1. Common Designs for Controlled Clinical Trials A.Parallel Group Trials 1.Simplest example - 2 groups, no stratification 2.Stratified design 3.Matched pairs 4.Factorial design B.Crossover Trials

2. Parallel Group Design y ij  T i  e  overall mean T i  effect of i th treatment (i= A,B) e ij  error term for j th patient in i th group e ij ~N(0,  2 ) y ij ~N(  T i,  2 )  2  s 2  e 2

3. Diastolic BP (mmHg)5836940.62 Serum cholesterol120040016000.75 (mg/dl) Viral load0.160.090.250.64 (log 10 copies/mL) Overnight urine 3256259500.34 excretion of Na + (meq/L) Carbohydrate intake 1102083180.35 (% of calories) Estimates of  and  for Selected Response Variables 2s2s 2e2e 2e2e  2s2s  2e2e  2s2s  2e2e  2s2s     Total 2s2s 

4. A Poorly Designed “Crossover” Trial: Why? 2n patientsDrug ADrug B...orLow Dose Also referred to as “changeover” or “switchover”

5. Two-Period Crossover Trial 2n - Randomize Patients Drug A Drug B Drug A WASHOUTWASHOUT 12 Period n n

6. Schematic design of study to determine effects of dietary sodium on blood pressure in normotensive adults Low Sodium Diet -804610 Weeks Group I (n=25) Group II (n=23) 100 mEq NaCI Placebo100 mEq NaCI Placebo WASH-OUT SNaP Hypertension 1991; 17:I21-I26.

7. Crossover Trial Advantages: Fewer patients required in most situations due to elimination of between subject component of variability, therefore: –recruitment may be easier –can be more easily done in single center –fewer patients exposed to experimental treatment Less data collection Disadvantages: Interaction due to: –differential carry-over effects of treatments –treatment x period interaction –differences between the two randomized groups (AB and BA) at baseline Patients must be observed longer Losses to follow-up/missing data Naturally occurring changes in underlying disease state

8. Situations where crossover design is most applicable: 1.Rapid response and response is transitory 2.Variability between patients is large compared to variability within patients 3.Steady state physiological condition; disease or condition cannot be cured 4.No residual or carry-over effects of treatment expected

9. Survey of Randomized Trials in 2000 116 of 526 published trials were crossover Median sample size for crossover trials = 15 78% of crossover studies involved 2 treatments 70% of crossover studies reported a washout period 17% reported a test for carry-over effect and 13% tested for period effects Mills EJ et al. Trials 2009

10. General Model for 2-Period Crossover Trial y ijk =  +  k + T u + v +  ij +  ijk  = overall mean  k = effect of kth period (k= 1,2) T u = effect of uth treatment (u= A, B) (direct effect) v = residual effect of treatment given in first period on second period response (v= A, B) ( v = 0 for first period measurements, i.e., when k= 1) Hills and Armitage Br J Clin Pharmac 1979; 8:7-20. Senn Stat Methods Med Res 1994;3:303-324.

11. 2e2e  2s2s    ij = random effect between subject of j th patient in group i (I or II); same in each period  N(0,  s 2 )  ijk = random within patient effect for k th period  N(0,  e 2 ) Motivation for crossover

12.  +  + T +  +  + T 2-Period Crossover Trial Fixed Effects 12 I (AB) II (BA) Group 1 A 1 B 2 B 2 A A B Period

13. Group I (AB) Patient (i = 1) Response - Patient j Period 1: y 1j1 =  +  1 + T A +  1j +  1j1 Period 2: y ij2 =  +  2 + T B + A +  1j +  1j2 Paired difference d I 1 = y 1j1 - y 1j2 = (T A - T B )- A + (  1  2 )  (  1j1  1j2 ) E(d I )= (T A - T B )- A + (  1  2 )

14. Group II (BA) Patient (i = 2) Response - Patient j I Period 1: y 2j'1 =  +  1 + T B +  +  1 Period 2: y 2j'2 =  +  2 + T A + B +  2j' +  2 Paired difference y 2j'1 - y 2j2 = (T B - T A )- B + (  1  2 )+(  2j'1  2 ) Consider difference in opposite direction and call it d ll d II = (T A - T B )+ B + (  2  1 )+(  2j'2  1 ) E(d II )= (T A - T B )+ B + (  2  1 )

15. d I +d II  2(T A - T B )- ( A - B ) d I  d II  2(  1  2 )- ( A + B ) d I +d II 2 estimates T A - T B if A = B d I -d II 2 estimates  1  2 if A = B  0

16. How Do We Convince Ourselves That A  B Consider the sum of the period 1 and period 2 responses. Group I (AB): E(Sum I )= 2  + (  1  2 ) + (T A + T B )+ A Group II (BA): E(Sum II )= 2  + (  1 +  2 ) + (T A + T B )+ B Note that Sum I  Sum II estimates A - B

17. Test for Carryover Has Low Power It is a between, instead of within, patient test (even when considering change from baseline, variance is 4 times larger) Although totals are used, direct information on carry over only comes from 2 nd period – effect is diluted with sum’s. If there is evidence of a carryover effect (p<0.10 or 0.15), Grizzle proposed that the 1 st period effects be used. This has been shown to be a sequential testing procedure that is not optimal. Some believe it is better not to carry out the test at all. Better be sure in the design that there is no carryover effect!

