Which one-sample statistical test for ordinal variables? I have one group of <25 subjects and multiple evaluation scores:
New images of every patients has been rated by a radiologist compared to old images (1 to 5 with 1 Markedly worse, 3 similar and 5 Markedly better)
I don't have the score for the old images only the new ones that have been compared with the old images. I am confused how to do any statistical test?
I only know that it will be a non-parametric test: maybe wilcox test?
Something like that for every variable :
wilcox.test(tab$variable1,mu=4,alternative = "two.sided",conf.int = T)

?
 A: Comments:
It seems unlikely that you will find persuasive evidence
about the improvement of patients based on such a small sample.
Suppose you have $n = 21$ ordinal values denoted $1$ through $5.$
From the bit of R code you show, I assume you want to test the
null hypothesis that the mean or median is 4 against the two-sided
alternative. It is not clear to me exactly what conclusions you
would like to draw from your data, why you have chosen $4$ instead of $3$ as the null value, or why you have chosen a two-sided
alternative. [In my simulated data, with six patients possibly worse, nine possibly better, and six relatively unchanged, there seems not much to discuss.]
Consider the data I have simulated
in R below:
set.seed(2021)
x = sample(1:5, 21, rep=T, p=c(1,1,2,3,3))
table(x)
x
1 2 3 4 5 
2 4 6 5 4 

Especially for sample sizes as small as $n = 21,$ the
one-sample Wilcoxon (signed-rank) test does not work
well if there are many ties in the data or if relatively many values
are at the value (here $4)$ specified in the null hypothesis.
wilcox.test(x, mu=4, sim=T)

        Wilcoxon signed rank test 
        with continuity correction

data:  x
V = 22, p-value = 0.0153
alternative hypothesis: true location is not equal to 4

Warning messages:
1: In wilcox.test.default(x, mu = 4, sim = T) :
  cannot compute exact p-value with ties
2: In wilcox.test.default(x, mu = 4, sim = T) :
  cannot compute exact p-value with zeroes

Even so, the approximate P-value of about $0.014 < 0.05 = 5\%$
may suggest that the population from which your observations were
taken is not centered at $4.$
A one-sided sign test would reject at the 5% level. The logic
is that you have $21-5 = 16$ observations that are not $4$ and
among those only $4$ exceed score $4.$ The P-value of the sign test
is the probability $0.0384 < 0.5 = 5\%$ of getting $4$ or fewer scores above $4$ under
the assumption that scores above and below $4$ are equally likely.
pbinom(4, 16, .5)
[1] 0.03840637

The core procedures in R do not include a sign test, but Minitab does sign tests. For my hypothetical data the output is shown below.
Sign Test for Median: x 

Sign test of median =  4.000 versus < 4.000

    N  Below  Equal  Above       P  Median
x  21     12      5      4  0.0384   3.000

If my small dataset were real, I would say there is some evidence,
but weak evidence that the new x-rays look better. However, I don't
believe any choices about medical practice should be based on fewer
than two dozen subjects.

Note: If you assume your data are numerical instead of ordinal, and
pretend that scores are somehow nearly normal, then a test of $H_0:\mu=4$ vs. $H_a: \mu \ne 4$ rejects below the 5% level. I mention this test (pretty much based on fantasy) just so you will know I thought of it.
t.test(x, mu=4)$p.val
[1] 0.01185307

