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.