EXTRSCORE= is the fractional score point value to subtract from perfect (maximum possible) scores, and to add to zero (minimum possible) scores, in order to estimate finite values for extreme scores (formerly MMADJ=). Look at the location of the E's in the tails of the test ogive in Table 20. If they look too far away, increase EXTRSC= by 0.1. If they look too bunched up, reduce EXTRSCORE= by 0.1.

 

The measure corresponding to an extreme (maximum or minimum) score is not estimable, but the measure corresponding to a score a little less than maximum, or a little more than minimum, is estimable, and is often a useful measure to report.

 

This is how Winsteps handles extreme scores. In Rasch theory, an extreme score (maximum possible score or minimum possible score) on a set of items corresponds to an infinite ability measure (theta). This is impractical and also misleading in most situations. So Winsteps takes the following action:

 

If the score is maximum possible, then Winsteps estimates the ability measure for the score (maximum possible - EXTRSCORE) where EXTRSCORE is a small adjustment to the score. Usually EXTRSCORE = 0.3 score-points. The reasonable range of EXTRSCORE is 0.1 to 0.5.

 

If the score is minimum possible, then Winsteps estimates the ability measure for the score (minimum possible + EXTRSC)

 

So in one line, this becomes:

Score for estimation = Maximum (Minimum(observed score, maximum possible score - EXTRSCORE) , minimum possible score + EXTRSCORE)

 

The actual ability measure corresponding to an adjusted extreme score or any other score depends on the spread of the items, but it is at least Ln((score for estimation - minimum possible score)/(maximum possible score - score for estimation)) away from the mean item difficulty.

 

Rasch programs differ in the way they estimate measures for extreme scores. Adjustment to the value of EXTRSC= can enable a close match to be made to the results produced by other programs.

 

There is no "correct" answer to the question: "How large should extreme score adjustments be?" The most conservative value, and that recommended by Joseph Berkson, is 0.5. Some work by John Tukey indicates that 0.167 is a reasonable value. The smaller you set EXTRSC=, the further away measures corresponding to extreme scores will be located from the other measures and the larger the S.E.s. The technique used here is Strategy 1 in www.rasch.org/rmt/rmt122h.htm.The default value in Bigsteps was 0.5.

 

Weighted items or persons: the extreme score adjustment is multiplied by the smallest non-zero item weight for items with extreme scores, and the smallest non-zero person weight for persons with extreme scores.

 

See also: G.H. Fischer , On the existence and uniqueness of maximum-likelihood estimates in the Rasch model. Psychometrika 46 (1981), pp. 59–77

 

 

Example 1: You wish to estimate conservative finite measures for extreme scores by subtracting 0.4 score points from each perfect score and adding 0.4 score points to each zero person score.

  EXTRSCORE=0.4

 

Example 2: With the standard value of EXTRSCORE=, this Table is produced:

+-------------------------------------------------------------------------+

|ENTRY    RAW                        |   INFIT  |  OUTFIT  |PTBSE|        |

|NUMBER  SCORE  COUNT  MEASURE  ERROR|MNSQ  ZSTD|MNSQ  ZSTD|CORR.| PERSON |

|------------------------------------+----------+----------+-----+--------|

|    46     60     20    7.23    1.88| MAXIMUM ESTIMATED MEASURE | XPQ003 |

|    94     62     21    5.83    1.12| .44   -.9| .08   -.6|  .62| XPQ011 |

|    86     18      6    5.11    1.90| MAXIMUM ESTIMATED MEASURE | XPQ009 |

|    64     50     17    4.94    1.09| .53   -.7| .13   -.6|  .60| XPQ006 |

 

Here, the 5.11 corresponds to a perfect score on 6 easier items. The 5.83 was obtained on 21 harder items (perhaps including the 6 easier items.) To adjust the "MAXIMUM ESTIMATED" to higher measures, lower the value of EXTRSCORE=, e.g., to EXTRSCORE=0.2

 

Example 3. EXTRSCORE= is applied to the response string as observed. So, imagine that Mary is administered 5 dichotomous items out of a pool of 100 items and scores R on the 5 items. Her estimate will be based on:

Score for estimation = Maximum (Minimum(R, 5 - EXTRSCORE) , 0 + EXTRSCORE)