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Clinica l Tria l Writing II
S a mple S ize Ca lcula tion a nd
              Ra ndomiza tion


      Liying XU (Tel: 22528716)
      CCTER
      CUHK
      31st July 2002
1
Sample Size Planning
1.1 Introduction
   Fundamental Points
       Clinical trials should have sufficient
        statistical power to detect difference
        between groups considered to be of
        clinical interest. Therefore calculation
        of sample size with provision for
        adequate levels of significance and power
        is a essential part of planning.
Five Key Questions Regarding the Sample
Size
   What is the main purpose of the trial?
   What is the principal measure of patients
    outcome?
   How will the data be analyzed to detect a
    treatment difference? (The test statistic: t-test ,
    X2 or CI.)
   What type of results does one anticipate with
    standard treatment?
   Ho and HA, How small a treatment difference is
    it important to detect and with what degree of
    certainty? ( δ, α and β.)
   How to deal with treatment withdraws and
    protocol violations. (Data set used.)
SSC: Only an Estimate
 Parameters used in calculation are
  estimates with uncertainty and often
  base on very small prior studies
 Population may be different
 Publication bias--overly optimistic
 Different inclusion and exclusion
  criteria
 Mathematical models approximation
What should be in the protocol?
   Sample size justification
     Methods of calculation
     Quantities used in calculation:

        • Variances
        • mean values
        • response rates
        • difference to be detected
Realistic and Conservative
   Overestimated size:
       unfeasible
       early termination

   Underestimated size
       justify an increase
       extension in follow-up
       incorrect conclusion (WORSE)
What is α (Type I error)?
 The probability of erroneously
  rejecting the null hypothesis
 (Put an useless medicine into the
  market!)
What is β (Type II error)?
 The probability of erroneously failing
  to reject the null hypothesis.
 (keep a good medicine away from
  patients!)
What is Power ?
   Power quantifies the ability of the
    study to find true differences of
    various values of δ.

   Power = 1- β=P (accept H1|H1 is true)
   ----the chance of correctly identify H1
    (correctly identify a better medicine)
What is δ?
   δ is the minimum difference between groups
    that is judged to be clinically important
       Minimal effect which has clinical relevance in
        the management of patients
       The anticipated effect of the new treatment
        (larger)
The Choice of α and β depend on:

   the medical and practical consequences of the two
    kinds of errors
   prior plausibility of the hypothesis
   the desired impact of the results
The Choice of α and β
   α=0.10 and β=0.2 for preliminary trials
    that are likely to be replicated.
   α=0.01 and β=0.05 for the trial that are
    unlikely replicated.
   α=β if both test and control treatments are
    new, about equal in cost, and there are
    good reasons to consider them both
    relatively safe.
The Choice of α and β
   α>β if there is no established control treatment
    and test treatment is relatively inexpensive, easy
    to apply and is not known to have any serious side
    effects.
   α<β (the most common approach 0.05 and
    0,2)if the control treatment is already widely used
    and is known to be reasonably safe and effective,
    whereas the test treatment is new,costly, and
    produces serious side effects.
1.2 SSC for Continuous Outcome
Variables
 H0: δ=µC-µI=0
 HA: δ=µC-µI≠0
 If the variance in known
 If
       z=
          (x −x )c   I

               1   1
           σ     +
               NC N I

        Z > Zα
   If           H0 will be rejected at the
    α level of significance.
   A total sample 2N would be needed to
    detect a true difference δ between µI and µC
    with power (1-β) and significant level α by
    formula:


              2N =
                      (        )
                     4 Zα + Z β σ 2
                                   2


                          δ2
Example 1
   An investigator wish to estimate the sample size
    necessary to detect a 10 mg/dl difference in
    cholesterol level in a diet intervention group
    compared to the control group. The variance from
    other data is estimated to be (50 mg/dl). For a two
    sided 5% significance level, Zα=1.96, and for 90%
    power, Zβ=1.282.

