An instrument used for measurements has a known measurement imprecision of $CV = 10 \%$. Thus, if the variable to be measured has a true value $x_{true}$, the range of possible measured values will be given by:
$x_{measured} \sim \mathcal{N}(x_{true},\,(0.1\cdot x_{true})^2)$
Say we have two measurements, $x_{1_{measured}} = 100$ and $x_{2_{measured}} = 50$. We can easily calculate the 95% CI for $x_{1_{true}}$ and $x_{2_{true}}$ in R:
cv <- .1
x1_measured <- 100; x2_measured <- 50
x1_true_ci <- x1_measured / qnorm(c(.975, .025), 1, cv)
x2_true_ci <- x2_measured / qnorm(c(.975, .025), 1, cv)
However, I would like to determine a 95% CI for the ratio $\frac{x_{2_{true}}}{x_{1_{true}}}$. My current approach is to simulate ratios and then use quantile() to find the 2.5% and 97.5% quantiles as follows:
ratio_sim <- (x2_measured / rnorm(1e6, 1, cv)) / (x1_measured / rnorm(1e6, 1, cv))
ratio_ci <- quantile(ratio_sim, c(.025, .975))
log(ratio_sim)appears to be close to normally distributed, although I would assume that it is actually a kind of ratio distribution? Anyway, calculating the 95% CI from log(ratio_sim) yields values very similar to the quantiles calculated previously:
ratio_ci2 <- qlnorm(c(.025, .975), mean(log(ratio_sim)), sd(log(ratio_sim)))
My question: is there a more direct, less computationally intensive way to estimate the 95 % CI for the ratio, without using simulation?
log(x1)andlog(x2)if all the information I have iscv <- .1; x1_measured <- 100; x2_measured <- 50? I can simulate values ofx1_trueby dividingx1_measuredbyrnorm(1e6, 1, cv)and then simply usevar(x1_true)to find the variance, but how do I find it without simulation? Won't the distribution ofx1_truebe a kind of reciprocal normal distribution? – Peder Holman Mar 09 '22 at 17:50