Say we have a normally distributed population with mean of 1000 and standard deviation of 10, of which I am taking a sample of 100.

My question is what would be the expected standard deviation of the sum of the elements of my sample? I believe that my expected value would be n*E[X] = 100 * 1000 = 100000.

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    $\begingroup$ Please explain what you mean by the "cumulative statistic of my sample": what is that? And could you clarify how you know "my sample mean will come out to 1000"? In most populations, most sample means will not equal the population mean, so are you making a false assumption or are you trying to tell us something special about your sampling process? $\endgroup$ – whuber Apr 10 '17 at 18:27
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    $\begingroup$ There is no reason why the sample mean should be the same as the population mean. Also the sample standard deviation will not necessarily be 10 divided by the square root of the sample size. Also how do you define the sample standard deviation? Sometimes n-1 is used where n is the sample size. That is done so that the sample variance will be unbiased. But the biased estimate using n in the denominator can also be used. You seem to be confusing the expected value of the sample estimates with their actual values. $\endgroup$ – Michael R. Chernick Apr 10 '17 at 18:33
  • $\begingroup$ For "cumulative statistic of my sample" I'm referring to the sum of each object in the sample, sorry for the confusion. All of the literature I've found on the topic lists how to get summary stats for a sample from a population, such as sample mean and variance, but I have yet to find how to create an estimate for variance if you are to sum across the elements of a sample from a normal distribution. $\endgroup$ – JVal Apr 10 '17 at 18:44
  • $\begingroup$ @JVal, can you cite some of your sources? It also doesn't make a lot of sense to try to find $\overline{x}$ and $s$ if you already know $\mu$ and $\sigma$. $\endgroup$ – Tavrock Apr 10 '17 at 18:57
  • $\begingroup$ I might not be using the right terms to describe what I'm looking for, sorry if this is causing confusion. In general I just want to find out how to construct an estimate of what the sum of elements of a sample from a population with known mean and standard deviation might be. I'm fairly confident that the expected value will be n*μ, but not sure what the variance around this expected value will be. $\endgroup$ – JVal Apr 10 '17 at 19:01

If $\mu=1000$ and $\sigma=10$, then you should expect $\overline{x}\approx \mu\pm t_{1-\alpha/2,n-1}\frac{s}{\sqrt{n}}$ and $s\approx \left\{\sigma\sqrt{\frac{n-1}{\chi^2_{\alpha/2,n-1}}},\sigma\sqrt{\frac{n-1}{\chi^2_{1-\alpha/2,n-1}}}\right\}$, where $n$ is the sample size. Any sample you take should have approximately the same characteristics, within expected limits, of the population from which they were taken.

Your cumulative expected value of the sample should be approximately the same as the cumulative expected value of the population.

  • $\begingroup$ Could you please explain what a "cumulative expected value" is? $\endgroup$ – whuber Apr 10 '17 at 18:30
  • $\begingroup$ Sorry for the confusion, for this I am referring to the sum of the elements of a sample from a normal distribution. In the example above it would be the sum of the 10 elements sampled from the population. $\endgroup$ – JVal Apr 10 '17 at 18:45
  • $\begingroup$ You seem to be confusing $\sigma_M$, the standard error of the mean, with $s$, the sample standard deviation. $$\sigma_M=\frac{\sigma}{\sqrt{N}}$$ while $$s=\sqrt{\frac{\sum\left(x-\overline{x}\right)^2}{n-1}}$$ $\endgroup$ – Tavrock Apr 10 '17 at 19:02
  • $\begingroup$ I believe I was... I re-phrased my question in a comment above, let me know if this is helpful. Approaching this with an example was probably not the best decision.... $\endgroup$ – JVal Apr 10 '17 at 19:04
  • $\begingroup$ @Tavrock, did you update your answer to answer what the expected standard deviation of the sum of the sample might be? If so, would you mind elaborating further on the formulas you provided? My intuition still tells me that the sum of a sample of n from population with N(μ,σ) would be n*μ; is this not correct? $\endgroup$ – JVal Apr 10 '17 at 20:39

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