No, that’s not right at all – the formula given by Aheto below relates to the difference of two means, which is something totally different.

I assume that ‘the concatenation of the two samples’ means that they are simply to be merged.

If so, then you need to get back to N1, T1 and SSq1 for sample 1, the quantities that the mean and SD and SE were calculated from.

Then similarly N2, T2 and SSq2 for sample 2.

Then use N=N1+N2, T=T1+T2, SSq=SSq1+SSq2.

Then use these to calculate the pooled mean, and the SD for the combined sample and hence the SE of the pooled mean.

Essentially, T1 = N1*mean1

SSq1 = N1*mean1^2 + (N1-1)*SD1^2

SD1 = SE1*sqrt(N1)

With corresponding formulae for sample 2, and then to go back from the cumulants for the pooled sample.

Best done in a spreadsheet, checking each stage carefully!

One test of correct programming is to use zeros for sample 2, you should then end up reproducing the mean and SE for sample 1.

Note that here, the SD and hence the SE for the mean of the combined sample has a contribution from mean1-mean2, in effect.

If the means of the two samples are substantially different, this difference will increase the SD for the pooled sample.

Think of a sample consisting of body weights of two different breeds of dogs, one very small breed and one very large breed.

In this situation (in which you obviously wouldn’t want to naively pool the two samples), the overall SD would come mainly from between-breeds differences, not from within-breeds differences.

This is of course the starting point for the concept of analysis of variance.

SO – always consider carefully what it all means!

If the two samples apparently differ, either in mean or in spread (as measured by the SD), then it’s pretty meaningless to pool the data for the two samples.

Also, depending on how N1 and N2 were arrived at, you could argue for using an estimated mean of (mean1 + mean2)/2, instead of (N1*mean1 + N2*mean2)/(N1+N2).

This will have a quite different standard error, based on the pooled SD, which is sqrt{(N1-1)*SD1^2 + (N2-1)*SD2*2)/(N1+N2-2)}.

In this situation, too, you would need to check that SD1 and SD2 were similar enough to warrant pooling.

Hope this helps.

Robert G. Newcombe PhD CStat FFPH HonMRCR

Professor of Biostatistics

Cochrane Institute of Primary Care and Public Health

School of Medicine

Cardiff University

4th floor, Neuadd Meirionnydd

Heath Park, Cardiff CF14 4YS

Tel: (+44) 29 2068 7260

My book Confidence Intervals for Proportions and Related Measures of Effect Size is now published.

Available at http://www.crcpress.com/product/isbn/9781439812785

See http://www.facebook.com/confidenceintervals

Home page https://sites.google.com/site/robertgnewcombe/

From: A UK-based worldwide e-mail broadcast system mailing list [mailto:[log in to unmask]] On Behalf Of John Sorkin
Sent: 29 December 2014 11:40
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I thank you for your suggestion, but I don't see how it helps answer my question.

John

From: A UK-based worldwide e-mail broadcast system mailing list [mailto:[log in to unmask]] On Behalf Of Justice Moses K. Aheto
Sent: 29 December 2014 11:26
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