When can we use the z-score? (Since it's impossible to know the population SD)

Question

Sometimes I see papers that report the z-score. I am confused because the z-score requires us to know the population standard deviation, which is impossible. What are some reasons people still report the z-score?? Can you give an example please? (The only thing I can think of is when the sample size is large enough so the t-distribution converges to a normal distribution. Also, if the sample is large enough, the sample SD will approximately converge to the population SD. However, even in that situation, shouldn't the researcher report the t-score just to be more sure and conservative?)

Thank you!

Answer 1

Remember that the T-statistic also requires your data to be normal, which is usually not true either. We end up using Z scores in large samples because of the central limit theorem, one version of which tells us that for arbitrary random variables $ \{X_i\}_{i=1}^n $ with finite variance and the same unknown distribution $F$, the test statistic becomes normally distributed i.e. $$ \sqrt{n}(\bar{X}_n - \mu_0)/s \stackrel{d}{\to} \mathcal{N}(0,1).$$ By abusing asymptotics, we basically end up using critical values like 1.96 (which is a z-critical value at the 5% level!) to reject the null that $\mathbb{E}[X] = \mu_0 $.

Answer 2

When comparing proportions, one way to compute a p-value and confidence interval is to compute a z-ratio also called a z-score. T ratios aren't used.

When comparing means, you are right, I think. The t ratio should be used.

Don't confuse the z-score with "z-factor", which is used when evaluating high-throughput screening of drugs, and has an entirely different meaning than the z-score.