   2N=4(1.96+1.282)2(50)2/102=1050
Example1a
Baseline Adjustment
   An investigator interested in the mean levels of
    change might want to test whether diet
    intervention lowers serum cholesterol from
    baseline levels when compare with a control.
   H0: =0 ∆ I
        ∆c −

   HA: ∆ c − ∆ I ≠0
   σ=20mg/dl, δ=10mg/dl
   2N=4(1.96+1.282)2(20)2/102=170
A Professional Statement
   A sample size of 85 in each group will
    have 90% power to detect a difference
    in means of 10.0 assuming that the
    common standard deviation is 20.0
    using a two group t-test with a 0.05
    two-sided significant level.
Values of f(α,β) to be used in formula
for sample size calculation
                                β(Type II error)
    α                    0.05    0.1        0.2    0.5
 (Type I          0.1    10.8    8.6        6.2    2.7
  error)          0.05   13.0   10.5        7.9    3.8
                  0.02   15.8   13.0        10.0   5.4
                  0.01   17.8   14.9        11.7   6.6

(Z   α +Z β   )
              2
                   = f (α β)
                         ,
1.3 SSC for a Binary Outcome
   Two independent samples

                                      1   1 
      Z = ( pC − pI ) / p (1 − p )  N  +    
                                      C   NI 
                                              
      p = ( rI + rC ) /( N I + N C )
p = ( pC + pI ) / 2

2 N = 4( Zα + Z β ) p (1 − p ) / ( pC − pI )
                     2                         2
Example 2
   Suppose the annual event rate in the
    control group is anticipated to be 20%. The
    investigator hopes that the intervention
    will reduce the annual rate to 15%. The
    study is planned so that each participant
    will be followed for 2 years. Therefore, if
    the assumption are accurate,
    approximately 40% of the participants in
    the control group and 30% of the
    participants in the intervention group will
    develop an event.
2 N = 4(1.96 + 1.282 ) (0.35)(0.65) / ( 0.4 − 0.3)
                       2                             2


= 956 ≈ 960
A Professional Statement
   A two group x2 test with a 0.05 two-
    sided significant level will have 90%
    power to detect the difference between a
    Group 1 proportion, P1,of 0.40 and a
    Group 2 proportion P2 of 0.30 (odds
    ratio of 0.643) when the sample size in
    each group is 480.
Table 1.3 Approximate total sample size for comparing
 various proportions in two groups with significance level (α)
 of 0.05
 and power(1-β) of 0.8 and 0.9

True proportions                  α=0.05(one-sided)      α=0.05(two-sided)
pC                 pI             1-β             1-β    1-β             1-β
Control group      Intervention   0.90            0.80   0.90            0.80
                   group
0.6                0.50           850             610    1040            780
                   0.40           210             160    260             200
                   0.30           90              70     120             90
                   0.20           50              40     60              50
0.50               0.40           850             610    1040            780
                   0.30           210             150    250             190
                   0.25           130             90     160             120
                   0.20           90              60     110             80
0.40               0.30           780             560    960             720
                   0.25           330             240    410             310
                   0.20           180             130    220             170
0.30               0.20           640             470    790             590
                   0.15           270             190    330             250
                   0.10           140             100    170             130
0.20               0.15           1980            1430   2430            1810
                   0.10           440             320    540             400
                   0.05           170             120    200             150
0.10               0.05           950             690    1170            870
From Table 1.3 You can see:
 δ↑→N↓
 The power 1- β↑→N ↑
 The α↓→N ↑
Paired Binary Outcome
   McNemar’s test


           Np   =
                  [Z   α   + Zβ    ]   2
                                           f
                               2
                           d
   d=difference in the proportion of successes
    (d=pI-pC)
   f=the portion of participants whose response is
    discordant (the pair of outcome are not the
    same)
Example 3
   Consider an eye study where one eye
    is treated for loss in visual acuity by a
    new laser procedure and the other
    eye is treated by standard therapy.
    The failure rate on the control, pC, is
    estimated to be 0.4, and the new
    procedure is projected to reduce the
    failure rate to 0.20. The discordant
    rate f is assumed to be 0.50.
 α=0.05
 The power 1- β=0.90
 f=0.5

