If I'm running a simple 1-sample t-test, and I have x-bar, s, n, and mu, where p-hat and s come from the following type of numbers:
0
0
0
0
0
0
0
0
0
.8
.8
.8
.8
1.2
1.2
1.2
1.6
1.6
1.6
2
2
2
2
2
I.e, from only 5 different possible "scores" on a "test", can I proceed as usual with the t-test? Is there a different procedure for data like these?
If the one-sample $t$-test you're looking at is vs 0, the fact that 15 of your 24 numbers are positive and the rest is not negative but 0 is already a quite strong indication that the true mean of the distribution your samples are from is larger than 0.
If you perform a one-sided one-sample $t$-test on these 24 numbers, the result is $p = 7.9 \cdot 10^{-6}$, far below the common significance level of 0.05. So even if the $t$-test is not exact in this case, it is quite unlikely that a correct test would give you a non-significant result.
A non-parametric alternative to the one-sample $t$-test is the sign-permutation test*: Compute the mean of the numbers, but also on numbers where the signs have been switched (+ to -, - to +). The $p$-value is then the fraction of permutation means which are larger than or equal to the actual mean. There are $2^{24} = 16777216$ such permutations. The result is $p = 3.05 \cdot 10^{-5}$. Not only is this a significant result, but it also agrees well with the result of the $t$-test, which indicates any violation of the normality assumption here is not very strong.
*) See Good, Permutation, Parametric and Bootstrap Tests of Hypotheses, 3rd ed., Springer 2005, section 3.2.1. The procedure can be traced back at least to Fisher, The Design of Experiments, Oliver & Boyd 1935, section 21, where he describes an alternative to the paired $t$-test that drops the assumption of normality and tests whether the paired samples come from the same distribution. As whuber pointed out, for the one-sample test the corresponding assumption is that the distribution is symmetric around 0 under the null hypothesis.
Update after the poster's comments: $t$-test vs 0.8 gives $p = 0.27$, clearly non-significant, so the question whether the test is correct here is not really relevant. Sign-permutation test vs 0.8 gives $p = 0.30$, again a decent agreement, which indicates that the $t$-test isn't too bad.
Generally I'd recommend here to just use the sign-permutation test. If you have more data, you have many more permutations, which means you cannot generate them all. In this case, use a randomly drawn subset of the permutations (a.k.a. "Monte Carlo").
I would be quite comfortable that using a t-test here especially since you have reported "many more items" in the sample. The Central Limit Theorem dictates that the mean will be very close to normally distributed and thus the t-test will be appropriate.
