standard deviation of a division

M

michu

New Member
#1
#1
Hi,

I have a rather easy question: I have a batch of different dried fruits, let's say n=100, with a diameter of 8±2 cm, 2 being the standard deviation. After throwing them in water the distribution of the batch's diameters is let's say 16±5 cm because of the swelling. This results in a size ratio of 2, but what about the standard deviation of this calculation?

I tried propagation of uncertainty with covariance=0, so:

σ(A/B) ≈A/B*SQRT( (a/A)^2 +(b/B)^2 ). a and b are the standard deviation of A and B, respectively.
Hi,

I have a rather easy question: I have a batch of different dried fruits, let's say n=100, with a diameter of 8±2 cm, 2 being the standard deviation. After throwing them in water the distribution of the batch's diameters is let's say 16±5 cm because of the swelling. This results in a size ratio of 2, but what about the standard deviation of this calculation?

I tried propagation of uncertainty with covariance=0, so:

σ(A/B) ≈A/B*SQRT( (a/A)^2 +(b/B)^2 ). a and b are the standard deviation of A and B, respectively.

When using my example I get 0.8, so is the size ratio of the two population 2±0.8? This seems not too bad given that the extreme ratios 21/6 and 11/10 are 3.5 and 1.1, respectively.

Thanks a lot for the feedback, curious to see whether this is good enough or if there are better means.
 
K

katxt

Active Member
#2
#2
The method you have used works well enough for small a/A and b/B but is only approximate.
You could try it by doing a simulation, either generating A and B using say a normal distribution or using resampling if you have the data.
Is it possible to identify individual fruit before and after soaking?
 
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M

michu

New Member
#3
#3
katxt said:
The method you have used works well enough for small a/A and b/B but is only approximate.
You could try it by doing a simulation, either generating A and B using say a normal distribution (this gives SD about 1.0) or using resampling if you have the data.
Is it possible to identify individual fruit before and after soaking?
Click to expand...
hi katxt, thanks a lot for your reply! our a/A and b/B are not that small, 25 and 31 %, respectively, so the approximation might not be great. However, A and B are not that far from having a normal distribution, so your idea seems cool! And I agree that observing the fruits one by one would be better, but this is not possible in that process.

So if you don't mind I would be very happy if you could emphasize a bit more on the method you proposed? Thanks a lot in advance and best regards,
M
 
K

katxt

Active Member
#4
#4
When I read your first post, I only looked at the formula and didn't check your calculations. Now that I've seen your latest post, I realize that you haven't taken the sample size n = 100 into account.
a and b are not the SDs. They are the uncertainties in the sample means (the standard errors) = SD/sqrt(n)
a = 5/sqrt(100) = 0.5 and a/A is 0.5/16 = 3.1% b = 2/sqrt(100) = 0.2 and b/B = 0.2/8 = 2.5% These are small enough that the approximation is fine. Completing the calculations, SD of A/B = turns out to be 0.08 rather than 0.8
The suggestions about resampling are still true, but unnecessary in this case. They would be useful with badly non normal data.
kat
 
K

katxt

Active Member
#6
#6
You have calculation involving uncertain variables and in the formulas a and b are estimates as to how uncertain they are. For your calculation (as I understand your posts) you measure 100 dried fruit and 100 fruit after soaking. You find the means, and divide. So the variables A and B are the sample means, and the a and b are the standard deviation of those means (not of the individual fruit).
The SD of an estimate from a sample is known as the standard error but it is the SD of the mean, not individual fruit.
If you measure a length once and get 120 mm +/- 2 mm then A = 120 and a is 2.
If you measure n = 100 weights and get a mean = 50 g with SD = 5 g then A is 50 and a = SD of A = SE = 5/sqrt(100) = 0.5
 
M

michu

New Member
#7
#7
Hi katxt,

Thanks a lot for that beginners course in statistics, I think all is clear and the standard error of my division of 0.08 makes sense!
Best,
Michu
 
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