According to this link:

The general rule of thumb for when to use a t score is when your sample size meets the following two requirements:

  • The sample size is below 30
  • The population standard deviation is unknown (estimated from your sample data)

In other words, you must know the standard deviation of the population and your sample size must be above 30 in order for you to be able to use the z-score. Otherwise, use the t-score.

Question: Suppose that our sample size is below 30 and that we do know our population standard deviation.

Why not use the z-score? I understand that according to the central limit theorem, with $n < 30$ so low we would have no expectation that our sample estimator would be normal. Does this have something to do with why we should prefer the t-score in this case?

  • 1
    $\begingroup$ In the past when one did not have anything except a table, it makes sense to have a rule "if n is above ..., use the z-table". Most (recent) statistics textbooks still assume people look up things in tables, instead of using software. $\endgroup$
    – vinnief
    Commented May 3, 2016 at 3:52
  • $\begingroup$ @vinnief: Unfortunately correct. Too many "recent" texts are mindless 'retread' editions of older texts, with no significant changes in content or approach, published more for commercial than for educational purposes. $\endgroup$
    – BruceET
    Commented May 3, 2016 at 16:12

1 Answer 1


The highlighted points do not tell the whole story. And your point is correct about always using z tests and confidence intervals when the population standard deviation $\sigma$ is known.

The distinction between t and z methods (for confidence intervals and tests of hypotheses) is very simple.

If the population standard deviation $\sigma$ is known, always use z.

If the population standard deviation $\sigma$ is not known, and needs to be estimated by the sample standard deviation $S$, always use t.

Notes: (1) It is true that for 95% CIs and tests at the 5% level with more than about 30 observations, you can substitute the value 1.96 (about 2) from a printed standard normal table (which cuts 2.5% from the upper tail of the distribution) for the corresponding t value, which will be about 2.

(2) However, for confidence levels other than 95% or significance levels other than 5%, the sample size $n$ where t values become near to the z value will be different from 30. So the 'rule of 30' does not work.

(3) Also, when using software, z tests always require you to input $\sigma$ and t tests always require you to input $S$--regardless of the size of $n$.

(4) Beware of information about statistics on unknown web sites, especially ones with interspersed ads that try to get you to drink (very expensive) beet juice to cure high blood pressure.


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