There are two obstacles to doing a Wilcoxon signed-rank test: (a) You have only 13 non-zero differences among 21. The 0 differences provide no evidence that Q1 and Q2 differ. You may hate to 'discard' these differences, but they were never really there. (b) There are many ties among the non-zero differences; only six unique differences among 13. Because the Wilcoxon test is based on ranks, the existence of
so many ties makes it difficult to find a P-value. I simply don't think a Wilcoxon signed rank test is useful for analyzing your data.
In principle, a simulated permutation test can obtain a reasonable approximation
of the P-value. However, that does not help find a significant difference for your data. P-values of permutation tests (one or two-sided) on your data are
much too big to lead to rejection.
The root difficulty is that 4 of the 13 nonzero differences are positive and 9 are negative, and that is too well balanced an outcome to lead to rejection. (If a fair coin is flipped 13 times, there
is better than 1 chance in 4 that it will show Heads either 4 or fewer times or 9 or more times.)
The bottom line is that your data are consistent with the null hypothesis that
Q1 and Q2 do not differ.
Addendum: Here is one possible permutation test for your data.
First, a paired t test gives test statistic $T = -0.8495$ with P-value $0.4057.$
One Sample t-test
t = -0.8495, df = 20, p-value = 0.4057
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
mean of x
The legendary robustness of the t test notwithstanding, a t test may not be appropriate for your data because they are so discrete and because they
fail a Shapiro-Wilk test of normality with P-value $0.02569.$
Shapiro-Wilk normality test
W = 0.89302, p-value = 0.02569
The usefulness of the t statistic as a measure of any shift from Q1 to Q2 is
not in question. However, because the data are not normal, it is doubtful
whether the distribution of the t statistic is Student's t distribution with 20 degrees of freedom.
A permutation test is based on the idea that if there is no shift in values from Q1 to Q2, we could change the signs of the differences without harm. If we
change these signs at random many times, and compute the t statistic for each
such 'permuted' sample, then we can approximate the 'permutation distribution' of the t statistic, and use that distribution to get a reliable P-value.
are too many possible permutations to consider them all by combinatorial methods, but simulating many cases gives a serviceable result. The result will be
slightly different on each run, but with 100,000 iterations not enough
different to affect the conclusion whether to reject the null hypothesis.
(Results from a program in R are shown below.)
The run below gave P-value 0.49, which is very much larger than 5%, so we
cannot reject the null hypothesis. [The P-value is not enormously different from the P-value of the t test, but enough different that doing the permutation test was worthwhile.]
d = c(-.5, -2, 0, 2.5, -.5, .5, 0, -.5, -.5, -1, 0, -1, 0 ,0, 1, 0, 0, .5, -1, 0, -1)
n = length(d)
set.seed(706); m = 10^5
t.obs = t.test(d)$stat; pv.obs = t.test(d)$p.val
t.prm = replicate(m, t.test(d*sample(c(-1,1), n, rep=T))$stat)
p.val = mean(abs(t.prm) >= abs(t.obs)); p.val
The permutation distribution of the t statistic is highly discrete, but
when plotted with only a few histogram bars it is seen not to differ greatly
from Student's t distribution with 20 degrees of freedom, the
density of which is shown in the figure below.
Reference: Eudey, et al. (2010) provides an
elementary introduction to permutation tests; Sect. 3 deals with paired