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I have 40 people that I measure at baseline, getting their mean level of X at time zero. I also calculate the standard deviation and standard error of the mean of X.

Then after 100 days I measure their levels of X again, and again calculate a mean, standard deviation, and standard error. I lost five people to follow up, so the N for this group is only 35.

     Time    N    MeanX    SD    SE
        0   40      6.9   5.2   0.8
      100   35      5.7   5.7   1.0

I am interested in the difference in mean levels of X between 100 days and baseline. So I can calculate this easily, as

5.7 - 6.9 = -1.2

My question is... For this value, -1.2, I would also like to know its standard deviation and standard error. Could someone tell me how to do this? I've found a few possible formulas on the internet, including one from a similar question on this site, for example squaring the standard deviations, dividing them by their n's, and then taking the square root, but I am not one hundred percent sure if this is what I want.

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    $\begingroup$ It seems you have paired data where the easiest approach would be to take differences for each individual and then do a one sample test on those differences. A complication with calculating the standard error of this quantity is that the two samples are not independent, but if you take differences you don't need to worry about this. $\endgroup$ – dsaxton Aug 8 '15 at 23:22
  • $\begingroup$ The trouble is I don't actually have the individual participant data - only the summary data shown above $\endgroup$ – Alexander Aug 8 '15 at 23:23
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    $\begingroup$ @dsaxton you appear to be assuming non-negative covariance (that is something, not nothing). So your bound could be anti-conservative. $\endgroup$ – Mark L. Stone Aug 8 '15 at 23:45
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    $\begingroup$ Oops yeah. Dumb mistake. In any case you might be able to reasonably assume positive covariance. (I'll delete my comment anyways so it isn't confusing.) $\endgroup$ – dsaxton Aug 8 '15 at 23:46
  • $\begingroup$ I've revised my answer to include the commentary. But be careful about being "fine with not showing a significant difference"; there can be costs to type II errors, too. $\endgroup$ – EdM Aug 9 '15 at 15:04
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Go back to first principles: the variance of a difference is the sum of the variances minus twice the covariance.

Here the variances would be the squares of the standard errors of the means. The covariance would be between means at time zero and means at 100 days over repeated instances of the same exercise (40 cases at time 0 and fewer cases selected from the same individuals at 100 days). Without information on individual cases and the process that led to the loss of cases, I don't see a way to determine the covariance.

If you assume zero covariance, the square root of the sums of the squares of the standard errors is 1.28 versus a difference of -1.2 between the means, so there would be no significant difference. As @dsaxton and @Glen_b point out, you would typically expect a positive covariance in this situation, which might diminish the variance of the difference, but you don't have the necessary data. And as @MarkL.Stone points out, a negative covariance can't be ruled out a priori, which would instead increase the variance of the difference.

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    $\begingroup$ Same comment as for @dsaxton . You appear to be assuming non-negative covariance.. So your bound could be anti-conservative. How is covariance = 0 a conservative assumption? $\endgroup$ – Mark L. Stone Aug 8 '15 at 23:47
  • $\begingroup$ So are you saying that my answer would be (0.8^2 + 1.0^2) - 2*0 = 1.64? I am fine with not showing a significant difference - I basically just need to do this as "properly" as I can given the limitations of my data $\endgroup$ – Alexander Aug 8 '15 at 23:51
  • $\begingroup$ @MarkL.Stone Positive covariance is expected because of the same subjects being in both groups. If there's a subject effect (a given subject tends to be different from the average level of "X" e.g. because of - genetics, environment, lifestyle etc) then their two values will tend to both be higher or both lower than the overall mean for those measures. Indeed the existence of a subject effect why we use paired tests when we have the pairing information. $\endgroup$ – Glen_b♦ Aug 9 '15 at 1:13
  • $\begingroup$ @MarkL.Stone ... (ctd) The Wikipedia page on pairing puts it this way: > "Thus the variance of $\bar{D}$ is lower if there is positive correlation within each pair. Such correlation is very common in the repeated measures setting, since many factors influencing the value being compared are unaffected by the treatment" $\endgroup$ – Glen_b♦ Aug 9 '15 at 1:17
  • $\begingroup$ @Glen_b This is basically the same as common random numbers in stochastic simulation (my thing, so to speak). (Some) common factors may indeed tend to result in positive correlation, but I do not believe they guarantee non-negativity of correlation, even if negative correlation is unlikely. So until I see proof that correlation can't be negative, I do not accept the assumption that it is nonnegative as being conservative. $\endgroup$ – Mark L. Stone Aug 9 '15 at 4:36

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