Comment:
A suggestion has been made to do a chi-squared
test on your $2 \times 6$ table of counts to see if the distributions for High and Low are
different. That is one reasonable possibility among several.
H = c(6, 4,14, 3, 3, 2)
L = c(0, 3,15,10, 6, 1)
TBL = rbind(H,L); TBL
[,1] [,2] [,3] [,4] [,5] [,6]
H 6 4 14 3 3 2
L 0 3 15 10 6 1
chisq.test(TBL)
Pearson's Chi-squared test
data: TBL
X-squared = 11.168, df = 5, p-value = 0.04815
Warning message:
In chisq.test(TBL) :
Chi-squared approximation may be incorrect
It has also been commented that the counts in some
of the cells are too sparse for a reliable P-value. Hence
the 'Warning'. (Some of the expected counts are less than 5.)
chisq.test(TBL)$exp
[,1] [,2] [,3] [,4] [,5] [,6]
H 2.865672 3.343284 13.85075 6.208955 4.298507 1.432836
L 3.134328 3.656716 15.14925 6.791045 4.701493 1.567164
Warning message:
In chisq.test(TBL) : Chi-squared approximation may be incorrect
You could try combining Classes 0 & 1 and Classes 4 & 5 to get a $2 \times 4$ table of counts and then do a chi-squared test on that table.
Alternatively, as implemented in R, it is possible to obtain a
more useful P-value by simulation, which is marginally
significant at the 5% level.
chisq.test(TBL, sim=T)
Pearson's Chi-squared test
with simulated p-value
(based on 2000 replicates)
data: TBL
X-squared = 11.168, df = NA, p-value = 0.04698
Moreover, as you suggest, your six 'Classes' are ordinal. So, you might want to consider looking here. Also, you might look at the
links in the margin of this page noted as 'Related'.
Your strategy should be to select the one test that you think
best matches the nature of your data and your objectives. (You should not try all
possible tests and pick the one with the smallest P-value.)