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Why not just testing alternative hypothesis? Why do we need null hypothesis?
For example, I am testing the effectiveness of a new drug. I can choose two groups: control and experimental. Based on the result I can say that whether the drug is working or not. Why do I need to state the null hypothesis in this case anyway?
Welcome to Cross Validated!
I'm sure someone could give a more canonical answer, but here's the conceptual gist of it.
Think of it this way: there is only one null hypothesis, right? The hypothesis that there is no difference between your two samples/populations.
However, how could you define your alternative hypothesis? There could be infinite alternative hypotheses. If you have a statistically significant difference between your control and experimental group, that could be due to:
I am not a researcher, and I don't know your experiment, so these are just guesses. But I hope you can see the point - there is only one null hypothesis (nothing is different between the groups) vs. an infinite number of possible explanations for an observed difference between the two groups. You assume that it's your intervention/drug that is the difference, but it could be anything! This is the conceptual reason why you don't test an alternative hypothesis - because which one would you test? What would be your assumed effect size? Too many (literally infinite) possibilities and assumptions. In vast majority of cases, the only thing you can actually test is "Is there some difference between these two groups?", and you set up the experiment as best you can so that if there is a difference, you can attribute it to the intervention (because you controlled for demographics, diet, behavioral factors etc.)
I hope this helped clarify your understanding!
In a test of hypothesis, the null hypothesis provides the distribution of the test statistic, and thus the critical value or the P-value, either of which can be used to decide whether to reject.
Three elementary examples:
For normal data with known $\sigma,$ we can test $H_0: \mu = 5$ against $H_a: \mu \ne 5.$ Then the test statistic is $Z = \frac{\bar X - 5}{\sigma/\sqrt{n}} \sim \mathsf{Norm}(0,1),$ so that we reject $H_0$ at the 5% level, if $|Z| \ge 1.96.$
In a chi-squared test of independence for a table of observed counts $X_{ij},$ the null hypothesis provides a formula for finding the expected counts $E_{ij}$ to evaluate the test statistic $Q = \sum_{ij} \frac{(X_{ij}-E_{ij})^2}{E_{ij}}$ distributed approximately as $\mathsf{Chisq}(\nu),$ where $\nu = (r-1)(c-1),$ if the table has $r$ rows and $c$ columns.
Suppose we have $X = 9$ successes in $n = 25$ Bernoulli trials with Success probability $p$ and want to test $H_0: p = 0.5$ against
$H_a: p < 0.5$ Then the test statistic for an
exact binomial test is $X \sim \mathsf{Binom}(25, 0.5).$ In R, the procedure binom.test is shown below,
where the P-value, not leading to rejection, is computed using this binomial distribution.
.
binom.test(9, 25, p=0.5, alt="less")
Exact binomial test
data: 9 and 25
number of successes = 9, number of trials = 25,
p-value = 0.1148
alternative hypothesis:
true probability of success is less than 0.5
... # confidence interval omitted
sample estimate:
probability of success
0.36
pbinom(9, 25, .5) # separate computation of P-value
[1] 0.1147615
because alternative hypothesis (Ha) will have a certain degree of correlation attached to variables, like strongly correlated or weakly correlated. Suppose you simply state Ha, then ppl will ask how strong is the correlation between variables. Suppose its weak correlation, again ppl will question, how much weaker? suppose Ha is very weak almost close to Zero, in such cases you'll need Null hypothesis (Ho) to make a distinction between Ha and Ho
