Required conditions for using a t-test The conditions that I have learned are as follows: 


*

*If the sample size less than 15 a t-test is permissible if the sample is roughly symmetric, single peak, and has no outliers.

*If the sample size at least 15 a t-test can be used omitting presence of outliers or strong skewness.

*With a larger sample the t-test  can be use even if skewed distribution if the sample is greater than 30, but  less than 10% of the population.  
Why can't you use a the t-test when the sample size is larger than 10% of the population size?  What happens then? Do you use the z-test? 
 A: First, you have to understand why there are two tests, for a same quantity. Let's say you have a sample $x_1, \dots, x_n$, drawn from an unknown distribution and you want to test if the mean of the distribution is zero or not.
So you compute the sample mean $\overline x = {1\over n} \sum_{i=1}^n x_i$. And you compute the sample variance $s^2 = {1\over n-1} \sum_{i=1}^n (x_i-\overline x)^2$. And finally, you reduce $\overline x$ by the standard error $s = \sqrt{s^2}$, considering ${\overline x \over s/\sqrt n}$.
There are two cases :


*

*the underlying distribution is normal ; then ${\overline x \over s/\sqrt n}$ is distributed like a $t$ distribution (if the mean is zero), and you use a $t$ test. This is an exact procedure.

*you don’t know whether the underlying distribution is normal or not. If $n$ is big enough, the central limit theorem tells you that ${\overline x \over s/\sqrt n}$ is approximately distributed like a standard normal distribution (if the mean is zero), and you use a $z$ test. This is an approximate procedure.
What you were stating are just guidelines to help you decide if the assumptions required for $t$ test are satisfied. 
I don’t get rule 3. For me, it is just false. If the distribution is skewed, it is not normal, and you have no reason to think that the $t$ test will perform better than the $z$ test.
A: You can actually use the t-test if you like -- it's just more conservative.  As your sample size grows larger, the Central Limit Theorem says that the distribution of your mean approaches a normal distribution, regardless of the underlying population distribution.  Therefore, you can use the Z-test, since that compares your statistic with a normal distribution.
A: I believe the reason for the third rule is in its need to adhere to CLT, and therefore be nearly normal. CLT states that a sampling distribution model is relatively normal for a large sampling frame, regardless of the distribution of the population, as long as the sampled individuals are independent.
This 10% rule is to protect the independence of the sampled individuals when sampling without replacement by sampling only a small fraction of the population, assuring that any relation can be generally minimized by randomization.
If you want to see the mechanics on why this percentage is chosen, University of Texas explains it more in depth here: https://web.ma.utexas.edu/users/mks/M358KInstr/TenPctCond.pdf [WHERE DOES THE 10% CONDITION COME FROM?][1],
But I sourced my general information off of "Stats, Modeling the World" 1st ed.
A: I don't see the necessity for the comparison. T-test and Z-test I believe, operate under different conditions. T-test is Parametric, while Z-test is one of the known four nonparametric equivalent. Please somebody should correct me if my assumption wrong.
