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?
 A: $\mathbf{\mu}$ 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 $\mu$.
A: This isn't generally applicable, but here's a specific example where the mean is knowable.
Suppose the population is "individuals' net profits in poker games over the last 6 months".
Poker doesn't create or destroy money, so if you add up everybody's winnings and losses the total will be zero. (Let's assume that we are ignoring things like entry fees.)
So the mean profit is zero but the variance is unknown.
