What sampling distribution does $\hat{p}$ in this scenario follow? Homework question, not sure whether I should be using a normal distribution, a binomial distribution or a hypergeometric distribution.
A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders.
This is exactly how the scenario is worded, no extra info.
I am asked to construct a 98% confidence interval for p, but I'm really bugged that sometimes my teacher instructs me to use mere approximations when inappropriate (using a normal distribution when n < 30 and $\sigma$ is unknown, for example), so I want to figure out what the actual distribution is here.
Sorry if this is an odd question, my knowledge of statistical theory is very fragmented and I think I'm mixing things up in my head.
 A: The estimated Success ratio for fulfilling the promise is $\hat p = 35/50 = 0.7,$ which
is considerably less than 100%. Using a 98% confidence level makes the confidence longer, so that its upper end is nearer to $1$ than you would get
with a 95$ CI. Obviously, the null hypothesis that the promise is kept all of the time is rejected. 
It is not clear what approach you are intended to take, but it seems clear
that the company has to make some improvements in performance to live up to its promise of 3-day shipping.
Exact binomial hypothesis test. If you consider 90% of the time to be tolerable,
then you might test $H_0: p = .9$ vs. $H_a: p < .9,$ rejecting for small
numbers of successes. The exact binomial test of this hypothesis uses the null distribution $\mathsf{Binom}(n=50, p=.9).$ The P-value $P(X \le 35)$ is much less than 5%, so the null hypothesis is overwhelmingly rejected. 
pbinom(35, 50, .9)
[1] 7.383869e-05

The relevant test in Minitab statistical software gives the following output:
Test and CI for One Proportion 

Test of p = 0.9 vs p < 0.9
                                             Exact
Sample   X   N  Sample p  98% Upper Bound  P-Value
1       35  50  0.700000         0.826072    0.000

The one-sided 98% confidence interval give the upper bound 0.826, which
suggests that the true fulfillment probability is below 83%. (The P-value 0.000 means $< 0.0005.)$
Wald confidence interval. A Wald 98% confidence interval (CI) for $p$ based on 35 Successes in 50 trials is
$$\hat p  \pm 2.326\sqrt{\hat p(1-\hat p)/n},$$
where $\hat p = 35/50 = .7.$ This computes to  $(0.542, 0.858).$
This style of CI uses a normal approximation and has been deprecated for
use with small samples.
Bayesian-based interval. Another style of (frequentist) 98% CI based on a Bayesian approach with a non-informative
prior distribution $\mathsf{Unif}(0,1) \equiv \mathsf{Beta}(1,1)$ cuts 1% of the probability from each tail of $\mathsf{Beta}(35+1, 15+1).$ Especially for
small numbers of trials, this interval has more reliable coverage probability than the Wald interval. [See this Q & A.] Using R statistical software, it computes to
$(0.515, 0.827).$
qbeta(c(.01,.99), 36,16)
[1] 0.5358620 0.8272392

Note: There are many styles of CIs for a binomial proportion. Perhaps see Wikipedia for an overview. 
