T-tests provide info about the margin of mean difference? I have 2 unpaired datasets where the difference of the sample means is quite substantial (with the naked eye). The t-test shows that the difference is significant (p-value=0.007) and I would like to know if I can use the t-test to claim that one dataset provides significantly better results than the other or do I have to say that t-tests prove the difference is significant and use confidence intervals (or some other method) to show that one performs better than the other one?
If we go by the description of the T-test on Wikipedia: "The t-test can be used, for example, to determine if the means of two sets of data are significantly different from each other".
Based on that I believe t-tests can be used only to show if the means are different, we cannot extract info about the margin of the difference or one dataset is better than the other.

Consider the two independent fictitious normal datasets below.
set.seed(910)
x1 = rnorm(100, 50, 7)
x2 = rnorm(90, 60, 8)

mean(x1)
[1] 49.86995
mean(x2)
[1] 60.04863

The sample means show $\bar X_1 < \bar X_2$ suggesting that $\mu_1$ may be 'significantly' smaller than $\mu_2.$ Welch two-sample t test in R shows that the null hypothesis $H_0: \mu_1 = \mu_2$ is rejected in
favor of $H_a: \mu_1 < \mu_2.$ with P-value near $0.$
More specifically, the one sided CI: shows 95% confidence that
$\mu_1 - \mu_2 \le -8.412.$
t.test(x1, x2, alt="less")

        Welch Two Sample t-test

data:  x1 and x2
t = -9.5241, df = 187.95, p-value < 2.2e-16
alternative hypothesis: 
 true difference in means is less than 0
95 percent confidence interval:
      -Inf -8.412074
sample estimates:
mean of x mean of y 
 49.86995  60.04863 

Nothing here is a 'proof' that $\mu_1 < \mu_2.$ Of course, because
I simulated these data, we know $\mu_1 - \mu_2 = -10.$ However,
in a real statistical application we would not have such exact information.
The best we can do is to
report that sample means show $\bar X_1 =49.87 < \bar X_2 = 60.05$
and to say we are
'reasonably sure' that $\mu_1 - \mu_2 \le -8.412.$