 PC=0.4 PI=0.2


Np =
     (1.96 + 1.282) ( 0.5) = 262 × 0.5 = 132
                   2


         ( 0.4 − 0.2) 2
1.4 Adjusting for Non-adherence
   Ro =drop out rate
   RI=drop in rate
         / (1 − RO − RI )
                            2
   N∗=N

   If RO=0.20, RI=0.05
   N ∗=1.78N
1.5 Adjusting the Multiple Comparison
   α’= α/k

   k= the number of multiple comparison
    variables
Table 1.4 Adjusting for Randomization Ratio

Randomization Ratio   Increase in total N
1:1                   0
1:2                   +12.5%
1:3                   +33%
1:4                   +56%
1:5                   +80%
1:6                   +100%
1.6 Adjusting for loss of follow up
   If p is the proportion of subjects lost to
    follow-up, the number of subjects must be
    increased by a factor of 1/(1-p).
1.7 Other Factors:
 the rate of attrition of subjects during
  a trial
 intermediate analyses
Sample size re-estimation
 Events rates are lower than
  anticipate
 Variability of larger than expected


 Without unbinding data and
 Making treatment comparisons
1.8 Power Calculation
(assuming we compare two medicines)
   Power Depends on 4 Elements:
      The real difference between the two
       medicines, δ
        • Big δ⇒big power
      The variation among individuals,σ

        • Small σ⇒big power
      The sample size, n

        • Large n⇒big power
      Type I error,α

        • Large α ⇒big power
Sensitivity of the sample size
estimate
   to a variety of deviations from these
    assumptions

   a power table
Table 1 Statistical Power of the Tanzania
Vitamin and HIV Infection Trial (N=960)

              Effect of B
              0%                  15%                 30%


Effect of A   Loss to follow up   Loss to follow up   Loss to follow up


              0%    20% 33%       0%    20%    33%    0%    20%    33%


30%           89%   82% 74%       85% 76%      68%    79% 69% 61%


25%           75% 65%    58%      69%   59% 52%       62% 52% 45%
Example 4
Regret for Low Power Due to Small
Sample?
   I have a set of data that the mean change
    between the 2 groups is significantly
    different (p<0.05). But when I put
    calculate the power it gives only 50%.
    How should I interpret this? Also, can
    someone kindly advise as whether it is
    meaningful (or pointless) to calculate the
    power when the result is statistically
    significant?
Books and Software
   Sample size tables for clinical
    studies (second edition)
   By David Machin, Michael Campbell Peter Fayers
    and Alain Pinol
   Blackwell Science 1997
 PASS 2000 available in CCTER
 nQuery 4.0 available in CCTER
2. Randomization
Randomization
 Definition:
   randomization  is a process by which each
    participant has the same chance of being
    assigned to either intervention or control.
Fundamental Point
   Randomization   trends to produce study
   groups comparable with respect to known
   and unknown risk factors, removes
   investigator bias in the allocation of
   participants, and guarantees that statistical
   tests will have valid significance levels.
Two Types of Bias in Randomization
   Selection bias
      occurs if the allocation process is predictable. If any
       bias exists as to what treatment particular types of
       participants should receive, then a selection bias
       might occur.
   Accidental bias
      can arise if the randomization procedure does not
       achieve balance on risk factors or prognostic
       covariates especially in small studies.
Fixed Allocation Randomization
 Fixedallocation randomization procedures
 assign the intervention to participants with
 a pre-specified probability, usually equal,
 and that allocation probability is not altered
 as the study processes
     • Simple randomization
     • Blocked randomization
     • Stratified randomization
Randomization Types
            Simple randomization
Simple Randomization
   Option 1: to toss an unbiased coin for a randomized
    trial with two treatment (call them A and B)
   Option 2: to use a random digit table. A randomization
    list may be generated by using the digits, one per
    treatment assignment, starting with the top row and
    working downwards:
   Option 3: to use a random number-producing
    algorithm, available on most digital computer systems.
Advantages
 Each treatment assignment is completely
 unpredictable, and probability theory
 guarantees that in the long run the numbers
 of patients on each treatment will not be
 radically different and easy to implement
Disadvantages
 Unequal groups
    one treatment is assigned more often than
     another
 Time imbalance or chronological bias
    One treatment is given with greater frequency
     at the beginning of a trial and another with
     greater frequency at the end of the trial.
 Simple randomization is not often used, even for
  large studies.
Randomization Types