18. Possible Reasons for Rejecting Hypothesis of Equal Carry-over Effects 1.True carry-over effect for A or B, or both 2.Psychological carry-over effect 3.Difference between treatments depends on pre- treatment level of response variable 4.Group I and Group II differ significantly by chance

19. Response Group II Group I B A B A 1 2 Period xx 1.Treatment effect 2.No period effect 3.No interaction

20. 1.Treatment effect 2.Period effect 3.No interaction Response Group II Group I A A B B 1 2 Period x x

21. Quantitative Interaction Response Group II Group I B A B A 1 2 Period

22. Qualitative Interaction Response Group II Group I B A B A 1 2 Period

23. Example from Senn Statistical Issues in Drug Development John Wiley & Sons, 1997 BA20002300300 AB23002000300 300 SequencePeriod 1Period 2 A-B Difference FEV 1.0 (mL) Beta-Agonist (A) vs. Placebo (B) for Patients with Asthma

24. Senn Example (cont.) BA23502650300 AB23002000300 300 SequencePeriod 1Period 2 A-B Difference Add 350 mL to BA Differences are recovered in spite of difference between sequences

25. Senn Example (cont.) BA23502750400 AB23002100200 300 SequencePeriod 1Period 2 A-B Difference Add 350 mL to BA; and Add a secular trend causing FEV 1.0 to be 100 mL HIGHER in 2nd period Difference in each sequence not recovered, but average is okay – crossover still works!

26. Senn Example (cont.) BA23502750400 AB23002400-100 150 SequencePeriod 1Period 2 A-B Difference Add 350 mL to BA; and Add a secular trend causing FEV 1.0 to be 100 higher in 2nd period; and Add a carry-over effect of A (still completely effective when B is given) Crossover does not work!

27. Variance Estimates Treatment Effect= d I + d II 2 Period Effect= d I - d II 2 Carryover Effect=Sum I - II Variance of (1) and (2) = 1 4  2 d I n I +  2 d II n        2 d  pooled variance = (n I - 1)  2 d I + (n II - 1)  2 d II n I + n - 2 Variance of (3) =  2 Sum I n I +  2 II n  2 Sum = pooled variance= (n I - 1)  2 Sum I + (n II - 1)  2 Sum II n I + n - 2 (1) (2) (3)

28. Hypothesis Testing 1) H o : no interaction t(n I + n II - 2)= SUM I - II ˆ  SUM 1 n I + 1 n II         1 2 2) H o : no treatment effect t(n I + n II - 2)= d I + d II 2 ˆ  d 2 1 n I + 1 n         1 2 3) H o : no period effect t(n I + n II - 2)= d I - d II 2 ˆ  d 2 1 n I + 1 n         1 2

29. Group I 185313 31410424 48088 697216 7116517 935-28 116066 130000 161312125 18102812 1975212 211313026 22810-218 2477014 259099 27106416 282204 Patient Accession No. Period 1 (A) Period 2 (B) A-B Difference Sum

30. Group II 21211-123 568214 8139-422 1088016 1289117 1448412 15814622 172426 208`13521 2397-216 26710317 2976-113 Patient Accession No. Period 1 (B) Period 2 (A) A-B Difference Sum

31. Group I (N = 17) Mean8.125.292.8213.41 SD3.844.253.477.32 SE0.841.78 Period 1 (A) Period 2 (B) A-B Difference Sum Group II (N = 12) Mean7.678.921.2516.58 SD2.992.812.994.98 SE0.861.44 Period 1 (B) Period 2 (A) A-B Difference Sum

32. 1) Determine pooled variance of sum H o : No interaction s SUM 2 = (n I - 1)s I 2 + (n II - 1)s II 2 n I + n - 2 = 16(7.32) 2 + 11(4.98) 2 27 = 41.9

33. 2) Calculate test statistic 3) Compare with t-tables with 27df, p = 0.20 t(n I + n II - 2)= SUM I - II s SUM 2 1 n I + 1 n II         = 16- 13.49 41.86 1 17  1 12 = 1.30

34. same procedure H o : No treatment difference s d 2 = (n I - 1)s I d 2 + (n II - 1)s II d 2 n I + n - 2 = 16(3.47) 2 + 11(2.99) 2 27 = 10.78

35. t(n I + n II - 2)= d I + d II 2         s d 2 1 n I + 1 n         = 2.82+ 1.24 2       10.78 2 1 17  1 12  3.28 p=0.0028

36. Similarly for H o : No Period Effect t(n I + n II - 2)= d I - d II 2         s d 2 1 n I + 1 n         = 2.82- 1.24 2       10.78 2 1 17  1 12  1.27 p=0.21

44. Two-Period Crossover Trial 2n - Randomize Patients Drug A Drug B Drug A WASHOUTWASHOUT 12 Period n n

45. Consider Period 1 Data and Estimated SE of Treatment Difference (Parallel Group Design) s p 2 = (3.84) 2 (16)+ (2.99) 2 (11) 27 = 12.38 SE= (12.38) 1 2 1 17  1 12 = 1.33 compared to 0.62 for crossover

46. Advantages of Baseline Measurements 1.Description of study participants at entry 2.Comparability of treatment groups: AB vs. BA 3.More powerful test for treatment x period interaction 4.Improved precision for estimating treatment differences (e.g., analysis of covariance or change from baseline when correlation >0.5.) 5.Subgroup analysis

47. Advantage of a 2nd “Baseline” during Washout Between Periods Differences between “baseline” measurements for Group I and Group II may provide support for unequal residual effects. NOTE! This comparison does not replace comparison of sums of observation at the end of periods for Group I and Group II that was previously discussed.

48. Design for Estimating Direct and Residual Treatment Effects Group123 IABC IIBCA IIICAB Period Square 1 Group123 IACB IIBAC IIICBA Period Square 2

49. Recommendations 1.Always measure initial baseline. 2.If washout is employed, measure B 2. 3.Take multiple measures during each treatment period, e.g., comparison of treatment A with placebo (P). Week of Study Group12345678 IPAAAPPPP IIPPPPPAAA B 1 B 2 Period 1Period 2