I have some mean amount of DNA from my ChIP assay (A), and some mean amount from the IgG (B).
Call fold enrichment C. C=A/B
I have standard deviations for both A and B. However, If I calculate the standard deviation for C according to basic propagation of error rules for sample quotients, my standard deviation becomes impossibly large.
I'm looking for a different way to calculate a standard deviation, standard error, relative error or confidence interval for fold enrichment (C). I've heard that jackknife resampling or bootstrapping a CI may be my best bet, but both of those methods seem silly given that I have only 3 technical replicates (n).
Many thanks for your help!
ANOVA won't work for finding the error in fold enrichment, although it will certainly be a good option for comparing my multiple samples once I have a value for the error in fold enrichment.
i also think you should try using ANOVA using any statistical package available to you
Carl:
I’m agree that ANOVA is not the solution, first looking at your question and second, the measured variables are different. Your C variable (fold enrichment) is a result for the relationship between A and B, for that reason its variability is affected like you said, tending to be larger. I think that with only n=3, you can’t do nothing in terms of statistical inferences (estimates or significant test) nor any resampling method (bootstrapping or jacknife).
Best regards
How exactly are you calculating the standard deviation of the quotient? While, naturally, the variance of a relationship between two random quantities will be larger than either individual variance, it seems unlikely to me that you would be getting "impossibly large" results for C unless the variances of A and/or B were already very large to begin with. Standard rules for error propagation state that the error of a quotient is simply the sum of the fractional errors of the components of that quotient.
What do the standard deviations of A and B look like?
I would suggest the following Taylor series approximation for the variance of C:
Numerator = E(A^2)(E(B))^2 - 2*E(A*B)*E(A)*E(B) + E(B^2)(E(A))^2
Denominator = [E(B)]^4
where E(x) = the average for the variable x.
Then use a VERY conservative critical value: alpha=0.05 with df=1: t_(alpha/2) = 12.71 (that seems really high, but with df = 3 - 2, you don't have a lot of wiggle room)
then you could produce a CI as:
E(A/B) ± t_(alpha/2) * sqrt(numer/denom)
This won't be the most elegant solution, but it would be mildly defensible with a very small sample size.
Hope this helps.
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