     Blocked randomization
Blocked Randomization
(permuted block randomization)
 Blocked randomization is to ensure exactly
  equal treatment numbers at certain equally
  spaced point in the sequence of patients
  assignments
 A table of random permutations is used
  containing, in random order, all possible
  combinations (permutations) of a small series of
  figures.
 Block size: 6,8,10,16,20.
Advantages
 The balance between the number of
 participants in each group is guaranteed
 during the course of randomization. The
 number in each group will never differ by
 more than b/2 when b is the length of the
 block.
Disadvantages
 Analysis may   be more complicated (in
  theory)
    Correct analysis could have bigger power

 Changing block size can avoid the
  randomization to be predictable
 Mid-block inequality might occur if the interim
  analysis is intended.
Randomization Types
           Stratified randomization



geographic
                                    U .S .                                                   E u ro p e
location

previous
exposure             Yes                           No                           Yes                           No

site
           l y m p h s k i n b re a s t l y m p h s k i n b re a st l y m p h s k i n b re a s t l y m p h s k i n b re a s t
Stratified Randomization
   Stratified randomization process involves
    measuring the level of the selected factors for
    participants, determining to which stratum each
    belongs, and performing the randomization within
    the stratum. Within each stratum, the
    randomization process itself could be simple
    randomization, but in practice most clinical trials
    use some blocked randomization strategy.
Table 3. Stratification Factors and Levels
(3×2×3=18 Strata)

Age           Sex         Smoking history

1. 40-49 yr   1.Male      1. Current smoker

2. 50-59 yr   2 Female    2. Ex-smoker

3. 60-69 yr               3. Never smoked
Table 4 Stratified Randomization with Block Size of Four
        Strat   Age      Sex      Smoking        Group assignment
         a
         1      40-49     M       Current          ABBA BABA..
         2      40-49     M         Ex             BABA BBAA..
         3      40-49     M       Never               Etc.
         4      40-49     F       Current
         5      40-49     F         Ex
         6      40-59     F       Never
         7      50-59     M       Current
         8      50-59     M         Ex
         9      50-59     M       Never
         10     50-59     F       Current
         11     50-59     F         Ex
         12     50-59     F       Never
                 etc.
Advantages
 Tomake two study groups appear
 comparable with regard to specified factors,
 the power of the study can be increased by
 taking the stratification into account in the
 analysis.
Disadvantages
 The prognostic factor used in stratified
 randomization may be unimportant and other
 factors may be identified later are of more
 importance
Mechanism
Trial Type                       Mechanism

No central registration office   Randomization list
                                 sealed envelops
Double blind drug trial          Pharmacist will be involved

Multi-centre trial               Central registration office

Single-centre trial              Independent person
                                 responsible for patients
                                 registration and randomization
An Example of Stratified Randomization
 Patients will be stratified according to the following
  criteria:
 1) Treatment center (Hospital A vs Hospital B vs
  Hospital C)
 2) N-stage(N2 vs N3)
 3) T-stage (T1-2 vs T3-4)
What should be in the protocol?
   A dynamic allocation scheme will be used to
    randomize patients in equal proportions within
    each of 12 strata. The scheme first creates time-
    ordered blocks of size divisible by three and then
    uses simple randomization to divide the patients
    in each block into three treatment arms, in equal
    proportion. The block sizes will be chosen
    randomly so that each block contains either 6 or
    9 patients.
Cont…
   This procedure helps to ensure both
    randomness and investigator blinding (the block
    sizes are known only to the statistician), as
    recommended by Freedman et al.
    Randomization will be generated by the
    consulting statistician in sealed envelopes,
    labeled by stratum, which will be unsealed after
    patient registration.
Adaptive Randomization
 Number  adaptive
   Biased coin method

