Transcript Lecture 13
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 Parallel Group Design y ij = m + Ti + eij m = overall mean th Ti = effect of i treatment (i = A, B) th th eij = error term for j patient in i group eij ~ N(0, s 2 ) y ij ~ N (m + Ti , s 2 ) 2 2 2 s = ss + se Estimates of s and s for Selected Response Variables 2 s 2 e s 2s s 2s Diastolic BP (mmHg) s 2e Total s 2s + s 2e s 2s + s 2e 58 36 94 0.62 Serum cholesterol (mg/dl) 1200 400 1600 0.75 Viral load (log10 copies/mL) 0.16 0.09 0.25 0.64 Overnight urine excretion of Na+ (meq/L) 325 625 950 0.34 Carbohydrate intake (% of calories) 110 208 318 0.35 A Poorly Designed “Crossover” Trial: Why? 2n patients Drug A or Low Dose Drug B or Low Dose Also referred to as “changeover” or “switchover” ... Two-Period Crossover Trial Period 1 Drug A n 2n - Randomize Patients 2 W A S H Drug B n Drug B O U T Drug A SNaP 100 mEq NaCI Placebo Group I (n=25) WASH-OUT Group II (n=23) Placebo 100 mEq NaCI Weeks -8 0 4 6 Low Sodium Diet Schematic design of study to determine effects of dietary sodium on blood pressure in normotensive adults Hypertension 1991; 17:I21-I26. 10 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 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 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 General Model for 2-Period Crossover Trial y ijk = m + pk + Tu + lv + g ij + e ijk m = overall mean pk = effect of kth period (k = 1, 2) Tu = effect of uth treatment (u = A, B) (direct effect) lv = residual effect of treatment given in first period on second period response (v = A, B) (l 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. g ij = random effect between subject of jth patient in group i (I or II); same in each period 2 N(0, ss ) e ijk = random within patient effect for kth period 2 N(0, se ) Motivation for crossover s 2s s 2e 2-Period Crossover Trial Fixed Effects Period 1 2 I (AB) m + p1 + TA m + p2 + TB+l A II (BA) m + p1 + TB m + p2 + TA+l B Group Group I (AB) Patient (i = 1) Response - Patient j Period 1: y1j1 = m + p 1 + TA + g 1j + e1j1 Period 2 : yij2 = m + p 2 + TB + l A + g 1j + e 1j2 Paired difference d1I = y1j1 - y1j2 = (TA - TB ) - l A + (p 1 - p 2 ) + (e1j1 - e 1j2 ) E(d I ) = (TA - TB ) - l A + (p 1 - p 2 ) Group II (BA) Patient (i = 2) Response - Patient jI Period 1: y 2j'1 = m + p 1 + TB + g 2j' + e 2j'1 Period 2 : y 2j'2 = m + p 2 + TA + lB + g 2j' + e 2j'2 Paired difference y 2j'1 - y 2j2 = (TB - TA ) - lB + (p 1 - p 2 )+ (e2j'1 - e 2j' 2 ) Consider difference in opposite direction and call it dll dII = (TA - TB ) + lB + (p 2 - p 1)+ (e 2j' 2 - e 2j'1) E(dII) = (TA - TB ) + lB + (p 2 - p 1) dI + dII = 2(TA - TB) - (l A - lB ) d I - dII = 2(p 1 - p 2 ) - ( lA + l B ) d I + dII 2 estimates TA - TB if lA = l B d I - dII 2 estimates p 1 - p 2 if l A = lB = 0 How Do We Convince Ourselves That lA=lB Consider the sum of the period 1 and period 2 responses. Group I (AB) : E(SumI ) = 2 m + (p 1 + p 2 ) + (TA + TB ) + l A Group II (BA) : E(SumII) = 2 m + (p 1 + p 2 ) + (TA + TB ) + lB Note that SumI - SumII estimates l A - lB 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 2nd 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 1st 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! 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 pretreatment level of response variable 4. Group I and Group II differ significantly by chance A Response A Group II x x B Group I B 1 2 Period 1. Treatment effect 2. No period effect 3. No interaction Response A x A x B B Group II Group I 1 2 Period 1. Treatment effect 2. Period effect 3. No interaction Quantitative Interaction A Response B 1 A Group II B Group I 2 Period Qualitative Interaction A B Response B A 1 2 Period Group I Group II Example from Senn Statistical Issues in Drug Development John Wiley & Sons, 1997 Beta-Agonist (A) vs. Placebo (B) for Patients