- #1
Simon666
- 93
- 0
Hello,
The standard deviation is calculated as:
http://www.mathsrevision.net/gcse/sdeviation2.gif
Now the problem I have is that how you calculate the standard deviation (more accurately?) over both samplesvif you have two samples of different size, n1 and n2, in which the level of the average µ1 and µ2 can change, but the distribution remains otherwise the same?
Does it make mathematical sense to calculate an "overall" standard deviation using both samples, supposing more means better accuracy? And how would this overall standard deviation then be calculated?
[Edit]Anyway, wikipedia[url] seems to suggest that the variance S(X1X2) = sqrt ( ( (n1-1)S²(X1) + (n2-1)S²(X2) ) / (n1 + n2 - 2) )
so for for example 3 samples of different size, is it then:
S(X1X2X3) = sqrt ( ( (n1-1)S²(X1) + (n2-1)S²(X2) + (n3-1)S²(X3) ) / (n1 + n2 + n3 - 3) ) ?
The standard deviation is calculated as:
http://www.mathsrevision.net/gcse/sdeviation2.gif
Now the problem I have is that how you calculate the standard deviation (more accurately?) over both samplesvif you have two samples of different size, n1 and n2, in which the level of the average µ1 and µ2 can change, but the distribution remains otherwise the same?
Does it make mathematical sense to calculate an "overall" standard deviation using both samples, supposing more means better accuracy? And how would this overall standard deviation then be calculated?
[Edit]Anyway, wikipedia[url] seems to suggest that the variance S(X1X2) = sqrt ( ( (n1-1)S²(X1) + (n2-1)S²(X2) ) / (n1 + n2 - 2) )
so for for example 3 samples of different size, is it then:
S(X1X2X3) = sqrt ( ( (n1-1)S²(X1) + (n2-1)S²(X2) + (n3-1)S²(X3) ) / (n1 + n2 + n3 - 3) ) ?
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