" can be statistically 
assessed as "s6 times sqrt(6)", when "s6" is known, isn' it? Then if you 
insert this "s6 times sqrt(6)" for "s1" into "your" formula, you end with 
"my" formula again, although you use the t-distribution.


And you wrote:

>Finally, perhaps I have managed to refute Jan's bleak picure, as he
wished, and shown that, in the hands of a competent observer, 95% of
the time a lunar will be accurate to within 25" of arc with 6 replicate
observations, and 99.9% of the time accurate to within 44" of arc.

I wonder, if this is true. I consider only 22 lunars for stars I have sent to 
you in my last posting (there is undoubtely a systematic error in 10 lunars 
working with the Sun - it would be unfair to use them against you, not to 
speak about 2 clear outliers) And I find 9 errors greater than 25" among them 
and 2 errors greater than 44".

This rather corresponds to the standard deviation of 30" (computed from these 
22 observations and rounded) and the (quasi)normal error distribution 
supposed by me: 15 errors lie below the standard deviation of 30", i.e. 68% 
of items (should be 68%)  and 12 errors lie below 20" ("probable error" 
exceeding 50% of all errors), i.e. 54% of items (should be 50%). Therefore, 
the error limit "2 times the standard deviation" for 95% of cases and "3 
times the standard deviation" for 99,7% of cases (that made me so sad) 
probably apply, approximately at least.


For your convenience, I repeat those 22 error values ascertained by Bolte for 
stars as the distance bodies :

 +18,-5,-15,+72,+16,+7,-14,+8,+54,+12,-47,-32,-34,-28,+39,-36,-7,+19,-27,+13,-5,+25 ;
13 westerly distances, 9 easterly distances, the mean +1.5" , the standard deviation 30" (rounded)


Yes, we fiddle with a too small sample to be statistically persuasive. If 
anybody has another and greater set of lunars taken *at`sea* and correlated 
with the verified GMT times, all this thread will become a modest preface 
only.



Jan Kalivoda

   

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