If the distributions are independent then: Var(conbined) = Sum [Var(dist i)].
But, if the distribution are not independent then: Var(combined) = Sum[Var(dist i)] + 2 Sum[Cov(Cov ij)], or Var(combined) = Sum [Var(dist i)] 2 [Corr ij(Var(dist i) * Var(dist j)]. Var is the square of the standard deviation of each distribution, Cov is the covariance of the combinations by two of all the distributions, and Corr is the correlation coefficient of the respective combinations. [See page 216 of William Feller's An Introduction to Probability Theory And It's Applications] Good luck, WDA end "David Robinson" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > I've searched for this, and failed to come up with an answer. It's not > a homework question, or even a purely accademic question - I'm trying > to combine psychoacoustic data from a number of sources, because I > need to calculate the percentage of the population who are able to > detect certain sounds. > > > Here's the problem: > > I have several measurements of the SAME QUANTITY from different tests. > For each test, I have: > > number of subjects, mean value, standard deviation (or standard error > of the mean.) > > (All standard errors will be converted to standard deviations - I know > they don't quite mean the same thing (even when divided by root N), > but I believe SD is more appropriate in this context). > > I want to combine all the data, to get a global mean, and a global > standard deviation. The results I want for these two values should be > the results I would get if I had ALL the original data from all the > tests, and simply calculated the mean and standard deviation over all > the data. > > (I do not have all the original data - some of it does not exist > anymore) > > I can calculate the global mean easily (n1*x_bar1 + n2*x_bar2 + > ...)/n_total > > I do not know how to calculate the global standard deviation. Please > can you help? > > (I don't understand it, but from what I have read I believe the usual > technique of simply adding the squares of the SDs is not appropriate > because I have (hopefully) correlated data, and different sample > sizes) > > Cheers, > David. > http://www.David.Robinson.org/ > > > P.S. The SD is AS IMPORTANT as the mean for this work, because I will > integrate the resulting gaussian to give a psychometric function. > (Well, I have the integrated form giving a psychometric function, and > I can put the numbers in!) I have some real measured psychometric > functions to compare with, and these match the psychometric formula > predictions (from mean and SD values) very well. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
