Is it even possible to compare two smples with the data I have? I'm pretty new to statistics. and recently I learned how to check if the samples are equal. I want to learn how to find out which of the samples is greater. I have some friends in college, that are good at statistics, and they said that's impossible to find it out with the data I have. For example, I have two drugs ($x_1$, $x_2$), and I have to find out which of them is more effective. Here is the data:
Sample mean:$\bar x_1=5.9$, $\bar x_2=4.1$
Sample standard deviation: $\sigma_1=4.1$, $\sigma_2=3.7$
Sample size: $n_1=42$, $n_2=47$
And I want to test:
$$H_0:\mu_1-\mu_2>0$$
$$H_a:\mu_1-\mu_2<0$$
 A: I have some quibbles with your notation:
(a) The sample standard deviations should be
denoted $S_1, S_2.$ Notations $\sigma_1$ and
$\sigma_2$ are for population standard deviations.
(b) It is pointless to test $H_0: \mu_1 - \mu_2 \ge 0$
against $H_1: \mu_1 - \mu_2 < 0,$ because you say
$\bar X_1 = 5.9 > \bar X_2 = 4.1.$ So you have no
evidence that $\mu_1 - \mu_2 < 0$ might be true.
I will test $H_1: \mu_1 - \mu_2 \le 0$ against
$H_1: \mu_1 - \mu_2 > 0.$  You know that the first
sample mean is greater than the second. The
question whether it is enough greater that the
difference would be called 'statistically significant'.
(3) Also, notice that the null hypothesis $H_0$ must always
contain an equal sign (maybe as $=, \le$ or $\ge).$
Finally, you have given summary data rather than the
actual observations. So I have no way to check whether
the data are approximately normally distributed, as
is required for the Welch two-sample t test to be valid.
So I will simply assume your data are nearly normal.
I am using a Welch two-sample t test instead of a 'pooled' two-sample t test because the two sample standard deviations you provide are different. Thus the population standard deviations may also be different, which would
violate an assumption of the 'pooled' test.
Here is printout from a recent release of Minitab
statistical software:
Two-Sample T-Test and CI 

Sample   N  Mean  StDev  SE Mean
1       42  5.90   4.10     0.63
2       47  4.10   3.70     0.54

Difference = μ (1) - μ (2)
Estimate for difference:  1.800
95% lower bound for difference:  0.417
T-Test of difference = 0 (vs >): 
  T-Value = 2.16  P-Value = 0.017  DF = 83

Because the P-value $0.017$ is smaller than $0.05 = 5\%,$
we reject the null hypothesis and conclude that $\bar X_1 = 5.9$ is significantly larger than
$\bar 4.1$ at the 5% level of significance. (You could also
declare significance at the 2% level.)
"Proving" that one drug is better than another
is a practical situation requires evidence from
thousands of randomly selected subjects and
significance levels much smaller than 5%.
If required to show computations by hand using
a calculator, you can look for the Welch t statistic in a statistics text (or online). Also notice that there is a special formula for
the degrees of freedom DF in a Welch t test;
it uses sample sizes and sample standard deviations.
