Why null hypothesis in t-tests can be numbers other than zero? This may sound stupid but please bear with me because it's hard to explain also hypothesis testing is a totally new concept to me and cannot really wrap my head around it
Concern:
Should not T-distributions be used ONLY to test $H_o: \Delta \mu = 0$ ?
My reasoning:
Because basically that's what these T-distributions are only able to test
Let me illustrate , say we ran a paired t-test  with 3 degrees of freedom and confidence interval of 95% and $H_o: \Delta \mu = 0$ , $H_a: \Delta \mu \ne 0$ and we get t-score = -3.908 which corresponds to p-value = 0.015 and according to my "understanding" that means 1.5% of times "our null hypothesis" which says that $\Delta \mu = 0$ is true so we can reject the null hypothesis with a 95% of confidence , right?
BUT and here's what i am trying to say:
We could have got the same T-score (because t-scores are simply the no. of standard deviations away from the hypothesized mean in the sampling distribution) but for a totally "different" null hypothesis (say $H_o: \Delta \mu = 3$) , so should not we construct a totally new T-distrubtion with it's own degress of freedom to test the new null hypothesis ($H_o: \Delta \mu = 3$) and same goes to all other null hypotheses, Why are we using a single disturbtion to test all possible null hypotheses ?
NOTE : I might be making wrong assumptions and of course feel free me to correct me but that's what i was able to grasp
Thanks in advance.
 A: We know that $\frac{\bar x-\mu}{s/\sqrt n}$ follows a t-distribution with n-1 degrees of freedom. Under the null hypothesis, we specify a particular value of $\mu$ that we think is the true population mean. Call it $\mu_0$. By extension, since this is assumed to be the true population mean, therefore $\frac{\bar x-\mu_0}{s/\sqrt n}$ follows a t distribution. If the sample mean $\bar x$ is far from the $\mu_0$, it is unlikely that we would have observed such an extreme sample mean under this t distribution, because the shape of the density is bell-shaped, meaning there is less density on the tails. It is not the fact that it is far from the $\mu_0$ that means that it is incorrect. The mean is only the expected value; it is possible that under another distribution, the farther you are away from $\mu_0$, the more density there is, and so the closer to $\mu_0$, the more unlikely you are to observe that value - for example, take a bimodal normal distribution. Anyway, since there is less density on the tails, we don't think that this distribution is correct. But what were we specifying for this distribution? Everything is fixed except for the mean. Therefore it must be the mean that is incorrect, and we reject the null mean.
There is no particular reason why you must test the hypothesis that $\mu_0=0$. It can be any value that you believe is the true population mean (see the very first sentence).
