Understanding how to find the $1$ in $100$ chance on the $t$-table Background
In my university class, we've been discussing the following experiment:

Consider an experiment on artificially raised salmon, with two treat-ments (one a control) and $20$ fish per treatment. Average weight gains(g) over the experimental period were $1210$ and $1320$ grams. The estimate of variation between fish within a group was $s = 135\mathrm{g}$. Did treatment improve growth rate?

My professor established the

*

*observed difference between group means $1320 - 1210 = 110\mathrm{g}$

*variation between two means expected solely from chance $135×\left(\frac{2}{20}\right)^{0.5}= 42.7$

*test statistic = $\frac{110}{42.7} = 2.58$
My work so far
I've wanted to work through the formula, to get better intuition on the process
$$\begin{align*}
t&=\frac{\overline{x}-\overline{y}}{\sqrt{S^2_p\left(\frac{1}{n_1}\frac{1}{n_2}\right)}}\\[5pt]
t&=\frac{1320-1210}{\sqrt{S^2_p\left(\frac{1}{20}+\frac{1}{20}\right)}}\\[5pt]
&=\frac{110}{^{135\sqrt{\frac{2}{20}}}}\\[5pt]
&\approx2.57667\dots \\[5pt]
\end{align*}$$
I understand that $2.58$ is statistically signifigant. My professor also stated the $t$-table shows this being $38$, and the chance of a value as large as $2.58$ is about $1$ in $100$. I've had no success getting this, using the following table, to derive the $1$ in $100$ chance.
How would this be derived from the $t$-table?
 A: You are doing a one-sided, two-sample pooled t test. You are
correct that under the null hypothesis (no difference) the test
statistic $T$ has Student's t distribution with DF $= 40-2=38.$
The critical value $c$ for a test at significance level $0.01 = 1\%$ cuts probability $0.01$ from the upper tail of Student's t distribution with 38 degrees of freedom. You would reject the null hypothesis (no change) against the alternative that the treatment improves growth at the 1% level of significance if $T \ge c.$
The table in your link does not
show values for DF = 38, skipping from DF = 35 to DF = 40, on account of the small change in values as DF increases in that part of the table. According to the table, the $c$ is between 2.44 and 2.42.
Using software, you could get the exact value. For example, R statistical software gives $c = 2.429.$ Because your observed value
of the test statistic is $T = 2.577 > 2.492,$ you would reject the null hypothesis at the 1% level of significance.
qt(.99, 38)
[1] 2.428568

The P-value of your test is the probability of getting a $T$-value greater than or equal to $2.577$ if the null hypothesis is true.
Printed tables of t distributions are seldom adequate to give
exact P-values, but from your table you can guess that the P-value must be between $0.01$ and 0.005.$
The exact P-value from R (where pt is the CDF of a t distribution) is $P(T \ge 2.577 | H_0) = 0.00598.$ Similar precision for P-values is provided as output from t tests in most statistical software, but this degree of precision is not necessary to know whether to reject at the 1% level of significance.
1 - pt(2.577, 38)
[1] 0.006986493

Here is a plot of the density function of Student's t distribution with DF = 38. The
solid vertical line is the observed value of the $T$ statistic
and the P-value is the area (probability) under the density curve to the right of that line.

R code for figure:
curve(dt(x,38), -3.5, 3.5, ylab="PDF", xlab="t", 
      main = "T(38) Density")
 abline(h = 0, col="green2")
 abline(v = 0, col="green2")
 abline(v = 2.577, lwd=2)

