# Standard deviation vs standard error of sample mean

Wondering if anyone can clarify. I have some results from lit of a sample mean estimate, and 95% CI's. I can use the CI's to calculate the standard error of the mean. If we I want to find the standard deviation and don't know the population estimates do I interpret the standard error of the sample mean as the standard deviation?

I only have this info to work with: mean=-0.32, 95% CI -0.60 to -0.05, n=3090).

Thanks

• what's "lit" in "results from lit"? – Glen_b Apr 4 '16 at 7:36

No, the standard error and standard deviations are very different (but related) beasts. The standard error is the standard deviation of the sampling distribution of means. So we imagine that we draw a sample of $N$ observations from the population over and over again, compute the mean in each of these samples and the distribution of these means is the sampling distribution. The standard deviation of that distribution is much smaller than the standard deviation of the distribution of observations in one sample. Moreover you would expect the sampling distribution to be more compact (smaller standard error) if you draw larger samples. This intuition is correct: to get the standard deviation you need to multiply the standard error with the square root of $N$.
Standard deviation gives an indication of how far a single unit will differ from the population mean. Mathematically, $\frac {1}{N-1}\sum_i (y_i-\mu)^2$ (with $\mu=\frac {1}{N}\sum_iy_i$ being the population mean). These become model expectations if you have an infinite population (such as a normal distribution).
The standard error in the mean gives an indication of how far the sample mean will differ from the population mean. Mathematically, $E (\overline {y}-\mu)^2$ where $\overline{y}=\frac {1}{n}\sum_{i=1}^ny_i$ and the expectation is over all possible samples.
• You talk about standard deviations but your formulas are for variances, or estimators of variances. You know the difference but I think this could confuse almost all people attracted to this thread. Needs clarification on notation: $N, n, E$ are assumed familiar. If you know the population of $N$ values then the variance is defined too and there is no call to estimate it. – Nick Cox Apr 4 '16 at 9:30