NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Real accuracy of the method of lunar distances
From: Fred Hebard
Date: 2004 Jan 1, 10:43 -0500
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