What is the advantage of a scoring scale with many levels? In my research I have different ordinal variables with 3 levels of scores. Each score represents the severity of a condition. In other studies I see that the same variables are scored on a scale with 6 levels.
What is the advantage or disadvantage of having a scale with a higher number of levels? Having a scale with more levels would allow to classify the variables with more detail, but how would this affect statistical analyses when testing hypotheses?
 A: Fewer bins means more information loss.
Consider my colleague, who is $50$. He read a report about covid prognoses in people in their $50$s. "Wait...I'm more like someone who is $49$ than someone who is $59$," he protested.
By binning into decades, there was an assumption that the difference between a $50$-year-old and a $59$-year-old would be less than the difference between a $50$-year-old and a $49$-year-old.
Had the prognoses been on five-year intervals (so $50-54$, then $55-59$, etc), there would not have been as much of an objection. Even better than five-year intervals would have been two-year intervals. Even better than two-year intervals would have been each age.
A: Here is an example of a problematic attempt to use a
nonparametric 2-sample Wilcoxon rank sum test with Likert-5
data. The test gives approximate P-value $0.025,$ which
suggests significance, but is not sufficiently accurate
to give a definitive result.
set.seed(2022)
x1 = sample(1:5, 30, rep=T, p=c(2,2,4,5,4))
summary(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.000   3.000   4.000   3.567   4.750   5.000 
x2 = sample(1:5, 40, rep=T, p=c(1,3,5,2,1))
summary(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.000   2.000   3.000   2.975   3.250   5.000 

boxplot(x1, x2, horizontal=T, col="skyblue2")


Summaries show different medians, but whether they are significantly
different at the 5% level is not clear.
wilcox.test(x1, x2)

        Wilcoxon rank sum test with continuity correction

data:  x1 and x2
W = 784, p-value = 0.02443
alternative hypothesis: 
 true location shift is not equal to 0

Warning message:
In wilcox.test.default(x1, x2) : 
  cannot compute exact p-value with ties 

This difficulty causes some researchers to wander onto
the slippery slope of taking Likert-5 data to be
numerical (instead of categorical) and using a two-sample
Welch t test.
t.test(x1, x2)

        Welch Two Sample t-test

data:  x1 and x2
t = 2.1484, df = 53.184, p-value = 0.03626
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
 0.03932023 1.14401311
sample estimates:
mean of x mean of y 
 3.566667  2.975000 

It would be easy to report an "accurate" P-value below 4%.
It would be harder to say whether this P-value is any more
meaningful than the admittedly flawed P-value from the 2-sample Wilcoxon test just above.