with Asthma FEV1.0 (mL) Period 1 Period 2 A-B Difference BA 2000 2300 300 AB 2300 2000 300 Sequence 300 Senn Example (cont.) • Add 350 mL to BA Sequence BA Period 1 2350 Period 2 2650 A-B Difference 300 AB 2300 2000 300 300 Differences are recovered in spite of difference between sequences Senn Example (cont.) • Add 350 mL to BA; and • Add a secular trend causing FEV1.0 to be 100 mL HIGHER in 2nd period Sequence BA Period 1 2350 Period 2 2750 A-B Difference 400 AB 2300 2100 200 300 Difference in each sequence not recovered, but average is okay – crossover still works! Senn Example (cont.) • Add 350 mL to BA; and • Add a secular trend causing FEV1.0 to be 100 higher in 2nd period; and • Add a carry-over effect of A (still completely effective when B is given) Sequence BA Period 1 2350 Period 2 2750 A-B Difference 400 AB 2300 2400 -100 150 Crossover does not work! Variance Estimates d I + d II 2 d I - d II Period Effect = 2 (1) Carryover Effect = Sum I - Sum II (3) Treatment Effect = (2) Variance of (1) and (2) d 1 d = s I + s II 4 nI nII 2 s 2 2 = pooled variance 2 2 (nI - 1)s d I + (nII - 1)s d II = nI + nII - 2 d Variance of (3) = s s 2 Sum 2 Sum I nI + s 2 Sum II nII = pooled variance = (nI - 1)s + (nII - 1)s nI + nII - 2 2 Sum I 2 Sum II Hypothesis Testing 1) Ho : no interaction t(nI + nII - 2) = SUMI - SUM II 1 1 1 2 ˆ sSUM n + n ÷ ÷ II I 2) Ho : no treatment effect dI + dII 2 t(nI + nII - 2) = 1 2 ˆ 1 1 sd + ÷ 2 nII ÷ nI 3) Ho : no period effect dI - dII t(nI + nII - 2) = 2 1 1 2 sˆ d + ÷ ÷ 1 2 nI nII Group I Patient Accession No. Period 1 (A) Period 2 (B) A-B Difference Sum 1 3 4 6 7 9 11 13 16 18 19 21 22 24 25 27 28 8 14 8 9 11 3 6 0 13 10 7 13 8 7 9 10 2 5 10 0 7 6 5 0 0 12 2 5 13 10 7 0 6 2 3 4 8 2 5 -2 6 0 1 8 2 0 -2 0 9 4 0 13 24 8 16 17 8 6 0 25 12 12 26 18 14 9 16 4 Group II Patient Accession No. Period 1 (B) Period 2 (A) A-B Difference Sum 2 12 11 -1 23 5 6 8 2 14 8 13 9 -4 22 10 8 8 0 16 12 8 9 1 17 14 4 8 4 12 15 8 14 6 22 17 2 4 2 6 20 8 `13 5 21 23 9 7 -2 16 26 7 10 3 17 29 7 6 -1 13 Group I (N = 17) Mean SD SE Period 1 (A) Period 2 (B) A-B Difference Sum 8.12 3.84 5.29 4.25 2.82 3.47 0.84 13.41 7.32 1.78 Group II (N = 12) Mean SD SE Period 1 (B) Period 2 (A) A-B Difference Sum 7.67 2.99 8.92 2.81 1.25 2.99 0.86 16.58 4.98 1.44 Ho : No interaction 1) Determine pooled variance of sum s2SUM (nI - 1)sI2 + (nII - 1)sII2 = nI + nII - 2 2 16(7.32) + 11(4.98) = 27 = 41.9 2 2) Calculate test statistic t(nI + nII - 2) = SUMI - SUMII 2 SUM s = 1 1 n + n ÷ ÷ II I 16 - 13.49 1 1 41.86 + 17 12 = 1.30 3) Compare with t-tables with 27df, p = 0.20 Ho : No treatment difference same procedure s 2 d = (nI - 1)s 2 Id 2 II d + (nII - 1)s nI + nII - 2 2 16(3.47) + 11(2.99) = 27 = 10.78 2 dI + dII 2 ÷ ÷ t(nI + nII - 2) = s d 1 1 + ÷ 2 nI nII ÷ = 2.82 + 1.24 ÷ 2 = 3.28 10.78 1 1 + 2 17 12 p = 0.0028 Similarly for Ho : No Period Effect dI - dII 2 ÷ ÷ t(nI + nII - 2) = s d 1 1 + ÷ 2 nI nII ÷ = 2.82 - 1.24 ÷ 2 = 1.27 10.78 1 1 + 2 17 12 p = 0.21 Two-Period Crossover Trial Period 1 Drug A n 2n - Randomize Patients 2 W A S H Drug B n Drug B O U T Drug A Consider Period 1 Data and Estimated SE of Treatment Difference (Parallel Group Design) 2 2 (3.84) (16) + (2.99) (11) 2 sp = 27 = 12.38 SE = (12.38) 1 2 1 1 + 17 12 = 1.33 compared to 0.62 for crossover 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 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. Design for Estimating Direct and Residual Treatment Effects Square 1 Group I II III Period 1 2 3 A B C B C A C A B Square 2 Group I II III Period 1 2 3 A C B B A C C B A Recommendations 1. 2. 3. Always measure initial baseline. If washout is employed, measure B2. Take multiple measures during each treatment period, e.g., comparison of treatment A with placebo (P). Group I II 1 P P B1 Week of Study Period 1 Period 2 2 3 4 5 6 7 8 A A A P P P P P P P P A A A B2