Two methods for estimate $s_{\bar{X}}$?

Question

When reading in my book Mathematical statistics and Data Analysis (Ross), I figured there where two different methods to compute the estimate $s_{\bar{X}}$, with $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$. The first is $$s_{\bar{X}} = \sqrt{\frac{\sum_{i=1}^n (X_i - \bar{X})^2}{n-1}}$$ and the second is $$s_{\bar{X}} = \sqrt{\frac{s^2}{n}(1-\frac{n}{N})}$$

In Example A in section 11.3 in the book, we have the following table of values: \begin{matrix} Before & After & Difference \\ 25 & 27 & 2 \\ 25 & 29 & 4 \\ 27 & 37 & 10 \\ 44 & 56 & 12 \\ 30 & 46 & 16 \\ 67 & 82 & 15 \\ 53 & 57 & 4 \\ 53 & 80 & 27 \\ 52 & 61 & 9 \\ 60 & 59 & -1 \\ 28 & 43 & 15 \end{matrix}

From the column of differences, $\bar{D}=10.27$ and they say $s_{\bar{D}}$ is 2.40. This answer is true for the second method of computing I described above. In this case, $N=\infty$ is used. When computing this with the first method, the answer would be 7.94.

Now my question is: when do I use the first method and when the second? Maybe the answer is simple, but I can't figure it out by myself.

Answer 1

The first equation is $not$ for standard error of $\bar X$, it is for the sample standard deviation $S$. The standard error of $\bar X$ is $S/\sqrt{n}.$ The second equation looks as if you are sampling $n$ individuals from a finite population of size $N$. Careful reading of the context of these equations will probably clarify things for you. Ross is pretty careful about specifying conditions.

Here are computations from R, verifying what you say about the column of differences:

 d = c(2,4,10,12,16,15,4,27,9,-1,15)
 mean(d)
 ## 10.27273
 sd(d)
 ## 7.9761  # S
 sd(d)/sqrt(length(d))
 ## 2.404885 # SE

In this case the standard error seems to be for sampling with replacement or for sampling without replacement from a population viewed as (essentially) infinite. When $n$ is much smaller than $N$ the second factor under the square root sign (second formula) is very nearly unity.