Final Exams and Final Grades


I finished grading the final exams and  calculating overall averages and letter grades.  The grades have been submitted through Spire, and should appear there after they have been processed by the registrar's office.

The average and mean grades on the final exam for all (86) students who took were nearly the same. As a percentile (out of 100) both were 57 (which was 68.5 out of 120 points.)  This is considerably lower than the mean/median grades on the first two exams (of 71 and 69), and would translate to an average letter grade of F.

Analysis of  the exam results: 
The class average does not tell the full story. There were a significant number of high grades (6 were between 92% and 99%, 21 were over 77%), and as there were a number of virtually perfect papers, it was clear that the low average score for the class was not caused by a single question about a topic everyone was unprepared for. The exam was comprehensive, and had questions about most of the major topics of the course, and it was not a "short" exam. A valid question is whether the exam was excessively long. The fact that there were 10 scores in the A to A- range, and 21 scores that were B- or better indicates that although the exam was a long  and comprehensive one, it was not excessive in the sense that even a very well-prepared student would be unable to finish it.

This was borne out in what I saw on many individual papers while grading. It was often the case that a significant segment of the class had difficulty in recognizing the type of equation in various questions and in using the appropriate methods of solution.   For example, many students simply didn't know what to do the the first problem on part II about a  linear first order equation (in fact, if you replace the "-2" with a minus "-3", the equation is identical to problem 1 on exam I) or failed to recognize  that the last question on part II (where the DE was given as p'/p=k(1000-p) ) was  separable (and nonlinear). This was also a minor "tweak" of  the separable equation, y'=(y+1)(y+2) on exam 1 (and is essentially the same DE  as  y'=y-2y^2 on 2a) of sample exam 1). Yet  a large
segment of the class had great difficulty here and elsewhere on the exam. While some students clearly struggled a bit with time as well as with the mathematics, the largest part of the problem by far was with students who were unable to recognize the type of equation various questions were asking about  and/or were unable to use this information to find an appropriate mathematical method of solution.

Why then were the results on the final so inconsistent with the results on the previous exams?

While grading, it was obvious that some students were much better prepared and had a much better recollection of the various methods and types of problems that we studied during the semester. This was also confirmed by a look at the  overall set of homework scores for the two sections. It was clear that only about two thirds of the class had turned in  the required 6 of 9 homework sets for grading, and a substantial number of students had even turned in considerably less than that. Students in their 2nd or 3rd year of college should know by now that cramming for exams just doesn't work in matheamtics- it takes lots of practice over the course of a semester to really master a subject. I decided to look at the data more closely, since I had been a bit worried from the start of the term whether dropping the 3 lowest  homework scores was too lenient a policy. The table below presents certain data about average performance on the final exam based upon the number of students ("#" in the second colum) who turned in at least N homework sets for grading(first column) -the homework  grade didn't matter if it was larger than 0. For each set of students, the  last two columns give the median ("M") and mean ("m") (percentile) scores on the final exam:

N ,# ,M ,m
9,11,68,68
8,34,68,67
7,50,68,65
6,68,62,62
5,76,60,60
4,79,58,59
3,87,57,55
2,92,56,53
1,94,55,52

Thus, of  the 11 students who turned in all 9 homework sets, the median and mean final exam scores of 68%,while of the set of 68 students who had turned in at least 6 homework sets, the final exam median and mean had both fallen to 62%, which was nevertheless significantly higher than the scores for at least 3 homework sets.
(In preparing the table, the students who had offficialy or unofficially dropped and didn't show up for the final were included, but all of them had done very few homeowrks.)

The table clarified for me the dramatic correlation between students who attempt most homework sets (regardless of the numerical score they get on their homeworks) and how well they do on a comprehensive final exam. The data clearly show that signficant  segments of the class that did well on exams I and II, did this by cramming (since they hadn't turned in much homework) and were unable to retain the material later on the final exam.There will of course be exceptions since these are only average results, however the overall pattern seen in the table show that both the class and the instructor (i.e. me) share some responsbility for the lower overall performance  on the final. It is impossible with 90 to 100 students to give students with valid reasons for not being able to turn in homeworks sets by the due date the opportunity to work on a makeup homework. The three dropped scores accounts for such missing grades, but also probably for too many others that could have been turned in but weren't. The
most important part of the table that the 68 students who turned in 6 or more of the 9  hw sets  had a median and mean of only 62% on the final, but for 50 students who had turned in at least 7 of the 9 sets, the median score ros e to 68% and the mean rose to 65%, while for students who had turned in at least 8 assignments,the mean rose to 67%.  While one could argue that in an ideal world, an ideal student would work out the problems on each homework whether or not it is turned in for a grade. One could also argue that an ideal professor would recognize that we don't live in ideal world, and that a significant segment of the class would  end up working out far fewer homework problems if they have their lowest three scores dropped.

While I've taught this course many times, the trend of having faculty teaching classes
with many more students than in previous years is a relatively recent and continuing one, and both the depatment and faculty are learning how to make do with the resources at hand (e.g. TA support.) In this case, it seems ovious that the class would have been significantly better off had I only dropped the lowest, or the lowest two homeowork scores, rather than the lowest 3. I won't make that mistake again.

What I did:
I decided that a reasonable compromise would be to introduce a uniform scaling of the  final exam to bring the average (mean and median) scores of the group of 68 students who turned in at least 6 homework sets to between 70 and 71%. In short I added 10  points to the (raw) final exam score of everyone in the class (regardless of how many homework sets turned in). On a percentile basis, this as 8.3 points to the 62% mean and median on row 4 of the table.

To make a (very) long story short, here is the formula I used to calculate your
overall course average:


                                                 (4 x (average of best 6 homework scores) +Mid I +  Mid 2 + (Final+10))

 overall average =                      ---------------------------------------------------------------------------------------------------------------------
                                                                                                                   4
Mid I and Mid II were out of 100 points and already percentile scores, the average homework score is out
of 20 points, and the final exam score was out of 120 points, so the above weighing is the same as the
one announed on the information sheet where your homework average would be 20% of your final grade,
each midterm woudl count 25%, and the final exam would be 30%. For midterm II and the final, I used the
scaled score whether or not it exceeded the total number of points on each exam.

A letter grade was then assigned according to the table announced at the start of the term
on the class information sheet:


A    90
A-   87
B+  83
B    80
B-  77
C+ 73
C   70
C-  67
D+ 63
D  60
F  59 and below

While final letter grades where computed from this table and from the formula above, I reviewed
each student's performance on homeworks and exams before entering their grade into Spire.
When circumstances warranted it, I raised several grades (usually to the next highest one on the table,
but I did not lower anyone's letter grade from the grade calculated from formula.

If you would like your numerical score on the final exam, please send me an
e-mail. Best wishes for a pleasant summer.
Robert Gardner