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Jan,

I am claiming that the quantity you are denoting s1 is _less_ than or
equal to s6, and that, more specifically, s1 is approximately equal to
s6.

Here's an analysis of variance table; the dependent variable is the
difference between the observed lunar distance and that calculated from
the time of the observation.  There are 32 sets of observations, with
six observations per set.  I use s1*s1 to denote the variance
corresponding to the standard deviation s1 and s6*s6 likewise.  I use
sa*sa to denote additional variance due to systemic differences between
sets of observations.  These systemic differences could be due to
sighting on the sun rather than a star or sighting east rather than
west, etc.  Anythings that would give significant differences between
sets of observations.  At this point they have not been isolated as
treatment variables so I'm treating them as random variables, but that
makes no difference anyway.  The point is they would _increase_ s6
above s1.

source of      degrees of  Sum of  Mean   F statistic   Expected Mean
variation       freedom   Squares Square                   Square
------------------------------------------------------------------------

between sets of                                             s6*s6 =
observations       31        a     a/31  (a/31)/(b/160)  s1*s1 + 6*sa*sa

within sets of
observations      160        b    b/160                     s1*s1

total             191       a+b

------------------------------------------------------------------------

Thus, s1 does not equal s6*(square root 6), but rather, as above, s6*s6
= s1*s1 + 6*sa*sa.  You are correct, in my opinion, that there are not
enough data to reach any conclusions about departures from the normal
distribution.  However, when you look at these errors, you should
center them by setting the mean to zero, to help nullify any systemic
errors.

It seems that the standard deviations of Bolte & Behrens observations
are less than about 30" of arc (Bolte's standard deviation was about
40" of arc if the two errant observations were included).  I looked at
my own observations and they had a root mean square error of 32" of arc
if I threw out two early sun lunars that were particularly awful, and
44" without.  I have only done about 18 lunars.  Although mine were
done on land, my telescope is only 2.5x, which makes it more difficult.
  The one lunar Frank Reed reported recently had a standard deviation of
about 20" of arc.  I do not have the data, but I expect that the C.
Gregory lunar distance reported by Kieran Kelly had a standard
deviation of 10" or 15" of arc.  On land, a tripod would be most
helpful in achieving high accuracy.

It would be interesting if Frank Reed could supply us with his
voluminous data so that the effects of looking east and west and the
effects of stars versus the sun could be assessed more thoroughly.  I
doubt that many more sea data will surface, as the accuracy of lunars
could only be assessed after they were obsolete!

Regards,

Fred

On Jan 2, 2004, at 2:29 PM, Jan Kalivoda wrote:

> 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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