Imagine we have two sets of data m1 and m2. m1 components have the dimension of [m^2/s] and m2 components are measured in [cm^2/s].

m1={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

m2={31, 32, 33, 34, 35, 36, 37, 38, 39, 40}

Assuming that B is the proportionality constant that is estimated as a straight line fitted to the points with the least square method, it can be estimated by

NonlinearModelFit[Transpose[{10^-4 m1, m2}], a x, {a}, x, 
 VarianceEstimatorFunction -> (Mean[#^2] &)]


But, the estimated error variance is NOT non-dimensional. How can I normalize the error variance scale?


Sqrt[NonlinearModelFit[Transpose[{10^-4 m1, m2}], a x, {a}, x, 
   VarianceEstimatorFunction -> (Mean[#^2] &)]["EstimatedVariance"]]


Sqrt[NonlinearModelFit[Transpose[{m1, 10^4 m2}], a x, {a}, x, 
   VarianceEstimatorFunction -> (Mean[#^2] &)]["EstimatedVariance"]]


It looks quite inconsistent if I write B=52857.14 +/- # [cm^2/s] while B has no dimensions!

  • 1
    $\begingroup$ Be careful! You are fitting a model without a constant term: it's a poor fit to these data. Also: what does it mean for you to "normalize" the error variance? After all, if I tell you it's two miles from my house to work and somebody else tells you it's 3.2 kilometers, we're perfectly consistent: it's merely a change of units. How should we reply if someone asks us to "normalize" how we report that distance? $\endgroup$ – whuber Oct 26 '11 at 14:54
  • $\begingroup$ Actually, i'm looking for a way to show the margin of error. The calculation actually relates to turbulence modeling and I cannot use a fit with general form of m1=A+B m2. On the other hand, it doesn't make sense to mention a dimension for margin of error in this case. $\endgroup$ – K-1 Oct 27 '11 at 11:08
  • $\begingroup$ Margin of error of what? Of predictions made from the model or in the coefficients? In either case there is a unit of measurement involved: in the former, the M of E is in the units of the response variable m2; in the latter, it is in the units of the proportionality constant B. $\endgroup$ – whuber Oct 27 '11 at 16:44

I am not familiar with the coding language you are using, but it seems, that you are fitting the linear regression model:


Or vice versa

$$m_2=A+B m_1$$

Since you measure both $m_1$ and $m_2$ in $cm^2/s$, then $A$ has the dimension $cm^2/s$ and $B$ is dimensionless. Hence any associated measures (such as variance and standard error) with $B$ are dimensionless too.

  • $\begingroup$ The language is Mathematica (except the double stars, "**", make no sense--they might have been added in an attempt to format the code for presentation here). $\endgroup$ – whuber Oct 26 '11 at 14:49
  • $\begingroup$ As it is shown the "margin of error" might have the dimension of cm^2/s or m^2/s. Simply, I calculated variance of error. $\endgroup$ – K-1 Oct 27 '11 at 11:13
  • $\begingroup$ Margin of error for what? If for $B$ as you write in your question, still you get that it is dimensionless, since $B$ is dimensionless. $\endgroup$ – mpiktas Oct 27 '11 at 11:21

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