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Selecting the Appropriate Statistical Distribution for a Primary Analysis

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  1. Selecting the Appropriate Statistical Distribution for a Primary Analysis P. Lachenbruch

  2. A Study of Xeroderma Pigmentosa (XP) • A characteristic of XP is the formation of Actinic Keratoses (AK s ) • Multiple lesions appear haphazardly on a patient’s back • The rate of appearance may not be the same for different patients

  3. Background • Analysis: Rank Sum test. • Late in study the Statistical Analysis Plan (SAP) was amended to use Poisson regression • Unclear if stepwise selection of covariates was planned a priori

  4. Study Results • Poisson regression analysis showed highly significant treatment difference (p=0.009) adjusting for baseline AK, age, and age x treatment interaction (stepwise selection) • All these effects were highly significant. • Substantial outlier problem

  5. Assumptions • Each patient has the same incidence rate,  per area unit. • Chance of more than one AK in small area unit is negligible. • Non-overlapping lesions are independent, that is, lesions occurring in one area of the body are not affected by those occurring in another area.

  6. Outliers • Outliers are observations that are jarringly different from the remainder of the data • May be multiple outliers • If frequency is large, this may be evidence that we have a mixture distribution. • Can substantially affect analysis

  7. Analyses Two-Sample Wilcoxon rank-sum (Mann-Whitney) test trt | obs rank sum expected --------+--------------------------------- 0 | 9 158 135 1 | 20 277 300 --------+--------------------------------- Combined| 29 435 435 unadjusted variance 450.00 adjustment for ties -15.07 ---------- adjusted variance 434.93 Ho: ak12tot(trt==0) = ak12tot(trt==1) z = 1.103 Prob > |z| = 0.2701

  8. Distribution of AK Data at Baseline (Stem and Leaf)(Yarosh et al, Lancet) Lead | Trailing digits 0* | 00000000000000000011223335 // 4* | 27 // 10* | 0  oops!

  9. Distribution of 12 Month AK Total Data (Stem and Leaf) . stem ak12tot,w(10) Lead| Trailing digits 0* | 000000001111222233457 1* | 00345 2* | 3* | 7 // 7* | 1 8* | 9 // 19*| 3  same patient - in placebo group

  10. Results of Poisson Analyses Poisson regression Number of obs = 29 LR chi2(3) = 1044.65 Prob > chi2 = 0.0000 Log likelihood = -127.46684 Pseudo R2 = 0.8038 ---------------------------------------------------------- ak12tot | Coef. Std. Err. z P>|z| [95% Conf. Interval] ---------+------------------------------------------------ age | .017 .0056 3.00 0.003 .0058 .0276 trt | .532 .167 3.20 0.001 .2061 .859 akb | .045 .0019 23.10 0.000 .0409 .0485 _cons | .658 .219 3.00 0.003 .2282 1.0878 ---------------------------------------------------------- • G-O-F in control group, 2 =1222.5 with 8 d.f. • G-O-F in treatment group, 2 =682.5 with 19 d.f.

  11. Permutation Test • Procedure: Scramble treatment codes and redo analysis. Repeat many (5,000?) times. • Count number of times the coefficient for treatment exceeds the observed value.

  12. Command and Output . permute trt "permpois trt ak12tot age akb" rtrt=rtrt rage=rage rakb=rakb ,reps(5000) d command: permpois trt ak12tot age akb statistics: rtrt = rtrt rage = rage rakb = rakb permute var: trt Monte Carlo permutation statistics Number of obs = 30 Replications = 5000 ---------------------------------------------------------- T | T(obs) c n p=c/n SE(p) -------------+-------------------------------------------- rtrt | .5324557 2660 5000 0.5320 0.0071 rage | .0167116 3577 5000 0.7154 0.0064 rakb | .0446938 1118 5000 0.2236 0.0059 ---------------------------------------------------------- Note: c = #{|T| >= |T(obs)|} I deleted the confidence intervals for the proportions

  13. Permutation Tests (2) • Poisson with 5000 Replications • Treatment: p = 0.57 • Age: p = 0.62 • AK Baseline: p = 0.28 • All significant results disappear

  14. Results of Poisson Analysis • Sponsor found that all terms were highly significant (including the treatment x age interaction). • We reproduced this analysis. • We also did a Poisson goodness-of-fit test that strongly rejected the assumption of a Poisson distribution. • What does a highly significant result mean when the model is wrong?

  15. Conclusions • The data are poorly fit by both Poisson and Negative Binomial distributions • Permutation tests suggest no treatment effect unless treatment by age interaction is included • Justification of interaction term by stepwise procedure is exploratory • Outliers are a problem and can affect the conclusions.

  16. Conclusions (2) • The results of the study are based on exploratory data analysis. • The analysis is based on wrong assumptions of the data. • Our analyses based on distribution free tests do not agree with the sponsor’s results. • The results based on appropriate assumptions do not support approval of the product.

  17. Suggestions • Conduct a phase II study to determine appropriate covariates. • Need to use appropriate inclusion / exclusion criteria. • Stratification. • a priori specification of full analysis

  18. Reference Yarosh D. et al., "Effect of topically applied T4 endonuclease V in liposomes on skin cancer in xeroderma pigmentosum: a randomised study" Lancet 357:926-929, 2001.

  19. The End

  20. Grid on “Back”

  21. +-------------------------+ | sex trt akb ak12tot| |-------------------------| | F 0 0 5 | | M 0 0 1 | | F 0 0 1 | | F 0 0 0 | | F 0 1 15 | |-------------------------| | M 0 0 3 | | F 0 100 193 | | M 0 0 2 | | M 0 2 13 | | M 1 47 71 | |-------------------------| | F 1 0 0 | | F 1 0 1 | | F 1 0 0 | | F 1 42 37 | | F 1 2 0 | |-------------------------| +-------------------------+ | sex trt akb ak12tot| +-------------------------+ | F 1 3 2 | | F 1 0 10 | | M 1 0 0 | | F 1 0 2 | | M 1 0 0 | |-------------------------| | F 1 0 0 | | F 1 3 10 | | F 1 1 0 | | F 1 0 4 | | F 1 5 3 | |-------------------------| | M 1 0 0 | | F 1 0 2 | | F 1 0 7 | | F 1 3 14 | | M . . . | +-------------------------+ The Data

  22. Descriptive Statistics (1)

  23. Descriptive Statistics (2)

  24. Negative Binomial Model • Need a model that allows for individual variability. • Negative binomial distribution assumes that each patient has Poisson, but incidence rate varies according to a gamma distribution. • Treatment: p = 0.64 • Age: p = 0.45 • AK Baseline: p = 0.0001 • Age x Treat: p <0.001 • Main effect of treatment is not interpretable. Need to look at effects separately by age.

  25. Negative Binomial Results • This model shows only that the baseline AK and age x treatment effects are significant factors. • It also gives a test for whether the data are Poisson; the test rejects the Poisson Distribution: p<0.0005 • A test based on chisquare test (obs - exp) suggests that these data are not negative binomial.