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I need to calculate the mean(+/-SD) final value of a variable from the mean(+/-SD) baseline value and the mean(+/-SD) change from baseline. So, for example:

  • Mean baseline weight is 40(+/-2) kg.
  • Mean change from baseline is +5(+/-1) kg.

I'd assume the correct final mean is 45kg, but how would I go about calculating a new SD?

I've googled for methods of calculating 'Pooled SD', but those only seem to apply to pooling same variables (like weights from different groups) and I don't think those apply here. This question seems the most similar to my question but also seems to try the exact opposite of what I'm trying to achieve.

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  • $\begingroup$ Is this homework? If so, it should have the homework tag. $\endgroup$ – Peter Flom Nov 8 '12 at 11:19
  • $\begingroup$ No, I want to recalculate missing data from articles, but that's only useful if I can also calculate a (reasonable) SD. $\endgroup$ – Joep Nov 8 '12 at 11:36
  • $\begingroup$ You cannot compute an SD without knowing the correlation between the two variables. $\endgroup$ – whuber Nov 8 '12 at 15:01
  • $\begingroup$ @whuber What do you mean by that? $\endgroup$ – Joep Nov 8 '12 at 15:41
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    $\begingroup$ The SD is the root of the variance. Let $X$ be the baseline weight and $Y$ the change from baseline. You need the variance of $X+Y$. It equals the variance of $X$ plus the variance of $Y$ plus twice the covariance of $X$ and $Y$. If there is nonzero correlation between $X$ and $Y$, that covariance will be nonzero and cannot be neglected. $\endgroup$ – whuber Nov 8 '12 at 15:45
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Impossible, as explained in Whuber's comment:

The SD is the root of the variance. Let $X$ be the baseline weight and $Y$ the change from baseline. You need the variance of $X+Y$. It equals the variance of $X$ plus the variance of $Y$ plus twice the covariance of $X$ and $Y$. If there is nonzero correlation between $X$ and $Y$, that covariance will be nonzero and cannot be neglected.

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  • $\begingroup$ Can you explain why this is the case? We want our answers to be a little more substantial than this (please see our FAQ, which contains info about answering Q's among other topics). If you're just going to agree w/ @whuber's comment, this itself is best put as a comment instead of an answer. $\endgroup$ – gung Nov 9 '12 at 13:23
  • $\begingroup$ I was just going to agree with Whuber's answer, but figured this would show the question as resolved. $\endgroup$ – Joep Nov 10 '12 at 15:12
  • $\begingroup$ Oh, sorry. I didn't recognize that you were the OP. $\endgroup$ – gung Nov 10 '12 at 15:18

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