How can we know population mean but not variance Sorry if this is too elementary, but I'm doing some statistics review for an interview but I can't seem to reconcile the premise of z/t-scores. In most the materials I read, it states that we should use the t-score when the population variance is unknown. But given the formula: T = (X – μ) / [ s/√(n) ], how would we know population mean (μ) but not variance?

$\mathbf{\mu}$ is the theorized value under the null hypothesis.
For the situation in general:
$$
H_0:\mu = \mu_0\\
H_a:\mu\ne\mu_0
$$
We do our usual fun of calculating the sample mean $\bar x$ and sample variance $s^2$, and then we calculate the t-statistic.
$$
t = \dfrac{\bar x - \mu_0}{\sqrt{s^2/n}}
$$
Now let's do an example. We have $n = 3$ observations, $3, 4, 5$, and we want to test if $\mu = 3$. (Exercise: The mean of these numbers is, clearly, four. How could the mean possible be three?)
$$
H_0:\mu = 3\\
H_a:\mu\ne 3
$$
Let's calculate $\bar x$ and $s^2$.
$$
\bar x = \dfrac{3 + 4 + 5}{3} = 4 \\
s^2 = \dfrac{(3 - 4)^2 + (4 - 4)^2 + (5 - 4)^2}{3 - 1} = 1
$$
Now calculate the t-stat.
$$
t =  \dfrac{\bar x - \mu_0}{\sqrt{s^2/n}} = \dfrac{4 - 3}{\sqrt{1/3}} = \sqrt{3}\approx 1.73
$$
Software agrees.
x <- c(3, 4, 5)
t.test(x, mu = 3)$statistic # I get 1.732051 

In summary, you assume $\mu$, rather than know $\mu$.
