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I’m working with summary statistics (means, standard errors, and sample sizes) for two variables, and I’ve derived:

  • SD of variable A from its standard error and sample size
  • The ratio (P = A / B ) (e.g. expressing A as a percentage of B)

I now want to estimate the standard deviation of this ratio. My current approach is: SD(P) ≈ SD(A) / mean(B)

Is this the correct approach or should I be dividing by the SD (B). i.e: SD(P) ≈ SD(A) / SD(B)

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    $\begingroup$ Neither would be correct. Under certain conditions, such as zero correlation between $A$ and $B,$ $P\gg 0,$ and zero probability that $B=0,$ you can approximate the SD of the ratio using the delta method. $\endgroup$ Commented 2 days ago
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    $\begingroup$ If I recall correctly, a number of posts on site already discuss variance or the standard error of ratio (not the same thing as the ratio of standard deviations); some discuss the case where A and B are independent (or at least, uncorrelated) and some discuss the case where A and B are dependent. Some may do both. Try a few site searches $\endgroup$ Commented 2 days ago

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Are both measured with error and both positive? In that case, a delta method approach is one option. Of course, the distribution for a ratio of two random variables can often only rather poorly describe by a normal distribution, working on a log-scale often works better.

A delta method approach gives you this, I believe:

$$\operatorname{SE}\left(\frac{A}{B}\right) \approx \frac{\mu_A}{\mu_B} \sqrt{\frac{\text{SE}_A^2}{\mu_A^2} + \frac{\text{SE}_B^2}{\mu_B^2} - \frac{2\rho \text{SE}_A \text{SE}_B}{\mu_A\mu_B}}$$ and if the two estimates are not correlated (i.e. $\rho = 0$ )

$$ \operatorname{SE}\left(\frac{A}{B}\right) \approx \frac{\mu_A}{\mu_B} \sqrt{ \frac{\text{SE}_A^2}{\mu_A^2} + \frac{\text{SE}_B^2}{\mu_B^2}}.$$

You can easily visualize what a distribution looks like, e.g. using R via hist(pmax(rnorm(n=1000, 10, 2.5), 1e-3) / pmax( rnorm(n=1000, 10, 2.5), 1e-3), breaks=100).

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  • $\begingroup$ Thanks the challenge is that I only have the summary statistics mean, SE and sample size, so do not know the co-variance between A and B. $\endgroup$ Commented Apr 28 at 10:06
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    $\begingroup$ But, A and B will be correlated. $\endgroup$ Commented Apr 28 at 10:09
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    $\begingroup$ You'll have a hard time doing anything if there is correlation, but you have no idea what it is (I guess you can plot the SE of A/B for all possible values). You might have to guess or elicit from experts or estimate from other data or something else... $\endgroup$ Commented 2 days ago
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    $\begingroup$ Fieller's Theorem is also an option - when you know the correlation. As @Björn implies, without knowledge of the correlation between A and B, you are on very shaky ground because, whilst the equations will still "work", your conclusions will be unreliable. $\endgroup$ Commented 2 days ago
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Are you really interested in the standard deviation or is your actual interest a confidence interval? In the latter case, you can obtain it with the MOVER method by Donner & Zhou (2012):

Donner, Allan, and G. Y. Zou. "Closed-form confidence intervals for functions of the normal mean and standard deviation." Statistical Methods in Medical Research 21.4 (2012): 347-359.

Section 2.2. gives the formula for the confidence interval for $\theta_1/\theta_2$ in terms of the point estimators $\hat{\theta}_i$, the limits of the two confidence intervals $(l_i, u_i)$ for $i=1,2$, and the estimated correlation $\hat{\rho}$ between $\theta_1$ and $\theta_2$.

There is even a ready to run R function pairwiseCI::MOVERR() available in the R package pairwiseCI. This function does not have an argument for $\hat{\rho}$, though, so it presumably makes the assumption $\rho=0$.

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If you know that $\rho \ne 0$ but don't know what it is, then you can only make some guess at the SE of the ratio, using the first formula in Bjorn's answer. You may be able to take some guess as to $\rho$ (e.g. it is positive) and this may help, but, in general, there's no exact answer. How badly this lack of knowledge affects your estimate will depend on

$\frac{SE_A SE_B}{\mu_a \mu_b}$ which you do know.

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