I am struggeling with the following problem: Let A, B, C be normally distributed random variables, the means and variances are not known. Assume further that we have a sample of each of the distributions of A, B, C. With these samples one could calculate a confidence interval for the mean for each of the three random variables A, B, C (for example using Student's t-distribution), for e.g. alpha=0.05.

Now consider the following quantity with the (true) means mean(A), mean(B), mean(C) of the distributions of the random variables A, B, C:

m= (mean(A) + mean(B) + mean(C))/3,

v = 3 * \sqrt{(m-mean(A))^2 + (m-mean(B))^2 + (m-mean(C))^2} / (2 * m) (the variation coefficient of the sample mean(A), mean(B), mean(C)).

How can one calculate a confidence interval for v, given the confidence intervals for mean(A), mean(B), mean(C)?

It would be great if anyone had an idea or hints for literature. Thanks!