Page 309 - DEDU504_EDUCATIONAL_MEASUREMENT_AND_EVALUATION_ENGLISH

Unit 26: Achievement Test: Concept, Types and Construction

Notes
R H – R L
DP =
N
Where, N = The number of students in the high or low class
R H = The number of students in the high class solving an item correctly

R L = The number of students in the low class solving an item correctly
Example
A test was administered on 200 students. (Using 27% method), the item number 23 was
correctly solved by 45 and 31 students respectively out of a group of 54 students each.
Find out the discriminating power of this item.
Solution
N= 54

R H = 45
R L = 31

R H – R L
So, DP =
N
45 – 31 14
= =
54 54
= .26
So, the discriminating power of the item number 23 is .26
Third Method : When some of the items of the test are not attempted, the following
formula is used to find out the discriminating power.

R– R – 1
DP = ⎛ H L ⎞
R T ⎜ R T ⎟ 1 –
⎝ N– NR T ⎠
T
In which, R = R + R
T H L
N T = N H + N L

NR T = NR H + NR L
N H = The number of students in the high class

N L = The number of students in the low class

NR H = The number of students in the high class, who did not attempt
NR L = The number of students in the low class, who did not attempt
This formula is used only when the value of R is more than R . If the value of R is less
L
H
H
than R , the following amended formula is used :
L
R– R + 1
DP = ⎛ H L ⎞
R T ⎜ R T ⎟ 1 –
⎝ N– NR T ⎠
T

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