 Baseline adaptive (MINIMIZATION)
 Outcome adaptive
Biased Coin Method
 Advantages
   Investigators  can not determine the next
    assignment by discovery the blocking
    factor.
 Disadvantages
   Complexity in use
   Statistical analysis cumbersome
Minimization
 Minimization  is an well -accepted statistical
  method to limit imbalance in relative small
  randomized clinical trials in conditions with
  known important prognostic baseline
  characteristics.
 It called minimization because imbalance in
  the distribution of prognostic factors are
  minimized
Table 1 Some baseline characteristics of patients in a controlled trial
          of mustine versus talc in the control of pleural effusions
          in patients with breast cancer (Frientiman et al, 1983)
                                   Treatment
                         Mustine (n=23)    Talc(n=23)
Mean age (SE)                50.3(1.5)           55.3(2.2)

Stage of disease:
1 or 2                          52%                 74%
3 or 4                          48%                 26%
Mean interval in             33.1(6.2)          60.4(13.1)
month between BC
diag. and effusion
diag. (SE)
Postmenopausal                  43%                 74%
Minimization Factors
 Age ( years)              <=50     Or   >50

 Stage of disease          1 or 2   Or   3 or 4

 Time between diagnosis    <=30     Or   >30
 of cancer and diagnosis
 of effusions(months)

 Menopausal                Pre      Or   Post
Table 2 Characteristics of the first 29 patients in a clinical
       trial using minimization to allocate treatment
                                     Mustine        Talc

     Age                <=50             7            6
                         >50             8            8

     Stage              1 or 2          11           11
                        3 or 4          4            3

     Time              <=30m             6           4
     Interval           >30m             9           10

     Menopausal          Pre             7            5
                         Post            8            9
Table 3      Calculation of imbalance in patient characteristics
          for allocating treatment to the thirtieth patient

                                         Mustine        Talc
                                         (n=15)         (n=14)
 Age                     >50             8              8
 Stage                   3 or 4          4              3
 Time interval           <=30m           6              4
 Postmenopausal                          8              9
                         Total           26             24
Advantages
 It can reduce the imbalance into the minimum
  level especially in small trial
 Computer Program available (called Mini) and
  also not difficult to perform ‘by hand’
 Minimization and stratification on the same
  prognostic factors produce similar levels of
  power, but minimization may add slightly more
  power if stratification does not include all of the
  covariance
Disadvantages
 It is a
        bit complicated process compare to the
  simple randomization
Practical Considerations
Study type                 Randomization
Large studies              Blocked
Large, Multicentre studies Stratified by centre

Small studies              Blocked and Stratified
                           by centre
Large number of            Minimization
Prognostic factors
Large studies              Stratified analysis
                           without stratified
                           randomization

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31st july talk (20021)

