For each image, I assume that one spot (left or right) truly is the brighter one, and that you know which spot that is. Then for
each image a person sees, you can get a score -2, -1, 0, 1, or, 2, where 2 is a strong correct response, -1 is a weak incorrect response, etc.
Summarizing the 40 responses for each of ten participants
you will have ten median responses between -5 and +5. Specifically,
one participant x may have 40 responses as below, with median $1.$
x
[1] 2 0 -1 3 2 1 -3 3 2 -3 2 -1 2 1 2 2 1 3 1 2
[21] -3 2 3 0 1 -2 2 1 1 2 1 1 3 -1 1 1 2 1 0 1
sort(x)
[1] -3 -3 -3 -2 -1 -1 -1 0 0 0 1 1 1 1 1 1 1 1 1 1
[21] 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3
median(x)
[1] 1
Then your ten subjects might have median scores s as follows.
s
[1] 2 -3 3 1 1 0 2 1 3 3
table(s)
s
-3 0 1 2 3
1 1 3 2 3
Among the ten scores s in my fictitious data, we have one participant with median below 0 and eight above 0.
Thus, a sign test on the nine relevant (non-0) scores
has P-value $P(S \ge 8 | H_0) = 0.0195 < 0.05 = 5\%,$ significant at the 5% level.
The null hypothesis is that participants can't distinguish
dark and light areas meaningfully, and have about a 50-50 chance
of getting a positive median score.
1 - pbinom(7, 9, .5)
[1] 0.01953125
sum(dbinom((8:9), 9, .5))
[1] 0.01953125
Notes: (1) A sign test based on ten participants does not
have good power, so even if some participants are good
at identifying dark vs. light regions, you may not detect
this effect with only ten participants.
(2) It would probably not be possible to use
a one-sample nonparametric Wilcoxon signed-rank test,
because, for small number of participants, this test
does not work well when there are ties, and we are very likely to have tied medians
in this experiment.
wilcox.test(s, mu=0)
Wilcoxon signed rank test
with continuity correction
data: s
V = 37.5, p-value = 0.08171
alternative hypothesis:
true location is not equal to 0
Warning messages:
1: In wilcox.test.default(s, mu = 0) :
cannot compute exact p-value with ties
2: In wilcox.test.default(s, mu = 0) :
cannot compute exact p-value with zeroes
(3) By contrast, one might look at mean responses for each of
the ten subjects and hope they are nearly enough normal
to do a one-sample t test. But this would require interpreting
categorical scores as numerical values. There is controversy
about this interpretation.
(4) Perhaps a better experimental design would be to have more
than ten participants and (if necessary) fewer than 40 images
per subject (e.g., maybe 20 participants, each looking at 20 images).
