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Educational Measurement and Evaluation


                   Notes                   When the value found out using these methods is 10 or more, the discriminating power
                                           of the item is considered satisfactory. The items with less than this discriminating power
                                           are not considered suitable for the test. If the value of DP comes in negative, the item is
                                           considered very poor.
                                           Example
                                           The analysis of obtained marks in an achievement test for item number 23 is as follows.
                                           Find out the discriminating power of the item.
                                              Class   Number of Students   Correct Solution  Unattempted

                                              High           54                  40              10
                                              Low            54                  25               5

                                           Solution
                                                 R = 40,       R = 25,       N = 54,
                                                  H
                                                                L
                                                                               H
                                                 N = 54,       NR = 10,      NR = 5
                                                                                L
                                                                  H
                                                  L
                                           Here the value of  R  is more than  R , so the first formula will be used.
                                                           H
                                                                         L
                                                            N = N  + N = 54 + 54 = 108
                                                             T   H
                                                            R = R  + R = 40 +  25 = 65
                                                             T   H    L
                                                          NR = NR    + NR = 10 + 5 = 15
                                                             T     H     L
                                                                    R– R – 1
                                           So,             DP =    ⎛  H  L     ⎞
                                                                 R T  ⎜  R T   ⎟ 1 –
                                                                   ⎝   N– NR  T  ⎠
                                                                        T

                                                                   40 – 25 – 1
                                                               =        65
                                                                 65 ⎛  ⎜    ⎞  ⎟ 1 –
                                                                   ⎝  108 – 15  ⎠

                                                                   14
                                                               =
                                                                  ⎛   ⎞ 28
                                                                  ⎜ 65  ⎟
                                                                  ⎝   ⎠ 93

                                                                  14
                                                               =   1820
                                                                  93

                                                                   14     14
                                                               =       =     = 3.17
                                                                 19.5698  4.42
                                           So, the discriminating power = 3.17
                                           Note :
                                           The following formulae are used to know the difficulty level and discriminating power
                                           of the teacher-made tests respectively :

                                                                 R  + R
                                                           DL =   H   L
                                                                   2N



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