0
$\begingroup$

I have conducted a measurement of sample on an instrument. I thus have the value of my samples, of a blank, and the standard deviation of said blank. I would like to execute operations on the value of my sample and need help understanding how standard deviation carries over across these operations. Here,± designates SD. Sample-1 and sample-2 are replicates.

Starting data:

Blank = 136 ± 1.37
Sample-1 = 374
Sample-2 = 394

I then take into account the blank and standard deviation by doing Adjusted sample = Blank - Sample. I get:

Sample-1 = 238 ± 1.37
Sample-2 = 258 ± 1.37

I then do a linear operation (e.g. multiply by k) to convert to different units on both the value and standard deviation (to account for sample volume, which is equal in this case).

Sample-1 = 24 ± 0.09
Sample-2 = 26 ± 0.09

Now I'd like to get to the true measurement of my sample, so I average the two replicates. In terms of SD, I do SD(Sample) = SD(Sample-1, Sample,2) + AVERAGE(SD(Sample-1) + SD(Sample-2). I get:

Sample = AV(24, 26) ± SD(24, 26) + AV(0.09, 0.09) 
Sample = 25 ± 1.5

I then convert the sample to a different unit again by a linear operation applied to the value and SD but lets disregard that here.

Now, I have a different measurement with its own standard deviation let's call it Count = 2.86 ± 6.76.

I would like to divide Sample/Count, here is where I think I make a mistake as I simply do SD(SAMPLE)/SD(Count). Is that correct?

Sample and Count are not technically independent but I think can be considered as such here.

$\endgroup$

1 Answer 1

0
$\begingroup$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.