Session 3 The proportional odds model and the Mann-Whitney test

Transcription

1 Session 3 The proportional odds model and the Mann-Whitney test 3.1 A unified approach to inference 3.2 Analysis via dichotomisation 3.3 Proportional odds 3.4 Relationship with the Mann-Whitney test Session 3 1

2 3.1 A unified approach to inference Data (assumed here to be discrete): x = (x 1,, x n ) A single unknown parameter: θ Likelihood: L(θ; x) = probability of x, given θ Log-likelihood: l(θ) = log L(θ; x) Session 3 2

3 Suppose that θ is small (it might be a treatment effect) Taylor s expansion: l l l l ( θ ) = (0) + θ (0) + 2 θ (0) + O( θ ) + θ θ 1 2 const. Z 2 V where Z = l (0), efficient score V = l (0), Fisher s information Session 3 3

4 For large samples and small θ Z ~ N(θV, V) approximately (Scharfstein et al., 1997) This is the basis for many common statistical tests: Pearson s Chi-squared test Armitage s trend test The logrank test The Mann-Whitney (or Wilcoxon) test and it leads to asymptotically efficient methods Session 3 4

5 Estimation of θ 1. Maximum likelihood estimate where ˆ θ l ( θ ˆ) = 0 2. Estimate based on score statistics Z V which has variance 1 V Session 3 5

6 Hypothesis testing To test the null hypothesis of H 0 : θ = 0 1. Likelihood ratio test 2. Score test 2 Z ~ V 3. Wald test χ 2 1 { ˆ } 2 1 W = 2log L(0) / L( θ) ~ χ θˆ s.e. ( θˆ ) 2 ~ χ 2 1 Session 3 6

7 For likelihoods with nuisance parameters: φ Replace log-likelihood with profile log-likelihood l( θ, φ) l( θ, φˆ ( θ)) where ˆφ(θ) is the maximum likelihood estimate of φ, given the value of θ l( θ, φˆ ( θ) ) is a function of θ only Session 3 7

8 Example: Binary data Control Group Treated Group Total Success s C s T s Failure f C f T f Total n C n T n Probability of success: p C and p T on C and T respectively log p (1 p ) T C θ = e p C(1 p T ) (log - odds ratio) Session 3 8

9 The unconditional likelihood of θ (comparison of binomial observations) leads to (see Session 2) Control Group Treated Group Total Success s C s T s Failure f C f T f Total n C n T n Z s f n s f n n sf = n T C C T C T = V 3 Session 3 9

10 The conditional likelihood of θ given s successes in total (hypergeometric distribution) leads to Control Group Treated Group Total Success s C s T s Failure f C f T f Total n C n T n Z = s f T C n s f C T n n sf V C T = n 2 (n 1) Session 3 10

11 3.2 Analysis via dichotomisation Head injury data from Example 1 GOS at 3 months GCS on entry Total 1. Good Recovery Moderate Disability Severe Disability Vegetative State Dead Total Session 3 11

12 1. Success is Good GOS GCS on entry Total Success Failure Total θ = log odds of success for (GCS = 6-8) versus (GCS = 3-5) Z 1 = 85.8, V 1 = 53.4 Estimate of θ = Z 1 /V 1 = 1.61 Score test: Z 12 /V 1 = (c.f. χ 2 on 1 df) Session 3 12

13 2. Success is Good or Moderate GOS GCS on entry Total Success Failure Total θ = log odds of success for (GCS = 6-8) versus (GCS = 3-5) Z 2 = 124.9, V 2 = 67.0 Estimate of θ = Z 2 /V 2 = 1.87 Score test: Z 22 /V 2 = (c.f. χ 2 on 1 df) Session 3 13

14 3. Failure is Vegetative or Dead GCS on entry Total Success Failure Total θ = log odds of success for (GCS = 6-8) versus (GCS = 3-5) Z 3 = 124.7, V 3 = 68.0 Estimate of θ = Z 3 /V 3 = 1.83 Score test: Z 32 /V 3 = (c.f. χ 2 on 1 df) Session 3 14

15 4. Failure is Dead GCS on entry Total Success Failure Total θ = log odds of success for (GCS = 6-8) versus (GCS = 3-5) Z 4 = 113.3, V 4 = 66.3 Estimate of θ = Z 4 /V 4 = 1.71 Score test: Z 42 /V 4 = (c.f. χ 2 on 1 df) Session 3 15

