Tests for comparing two groups of ordinal discrete data I have two groups of an ordinal discrete variable (rating grade takes values from 1 to 14 but cannot have values like 1.5, 2.5 etc.) from two different models. Which test is appropriate for identifying how similar / divergent the outputs are? Model 1 is the proposed model to be used and Model 2 is a challenger model.
E.g.
Model 1 output: 1 3 2 4 6 5 7 (for a sample of 7 cases, the scale indicating quality)
Model 2 output: 2 4 3 5 4 6(for same set of 7 sample cases)
The developer used Spearman correlation to quantify correlation between two groups but I am not convinced if this is the right measure(may be I am wrong).I was thinking more like chi-square independence test,Kruskall wallis test?
Best Regards,
 A: Depending on what you want to know, it may be appropriate to look at
correlation because you use the same $n = 7$ 'cases' for each Model.
It would be surprising if the same sample of cases did not show some
positive association between the two models.
If you want to know the difference between the two models then you
should look at the seven differences $D_i = X_{1i} - X_{2i}, i = 1,2,\dots,7,$
between quality measurements.
If the sample distribution of the seven $D_i$ is roughly symmetrical, then
a Wilcoxon signed rank test might tell you whether median difference is significantly different from $0.$ However, $n = 7$ is near
the lowest possible sample size to get useful results from a nonparametric
rank-based test such as the Wilcoxon SR.
Specifically, $n = 7$ might be too few observations to detect
a real difference from population median $\eta=0,$ especially if several of the $D_i = 0$ or if
there are ties among the absolute differences $|D_i|.$
Suppose the differences are as shown below. Then a two-sided Wilcoxon test
testing $H_0: \eta = 0$ against $H_a: \eta \ne 0$ gives a
P-value $0.078,$ which does not show a significant difference from $0$
at the 5% level--even though six of the seven differences are positive.
d = c(-3, 1, 2, 4, 5, 6, 8) 
wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 25, p-value = 0.07813
alternative hypothesis: true location is not equal to 0

However, a one-sided test of $H_0: \eta \le 0$ against $H_a: \eta > 0$
does reject the null hypothesis at the 5% level. (Same data as above.)
wilcox.test(d, alt="g")

        Wilcoxon signed rank test

data:  d
V = 25, p-value = 0.03906
alternative hypothesis: true location is greater than 0

If the $D_i$ are all positive (or all negative) and unique, then a two-sided test rejects
$H_0: \eta = 0.$
d = c(1, 2, 3, 5, 6, 8, 9)
wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 28, p-value = 0.01563
alternative hypothesis: true location is not equal to 0

However, the Wilcoxon SR test does not give an exact P-value
when there ties or $0$s in the data.
d = c(1, 2, 3, 5, 6, 9, 9)
wilcox.test(d)

     Wilcoxon signed rank test 
     with continuity correction

data:  d
V = 28, p-value = 0.02225
alternative hypothesis: true location is not equal to 0

Warning message:
In wilcox.test.default(d) : cannot compute exact p-value with ties

[The 'continuity correction' in the title line is a clue that the P-value
is estimated.]
If you were to treat your ordinal data as numerical and use
t tests, then the difficulty with $0$'s and ties goes away. (I'm not not recommending that you use t tests because I don't know what distortions might arise, pretending your data categorical data are numerical.) However, with only
seven observations, a
t test may not detect a true non-0 difference. Results for a two-sided t test are shown below, for data with five of seven positive differences.
The P-value $0.22 > 0.05 = 5\%.$
d = c(-2, 0, 1, 1, 1, 2, 2)
t.test(d)

        One Sample t-test

data:  d
t = 1.3693, df = 6, p-value = 0.2199
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 -0.5621227  1.9906941
sample estimates:
mean of x 
0.7142857 

Note: Suppose that differences are roughly normally distributed
with stamdard deviation $\sigma = 2.$ Then if you want to have power
80% of detecting a difference in quality scores of $\Delta = 1$ or $2,$ then
a 'power and sample size' procedure suggests that you'd need $n = 34$ or $10$ pairs, respectively. [Output from Minitab.]
Power and Sample Size 

Paired t Test

Testing mean paired difference = 0 (versus ≠ 0)
Calculating power for mean paired difference = difference
α = 0.05  Assumed standard deviation of paired differences = 2

            Sample  Target
Difference    Size   Power  Actual Power
         1      34     0.8      0.807778
         2      10     0.8      0.803097

Because Wilcoxon SR tests have slightly lower power than t tests (in comparable situations),
you would need slightly larger sample sizes for Wilcoxon tests to
be useful.
