There are some subtleties depending on how your 95% confidence interval was computed, but typically:
CI = tinv(.975,N-1)*std(x)/sqrt(N)
T = mean(x)/std(x)*sqrt(N);
so, if you have the sample mean M (M=mean(x)) and the confience interval CI, you can compute T as:
T = M/CI*tinv(.975,N-1);
If you do not know the sample size N, you may assume N is relatively large and approximate the cumulative T distribution by a cumulative Normal distribution to get:
T = M/CI*1.96;
The subtlety involves how your 95% CI was originally computed. The equation above assumes that this was a two-sided 95% interval (meaning that the population mean of x is expected to lie between M-CI and M+CI with 95% chance). Some times people use one-sided 95% intervals instead (meaning that the population mean of x is expected to be greater than M-CI with 95% chance) and that would mean that in the equations above you would need to change tinv(.975,N-1) to tinv(.95,N-1), and in the last equation change 1.96 to 1.64.
