Of course, it would be better if you had repeated this experiment
for several steel specimens. But it is almost always desirable to have more
data. The question here is what you can do with the data you have.
It seems you had $n_1 = 10$ ferrite determinations before heat
treatment with $\bar X_1 = 25$ and $s_1 = 1.2.$ Then after
heat treatment, you made $n_2 = 10$ comparable determinations
with $\bar X_2 = 23$ and $s_2 = 1.0.$
Then, assuming the data are not markedly far from normal, a two-sample t test is appropriate for assessing
whether the decrease in ferrite content is statistically significant.
You can look at the formulas in Sect. 188.8.131.52 of the NIST Engineering Handbook (search
two sample t NIST online). For your data, I did the computation using Minitab
statistical software, as follows:
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 10 25.00 1.20 0.38
2 10 23.00 1.00 0.32
Difference = mu (1) - mu (2)
Estimate for difference: 2.000
95% CI for difference: (0.962, 3.038)
T-Test of difference = 0 (vs not =):
T-Value = 4.05 P-Value = 0.001 DF = 18
Both use Pooled StDev = 1.1045
I used the 'pooled' version of the test, in which one assumes
that the population variances are equal.
The results show that the decrease in ferrite content is
statistically significant: (1) The P-value 0.001 indicates
that if there truly was no change, there would be only
about one chance in a thousand seeing a decrease as large as 2.
(2) A 95% confidence interval for the true decrease is
$(0.962, 3.038)$, which does not include $0$ difference
as believable value. (3) Also, not shown in the Minitab printout:
a 'critical value' is 2.101; if the absolute value $|T| = 4.05 > 2.101$, that is evidence at the 5% significance level that
the two population means differ. (Any one of these three
interpretations, suffices to answer your question.)
If you had provided data for ferrite measurements at higher
temperatures or for longer heating times, one could use a
more complicated ANOVA (analysis of variance) or regression design to
decide whether any significant differences exist, and (to an
extent) also to say which heat treatments differ from which
other ones. You seem to feel confident, on this occasion, that these effects are obviously real. But bear in mind for future reference
that it is possible to deal with more than one such comparison
at a time in a statistical model.
Notes (sort of like lawyers' fine print):
(1) One might have used the Welch (separate variances)
version of the two-sample t test, also referenced in the NIST
handbook. It would give the same value of $T,$ used degrees of
freedom $df = 17,$ and given a P-value that is not noticeably
different. Here, the CI is $(0.958, 3.042)$ also essentially
the same as above, and the critical value is 2.110.
(2) If, before seeing the data, you expected the ferrite
content to decrease after the heat treatment, you could have
used a one-sided test, resulting in an even smaller P-value.
(3) Of course, one could quibble and say that it is the
passage of time, or some other reason other than heat treatment
that caused the decrease in ferrite content (drift in accuracy of ferrite determination? humidity? lunar
phase?). But the time
order of untreated preceding treated is inevitable, and the
objection seems silly.
(4) I see no justifiable way to
use a paired t test here because I see no natural pairing
of individual before-after measurements. Perhaps you could
run a paired experiment if it were possible to measure the
same exact pieces of steel before and after, but I don't
think that is what you did.
(5) It would have been
preferable to see the 20 individual measurements. That would
have made it possible to do certain diagnostic tests. If you
have these individual measurements available, edit then into
your question and I will have a look. But the effect is
sufficiently large that I do not foresee any change in