# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Resume of "Averaging"**

**From:**Alexandre Eremenko

**Date:**2004 Nov 5, 21:57 -0500

Dear Herbert, You are not my student, so please do not call me "Professor Eremenko":-) I think it is a good custom in this list to address each other by the first name. On Fri, 5 Nov 2004, Herbert Prinz wrote: > undergraduate and graduate students at Perdue University, PUrdue University. > average of the average of two data sets equals, in general, the average > of the union set of the two data sets? No, this I don't teach because this is incorrect. Could you explain how is this related to our discussion? > This is, to wit, in simplified form, > the question which is at the basis > of my most important objection to the method of averaging individual > "runs" of sights. I have asked Professor Eremenko several times to > address the problem, the last time on Nov 3. So far, I saw in your messages 2 objections that I understand. The first one indeed was important from my point of view, so I made some efforts to verify whether it indeed applies. 1. Non-linearity problem. (Do you agree now that this is not a problem?) 2. Your statement on November 3 that fitting a linear function by the least square method gives something else than a simple averaging. (You do not think so anymore?) 3. The objection which I do not understand at all, about averaging of the averages of several groups. In the end of your message of November 3, you said "One problem with averaging groups of sights individually before reducing" What groups of sights are you talking about? It seems to me that we were discussing the following specific situation: on an interval of 5 minutes we measure altitude of the same body several times. We obtain altitudes, say a_1,a_2,a_3 taken at the moments t_1,t_2,t_3. What do you call "groups of sights" here? I see just ONE group of 3 sights. And we want to obtain one position line from this group. This is what I was talking about all the time. And you? Alex.