  • 1. Clinica l Tria l Writing II S a mple S ize Ca lcula tion a nd Ra ndomiza tion Liying XU (Tel: 22528716) CCTER CUHK 31st July 2002
  • 3. 1.1 Introduction  Fundamental Points  Clinical trials should have sufficient statistical power to detect difference between groups considered to be of clinical interest. Therefore calculation of sample size with provision for adequate levels of significance and power is a essential part of planning.
  • 4. Five Key Questions Regarding the Sample Size  What is the main purpose of the trial?  What is the principal measure of patients outcome?  How will the data be analyzed to detect a treatment difference? (The test statistic: t-test , X2 or CI.)  What type of results does one anticipate with standard treatment?  Ho and HA, How small a treatment difference is it important to detect and with what degree of certainty? ( δ, α and β.)  How to deal with treatment withdraws and protocol violations. (Data set used.)
  • 5. SSC: Only an Estimate  Parameters used in calculation are estimates with uncertainty and often base on very small prior studies  Population may be different  Publication bias--overly optimistic  Different inclusion and exclusion criteria  Mathematical models approximation
  • 6. What should be in the protocol?  Sample size justification  Methods of calculation  Quantities used in calculation: • Variances • mean values • response rates • difference to be detected
  • 7. Realistic and Conservative  Overestimated size:  unfeasible  early termination  Underestimated size  justify an increase  extension in follow-up  incorrect conclusion (WORSE)
  • 8. What is α (Type I error)?  The probability of erroneously rejecting the null hypothesis  (Put an useless medicine into the market!)
  • 9. What is β (Type II error)?  The probability of erroneously failing to reject the null hypothesis.  (keep a good medicine away from patients!)
  • 10. What is Power ?  Power quantifies the ability of the study to find true differences of various values of δ.  Power = 1- β=P (accept H1|H1 is true)  ----the chance of correctly identify H1 (correctly identify a better medicine)
  • 11. What is δ?  δ is the minimum difference between groups that is judged to be clinically important  Minimal effect which has clinical relevance in the management of patients  The anticipated effect of the new treatment (larger)
  • 12. The Choice of α and β depend on:  the medical and practical consequences of the two kinds of errors  prior plausibility of the hypothesis  the desired impact of the results
  • 13. The Choice of α and β  α=0.10 and β=0.2 for preliminary trials that are likely to be replicated.  α=0.01 and β=0.05 for the trial that are unlikely replicated.  α=β if both test and control treatments are new, about equal in cost, and there are good reasons to consider them both relatively safe.
  • 14. The Choice of α and β  α>β if there is no established control treatment and test treatment is relatively inexpensive, easy to apply and is not known to have any serious side effects.  α<β (the most common approach 0.05 and 0,2)if the control treatment is already widely used and is known to be reasonably safe and effective, whereas the test treatment is new,costly, and produces serious side effects.
  • 15. 1.2 SSC for Continuous Outcome Variables  H0: δ=µC-µI=0  HA: δ=µC-µI≠0  If the variance in known  If z= (x −x )c I 1 1 σ + NC N I Z > Zα  If H0 will be rejected at the α level of significance.
  • 16. A total sample 2N would be needed to detect a true difference δ between µI and µC with power (1-β) and significant level α by formula: 2N = ( ) 4 Zα + Z β σ 2 2 δ2
  • 17. Example 1  An investigator wish to estimate the sample size necessary to detect a 10 mg/dl difference in cholesterol level in a diet intervention group compared to the control group. The variance from other data is estimated to be (50 mg/dl). For a two sided 5% significance level, Zα=1.96, and for 90% power, Zβ=1.282.  2N=4(1.96+1.282)2(50)2/102=1050
  • 18. Example1a Baseline Adjustment  An investigator interested in the mean levels of change might want to test whether diet intervention lowers serum cholesterol from baseline levels when compare with a control.  H0: =0 ∆ I ∆c −  HA: ∆ c − ∆ I ≠0  σ=20mg/dl, δ=10mg/dl  2N=4(1.96+1.282)2(20)2/102=170
  • 19. A Professional Statement  A sample size of 85 in each group will have 90% power to detect a difference in means of 10.0 assuming that the common standard deviation is 20.0 using a two group t-test with a 0.05 two-sided significant level.
