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Re: statistics questions
*To*: mathgroup at smc.vnet.net
*Subject*: [mg62404] Re: statistics questions
*From*: "Ray Koopman" <koopman at sfu.ca>
*Date*: Wed, 23 Nov 2005 01:12:24 -0500 (EST)
*References*: <dkcocf$qeb$1@smc.vnet.net><dls7fb$3fk$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Chris Chiasson wrote:
> I guess what I mean by "mean response" is that regression generates a
> least squares fit function for y, which then has a confidence interval
> associated with it:
> regression function + or - t[v,ci]*sqrt(variance/n)
>
> I think if y is supposed to be a constant value, then least squares
> regression is equivalent to the mean of the data.
>
> Anywho, does anyone know of a way to obtain the regression function
> plus its confidence interval directly?
>
> Am I just using statistics incorrectly here?
>
> Thanks for your response, Bill Rowe.
>
> Regards,
> --
> http://chrischiasson.com/contact/chris_chiasson
The usual linear model is Y = X.beta + e, where X is a given n by k
matrix with rank k, beta is an unknown parameter vector, and the
elements of e are iid normal random variables with mean 0 and unknown
variance v. Let c = Inverse[Transpose[X].X], let y be an observed Y,
let b = c.Transpose[X].y, and let u = (y-X.b).(y-X.b)/(n-k). Then
(b-beta).c.(b-beta)/u is distributed as k*F(k,n-k), and this can be
used to get a confidence region for beta: any vector beta is in the
confidence region if (b-beta).c.(b-beta) < u*k*F(k,n-k,conlev),
where conlev is the confidence level. If k = 2 and one column of X
is constant then the elements of beta represent an intercept and a
slope, and we have a confidence region for the true regression line.
For nonlinear models, the usual approach is to get a matrix A that
is some sort of asymptotic estimate of the inverse of the covariance
matrix of b, and to define the region as (b-beta).A.(b-beta) <
k*F(k,n-k,conlev) or (b-beta).A.(b-beta) < ChiSquare(k,conlev).
Such regions tend to be optimistic; i.e., too small.
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