Interesting thread! Demetri suggested one possibility for analyzing your data. But I wonder if something like what I outline below might get you closer to what you want?
It seems like you would like to see if A and B are rated higher than C and D. So why not compute the average rating given by each participant to the A and B images and the average rating ...
The unfortunate fact is that this can not likely be done with an easy statistical test like a chi-square or t test. That you've chosen an ordinal outcome (not a count, as you say) and have experimental subjects rank multiple images will require us to use an ordinal mixed effects model.
Thankfully, the ordinal package in R can do this. However, the model ...
The chi-square test tests against the alternative that the distributions are different; the Mann-Whitney test tests against the alternative that one group is stochastically larger than the other. It depends on what you are interested in (i.e., if eventually you'd like to know whether one group is generally rating higher than the other, or whether they are ...
You can say that in your data, on average, belonging to the fifth quantile of parental income instead of the second leads to an increase in the score equal to:
7.49 - (-0.95) = 8.44. And this difference was statistically significant at an alpha level of 0.01.
Regarding your second question:
given your model, the difference in score between two people of the ...
With such large samples, do you think results of a statistical test adds much value?
Formally, you could used prop.test in R (or essentially equivalently, a chi-squared test on a 2-by-2 table).
prop.test(c(23*10^6,19*10^6), c(111*10^6,126*10^6), cor=F)
2-sample test for equality of proportions
without continuity correction
data: c(23 * 10^...
One possibility is to use Monte Carlo, that is, a chi-squared test with a simulated p value. I will show how to do that in R:
mytab <- cbind( control=c(0, 1, 0, 1, 1, 1, 3, 3, 8, 10, 3),
pilot=c(1, 3, 2, 2, 2, 5, 3, 5, 1, 4, 2) )
chisq.test( mytab, sim=TRUE, B=20000 )
Pearson's Chi-squared test with simulated p-value (based on 20000