Confidence interval - To use standard error or standard deviation? We are given a few true-false questions. I would like to know if my reasoning is correct. 
25 readings are made on the elevation of a mountain peak. These averaged out to 81,000 inches, and their SD was 30 inches.
Therefore, it works out that the standard error of the readings is $\frac{30}{\sqrt{25}}$ inches = 6 inches.
Given statement: the elevation of the mountain peak is estimated as 81,000 inches and this estimate is likely to be off by 6 inches (SE).
Statement I have a problem with: There is about a 95% chance that the next reading will be in the range 81,000 $\pm$ 12 inches
This statement is false, because according to my tutor, the range should not be 81,000 $\pm$ 12 inches, but instead $\pm$ 1.96 x 30 inches $\approx$ 60 inches.
My intuition (after answer was given) is that the SD of 30 given is already considered to be the SE, because it is the standard deviation of the mean itself. If we use 12 inches, we are actually using the standard error of the readings instead. 
But in this case, I have a question: then why would we still want to find the SE of the readings in the first place? Why not just use the SD of the readings (which estimates $\bar{X}$ in the first place)?
If possible, I would still like a more detailed explanation than what I have. 
 A: Your tutor is incorrect.  The standard deviation of the sample estimates the variability of the underlying population distribution--in this case, the elevation measurement.  The standard error reflects the variation about the observed sample mean, and thus is a measure of the uncertainty of the sample mean.
Let's put it a bit more simply.  If you take $100$ times more measurements (i.e. $n = 2500$), your estimate of the standard deviation would remain largely unchanged, because there is intrinsic variability in the precision of the measurement, consequently there is a natural variation from one observation to the next that is not reduced in your sample simply by taking more observations.  This variation is a feature of the population.  Thus, if you were to use the standard deviation to create your interval estimate, it would not have the property of decreasing in size with increasing sample size.  Its expected width would remain constant irrespective of the number of observations you make.
However, with increasing sample size, the standard error of the mean would most certainly be smaller, and should decrease as you gather more information about the population distribution, which is then reflected by a correspondingly narrower confidence interval for the mean.

Perhaps a simple example is in order.  Suppose that you work at a factory that produces ceramic floor tiles.  The machine that makes these tiles is supposed to be calibrated so that the completed tiles are 10 inches on a side.  As the quality control technician, you have conducted past measurements, and have observed that due to naturally occurring variability in materials composition and other uncontrolled variables, the machine produces tiles that have a standard deviation of $\sigma = 0.03$ inches.  So in order to determine whether the machine is out of calibration, you take a random sample of $n = 9$ tiles and compute a sample mean length of $\bar x = 9.99$ inches.
What is a $95\%$ confidence interval for the true mean dimension of the tiles made by the machine?  Your point estimate is simply $\bar x = 9.99$, and you are given the standard deviation $\sigma = 0.03$.  But based on the sample size, your standard error is $$SE = \frac{\sigma}{\sqrt{n}} = \frac{0.03}{\sqrt{9}} = 0.01.$$  The critical value is based on a normal distribution assumption, so $z_{\alpha/2} \approx 1.96$, and the margin of error is $$ME = z_{\alpha/2} \cdot SE \approx 1.96 \cdot 0.01 = 0.0196.$$  It follows that the CI is $$[9.9704, 10.0096].$$  Since this interval contains $10$, there is insufficient evidence from the sample you drew to suggest that the machine is out of calibration.
But what if the sample size had been, say, $n = 900$ rather than $n = 9$, all else being the same?  The fact that you measured $900$ tiles, and found that the average tile dimension was $9.9$ inches, should be cause for concern that the machine is very slightly out of calibration.  Indeed, this is reflected in the formula:  the standard deviation remains $0.03$, but now $$SE = \frac{\sigma}{\sqrt{n}} = \frac{0.03}{\sqrt{900}} = 0.001,$$ reflecting an order of magnitude more precision in your estimate of the sample mean due to the fact that you sampled $100$ times more tiles.  The resulting $95\%$ confidence interval is correspondingly narrower:  $$CI = [9.98804,9.99196]$$ meaning that this larger sample gives enough evidence to suggest that the machine is not calibrated to produce tiles exactly 10 inches on a side.  Note that the variability in the machine has not changed in either case, but because you've taken more data on the produced output, you have more information about the true mean tile length.
