is there any way to calculate confidence intervals (CI) of a ratio? I need to calculate a ratio of 2 values and each value has its own  95% confidence intervals. The question is: how can I calculate the  95% confidence intervals of the ratio? 
 A: In general, the variance of the ratio of two random variables can be calculated using the delta method. Specifically, $$\mathrm{var}\left(\frac{X}{Y}\right) \approx \frac{\mathrm{var}[X]}{E[Y]^2} - \frac{2E[X]}{E[Y]^3}\mathrm{cov}[X,Y] + \frac{E[X]^2}{E[Y]^4}\mathrm{var}[Y] $$
You don't give specifics about the two random variables you want to find the ratio of, except that you have the values and the confidence intervals. If you know that the values are the maximum likelihood estimates of the two values and you are willing to take a ride on the assumption train to the mythical land of Asymtopia, then you can treat those values as normally distributed. Specifically, the point estimates can be treated as $E[X]$ and $E[Y]$. You can find $\mathrm{var}[X]$ and $\mathrm{var}[Y]$ by inverting the confidence intervals
$$\text{(e.g. }\ UCL_X - E[X] = E[X] - LCL_X = 1.96\frac{\sigma}{\sqrt{n}},\text{ solve for }\ \sigma^2 = \mathrm{var}[X]).$$
In for a penny in for a pound, so you might as well assume that $\mathrm{cov}[X,Y] = 0$, unless you have good reason to suspect otherwise. After calculating this approximate value of $\mathrm{var}[\frac{X}{Y}]$ it is easy to come up with a confidence interval. So with a small amount of math and very large dose of assumptions you can arrive at a confidence interval for the ratio.
User beware, Asymtopia is a mythical land for a reason. Some very helpful approximating gnomes certainly live there, but there is also a sizeable population of filthy, devious, lying uncertainty trolls that feast on half-baked assumptions and drink the blood of the credulous.
A: The other answer offers a nice approach, when the two intervals are created using standard normality assumptions. In the more general case, it may not be possible to find the variance of each estimate given only the CI.
More generally, consider two parameters $\theta_1$ and $\theta_2$ with $(1-\alpha)\times 100\%$ confidence intervals $(L_1, U_1)$ and $(L_2, U_2)$ respectively. The set
$$\mathcal C = \{(a,b)|L_1 < a < U_1, \ L_2 < b < U_2\}$$
is a joint confidence region with confidence level $1-\tilde\alpha$. By the Bonferroni inequality, this joint confidence level is at least $1-2\alpha$. If the two confidence intervals are independent, then the joint confidence region is equal to $(1-\alpha)^2$. For example, if the original CI's have a $95\%$ confidence level, then the familywise confidence level is at least $90\%$ (and is exactly $90.25\%$ in the case of independence).
Since the bivariate parameter $(\theta_1, \theta_2)$ will belong to the set $C$ with probability $1-\tilde\alpha$ (as usual, this statement must be interpreted with care), confidence intervals for any function $g(\theta_1, \theta_2)$ can be easily derived. In general, the CI is given by $(\tilde L, \tilde U)$ where
$$\tilde L = \inf_{(\theta_1, \theta_2) \in \mathcal C} g(\theta_1, \theta_2)$$
$$\tilde U = \sup_{(\theta_1, \theta_2) \in \mathcal C} g(\theta_1, \theta_2).$$
For ratios of the form $\theta_1/\theta_2$, we get
$$\tilde L = \frac{L_1}{U_2} \quad\quad\text{and}\quad\quad \tilde U = \frac{U_1}{L_2}.$$

To summarize, if $\theta_1$ and $\theta_2$ have CIs $(L_1, U_1)$ and $(L_2, U_2)$ respectively (with confidence level $1-\alpha$), then $(L_1/U_2, U_1/L_2)$ is a valid confidence interval for $\theta_1/\theta_2$ with a confidence level which is at least $1-2\alpha$.
Note: The exposition here assumes that all values are positive. If not, a confidence interval can still be obtained using the more general version, although one of the confidence bounds may be $\pm \infty$.
