There is another way to rank the difficulty of test questions using Networks: Each student gives scores for each problem, then w

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3 years ago

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There is another way to rank the difficulty of test questions using Networks: Each student gives scores for each problem, then w

e construct a network whose nodes/vertices are the problems of the exam. There is an arc from problem A to B for student k that did better in problem B than in problem A, (i.e., if sA, sB are the scores of those problems, sB > sA). The weight of the arc yk associated to student k equals (approximately) the difference of score the student received in those two problems sB −sA = yk. Explain why the Massey’s method we saw in class can be used for rating exam problems by difficulty.

Mathematics

1 answer:

Vlad [161] · Answer 1 · 2021-12-30T20:27:37+03:00

Answer:

Step-by-step explanation:

Truly, on a single student level, the Massey method can be applied i.e. problems are ranked according to their level of difficulty for a single student.

Let us say there are 4 problems P1, P2, P3, P4 set in an exam where a student has to solve any two.

Assuming student 1 solves and scores more in P1 than in P2, r1>r2

student 2 more in P1 than P4 r1>r4

student 3 more in P2 than P4 r2>r4

student 4 more in P4 than P3 r4>r3

student 5 more in P2 than P3 r2>r3

and such many more cases.

then the Massey's model can be written as

P1 P2 P3 P4

\begin{bmatrix} 1 &-1 &0 &0\\ 1 &-1 &0 &0 \\ 0 &1 & 0 &-1 \\ 0 & 0 &-1 & 1 \\ 0 & 1 & -1 & 0 \end{bmatrix}.

\times \begin{bmatrix}r1 \\r2 \\r3 \\r4 \end{bmatrix} =

\begin{bmatrix}y_{1} \\ y_{2} \\ y_{3} \\ y_{4}\\ y_{5}\end{bmatrix}.

If Pi gives more score than Pj, the entry below Pi shall be 1 and below Pj shall be -1.

The rest of the entries shall be 0.

r1,r2,r3,r4,r5 are ranks .

There may be more rows depending upon more combination of 2 problems.

Should in case more than one student solve the same combination and get scores in the same order, for instance, 12 students solve P1 and P2, for 7 of them r1>r2, then their scores can be averaged.

Normally, such systems are overdetermined, i.e., more rows than the number of unknowns, then it is solved by the Least square method ., as there shall be no a solution with least error is found out.

The overall ranking, in reverse order, shall give the ranking according to increasing difficulty.