Re: statistical comparison of parameters from two applications of NonlinearModelFit

*To*: mathgroup at smc.vnet.net*Subject*: [mg116139] Re: statistical comparison of parameters from two applications of NonlinearModelFit*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 3 Feb 2011 05:30:12 -0500 (EST)

For nonlinear models, it seems rash beyond words, to me, to assume "normality, homoscedasticity, and independence". Hence, I doubt most of the statistics returned mean much of anything. The same is true in linear modeling, but you have a better chance of testing the assumptions. Bobby On Wed, 02 Feb 2011 05:08:31 -0600, Ray Koopman <koopman at sfu.ca> wrote: > On Feb 1, 3:54 am, dantimatter <goo... at dantimatter.com> wrote: >> Hello Everyone, >> >> First, I must apologize if the following questions are silly; >> I am neither a statistician nor a particularly good Mathematica >> user (although I do occasionally pretend to be one). And now, >> on the question: >> >> I've got some data that I try to fit with NonlinearModelFit: >> >> time1 = {0, 100, 200, 300, 400, 500, 600, 700, 800}; >> data1 = {1, 0.92, 0.79, 0.81, 0.81, 0.78, 0.72, 0.75, 0.68}; >> >> nlm1= NonlinearModelFit[Transpose[{time1, data1}], >> a1*Exp[-b1*x] + c1, {{a1, .5}, {b1, 0.001}, {c1, 0.5}}, x, >> ConfidenceLevel -> .95] >> >> with results: >> >> In[11] := nlm1["BestFitParameters"] >> nlm1["ParameterConfidenceIntervals"] >> nlm1["ParameterErrors"] >> >> Out[11] = {a1 -> 0.295575, b1 -> 0.00334141, c1 -> 0.696761} >> Out[12] = {{0.179807, 0.411344}, {-0.000284765, 0.00696758}, >> {0.582507, 0.811014}} >> Out[13] = {0.0473121, 0.00148194, 0.0466929} >> >> I then do the same thing on a second data set: >> >> time2 = {0, 100, 200, 300, 400, 500}; >> data2 = {0.8, 0.73, 0.65, 0.61, 0.58, 0.57}; >> >> nlm2= NonlinearModelFit[Transpose[{time2, data2}], >> a2*Exp[-b2*x] + c2, {{a2, .5}, {b2, 0.001}, {c2, 0.5}}, x, >> ConfidenceLevel -> .95] >> >> In[15]:= nlm2["BestFitParameters"] >> nlm2["ParameterConfidenceIntervals"] >> nlm2["ParameterErrors"] >> >> Out[15] = {a2 -> 0.287834, b2 -> 0.00362383, c2 -> 0.516796} >> Out[16] = {{0.204254, 0.371413}, {0.00121124, 0.00603641}, >> {0.427572, 0.60602}} >> Out[17] = {0.0262627, 0.000758091, 0.0280362} >> >> Based on our understanding of the real system, we actually expect >> b2 <= b1, but this not what the fitting tells us (of course, there's >> a bit of error in the extracted parameters). >> >> So what I'd really like to know is: >> >> (1) how can I make sense of these "ParameterConfidenceIntervals" >> and "ParameterErrors" (how are they determined, and what is meant >> by them, beyond their brief description in >> http://reference.wolfram.com/mathematica/tutorial/StatisticalModelAnalysis.html >> )? >> >> and >> >> (2) is there a way to establish some kind of upper limit on the >> detectability of the difference between b2 and b1? >> >> Any thoughts you might have would be most appreciated. >> >> Many thanks! > > The parameter estimates are the values of the parameters (a, b, c) > that minimize the sum of squared differences between the observed > data and the values predicted from the model. > > Each ParameterError is an estimate of the standard error of > the corresponding parameter estimate. > > Each ParameterConficenceInterval is an interval that you are > (100*ConfidenceLevel)% certain contains the true value of the > corresponding parameter *if the model is complete (all the variables > that matter are included), the form of the model function is correct, > and the distributional assumptions about the errors (normality, > homoscedasticity, independence) are correct*. > > Each confidence level is constructed as > estimated_parameter +/- ( estimated_standard_error * > Quantile[StudentTDistribution[degrees_of_freedom], > (1 + ConfidenceLevel)/2] ), > where degrees of freedom = > #_of_data_points - #_of_estimated_parameters. > > Each estimated standard error is the square root of the product of two > factors -- one that is common to all the estimates, and one that is > specific to each estimate. > > The common factor is the mean square error of approximation: > the sum of squares that the parameter estimates minimize, > divided by the degrees of freedom. > > The specific factors are more complicated, so I'll say only that: > 1. They do not directly reflect the size of the data-approximation > errors; > 2. They are strongly tied to the values of the independent variable > (your 'time'); > 3. Ceteris paribus, they are inversely proportional to the number of > data points. > > Now we're ready to talk about detecting a difference between b1 and > b2. The formal test of the hypothesis that there is no difference > is a student t: (b1 - b2)/Sqrt[ASE[b1]^2 + ASE[b2]^2] = -.170 . > The degrees of freedom are data dependent, but in your case the t > is so small that it will never be significant, no matter what the > degrees of freedom are. > > To achieve significance, the t would have to be at least 12 times > its current value, which means you would need 144 times as many > sample points as you currently have. (Note: I've cut a lot of > corners here. A proper analysis would be much messier but would > lead to the same conclusion -- you need a whole lot more data, > on the order of 200 times as much as you have shown us.) > > Finally, here's some code that should reproduce some of the > NonlinearModelFit output: > > r1 = FindFit[Transpose@{time1,data1}, a1*Exp[-b1*x]+c1, > {{a1,.5},{b1,.001},{c1,.5}}, x] > f1 = a1*Exp[-b1*time1]+c1; > e1 = f1 - data1; > df1 = Length@data1 - 3; > var1 = (e1.e1/.r1)/df1 (* error variance *) > cov1 = var1*Inverse[Transpose at #.#]&@Outer[D[##]&,f1,{a1,b1,c1}]/.r1; > ase1 = Sqrt@Diagonal@cov1 (* ASEs *) > cov1/Outer[Times,ase1,ase1] (* correlations *) > ({a1,b1,c1}/.r1) + Outer[Times,ase1, (* CIs *) > {-1,1}]*Quantile[StudentTDistribution[df1],.975] > -- DrMajorBob at yahoo.com

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