The 63% Distractor: What to Do With a Flawed Item
You teach physics to 41 second-year students. In the mid-term, Item 14 asked: a ball is thrown horizontally at 15 m/s from a 20 m cliff; how far from the base does it land? (g = 10 m/s²). The keyed answer is 30 m. Twenty-six students chose 60 m, including seven of the eight students who scored above 80% on the paper. The item's discrimination index is 0.05 and the class mean is 14.2 out of 25. While checking scripts you discover that the one-page revision handout you distributed two weeks earlier printed the time-of-flight formula as t = 2h/g, the square-root sign having been lost during formatting. The handout was never corrected on the board and no student queried it. The head of department wants the item voided and the marks awarded to everyone.
1. What do the response statistics actually tell you about Item 14, and should it be voided, credited to all, or retained with the original key? Defend your ruling using the numbers given.
Read the worked answer
Read the pattern before the pressure.
- 26/41 = 63% chose 60 m. Note that 60 = 15 × (2h/g) = 15 × 4: the distractor is the fingerprint of the missing square root, not of random guessing.
- Discrimination of 0.05, with 7 of the 8 strongest students choosing the distractor, means the item is currently measuring exposure to your handout, not projectile understanding. A sound item separates strong from weak; this one does not.
- But 15 students did avoid it, so the item is not unanswerable and the physics is correct.
Ruling: credit both 30 m and 60 m. Giving everyone the mark punishes the 15 who applied the correct formula; keeping only 30 m penalises students for relying on teacher-supplied material, which is a legitimate reliance interest. Crediting both, excluding the item from the discrimination analysis, and recording the reason in the moderation note is the proportionate response.
Duty afterwards: correct the handout in writing, tell the class plainly what happened, and do not let the episode become a precedent for re-crediting every difficult item.
2. Design a 15-minute corrective sequence for the next period and one replacement item, including the two distractors that would diagnose the same misconception.
Read the worked answer
Sequence (15 minutes).
- Board the two derivations side by side: h = ½gt² gives t = √(2h/g) ≈ 2 s, while t = 2h/g gives 4 s. Ask what unit the second expression produces; seconds-squared per metre appears and the error collapses.
- Concept check, no numbers: if the cliff were 80 m instead of 20 m, by what factor does the flight time change? Correct answer 2; the 2h/g misconception predicts 4. This separates rule-users from formula-reciters.
- Set the replacement item as a short exit ticket.
Replacement item. A ball is thrown horizontally at 12 m/s from a 45 m cliff (g = 10 m/s²). Key: t = 3 s, x = 36 m.
- Distractor 108 m: student uses t = 2h/g = 9 s.
- Distractor 12 m: student uses t = √(h/g) = 1 s, dropping the 2 entirely.
The two distractors target the same square-root misconception from opposite directions, so a wrong key will tell you which flavour of error you are still teaching. Re-test as a low-stakes three-item check within a week, and judge success at class level (80% mastery), not student by student.