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How to get linear fit standard deviation?

Asked by Adam Cervenka on 11 Aug 2018 at 15:16
Latest activity Commented on by David Goodmanson about 12 hours ago

After linear fitting I need to get value of standard error. When I use this

A = [x(:),ones(length(x),1)];
[u,std_u] = lscov(A,y(:));
for x=[2550	2450	2352	2256	2162]
for y=[1,93	2,11	2,39	2,44	2,63]

I get right values.

But when I use the same code where

x = [5,13e-19	5,55e-19	5,9e-19	6,370e-19	6,77e-19]
y = [40,40	39,97	39,21	38,67	38,75]

std_u and u value are 0 but they shouldn't be.

Did I use the wrong function? How can I get these values?


The documentation section on Accuracy of Floating-Point Data (link) might be helpful.

So what should I do now?

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1 Answer

Answer by David Goodmanson on 13 Aug 2018 at 1:11
 Accepted Answer

Hi Adam,

Time to rescale.

S = 1e18;     % scale factor to bring x into the same ballpark as the second column of A
x = S*[5.13e-19	5.55e-19	5.9e-19	6.370e-19	6.77e-19];
y = [40.40	39.97	39.21	38.67	38.75];
A = [x(:),ones(length(x),1)];
[u,std_u] = lscov(A,y(:))
u=                           (rescaled)
std_u =

In terms of the rescaled x you get a good result.

The problem being solved is basically ux = y, or (u/S)(Sx) = y. So for the rescaled problem, u -> u/S.

To get back to the original problem you have to multiply the rescaled u by S. But since only the first column of A was rescaled, only the first element of u and std_u have to be changed. Then

u(1) =  -1.1156e+19          (original)
u(2) = 46.0310
std_u(1) = 2.0929e+18
std_u(2) = 1.2499

Lots of scale factors that are within a few powers of 10 of S=1e19 are possible, and they don't change the 'original' results above (by any significant amount). But all in all, it makes at least as much sense to stay with the rescaled problam, because then the elements of u and std_u are of comparable size. I used a scale factor of 1e18 instead of 1e19 because for example if x were in meters, then the rescaled x is in the SI unit of attometers.


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Not that I know of. Is there evidence that these numbers are inaccurate?

No, but I need values with more decimal places (about ten).

You could use 'format long' and get more digits, which are there but not being displayed. However, your inputs x and y appear to only have four significant figures. Unless your x and y are good to many more sig figs than they appear to be (which I guess is possible), any more than four sig figs in the result is unjustified.

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