Assume $n = 25$ observations from a normal distribution have a sample mean $\bar{x} = 21.2$ and $s_{n} = 0.5$. We wish to test whether this observed mean is lower than a population mean with $\mu_{old} = 20$. Intuitively, we know the test should be rejected, since $21.2 > 20$, but I was asked to compute the $p$-value of the test (without using software). But first, let us observe a few facts.
The null hypothesis is $H_0 : \mu_{old} = \mu_{new}$. The alternative hypothesis is $H_a : \mu_{old} > \mu_{new}$. It is known that
\begin{align*} T = 5 \times \frac{\overline{x} - \mu_{old}}{\sigma_{old}} \sim t_{24} \end{align*}
The rejection region is $t \geq t_{\alpha; 24}$. With $\alpha = 0.01$, $t_{0.01;24} = 2.4922$. Observe that
\begin{align*} T = 5 \times \frac{21.2 - 20}{0.5} = 12 \end{align*}
Since $ 12 \not\leq 2.4922$ we do not reject the null hypothesis. But what is the $p$-value of our result? I know conceptually
\begin{align*} 1 - P(T < 12) &= p-\text{value} \end{align*}
This is, the probability of observing a $T$ as extreme as $T = 12$ or more, if the null hypothesis were true. I suspect this value is close to 1. But still, I was asked to compute it without using statistical softwares. How can one compute this? The CDF of a $T$-student distribution doesn't seem to be the simplest path.