How to compare two performances? The Merry Men of Sherwood Forest dispute again over archery proficiency. Little John and Will Scarlet shoot 10 arrows at a target. To make the comparison fair, they shoot the same 10 arrows in the same order.
The Data | |||||||||||
Arrow: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Score |
John | Hit | H | H | H | M | M | H | H | H | H | 8 |
Will | Miss | H | M | H | H | M | M | M | H | M | 4 |
Robin Hood insists "Those 10 arrows were crafted to be equivalent. The comparison should be:"
Robin Hood's Analysis | ||||
Hits | Misses | Odds of Success | Log-Odds | |
John | 8 | 2 | 8/2 | 1.4±.8 |
Will | 4 | 6 | 4/6 | -.4±.6 |
John ----- Will |
8/2 -- 4/6 |
1.0±1.0 |
"John is 6 times better than Will!"
Maid Marian is skeptical. "No two arrows are identical. Some fly true. Some less so. If John and Will miss the target with the same arrow, perhaps that arrow is defective. If John and Will hit the target with the same arrow, perhaps that arrow is exceptionally well made. The comparison should be:"
Maid Marian's Analysis | |||
John' Hits | John's Misses | ||
Will's Hits | - | 1 | |
Will's Misses | 5 | - | |
John ---- = Will |
(John's Hits & Will's Misses) -------------- (Will's Hits & John's Misses) |
Odds: 5/1 |
Log-Odds: 1.6±1.1 |
"John is still better than Will, but only 5 times."
Robin Hood objects: "But you are throwing away almost half of the data! My answer must be better."
Friar Tuck intercedes. "Those answers mean the same! You have forgotten that meaning is in the probabilities, not data! Write both statements in probability terms counting only the arrows you counted! Let P_{J} be John's probability of hitting the target. P_{W} be Will's. Here is Robin Hood's computation, considering the arrows to be equivalent:"
Friar Tuck's Version of Robin Hood's Analysis | ||||
Hits | Misses | Odds of Success on Target | ||
John | P_{J} | 1-P_{J} | P_{J} / (1-P_{J}) | |
Will | P_{W} | 1-P_{W} | P_{W} / (1-P_{W}) | |
John ---- Will |
P_{J} / (1-P_{J}) = --------------- P_{W} / (1-P_{W}) |
P_{J} * (1-P_{W}) = --------------- P_{W} * (1-P_{J}) |
"and here is Maid Marian's, considering the arrows to be heterogeneous:"
Friar Tuck's Version of Maid Marian's Analysis | ||
John' Hits | John's Misses | |
Will's Hits | P_{W} * (1-P_{J}) | |
Will's Misses | P_{J} * (1-P_{W}) | |
John (John's Hits & Will's Misses) P_{J} * (1-P_{W}) ------ = ------------------------------------- = --------------- Will (Will's Hits & John's Misses) P_{W} * (1-P_{J}) |
"Algebraically, both numbers are attempts to estimate the same unknowable truth! Now look at those numbers in log-odds terms. The difference between the measures is 1.6 - 1.0 = 0.6, but the precision of a measure is ±1.0 at best. Numerically, we are talking about the same number. 5 times better or 6? All we are arguing over is the imprecision in the estimates. These results provide no firm evidence about the quality of the arrows one way or the other, so we don't know which estimate is better, but we do know that regarding the arrows as heterogeneous is the more conservative option. So, Maid Marian's 5 times is the more defensible result!"
John Michael Linacre
A Pointed Argument in Sherwood Forest. Linacre J.M. … Rasch Measurement Transactions, 1997, 11:3 p. 574
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