Do Z-test and T-test assume normality? I’ve been digging into the internet trying to find an answer for this question. Notwithstanding, all that I’ve found were empty statements like: “z-tests depends on the population to be normally distributed” or “with t-tests you don’t have to concern if the population is normally distributed or not”. No proof was given, though.
After all, do z and t tests assume that the population is normally distributed? How do you prove it? 
Ps: Just to be clear, I'm not trying to prove anything. I just couldn't find any deep discussion or proof of it in any book or paper. Feel free to suggest papers or books in which this subject is discussed, I'll be glad to read them all.
 A: Yes, both tests are designed using assumption that the underlying distribution is normal. It may not be obvious just from looking at the test statistics, but the calculations that go into calculating p-values are heavily dependent on normality.
Look for Mann-Whitney test, as a nonparametric alternative.
A: Coolserdash's answer is correct.
Ill just add that you see a lot of stuff like "The Z-test requires that your data is normally distributed, etc".
This is not actually correct. The Z test requires that your data generating process comes from a probability distribution that has finite variance.
If that is the case, then you can show that the CLT can be used to show asymptotic convergence of the Z statistic (i.e. the standardized sample mean) to the normal distribution. Thus the Z statistic is normal although the underlying data itself does not have to be (i.e the data could be poisson, etc)
As coolserdash pointed out, on the other hand, the t-test does require that the data be drawn from a normal distribution as the derivation of the t-distribution requires that the sample mean and variance are independent. It can be shown that this is only true under assumption that the generating process for the underlying data is normal.
A: I think both test assume normality.
The only difference is that in Z-test, we assume we know the true standard deviation $\sigma$.
However, for t-test, we don't know the true s.d and we use sample standard deviation $\hat{\sigma}=\sqrt{\frac{\sum_{i=1}^{n}(\bar{\mu}-x_i)^2}{n-1}}$.
In Z-test and t-test, we assume $x_i$ follows $N(\mu,\sigma^2)$.
Then $\sum_{i=1}^{n}(\frac{\bar{\mu}-x_i}{\sigma})^2$ follows $\chi^2_{n-1}$-distribution.
Then $\frac{\bar{\mu}-\mu}{{\sigma}}$ follows $N(0,1)$ and $(n-1)\hat{\sigma}^2/\sigma^2$ follows $\chi^2_{n-1}$.
Hence,the t-statistics=$\frac{\bar{\mu}-\mu}{\hat{\sigma}}=(\frac{\bar{\mu}-\mu}{{\sigma}})/\sqrt{\hat{\sigma}^2/\sigma^2}$ follows $t_{n-1}$ since $\frac{X}{\sqrt{Y/K}}$ follows $t_K$ given that $X$ follows $N(0,1)$ and $Y$ follows $\chi^2_K$ and $X,Y$ are independent.
