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STATISTISTICAL METHODS FOR RANKING

STATISTISTICAL METHODS FOR RANKING. Sture Holm FMS Jubileumskonferens på Utö 25 oktober 2012. A COMMON TYPE OF PRESENTATION Comparison between care centers. Here about 200 care centers 40 000 involved patients, 56 percent answers In central part often 1 unit differences

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STATISTISTICAL METHODS FOR RANKING

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  1. STATISTISTICAL METHODS FOR RANKING Sture Holm FMS Jubileumskonferens på Utö 25 oktober 2012

  2. A COMMON TYPE OF PRESENTATIONComparisonbetweencare centers

  3. Here about 200 care centers • 40 000 involved patients, 56 percent answers • In central part often 1 unit differences • Weighting of 6 ”items” med 50-steps (from 0 to 600 ”points” for the individuals) • NO MEASURES OF DISPERSION GIVEN !! • ”Manufactured” by consultants ”Indikator”

  4. A typical exempleSuccess rates in units

  5. Howcanwe make correctstatisticalstatements on the ranks ? • As I see it there are two steps: • A suitablestatisticalmethod for pairwisecomparisonbetweencases. This is standard technique i statistics. • To useresults in this first step for making a proper statisticalstatement on the rank of a unit with a confidence interval.

  6. Step 1. Comparison within pairs • Different problems requires different basic methods, for example: • Wilcoxon test methods • Binomial comparison • Normal approximation • OBSERVE: MOSTLY THERE ARE DIFFERENT SAMPLE SIZES IN THE UNITS

  7. Step 2: • Principle: To use one-sided p values for the pairs in order to get the number of pairs, that can be considered ”significant” on each side. (within the half over all risk). • But how ? • To use mutiple test methods • Will give choosen confidence degree

  8. Simplest: Holm-Bonferroni • Bonferroni: Devide the risk (e.g. 2.5 %) with the number of cases, and use it individually. • + Holm (SJS 1979): Continue with new steps where the number is the remaining non-rejected. Go on as long as there are new rejections. • ”Translate” to intervals for rank.

  9. Illustration of multiple test

  10. The simple binomial example • Consider for instance unit E. Test hypothesis that all other units are at least as good with multiple level 2.5 % and find the ones which can be declared worse than E. • Repeat on other side

  11. The estimates

  12. Estimated order: • I, A, H, B, E, G, F, J, D, C • The ”central” B and E have no or almost no information on rank • I, H, A are positioned low [1;4] • G, F, J, D, C are positioned high [4;10]

  13. An exemple with scales • A: 35, 44, 51, 88, 46 • B: 19, 29, 34, 41, 30 • C: 43, 52, 54, 99, 56 • D: 42, 50, 70, 117, 61 • E: 13, 37, 49, 85, 45 • F: 15, 21, 28, 71, 35 • G: 16, 40, 46, 86, 34 • H: 19, 36, 39, 73, 31 • I: 17, 37, 55, 126, 65 • J: 16, 15, 18, 50, 20

  14. Confidence intervals for ranks

  15. Estimated order • B, C, A, H, D, G, J, E, F, I • For the extremes B and I (153 resp. 300 obs.):

  16. Advantages and disadvantages • Correct general method for different situations with chosen confidence degree. • A little inefficient since it is based on the Boole inequality, which is very good for negatively dependent and independent cases, but not so good for positively dependent cases. We have positive dependence here since there is comparison with the same case.

  17. Is there a big loss ? • Here follows a number of cases with 50 % correlation exact normal bound correspoding Bonferroni bound. Total risk 1 %. • Is it worthwhile to try to be exact ?

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