# Should I use Student's t-distribution to evaluate my measurement's confidence interval?

I'm an experimental physicist who mainly needs statistics for the calculation of uncertainties/confidence intervals. Since my results are usually normally distributed, I simply take $$N$$ measurements of the same signal and then calculate the sample standard deviation, i.e. $$\sigma = \sqrt{\frac{\sum_{i=1}^N (x_i - \bar{x})^2}{N-1}} \,.$$ I usually then assume that my measurement uncertainty is $$\Delta y = \sigma$$.

However, there are people who say that a correction is necessary here. And I know of Student's t-distribution, which is a correction of the normal distribution for small sample sizes. Some people suggest that this should be used if you have only a small amount of measurements, which would be true in my case because I usually have $$N=10$$.

In practice, I have never seen anybody use this correction, I do not frequently read about it in papers, and in my own statistics courses I have never been told to do it. I would not know exactly how to use it, except that the Wikipedia entry has a bit about confidence intervals, including this:

Therefore, the interval whose endpoints are $$\bar{X_n} \pm A\frac{S_n}{\sqrt{n}}$$ is a 90% confidence interval for μ.

This is difficult to parse for me. As far as I understand it, A is (in this example) the tabulated correction factor that gives the two-sided 90% confidence interval, which is $$A=1.833$$ for $$N=10$$ because I have 9 degrees of freedom. $$S_n$$ seems to be the standard deviation. So if I say that my uncertainty is half the confidence interval, then the total correction factor is $$1.833/\sqrt{10} \approx 0.58$$, and $$\Delta y = 0.58 \sigma$$.

So is this something I should do with my data?

I am unsure because there is so little information on it, and nobody else seems to be doing it. Actually, a different correction method I have heard frequently about is to divide my standard deviation by a factor of $$\sqrt{N}$$, i.e. $$\Delta y = \frac{1}{\sqrt{N}} \sigma \,,$$ which would be $$\Delta y = 0.32 \sigma$$ for $$N=10$$. The usual explanation for this is that the likelihood of the mean value being the correct value should increase if $$N$$ increases. That makes some sense, but I have never seen a statistical explanation for this correction factor. And the difference to the correction factor from Student's t-distribution is significant. So which one is correct?