# T-distribution or Z-distribution?

I have one dataset with 800+ records. This dataset contains two groups of studies (e.g. Group A and B). With the sample size of 261 and 581.

I have generated the Desc stats using Excel tools.

Both the Group A and B is left-skewed. With the mean of 4.3 and 4.6 and median of 5, based on Ordinal Variable (E.g. Rating of Customer Satisfaction, 1-5) Since the distribution is left-skewed, do I use the mean or median?

With population SD unknown and large sample size, it means z-distribution. Am I right?

I think what I am confused with right now is whether to further sample my data, or if the data provided is already a population and not a sample.

Can someone enlighten me? Thanks!

There are multiple distinct questions here

Whether you use the mean or the median depends on which of the two you want to measure.
Both may be reasonable regardless of skew (If you try to model a yes/no event that is highly unlikely the median won't get you very far).
Using the median will require you to use different methods such as Mood's median test, Mann-Whitney U test or Wicoxon test.

Your overall distribution not being perfectly normal does not immediately present an issue to z- and t-tests since they assume the distribution of the test statistic being normally distributed not directly the underlying variable. Your population needs some very strong skew before you start getting skew in the sampling distribution.

If you go with the mean: the z-test is just the special case of the t-statistic for an infinitely large sample size. At your sample size it doesn't really matter although using a t-test is common regardless of sample size.

Is your data everyone in the population? Is there no other human or other that can participate in your experiment?
If there are some left then your data are a sample.
If there are only a tiny amount left your data are also still a sample but they may not be considerable independent anymore.