1. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 91 Lecture 9 Bayesian approaches for quantitative responses 9.1 Proof-of-concept studies outside oncology 9.2 A frequentist approach 9.3 A Bayesian approach 9.4 Bayesian sample size calculation 9.5 Extensions and alternatives

2. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 92 9.1 Proof-of-concept studies outside oncology General characteristics: The experimental treatment group is usually compared with a control group The measure of efficacy usually has a continuous distribution The conditions studied are often not life-threatening eg: obesity, depression

3. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 93 Example: A CETP-inhibition trial CETP (plasma Cholesteryl Ester Transfer Protein) helps to transport cholesterol from tissues to the liver Low levels of CETP tend to be associated with high levels of HDL-Cholesterol in the blood, which in turn decreases the risk of coronary heart disease (CHD) CETP inhibitors may thus increase HDL-C and decrease the risk of CHD

4. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 94 CETP inhibition is assessed via the reduction of CETP between baseline and 24 hours, expressed as a proportion of the baseline value The trial will compare the proportionate CETP-inhibition (Y) between a control (C) and an experimental group (E) For the h th of n patients on treatment j, Y hj will be modelled as normally distributed: Y hj ~ N(  j,  1 ) h = 1,…, n j ; j = E, C  In the Bayesian approach it is convenient to parameterise in terms of precision  1/   2 ) rather than variance

5. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 95 It is anticipated that in the control group we will have  C = 0  on average,  no change in CETP For the experimental group a value of  E = 0.20 would be worth detecting The standard deviation  will be treated as known,  = 0.20 or precision, = 25

6. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 96 9.2 A frequentist approach To determine whether to PROCEED with the experimental treatment or not, test the hypotheses H 0 :  = 0 and H 1 :  > 0, where  =  E   C. The power requirement will be set so that P( PROCEED ;  = 0) =  P( PROCEED ;  =  R ) ≥ 1 –  for  = 0.05, 1 –  = 0.9 and  R = 0.2

7. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 97 Sample size where z 1  is the 100  % point of the standard normal distribution, and the allocation ratio (E:C) is R:1 If then PROCEED to phase III

8. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 98 For  = 0.05 and 1 –  = 0.90, z 1  = 1.645 and z 1  = 1.282 We have  R = 0.2, = 25, and assume unequal allocation 3:1 (R = 3) in favour of E Total sample size (to give integer values) 48

9. MPS/MSc in Statistics Adaptive & Bayesian - Lect 9 9 Impose normal priors:  typically  C0 >  E0 The data will comprise: n j observations on treatment j with mean 9.3 A Bayesian approach

10. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 910 Posteriors can be found as: where and (9.1) (9.2)

11. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 911 Posterior mean  j1 is a weighted average of (from the data) and  j0 (from the prior) if n j = 0,  j ~ N  the prior as n j  ,  j ~ N as  j0  0,  j ~ N  non informative prior

12. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 912 Posterior treatment effect The posterior for  =  E –  C is We will PROCEED if which occurs if (9.3)

13. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 913 Non-informative prior PROCEED if This is the same rule as in the frequentist case (slide 9-7) The frequentist rule  the Bayesian rule for a non- informative prior

14. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 914 Back to general prior: Predictive distribution Here we are, at the design stage What is going to look like? Given the value of  j, but  j is itself variable: 

15. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 915 Combining these two sources of variation gives the predictive distribution From (9.1) and (9.2),

16. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 916 So the predictive distribution of  j1 is

17. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 917 As the predictive distribution of the test statistic Z* is where  0 =  E0 –  C0

18. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 918 Probability of proceeding At the design stage, we can calculate P ( PROCEED ) For non-informative priors, (9.4)

19. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 919 Whitehead et al. (2008) As for the case of binary data in Lecture 8, find n and k such that and where  is a large value and  R > 0 is a promising treatment difference 9.4 Bayesian sample size calculation

20. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 920 The posterior distribution of  is so that Hence

21. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 921 so that It follows that we need (9.5)

22. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 922 As It follows that we need (9.6) The condition is that the design will either give convincing evidence that the drug is promising or convincing evidence that it is not a breakthrough

23. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 923 If Z* = k, then the posterior density of  must be:

24. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 924 For a non-informative prior,, j = E, C Thus, if n E = Rn/(R + 1) and n C = n/(R + 1), the formula becomes which is the same as the frequentist sample size formula on slide 9-7

25. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 925 Example For the sample size formula becomes With and a 3:1 allocation ratio, the sample size formula becomes

26. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 926 For equality, this is a quadratic in n yielding n = 14.86  take n = 16, with n E = 12 and n C = 4 The frequentist sample size was 48, showing the reliance on prior information in the Bayesian formulation The Bayesian approach also gains as n E + 4 = 16 and n C + 16 = 20 are more equal to one another than n E and n C  the larger sample is taken from the lesser known population See SAS program Bayes nor ss provided

27. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 927 9.4 Extensions and alternatives (a) Extension to unknown variance Prior: ~ Gamma (  j0,  j0 ), j = E, C Now we find n and k such that and with probability 

28. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 928 It can be shown that, the sample size must satisfy where t d,  is the 100  % point of the t-distribution on d degrees-of- freedom and beta a,b,  is the 100  % point of the beta distribution parameters (a, b)

29. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 929 (b) Alternative method based on predictive power Find n such that P ( PROCEED ) =  where the probability is given by (9.4) and  is a large value This probability should not be set as high as traditional power specifications, as it averages over both good and bad values of the parameter  No satisfactory sample size exists if the prior is sufficiently pessimistic

30. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 930 For = 25,  E0 = 100,  C0 = 400, we have: To achieve  = 0.90, take n = 88 (n E = 66, n C = 22) O’Hagan and Stevens (2001, 2002) describe this approach, and also suggest the use of a non-informative prior for analysis (Bayesian design of a frequentist procedure) See also Section 6.5.3 of Spiegelhalter et al (2004)