Confusion in Null hypothesis and Alternative Hypothesis

Question

Is it always true: The claim goes to the alternative hypothesis and the opposite of the claim goes to the Null hypothesis! If not then how to identify the case?

Answer 1

In contemporary frequentist hypothesis testing, we typically see four kinds of null hypothesis (the symbol $`\theta\text{'}$ means "some population statistic we are making inference about", for example, a population mean, $\mu$, or a difference in population means $\mu_1 - \mu_2$):

$$\begin{align}\text{H}_{0}\text{: }&\theta \ge c\text{; with }\text{H}_{\text{A}}\text{: }\theta< c\phantom{\text{---}}\text{ where $c$ is some constant}\\\\ \text{H}_{0}\text{: }& \theta \le c\text{; with }\text{H}_{\text{A}}\text{: }\theta> c \\\\\text{H}_{0}\text{: }&\theta = 0\text{; with }\text{H}_{\text{A}}\text{: }\theta \ne 0\phantom{\text{---}}\text{(i.e. the ‘two-sided’ test for difference)} \\\\\text{H}_{0}\text{: }& |\theta|\ge c\text{; with }\text{H}_{\text{A}}\text{: }|\theta|<c\phantom{\text{---}}\text{(i.e. the test for equivalence within $(-c, c)$)}\end{align}$$

When we test such null hypotheses, if we rejected the null ($\text{H}_{0}$), then we found enough evidence that the alternative hypothesis was true ($\text{H}_{\text{A}}$).

If we failed to reject the null, then we failed to find evidence that the alternative hypothesis was true.

There's another form of hypothesis using the Neyman-Pearson Lemma (NPL) which works a little differently, and instead asks given our observations and some distributional assumptions, which hypothesis is more probable $\text{H}_{1}\text{: }\theta = c_1$ or $\text{H}_{2}\text{: }\theta = c_2$? Unlike the four contemporary frequentist null hypotheses above, the NPL does not make null hypotheses about complementary sets that together span the entire sample space for $\theta$ (someone more confident on NPL please correct me on that :).