Advanced Math/% Improvement
QUESTION: I'm teaching a class and I would like to take a baseline and measure the percent improvement. So let's say the baseline is a set of homework assignments and the student only scores 5% correctly. Three months later the student takes the same homework assignment and scores 100% correctly. So would I score the improvement as follows:
(100-5)/5 = (95/5) = 19X so the improvement is 1900%
But what if the student scores 0% initially? Then we'd have:
(100-0/0) = (100/0) = But we can't divide by 0. How can we handle the situation here?
ANSWER: This does not make sense -- a percentage score is a measure of correctness, a rate of correct answers, basically an average or expected number of correct responses out of 100.
You can measure the absolute improvement, i.e. 100% - 5% = 95%.
I would strongly recommend against attempting to find the relative
improvement of something that is itself already a unitless average/rate. If you had a student who went from 5% to 15%, and a student who went from 70% to 100%, you would give them "200% improvement" and "43% improvement" respectively. Do you really think the first one had "more" improvement?
If you really want a better mathematical answer than "just use the difference" (i.e. 100%-5%=95%), you could consider instead a ten question assessment, where 70% is considered passing. If a student has average 100p%, ie. "p" is the probability this student gets any single question correct, where 0<p<1, then what is the probability the student passes the assessment?
f(p) = p^10 + 10 p^9 (1-p) + 45 p^8 (1-p)^2 + 120 p^7 (1-p)^3
You can then examine the difference in this score instead:
f(pnew) - f(pold)
f(100%) - f(5%) = f(1) - f(0.05) = 0.9999999.... ≈ 100% (this is incredible
f(15%) - f(5%) = f(0.15) - f(0.05) = 0.000134498... = 0.01% (not much improvement)
f(100%) - f(75%) = f(1) - f(0.75) = 0.224125... = 22.4% (a fairly significant improvement)
f(90%) - f(70%) = f(0.9) - f(0.7) = 0.337594... = 33.8% (huge improvement)
And in case of 0%, nothing really changes except you might note f(0)=0.
f(85%) - f(0%) = f(0.85) - f(0) = 0.95003... = 95% (incredible improvement)
These metrics / statistics are all subject to their utility. I suspect, however, that what I am proposing (either the normal difference 100% - 5% or this f(p) - f(q) idea) is much more useful than your strange percent-of-percent measure.
---------- FOLLOW-UP ----------
QUESTION: HI this is a question regarding implementation: Typical when I administer homework assignments, I provide help to the children. Now, if I am going to baseline/benchmark them, shouldn't I *not* provide any help as that will bias their initial scores? Then later, I administer the exact same homework to see whether they've improved?
It really depends on what you want the initial assignment to be. Ask yourself what is the purpose of that initial assignment. If you want it to be practice where they learn a little bit in their first try at the material, then you might as well help them to increase their learning, retention, etc.
(I am also assuming this assignment is not a significant grade like a final exam or term project, since it would be difficult to justify withholding assistance to students for a major assessment.)
On the other hand, if the goal is to establish a baseline and get a good idea of where each student stands, individually and without your help, then you can refrain from helping them with this initial benchmark assignment. You may want to let them know that this is the purpose of the assignment too -- take a moment to discourage them from working together, looking up solutions online, asking a parent/older student, etc.