# Standard error of mean - two methods don't give the same answer

Suppose I have repeated measurements of the same sample done by different people. Each person has different number of repeats. Here are the measurement data:

Person #1: 2, 3, 2

Person #2: 3, 3, 6

Person #3: 2, 5, 6, 4

Person #4: 2, 3, 4, 5

Person #5: 3, 2

I can group all these data together, and get the mean 3.4375, and its corresponding standard error $s/\sqrt{16}$ where s is just the standard deviation of all the data. The standard error turns out to be 0.353.

Another way to estimate the mean for this data set is: first calculate the mean for each person $\bar{x}_i$, and then pool the means $\Sigma_i n_i\bar{x}_i/\Sigma_i n_i$. This of course gives me the same mean at 3.4375. But if I estimate the standard error of mean using this pooled mean formula, it would be $\sqrt{\Sigma_i n_i^2var(\bar{x}_i)/(\Sigma_in_i)^2}$ where $var(\bar{x}_i)=var(x_i)/n_i$. But this gives me a standard error of 0.338.

Shouldn't I expect the standard errors to be the same using these two different mean calculation methods?

• You seem to be on the threshold of rediscovering Analysis of Variance :-). For some intuition, consider what would happen if the data for Person #5 were 103, 102. Could you interpret the two versions of the SE in that case? – whuber Jun 9 '17 at 21:52
• No, you shouldn't. Except in very special circumstances. – David Smith Jun 9 '17 at 22:02
• @whuber Thanks for pointing that out. So I think it boils down if all the persons data come from the same population or each comes from a separate population. If the latter is true, then the underlying assumption behind my standard error calculation is invalid. Correct? – hooyeh Jun 13 '17 at 21:14
• I don't know about invalid, but evidently it results in some combination of two components of variance: the within-group variance and the between-group variance. That makes it difficult to interpret or assign any meaning to. – whuber Jun 13 '17 at 21:18