Confidence Intervals for a relative risk where i lack the underlying data I have two estimated rates and their 95% confidence intervals but not the underlying data.  I take the ratio of the two to get a relative risk, but how do I determine the 95% confidence intervals for that relative risk?
Example:  At baseline I have an estimated rate of 38.3 (95% CI, 36.0-40.8) and in the study group I have an estimated rate of 45.2 (95% CI, 43.2-47.2).  The relative risk is 1.18, but what are the 95% confidence limits.  I do not have the underlying data upon which the rates were calculated. 
 A: So there are methods for propagating error through arithmetic opperations that are used in many areas of science. (for example look at this.)
As the relative risk is
$$R = \frac{X}{Y}, \text{with std. of } \sigma_x , \sigma_y$$
We are interested in error through the divisor operator which is 
$$\sigma_R = R\sqrt{\big(\frac{\sigma_x}{X}\big)^2 + \big(\frac{\sigma_y}{Y}\big)^2}$$
However, if X and Y are gaussian distributed R will not be. 
Edit:
Example

Example: At baseline I have an estimated rate of 38.3 (95% CI, 36.0-40.8) and in the study group I have an estimated rate of 45.2 (95% CI, 43.2-47.2). The relative risk is 1.18, but what are the 95% confidence limits. I do not have the underlying data upon which the rates were calculated.

$$R = 1.18$$
95% confidence interval is roughly $2\sigma$ on either side, so $\sigma_b = 1.25$ and $\sigma_s = 1$. So then
$$\sigma_R = 1.18\sqrt{\big(\frac{1.25}{38.3}\big)^2 + \big(\frac{1}{45.2}\big)^2}$$
$$\sigma_R = 0.053$$
So the relative risk is approximately $1.18\pm0.1$ with 95% confidence.
A: An additional approach would be to assume that the sampling distribution of log relative risk is normally distributed. Although the same is true for the relative risk, the sample size needs to be larger for this fact to be helpful in practice. In fact, it is often the case that the most commonly used formula for relative risk confidence intervals (Wald CIs) makes this assumption about the log relative risk. Using OP's example, we have:
set.seed(12345) # For reproducibility
rr.a <- 38.3
rr.b <- 45.2
ci.a <- c(36, 40.8)
ci.b <- c(43.2, 47.2)

We can calculate the standard deviations on the log scale. They would be the difference between the log confidence intervals divided by about 4. qnorm is the quantile function of the normal distribution.
l.sd.a <- diff(log(ci.a)) / qnorm(.975) / 2
l.sd.b <- diff(log(ci.b)) / qnorm(.975) / 2

We next assume that the log relative risks are normal with mean as the estimated log relative risk and standard deviations as the values above. We can draw a large number of samples assuming both distributions then take the difference between both samples.
nboots <- 1e4
log.rr.s <- rnorm(nboots, log(rr.b), l.sd.b) - rnorm(nboots, log(rr.a), l.sd.a)

And for results, the exponentiated 2.5th and 97.5th percentiles:
exp(quantile(log.rr.s, c(.025, .975)))
#     2.5%    97.5% 
# 1.094151 1.274278 

In OP's specific example, it probably does not make a difference since the CIs for the RR are themselves symmetric. This is more likely when the RRs are huge numbers as in OP's example.
