-modal.
Duncan Murdoch
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to each alternative is
the key idea.
Duncan Murdoch
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On 2 Feb 2001 01:12:59 -0800, [EMAIL PROTECTED] (Will Hopkins) wrote:
I've been involved in off-list discussion with Duncan Murdoch. At one
stage there I was about to retire in disgrace. But sighs of relief... his
objection is Bayesian.
Just to clarify, I don't think this is a valid
Serve me right for not checking my work. Here's a correction:
On Wed, 31 Jan 2001 12:37:13 GMT, [EMAIL PROTECTED] (Duncan Murdoch)
wrote:
To find it, do this. Suppose the quadratic curve is A x^2 + B x + C.
Then you've got three equations:
A (-1)^2 + B (-1) + C = 2.05
A (0)^2 + B (0) + C
probably want to use "weighted least squares", with the weights
equal to the group sizes.
Duncan Murdoch
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e question, but if you only have 3
groups, your estimates will interpolate the group means. Just solve
the 3 linear equations in 3 unknowns, using whatever equation solving
method you like.
Duncan Murdoch
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was used in every
state? He got a lot of votes from California based on a very close
popular vote.
Duncan Murdoch
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candidates, but only chose to vote for 2.
I would hope that the recount is an attempt to correct counting
errors, not to create votes where there were none.
Duncan Murdoch
On Sun, 19 Nov 2000 09:32:31 -0500, Bob Wheeler [EMAIL PROTECTED]
wrote:
The model is simplified, but I assume that (B)
votes
the range
(largest value minus smallest value).
That's true for the n denominator ("population standard deviation"),
but not for n-1 ("sample standard deviation"). For example, if your
sample is just the two points 0 and 1, the sample standard deviation
is 0.71, and t
