Can a 2 paired sample sets of size 15 (30 readings in total) be assumed to have normal distribution when finding confidence intervals? I've seen that for large sample sizes (about 30 or more) the distribution tends to a normal distribution, but what if I have 2 sets of 15 which are paired? Would that count as a normal distribution, or would I need to use a t-distribution to approximate confidence intervals for the difference between their means?
Thanks in advance.
 A: The "rule" you are using is baed on the following: The value 1.96 cuts 2.5% from the upper tail of standard normal. For a t test with $n \ge 30$ we have at least 29 degrees of freedom. The exact number that cuts 2.5% from the upper tail of $\mathsf{T}(29)$ is $2.04523 \approx 1.96.$ (Approximate in the sense that both 2.045 and 1.96 round to 2.0.) I stress that this works well only for 2-sided tests at the 5% level.
[For a two-sided test at the 1% level, you need to cut .5% of the area from
the right tail. For std. normal the cutoff point is 2.576; for $\mathsf{T}(df = 29),$ the cutoff point is 2.756. And 2.576 is not approximately 2.756. 
For $\mathsf{T}(df = 69),$ the cutoff point is 2.649. So it takes 70
observations to match the normal value (in the sense that both cutoffs round
to 2.6).]
In your paired t test you have df = 14, so you can't use the same 'rule'--
not even at the 5% level. 
If your textbook and instructor are using a "rule of 30" to decide between
z-tests and t-tests, then be a diplomat and go along with it. Over time
you may forget some things from your stat course. Maybe schedule the "rule
of 30" to be one of the first things to forget.
Also, in applying the Central Limit Theorem generally, $n = 30$ is not a good rule of thumb for 'convergence'. Means of data from some distributions are approximately normal for $n = 10,$
and for other distributions (especially skewed ones, such as exponential) $n = 50$ really isn't enough.
Accurate rule for deciding between z- and t-procedures (tests or confidence intervals): (a) If the population standard
deviation $\sigma$ is known, then use z. (b) If the population standard deviation $\sigma$ is unknown and estimated by the sample standard deviation $S$, then use t. 
