Why not just testing alternative hypothesis? Why do we need null hypothesis? 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?

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

