Testee A: 1 1 1 1 1 = 5 points
Testee B: 1 0 1 1 1 = 4 points
Testee C: 1 0 0 1 1 = 3 points
Testee D: 0 1 1 1 0 = 3 points
Testee E: 1 0 1 0 0 = 2 points
So, item 1 is solved by 4 testees: A, B, C, E. This means that S=4 according
to http://ivec.ultimaiq.net/quality.htm
Among top 4(S) testees item 1 is solved by A, B and C, so R=3.
The quality of item 1 is so R/S=3/4=75%.
Item 5 is problematic here. It is solved by A, B and C, which means S=3.
But top 3 testees do not exist because testees C and D share 3 points each.
So, I will take S=4 here, and then again R=3, and again 3/4=75%.
Item 2: 1/2=50%
Item 3: 3/4=75%
Item 4: 4/4=100%
Try to clarify this to yourself! If S is not an integer, round to the nearest integer.
So we have 4 good items (1,3,4,5) and 1 average (2).
Solvability of item is just an average score. For item 1 it is 4/5=0.8,
and for all items we have 0.8, 0.4, 0.6, 0.8, 0.6 in order.
Sorting by solvability we have the following table (I call it table A below):
item |
solvability |
quality |
4 |
0.8 |
100% |
1 |
0.8 |
75% |
3 |
0.6 |
75% |
5 |
0.6 |
75% |
2 |
0.4 |
50% |
Now, let us assume that we have the following norm:
raw score |
IQ |
1 |
130 |
2 |
140 |
3 |
150 |
4 |
160 |
5 |
170 |
IQ 130-150 range is raw score 1-3 range, so for quality in that range we take
only the first three rows from table A:
(100%+75%+75%)/3=83.3%.
140-160 range is raw 2-4, so we take rows 2-4 from table A:
(75%+75%+75%)/3=75%.
Quality in the 150-170 range is thus (75%+75%+50%)/3=66.7%.