  • 20. Values of f(α,β) to be used in formula for sample size calculation β(Type II error) α 0.05 0.1 0.2 0.5 (Type I 0.1 10.8 8.6 6.2 2.7 error) 0.05 13.0 10.5 7.9 3.8 0.02 15.8 13.0 10.0 5.4 0.01 17.8 14.9 11.7 6.6 (Z α +Z β ) 2 = f (α β) ,
  • 21. 1.3 SSC for a Binary Outcome  Two independent samples  1 1  Z = ( pC − pI ) / p (1 − p )  N +   C NI   p = ( rI + rC ) /( N I + N C )
  • 22. p = ( pC + pI ) / 2 2 N = 4( Zα + Z β ) p (1 − p ) / ( pC − pI ) 2 2
  • 23. Example 2  Suppose the annual event rate in the control group is anticipated to be 20%. The investigator hopes that the intervention will reduce the annual rate to 15%. The study is planned so that each participant will be followed for 2 years. Therefore, if the assumption are accurate, approximately 40% of the participants in the control group and 30% of the participants in the intervention group will develop an event.
  • 24. 2 N = 4(1.96 + 1.282 ) (0.35)(0.65) / ( 0.4 − 0.3) 2 2 = 956 ≈ 960
  • 25. A Professional Statement  A two group x2 test with a 0.05 two- sided significant level will have 90% power to detect the difference between a Group 1 proportion, P1,of 0.40 and a Group 2 proportion P2 of 0.30 (odds ratio of 0.643) when the sample size in each group is 480.
  • 26. Table 1.3 Approximate total sample size for comparing various proportions in two groups with significance level (α) of 0.05 and power(1-β) of 0.8 and 0.9 True proportions α=0.05(one-sided) α=0.05(two-sided) pC pI 1-β 1-β 1-β 1-β Control group Intervention 0.90 0.80 0.90 0.80 group 0.6 0.50 850 610 1040 780 0.40 210 160 260 200 0.30 90 70 120 90 0.20 50 40 60 50 0.50 0.40 850 610 1040 780 0.30 210 150 250 190 0.25 130 90 160 120 0.20 90 60 110 80 0.40 0.30 780 560 960 720 0.25 330 240 410 310 0.20 180 130 220 170 0.30 0.20 640 470 790 590 0.15 270 190 330 250 0.10 140 100 170 130 0.20 0.15 1980 1430 2430 1810 0.10 440 320 540 400 0.05 170 120 200 150 0.10 0.05 950 690 1170 870
  • 27. From Table 1.3 You can see:  δ↑→N↓  The power 1- β↑→N ↑  The α↓→N ↑
  • 28. Paired Binary Outcome  McNemar’s test Np = [Z α + Zβ ] 2 f 2 d  d=difference in the proportion of successes (d=pI-pC)  f=the portion of participants whose response is discordant (the pair of outcome are not the same)
  • 29. Example 3  Consider an eye study where one eye is treated for loss in visual acuity by a new laser procedure and the other eye is treated by standard therapy. The failure rate on the control, pC, is estimated to be 0.4, and the new procedure is projected to reduce the failure rate to 0.20. The discordant rate f is assumed to be 0.50.
  • 30.  α=0.05  The power 1- β=0.90  f=0.5  PC=0.4 PI=0.2 Np = (1.96 + 1.282) ( 0.5) = 262 × 0.5 = 132 2 ( 0.4 − 0.2) 2
  • 31. 1.4 Adjusting for Non-adherence  Ro =drop out rate  RI=drop in rate / (1 − RO − RI ) 2  N∗=N  If RO=0.20, RI=0.05  N ∗=1.78N
  • 32. 1.5 Adjusting the Multiple Comparison  α’= α/k  k= the number of multiple comparison variables
  • 33. Table 1.4 Adjusting for Randomization Ratio Randomization Ratio Increase in total N 1:1 0 1:2 +12.5% 1:3 +33% 1:4 +56% 1:5 +80% 1:6 +100%
  • 34. 1.6 Adjusting for loss of follow up  If p is the proportion of subjects lost to follow-up, the number of subjects must be increased by a factor of 1/(1-p).
  • 35. 1.7 Other Factors:  the rate of attrition of subjects during a trial  intermediate analyses
  • 36. Sample size re-estimation  Events rates are lower than anticipate  Variability of larger than expected  Without unbinding data and  Making treatment comparisons
  • 37. 1.8 Power Calculation (assuming we compare two medicines)  Power Depends on 4 Elements:  The real difference between the two medicines, δ • Big δ⇒big power  The variation among individuals,σ • Small σ⇒big power  The sample size, n • Large n⇒big power  Type I error,α • Large α ⇒big power
  • 38. Sensitivity of the sample size estimate  to a variety of deviations from these assumptions  a power table
  • 39. Table 1 Statistical Power of the Tanzania Vitamin and HIV Infection Trial (N=960) Effect of B 0% 15% 30% Effect of A Loss to follow up Loss to follow up Loss to follow up 0% 20% 33% 0% 20% 33% 0% 20% 33% 30% 89% 82% 74% 85% 76% 68% 79% 69% 61% 25% 75% 65% 58% 69% 59% 52% 62% 52% 45%
