To: WSFAlist at keithlynch.net
Date: Wed, 19 Mar 2003 03:46:38 -0500
Subject: [WSFA] Re: Predicting who will be at WSFA meetings
From: ronkean at juno.com
Reply-To: WSFA members <WSFAlist at keithlynch.net>

On Sat, 15 Mar 2003 14:54:27 -0500 (EST) "Keith F. Lynch"
<kfl at KeithLynch.net> writes:
...  With 143 people who have attended
> three or more of 192 meetings I have data for, a total of 5149 data
> points, I finally have a fair amount of data to play with.

You presumably mean that 5149 is the cumulative recorded attendance over
that time period, (but just among those 143 people), much the same way
that an airline might report carrying a million passengers in a year,
even though many of those million were repeat customers during the year.
It occurs to me that an instance of non-attendance is also a data point,
so there would be 27,456 (143 x 192) data points in all, at least in the
sense that a tabular representation of the data would have 27,456 cells
in the table.  Also, it seems that the average attendee who had been to
at least three meetings during the period, went to only about 19% of the
meetings (5149 / 27,456).

Doubtless you will be able to construct some formula which has predictive
value, one that has inputs such as past attendance with time weighting,
weather and traffic conditions, meeting location, competing events, time
of year, time of the month and proximity to major holidays, etc.  But
each attendee is different; for example one person might be more likely
to come to a meeting when it is raining, and another may be less likely
show up when it is raining.  So it seems like you would have to come up
with 143 formulas.  Perhaps a good way to tackle the problem would be to
first develop a single formula to predict the number of attendees at a
meeting.  With the large amount of data overall, that analysis would be
most sensitive to subtle factors, much more so than the data for any one
individual, and so would be a good way to identify and quantify at least
some of the relevant factors.

Ron Kean

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