Here's an anonymous function that computes the 95% CI based on the tinv method which requires that your data approximately form a normal distirbution. See this link for more information on this function.
% x is a vector, matrix, or any numeric array of data. NaNs are ignored.
% p is a the confident level (ie, 95 for 95% CI)
% The output is 1x2 vector showing the [lower,upper] interval values.