t-test vs z-test, when and why

Question

You want to test whether a population mean equals $\mu_0$ based on an i.i.d. sample.

When do you use a z-test versus a t-test, why does the t-distribution have heavier tails than the normal, and when does the distinction stop mattering?

Accepted Answer

The two statistics. Z = \frac{\bar{X} - \mu0}{\sigma/\sqrt{n}} \qquad 	ext{vs.} \qquad T = \frac{\bar{X} - \mu0}{S/\sqrt{n}}. The z-test assumes the population standard deviation \sigma is known; the t-test replaces it with the sample estimate S. For normal data, Z \sim \mathcal{N}(0,1) exactly, while T follows a Student-t distribution with n-1 degrees of freedom. Why heavier tails. T contains an extra source of randomness: the denominator. In samples where S happens to come out small, T gets inflated, so extreme values of T are more likely than extreme values of Z. Formally, T = Z / \sqrt{\…

t-test vs z-test, when and why

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