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18 Analyses each indicate that GCS 6-8 is preferable to GCS 3-5 Magnitude of advantage, on the log-odds ratio scale, is consistent How can these four analyses be combined? Session 3 18

19 3.3 Proportional odds Notation Category Control Group Treated Group Total C 1 n 1C n 1T n 1 C 2 n 2C n 2T n 2 M M M M C m n mc n mt n m Total n C n T n Session 3 19

20 Let L kt = n 1T + + n (k-1)t, U kt = n (k+1)t + + n mt L kc = n 1C + + n (k-1)c, U kc = n (k+1)c + + n mc Thus, if Success is {C 1,, C k }, the derived 2 2 table is Control Group Treated Group Total Success L (k+1)c L (k+1)t s Failure U kc U kt f Total n C n T n Session 3 20

21 Let p kc = P(C k ; Control Group) Q kc = P(C k or Better; Control Group) = p 1C + + p kc, k = 1,, m; so that Q mc = 1 p kt and Q kt are defined similarly for treated group and Q kt (1 Q kc) θ k = log Q kc (1 Q kt ) k = 1,, m 1 Session 3 21

22 θ k is the log-odds ratio of Success where Success is {C 1,,C k } The proportional odds assumption is θ 1 = θ 2 = = θ m 1 = θ The common value, θ, is a measure of the advantage of being in the Treated Group > 0 Treated Group better θ = 0 no difference < 0 Treated Group worse Session 3 22

23 Score and information Using a marginal likelihood based on the ranks, with allowance for ties (Jones and Whitehead (1979)) the efficient score for θ is m 1 Z = n kc(lkt U kt ) n + 1 k = 1 and Fisher s information is m ntncn nk V = 1 2 3(n + 1) k= 1 n 3 Session 3 23

24 Application to Head Injury data 1 Z = {73 ( ) ( ) + 79 ( ) + 37 ( ) ( )} = V = = ( ) = 83.0 Session 3 24

25 Hence 2 Z V = 2 larger than any individual 2 2 table, and c.f. very highly significant χ 1 Estimate of θ = Z V = 1.74 between the values from the 2 2 tables 95% confidence interval for θ is Z V ± 1.96 V = (1.52, 1.96) Note: The 2 2 tables contain 64%, 81%, 82% and 80% of total information respectively Session 3 25

26 The 2 2 table as a special case (m = 2) Control Group Treated Group Total Success s C s T s Failure f C f T f Total n C n T n 1 s f s f Z = sc ( 0 ft ) + fc ( st 0) = n + 1 n + 1 { } T C C T Note: n+1 instead of n in denominator Session 3 26

27 The 2 2 table as a special case (m = 2) Control Group Treated Group Total Success s C s T s Failure f C f T f Total n C n T n 3 3 ncn T s f ncntsf V = 1 2 = 2 3(n + 1) n n (n + 1) n Note: (n+1) 2 n instead of (n) 2 (n-1) in denominator Session 3 27

28 3.4 Relationship with the Mann-Whitney test (Wilcoxon, 1945; Mann and Whitney, 1947) Samples: x 1,..., x a (low values y 1,..., y b are good) Scores: Mann-Whitney statistic: 1if xi < y d = 0 if x = y + 1if xi > y ij i j M a b i= 1 j= 1 ij (other variations exist) j j = d var M = ab( a + b + 1) 3 Mann-Whitney test: 2 2 M var M c.f. χ 1 Session 3 28

29 Mann-Whitney test with ties Values : u 1 u 2... u m Total x s n 1C n 2C n mc n C y s n 1T n 2T n mt n T Total n 1 n 2 n m n M = (n + 1)Z Siegel (1957) gives variance of M with ties as m ( ) ( 3 ) ( ) C T C T j j j= 1 { } var M = n n n n n n n 3n n 1 n n n n n n j ( ) m m C T 3 3 = j + 3n n 1 j= 1 j= 1 Session 3 29

30 3 3 m n C n Tn j = 1 (n + 1) 3n(n 1) j 1 n = n 2 V Thus, the score test under the proportional odds model is the Mann-Whitney test Session 3 30