Practical implication of failing to reject a null hypothesis Consider a scenario where you are trying to measure a dosage of a medicine. The machine is calibrated to fill a mean dosage of 50mg. But for a reason you believe that machine's calibration is off. For the above mentioned example null and alternative hypothesis will be as follows: 
null: On the average, the dosage sold is 50 mg 
alternative: On the average, the dosage sold is not 50 mg 
Upon conducting the experiment p-value is greater than say 0.05 for 95% significance level. The conclusion of this test would then be "fail to reject null hypothesis." But practically can I say that the machine is calibrated and is filling a mean dosage of 50mg?

Here is a discussion of a simple version of your question. Suppose the dosages are normal. And you have looked
at 30 bottles sampled before the faulty filling may have begun and 30 after.
Then 'Before' dosages may have been
$X_i \sim\mathsf{Norm}(50, 2),$ $i=1,2,\dots,30,$
and 'After' dosages
$Y_i \sim\mathsf{Norm}(52, 2),$ $i =1,2,\dots,30,$
so that the actual over-filling amount we wish to detect is 2g. Then
a Welch two-sample t test will detect the fault in about 97% of such experiments. That is, the power of my proposed test is about 97%, as found by looking at 100,000 simulated Welch t tests below.
set.seed(2022)
pv = replicate(10^5, t.test(rnorm(30, 50, 2),
                      rnorm(30, 52, 2))$p.val)
mean(pv <= .05)
[1] 0.96828

Notes: (1) I used the Welch t test because I can only
guess whether the standard deviation of the filling amounts is the same before and after the difficulty
arose. (2) I am interested in detecting an effect
of size 2 (which is the standard deviation of filling amounts). (3) The null hypothesis is rejected if the P-value of the test is smaller than $0.05.$ (4) The numeric vector pv contains 100,000 P-values of Welch tests; the logical vector pv <= .05 contains 100,000 TRUEs and FALSEs; and
its mean is its proportion of TRUEs.
