Test Item Analysis for its Effectiveness

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In this blog I introduce the following key terms used for finding the level of questions set by an examiner in an examination and the perception of the students about it. These terms are the facilitation value and the discrimination value.
1.       Introduction
A.      Facilitation Value: The facilitation value of a question indicates how easy the question is. The facilitation value is calculated by using the following formula.
FV= (RU + RL)(2xN*marks of test item)
Where N= no. of students in a group. The value is chosen as 30% of the total students from top scorer and 30% of students from low scorers.
RU = sum of marks obtained by upper group
RL = sum of marks obtained by lower group
When FV=1 means that all students in lower group and all students in upper group answered the test item. So it can be so assumed that the test item was very easy.
 When FV=0, means nobody answered the question, it means that either the question was so tough that no one was able to answer the question or else there was a problem in understanding the topic by the students which further mean that there might be problem in teaching learning methodology adopted by the faculty.
A.      Discrimination Index: Discrimination value is used in used to discriminate among good and the bad students. This value also indicate how good the test item was in deciding among the good and the bad students.
The formula for finding the decimation value is:
DI = (RU – RL) / (N*marks of the test item)
DI=1 when only the upper group answered the question
DI=0 when equal number of good and equal number of weak student answered the test ited and scored the same marks. So the test item is so ambiguous that it is not able to distinguish among the good and weak students.
DI=-1; This is when only the weak students answered the test item but all the good students failed to answer the test item.  It may be that weaker students answered by cheating or by some other means.
When FV=1, DI is definitely ‘0’ but when DI=1, FV is 0.5.
2.       Analysis of the Test Item
For the purpose of calculations, a class size of twenty student and their performance in some course is taken. The score of a class obtained by totawas taken
Test Item-->
1A
1B
1C
1D
1E
2A
2B
Total
Student Roll No.
2
2
2
2
2
5
5
Total
1
2
2
2
2
2
5
4
19
2
2
2
2
2
2
4
4
18
3
2
1.5
1.5
2
2
5
3
17
4
2
2
1.5
2
1.5
4
4
17
5
2
2
2
2
2
3
3
16
6
2
2
0.5
1.5
2
4
4
16
7
2
2

2
2
5
3
16
8
2
1
2
2
2
2.5
4
15.5
9
2
2

1.5
0.5
5
4
15
10
2
2
1.5
1.5
2
3
3
15
11
1.5
2
2
1.5
1.5
3
3.5
15
12
2
2
2
2
1
2.5
3.5
15
13
2
1
0
1.5
1.5
5
3.5
14.5
14
1.5
2
0.5
2
1.5
2.5
4
14
15
1.5
1.5
0
2
2
2.5
3.5
13
16
0
2
2
2
2
3
2
13
17
1
2
0.5
1
2
2.5
4
13
18

2

1.5
2
2
5
12.5
19

2
1.5
2
2
4

11.5
20
2
2
2
2
2


10
Index
1A
1B
1C
1D
1E
2A
2B

FV
0.69
0.96
0.65
0.92
0.98
0.65
0.61

DI
0.63
0
0.29
0.08
-0.04
0.37
0.25







Conclusion:
As per the observation from the above representation, it can be concluded that some test items were very easy; some test items were very ambiguous as they were not able to discriminate among the good and weak students. The table below summarizes these facts.
Test Item-->
1A
1B
1C
1D
1E
2A
2B
How easy is the question
Not very Easy
Quite easy
Not very Easy
Quite easy
Very Easy
Not very easy
Not very easy
How Good the Question is in differentianing Good and weak students
Almost equal number of students in both upper and lower group answered
Very Tough Question
Quite tough question
Quite Tough question
Ambiguous question No able to discriminate among good and weak students
Not Very Good as it does not clealy discrimate good and weak students
Not Very Good as it does not clealy discrimate good and weak students



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