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.