Re: Standard deviations and Confidence intervals with respect to errors

*To*: mathgroup at smc.vnet.net*Subject*: [mg101821] Re: Standard deviations and Confidence intervals with respect to errors*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Sat, 18 Jul 2009 04:50:51 -0400 (EDT)

On 7/16/09 at 8:20 AM, lehin.p at gmail.com (Alexey) wrote: >Hello, I have a set of independent observations of two values (Y,X) >with known (and different) standard errors in both values for each >observation. These valueses are linearly dependent: Yi=a*Xi+b I need >to calculate the unknown parameters {a,b} and confidence intervals >with respect to known standard errors of observations. As I >understand, the standard Mathematica's function LinearModelFit is >designed only for the case when all observations have equal standard >deviation and are normally distributed. These are the standard assumptions for regression analysis. The routines available in Mathematica version 7 do allow for some deviation from these assumptions. Specifically, you can use the Weights option in LinearModelFit to adjust for variations in the standard deviation between observations. The usual procedure is to specify the weight for each observation as 1/variance for that observation. For even more flexibility you can use GeneralizedLinearModelFit which allows for probability distributions for the error term different than a normal distribution. This function will also do a weighted regression to allow for variations in the standard deviation of observations. However, none of the built-in routines are intended to address the case where there are errors in X as well as errors in Y. If you really do have significant errors in X you will need to write your own fitting routine in order to get statistically valid confidence limits. >In my case it is known only that all observations have the same >distribution that is supposed to be nearly-normal. But standard >deviations for observations are significantly different and known. Then it is quite possible the values computed using LinearModelFit will be adequate. Surely, it is not reasonable to expend considerable effort to meet some standard for mathematical rigor if the end result is say a difference in the third significant digit when your data is say only accurate to two significant digits. If the error distribution is "nearly normal" it could well be there is nothing to be gained by being more rigorous. What level of rigor do you really need? What is your end goal? >And another question: how in this case the term "standard deviation" >may be defined and calculated? If you are simply asking how standard deviation is defined and can be computed, then standard deviation is the square root of variance. Look up Variance using the Documentation Center. In the more information entry are details of how variance is computed. Or in standard statistics language variance is the expected value of (x-mu)^2 where x is the data sample and mu is the mean of the data sample. If this doesn't address your question about standard deviation, you will need to re-phrase your question or expand on it. If I have answered your question, that suggests to me you don't have much knowledge of statistical computations/theory. If that is true, trying to compute confidence limits where there are significant errors in both x and y that are not from a normal distribution is very much akin to trying to run a marathon when you haven't learned to walk. In which case, you would be best advised to study some good texts on statistics that should be available in your local library. The documentation for various statistical routines in Mathematica is no substitute for a good text.

**Re: False divergence of the NDSolve solution: how to avoid**

**Re: Dynamic freeze**

**Standard deviations and Confidence intervals with respect to errors**

**how to use vb.net/netlink to export a dxf file**