How can I write down a significative statistic test? I'm studying statistics and thinking about some exercise to do, I came up with the idea to write them down my own. However, I'm coming across some problems, for which I'm asking for help.
I have 35 subjects with these scores of a random test, divided between pre and post. I'm trying to fit them in a normal distribution, having a not so different variance between them, and a significative shapiro wilk test. I can't understand how to change them in order to get my results.... any help? Thanks in advance
pre test;post test 
45;52
45;52
45;52
45;52
45;52
45;52
44;50
44;50
44;50
44;50
44;50
46;53
46;53
46;53
46;53
46;53
43;49
43;49
43;49
43;49
47;54
47;54
47;54
47;54
42;48
42;48
42;48
48;55
48;55
48;55
41;47
41;47
49;56
49;56


I suppose you want to know if there is a difference between pre-test and post-test values. Unless I have misread your data, for every one of the 35 subjects, the post-test score is higher. So you can use a simple sign test, based on a binomial distribution.
Under the null hypothesis that there is no difference between pre and post-test, the probability that the post-test score was always larger the the pre-test score is $(1/2)^{35} \approx 2.9 \times 10^{-11} < 0.01 = 1\%.$
So you can reject the null hypothesis in favor of the alternative that post-test scores are larger at the 1% level of significance. Computation in R:
.5^35
[1] 2.910383e-11

Formal sign test. For a slightly less-obvious case, consider 35 fictitious "post minus pre" differences as follows:
dif
 [1]  2  1  3  4  2  1  4  2  3  3  3  1  1 -1  4  2  1  1  2  1
[21]  2  2  1  3  3  2  2  0  1  0  0  2  1  2  3

In summary, there are 32 non-zero differences, of which 31
are positive.
table(dif)
dif
-1  0  1  2  3  4 
 1  3 10 11  7  3 

Thus, the P-value for a one-sided sign test would be nearly $0,$ computed in R as follows. Again, the null hypothesis is rejected.
pbinom(1, 31, .5)
[1] 1.490116e-08

Wilcoxon signed rank test. If you have no tied or zero differences, you may want to try a nonparametric one-sided Wilcoxon signed-rank test.
For my fictitious data I get a warning message that the
P-value may not be accurate:
wilcox.test(dif, alt="g", cor=F)

        Wilcoxon signed rank test

data:  dif
V = 522, p-value = 5.361e-07
alternative hypothesis: 
 true location is greater than 0

Warning messages:
1: In wilcox.test.default(dif, alt = "g", cor = F) :
   cannot compute exact p-value with ties
2: In wilcox.test.default(dif, alt = "g", cor = F) :
   cannot compute exact p-value with zeroes

Below we show that a permutation test can find an accurate P-value, even when ties and 0's are present. You say that the your data are not normal; according to the Shapiro-Wilk test, neither are my fictitious data dif. So one may wonder if the test
statistic of a t test has Student's t distribution, and so doubt the resulting P-value. [Notice that It is the differences that would need to be normal, not necessarily the pre or post scores.]
shapiro.test(dif)

        Shapiro-Wilk normality test

data:  dif
W = 0.93772, p-value = 0.04772

However, the t statistic seems a reasonable way to
express how far from the mean of the differences is from $0.$ Thus, we use the t statistic as a 'metric' in a permutation test. If pre and post scores are the same
we are just as likely to get positive and negative differences. [Some elementary texts seem to say it is OK to do a t test anytime the sample size exceeds 30, but this is not good advice.]
Permutation test. We can get a good idea of the distribution of the t statistic under the null hypothesis by repeatedly and
randomly switching the signs of the differences. If
t statistics for differences with switched signs tend to
be smaller than the t statitistic for dif, that is a good indication that $H_0$ is not true.
The permutation distribution of the t statistic under $h_0$ is approximated by the R simulation below.
t.obs = t.test(dif, alt="g")$stat; t.obs
       t 
9.013927 
t.prm = replicate(10^5, t.test( 
  dif*sample(c(-1,1),35,rep=T), alo="g" )$stat )
mean(t.prm > t.obs)
[1] 0

So the P-value is essentially $0$ and $H_0$ is rejected.
The figure below shows the simulated permutation distribution of the t statistic.

hdr = "Simulated Permutation Dist'n of t Statistics"
hist(t.prm, prob=T, xlim=c(-10,10), col="skyblue2", main=hdr)
abline(v = t.obs, lwd=2, lty="dotted", col="red")

