When we approximate a parameter of a population(let’s say μ) and the size of our sample is small( for example:n=9) Given the standard deviation of the popluation( not the sample). Do we use the Z-distribution(normal) or the T-distribution to approximate the parameter?
I am stuck here since I know that we use the t-distribution for a small sample given the standard deviation of the sample, and we use the Z-test with a large sample since we can approximate its distribution to be normal by the central limit theorem(with either, the standard deviation of the population or of the sample given).
According to the solution of a certain problem, the z-distribution is used in the latter case.Can you please explain why?

Data must be normal (or nearly normal) in order to use either z or t test (or confidence interval). Then the choice between z and t is easy:
The sample size $n$ has nothing to do with this choice between z and t.
Note: If (i) the significance level of a test is $\alpha = 0.05$ or a confidence interval has confidence level 95%, (ii) the sample standard deviation is unknown, and (iii) the sample size is $n \ge 30,$ then some textbooks point out that you can use a cut-off value from z tables instead of t tables as an approximation.
Properly used, the approximation usually causes no harm. (In the example, 2.032 and 1.96 both round to 2.0.)
The only real problem with this is that some people have trouble remembering all three conditions (i), (ii), and (iii). Then they get to thinking that t methods are only good for small samples. Or they use z methods for anything where $n \ge 30.$