2

According to this post,

The choice between a t-test and a z-test, therefore, depends on whether or not σ is known prior to collecting the data.

In classical t-test example, one sample t-test can be used in quality control. We can get a sample of product and compare sample mean to design value.

But, I do not understand how can we use z-test in real data. If we already know $\sigma$ (not estimated from data) what we test for?

Community
  • 1
Haitao Du
  • 32,885
  • 17
  • 118
  • 213
  • 1
    You might not know the means.... We use the test to test for difference in means. Does that answer your question? – SmallChess Apr 09 '17 at 04:16
  • @SmallChess, so it still can be applied to "quality control" example, where we measure products and calculate sample mean, but we use population variance to normalize? – Haitao Du Apr 09 '17 at 04:25

1 Answers1

1

You declare that $\sigma$ and $\mu$ are both unknow in "real world". Ok, I don't want to refer to any "real" case where the $\sigma$ is known. But in any case where I run t-test, I can set many assumption (not the "hypothesis" in hypothesis testing, the "assumption" used by me is prior on it ) $\sigma_1,\sigma_2,...$ and run z-test. Since the different $\sigma_i$ get different z-test result. I can say that in this $\sigma_i$ assumption, I reject (or not reject) the null hypothesis.

Then I run the z-test on whole $R$, and get partition $R = R_1 \cup R_2, R_1 \cap R_2 = \emptyset$, where $R_1$ include $\sigma_i$ such that cause rejecting null hypothesis and $R_2$ include $\sigma_i$ such that cause not rejecting null hypothesis. I think of that it make sense to obtain the partition. It summary all "If ... then ..." result from any "real" $\sigma_i$. And it are not effected by $\sigma$ estimation error.