Suppose I have the following $n = 20$ observations sampled
at random, as listed and summarized below:
 37.0 38.4 46.2 57.1 40.2 39.0 54.7 54.0 52.5 37.1
 49.6 45.7 43.6 41.1 49.7 56.3 41.8 53.8 48.1 38.3
Min. 1st Qu. Median Mean 3rd Qu. Max.
37.00 39.90 45.95 46.21 52.83 57.10
I suppose that the population from which these data were randomly sampled is normal, and I want to test $H_0: \mu = 50$ against $H_a: \mu \ne 50,$ where $\mu$ is the population mean.
The sample mean is $\bar X = 46.2,$ which is somewhat below $50.$ I use a t test to find out if $\bar X$ is 'significantly' different from $50.$
One Sample t-test
t = -2.4495, df = 19, p-value = 0.02417
alternative hypothesis: true mean is not equal to 50
95 percent confidence interval:
mean of x
I want to do this test at the 5% level of significance.
If $\mu$ really is $50,$ I want the probability of rejecting $H_0$ in favor of the alternative $H_a$ (a mistake) to be small: below 5%.
The t statistic, $T = -2.4495$ above, is distributed according to Student's t distribution with degrees of freedom
$\nu = n - 1$ $= 20 - 1 = 19.$ That distribution puts 95% of
its probability between $\pm 2.228$ (vertical red dotted lines in the plot below). These are the upper and lower 'critical values'.
The R code below finds the upper critical value. You can also find this critical value in printed tables of Student's t
The observed value of the t statistic, $T = -2.4495,$ shown in the plot as a vertical blue line, lies beyond the lower critical value. So we say that the null hypothesis $H_0$ is rejected at the 5% level of significance.
A second way to see that we reject $H_0$ at the 5% level is
shown in the output from the t test above. If $H_0$ is true, then the probability
$$P(|T| \ge -2.4495)= P(T \le -2.4495) + P(T \ge 2.4495) = 0.2417.$$
If $H_0$ is true, this is the probability that the $T$-statistic lies farther from 0 (in either direction) than the observed
value $T = -2.4495.$ We know to reject $H_0$ at the 5% level
of significance because the P-value is smaller than 0.05.
2 * pt(-2.4495, 19)
Generally speaking, there is not sufficient information in printed tables of Student's t distribution to find an exact P-value. So you will ordinarily have to rely on computer output
(or results from a statistical calculator) to compute P-values.
Using p-values has one advantage. If I want to test the null
hypothesis at the 1% level, I know that I cannot reject $H_0$ at the 1% level because the P-value exceeds 0.01. (So it is not
necessary to 'tell' a computer program what your level of significance is. Once you get the P-value you can test at
any desired level of significance.)
Note: I was not surprised that the t test rejected the null hypothesis because I simulated the 20 observations in R, sampling from a normal distribution with $\mu = 47.$
x = round(rnorm(20, 47, 7),1)