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Can you help me understand, or refer me to a source about, what is the sampling distribution of the sum of scores? I know how to get the sampling distribution of the mean to construct a t-test, etc. More specifically how does the variability of the sampling distribution of sums of scores relates to the variability of the underlying variable in the population. e.g. People pass sand bags to each other. I measure the total time for a sample of n=10 people to pass the bag, and calculate an average. If I repeat this sampling I will have a standard deviation of the sample mean, and then I can calculate the variability in the population. Assuming a relatively normal process, even though I am measuring time, etc. The problem is if I measure the same sample of 10 people passing bags every time. The samples are related. Any advice?

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    $\begingroup$ How do you work out the sampling distribution of a mean without knowing what the distribution of a sum is first? When you say 'the samples are related' ... do you mean 'the observations are related', or do you have multiple samples each containing many observations? $\endgroup$
    – Glen_b
    Aug 9 '13 at 1:17
  • $\begingroup$ I am interested in the case of a specific group of people engaged in a task on which I measure the total time to perform that task. That is what I mean by, the samples are related. Regarding my comment on calculating sampling distribution of the mean, I was trying to say something about my background not that I know how to calculate it for this specific case. $\endgroup$
    – Adam SA
    Aug 13 '13 at 7:22
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You suggest you're happy with assuming approximate normality. Then it's straightforward - a linear combination of multivariate normal variables is itself normal. The mean and variance follow from elementary properties of mean and variance.

If $X\sim N(\mathbf{\mu},\Sigma)$ then $a'X \sim N(a'\mathbf{\mu},a'\Sigma a)$.

In the case of a sum, $a = \bf 1$. That is, the sum is normal, where the mean of the sum is the sum of the means and the variance of the sum is the sum of all variances + twice the sum of all pairwise covariances.

Additionally, if the series are no too dependent and are iid, or independent and none of them have too large a variance relative to all the others, the CLT definitely applies, so if there are enough terms, you should have that $\sqrt n \bar x$ is normal. Note that $\sqrt n \bar x = \frac{X_1+\dots+X_n}{\sqrt n} $.

From there, if you think the $n$ is big enough to use the normal approximation at some point, you can back out one for the sum as well (the CLT - being about limits - doesn't apply to an unscaled sum, but the quality of the approximation of the cdf at some specific $n$ carries over to the sum). However the sample sizes required for these to kick in may be quite large.

Beyond that, convergence to normality of appropriately scaled sums can occur in a wide variety of situations. Again, sample sizes may need to be large.

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