How to calculate 95%CI of basic reproduction number from an available (linear) equation? I am working on my thesis and Prof. asked me to add 95%CI on the basic reproduction number R0, but I don't know how to do in R.
As I have an equation:
R0 = Re/(1-p*VE)
where: R0 is the basic reproduction number to be estimated; Re: Effective reproduction number (I have already estimated at 1.27 (95%CI: 1.22 - 1.33)); p: vaccination coverage (p = 0.946);  VE: vaccine effectiveness (VE = 80.19% (95%CI: 70.41-89.98))
I have not much exposed with statistic, please help me with this.
 A: From what you say in your Comment, I am not sure what the distribution of
R0 might be. Certainly, I see no reason to suppose the distribution would
be symmetrical.
Suppose you have $n = 30$ observations as in my fictitious data vector x, summarized below:
summary(x)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
 0.03995  2.73616  7.92256  9.38491 16.43023 30.52458 
boxplot(x, col="skyblue2", horizontal=T)


Data are obviously skewed, and a normal probability plot below
is far from linear, so t and Wilcoxon CIs for (population mean
and median, respectively) are inappropriate.
qqline(x, datax=T, col="blue")


So I will make a simple quantile 95% confidence interval for the population mean $\mu,$
which makes no assumptions about the population except that it does have a mean.
The procedure is to taka a relatively large number of re-samples of size $n=30$ with replacement from x, find the mean a.re of each re-sample, and take quantiles
$.025$ and $.975$ to be the endpoints of a 95% bootstrap confidence interval $(6.57, 12.35)$ for $\mu.$
set.seed(2021)
a.re = replicate(10^4, mean(sample(x, 30, rep=T)))
quantile(a, c(.025,.975))
     2.5%     97.5% 
 6.571277 12.353882 

hdr = "Bootstrap Dist'n of Resampled Mean"
hist(a.re, prob=T, col="skyblue2", main=hdr)
 abline(v = c(6.57, 12.35), col="maroon", lwd=2, lty="dotted")


This is a random procedure, so I have shown the 'seed' used in the procedure.
If you use the same seed and data I did, you will get the same result each time
you run the program. Also, two additional runs with different seeds gave
CI endpoints that are the same when rounded to two places.

There are
many different kinds of bootstrap CIs. This particular method is easy to
understand and I have gotten reasonable results using it. Because I simulated the fictitious data I know that the population mean is $\mu = 10,$ which is included
in the resulting CI. I also know that the fictitious data are exponentially distributed. The procedure below is based on the exponential distribution and is
known to be an appropriate one for such data. [The the quantity $\bar X/\mu \sim \mathsf{Gamma}(n,n)$ for a sample of size $n.]$ The resulting CI $(6.76, 13.91)$ is similar to the bootstrap CI.]

mean(x)/qgamma(c(.975,.025), 30, 30)
[1]  6.760025 13.909833

Notes: (1) If your R0 data are extremely skewed, or if you have less than a dozen values, then please leave a Comment and one of us may have additional suggestions.
(2) The fictitious data used above was sampled in R, using the following code. Use the same seed and you will get the same x.
set.seed(1234)
x = rexp(30, .1)

