Example 7: A partial credit analysis

A 30 item MCQ Arithmetic Test is to be analyzed in which credit is given for partial solutions to the questions. Each item is conceptualized to have its own response structure, as in the Masters' Partial Credit model. Estimation for the Partial Credit model is described in RSA, p. 87.

In this example, item 1 has 3 scoring levels. "C" is correct, worth 2 points. "A" and "D" are partially correct, worth 1 point. "B" is incorrect, worth 0 points. CODES= identifies all possible valid responses. In this example, KEY1= identifies responses worth 2 points. KEY2= and KEY3= identify responses worth 1 point. The values of KEY1=, KEY2= and KEY3= are set by KEYSCR=. So for item 1, KEY1=C..., KEY2=A..., and KEY3=D. Response B is not in a KEY= and so is scored 0. Here, invalid responses are treated as not-administered. If invalid responses are to be treated as "wrong", specify MISSCORE=0.

; This file is EXAM7.TXT

TITLE="A Partial Credit Analysis"        page heading

NAME1=1                ; Person-id starts in column 1

ITEM1=23                ; Item responses start in column 23

NI=30                        ; There are 30 items

; USING KEY1= (an alternative is to use IREFER=)

CODES=ABCD                ; Scores entered as A through D

;    0        1         2         3

;    123456789012345678901234567890

KEY1=CDABCDBDABCADCBDBCADBABDDCDABC        ; Fully correct

KEY2=ABCDABCCBADDABACABBBACCCCDABAB        ; Partially correct

KEY3=DABABAAACC*CCABAC*************        ; Some partially correct

                       ; if no matching response, use a character not in CODES=, e.g., *

                       ; the keys are matched in sequence, "B" for item 15 matches Key1=, and Key3= is ignored

KEYSCR=211                ; KEY1 fully correct (2 points),

                       ; KEY2, KEY3 partially correct (1 point each)

ISGROUPS=0                ; Each item is its own grouping, i.e., the Partial Credit model

MODELS=R                ; Each item has its own Andrich rating scale

STKEEP=Y                ; Keep all intermediate categories in analysis, even if never observed

CURVES=111                ; Print all 3 item curves in Tables 2 and 21

CATREF=2                ; Use category 2 for ordering in Table 2

&END

0.7  A1                ; item labels

1.1  A2

 .

3.1  A29

2.7  A30

END NAMES

090111000102   10001  BDABADCDACCDCCADBCBDBCDDACADDC

090111000202   10002  BDCDCDCDADCBCCBDBDCABCBDACDABC

 .

090111005302   10053  BDABCDDDCCCDDCDBACBABABCMCDBDC

090111005402   10054  BDADCDCCDACBCCABBDADBBCBBDDDDC