Standard deviation calculation 1

[Pucp]

Hi

I have measured a dimension of particles (L) as a function of time. I measured several particles for each time. Then I got for each time a mean value and a standard deviation. No problem with that. Then I plot my data (mean value of L and time) in the form:

L[sup]2[/sup] - Lo[sup]2[/sup] (L-squared minus L-initial squared) vs. time

My question is: What should I do with the standard deviation, how to operate it in order to plot it as bars in the same graph??.

Thanks in advance

 

[Denial]

Is there some fundamental a priori reason you are dealing with "L-squared" rather than simply "L"?

Why not use simple linear regression of L (or L-squared) against time? The fact that your sample values are clustered on a small number of time values does not invalidate the approach.

 

[Pucp]

Dear Denial

Thanks for the reply. I am dealing with L[sup]2[/sup] vs. time because I am trying to show if my data follows the solution of the differential equation for diffusion (R[sup]2[/sup] = constant * time).

You can use L vs. squared root of time as an option to show the same but I think that is when you assume that your initial radius is zero. In my data for the time zero I have already an initial size (Lo) and that is why I am plotting L[sup]2[/sup] - Lo[sup]2[/sup] vs. time.

I still do not know what to do with the standard deviation..

 

[IRstuff]

Why wouldn't you simply plot them as boundaries to your mean value line? There are lots of constructs you can use.

TTFN

FAQ731-376

 

[btrueblood]

Presumably you have a std. dev. associated with your mean L. Add the std. dev. (sigma) to L, and calculate your L^2-Lo^2 term, i.e. (L+sigma)^2 - Lo^2 giving you the upper standard deviation of the term; similarly (L-sigma)^2 - Lo^2 gives the lower std. deviation limit.

But - are your L's truly normally distributed, i.e. can you plot a bell-shaped curve of L vs. no. of particles with dimension L? I'm asssuming you have some type of chemical or abrasive reaction occurring, where the particle dimension is diminishing or growing with time. In the former case, the particle dimension may diminish to zero, and a calculation of L-sigma may give a negative number...clearly not a real solution, thus using Gaussian probability math may give you goofy solutions if you extrapolate from your base data to endpoint conditions.

The distribution for the particles is likely more of a gamma function, which is a pain to calculate probability distributions (?not sure of that term, probability vs. frequency curve is what I'm talking about) with. You might have better luck fitting a Weibull curve to the particle distribution data at any given time; the Weibull function has more tractable math for probability distributions.

 

[Denial]

So you have a hypothesis [(L[sup]2[/sup]) proportional to time], and you want to see whether your recorded data supports it.

I am not formally trained as a statistician, but seem to end up doing a bit of stats work for my employer because nobody else is silly enough to put their hand up. So don't put too much faith in what I say. With that proviso, let me spell out how I would start to investigate your problem.

I would regress L[sup]2[/sup] against time, ie try a model of the form
L[sup]2[/sup]=a*t+b.
Among the standard results that would come out from this would be an estimate for the value of b, and confidence levels for this estimate. (This implicitly incorporates the standard deviation that you "still do not know what to do with".)

You are now in a position to decide whether you can accept your hypothesis that b=0. If you decide that you can accept it, then you can determine your proportionality constant (a) by doing a second regression, this time using a model that presupposes b=0, ie using a model of the form
L[sup]2[/sup]=a*t.

 

[Pucp]

Thanks a lot guys. I will follow btrueblood advice.

Regards

Ian