  • 40. Example 4 Regret for Low Power Due to Small Sample?  I have a set of data that the mean change between the 2 groups is significantly different (p<0.05). But when I put calculate the power it gives only 50%. How should I interpret this? Also, can someone kindly advise as whether it is meaningful (or pointless) to calculate the power when the result is statistically significant?
  • 41. Books and Software  Sample size tables for clinical studies (second edition)  By David Machin, Michael Campbell Peter Fayers and Alain Pinol  Blackwell Science 1997  PASS 2000 available in CCTER  nQuery 4.0 available in CCTER
  • 43. Randomization  Definition:  randomization is a process by which each participant has the same chance of being assigned to either intervention or control.
  • 44. Fundamental Point  Randomization trends to produce study groups comparable with respect to known and unknown risk factors, removes investigator bias in the allocation of participants, and guarantees that statistical tests will have valid significance levels.
  • 45. Two Types of Bias in Randomization  Selection bias  occurs if the allocation process is predictable. If any bias exists as to what treatment particular types of participants should receive, then a selection bias might occur.  Accidental bias  can arise if the randomization procedure does not achieve balance on risk factors or prognostic covariates especially in small studies.
  • 46. Fixed Allocation Randomization  Fixedallocation randomization procedures assign the intervention to participants with a pre-specified probability, usually equal, and that allocation probability is not altered as the study processes • Simple randomization • Blocked randomization • Stratified randomization
  • 47. Randomization Types  Simple randomization
  • 48. Simple Randomization  Option 1: to toss an unbiased coin for a randomized trial with two treatment (call them A and B)  Option 2: to use a random digit table. A randomization list may be generated by using the digits, one per treatment assignment, starting with the top row and working downwards:  Option 3: to use a random number-producing algorithm, available on most digital computer systems.
  • 49. Advantages  Each treatment assignment is completely unpredictable, and probability theory guarantees that in the long run the numbers of patients on each treatment will not be radically different and easy to implement
  • 50. Disadvantages  Unequal groups  one treatment is assigned more often than another  Time imbalance or chronological bias  One treatment is given with greater frequency at the beginning of a trial and another with greater frequency at the end of the trial.  Simple randomization is not often used, even for large studies.
  • 51. Randomization Types  Blocked randomization
  • 52. Blocked Randomization (permuted block randomization)  Blocked randomization is to ensure exactly equal treatment numbers at certain equally spaced point in the sequence of patients assignments  A table of random permutations is used containing, in random order, all possible combinations (permutations) of a small series of figures.  Block size: 6,8,10,16,20.
  • 53. Advantages  The balance between the number of participants in each group is guaranteed during the course of randomization. The number in each group will never differ by more than b/2 when b is the length of the block.
  • 54. Disadvantages  Analysis may be more complicated (in theory)  Correct analysis could have bigger power  Changing block size can avoid the randomization to be predictable  Mid-block inequality might occur if the interim analysis is intended.
  • 55. Randomization Types  Stratified randomization geographic U .S . E u ro p e location previous exposure Yes No Yes No site l y m p h s k i n b re a s t l y m p h s k i n b re a st l y m p h s k i n b re a s t l y m p h s k i n b re a s t
  • 56. Stratified Randomization  Stratified randomization process involves measuring the level of the selected factors for participants, determining to which stratum each belongs, and performing the randomization within the stratum. Within each stratum, the randomization process itself could be simple randomization, but in practice most clinical trials use some blocked randomization strategy.
  • 57. Table 3. Stratification Factors and Levels (3×2×3=18 Strata) Age Sex Smoking history 1. 40-49 yr 1.Male 1. Current smoker 2. 50-59 yr 2 Female 2. Ex-smoker 3. 60-69 yr 3. Never smoked
  • 58. Table 4 Stratified Randomization with Block Size of Four Strat Age Sex Smoking Group assignment a 1 40-49 M Current ABBA BABA.. 2 40-49 M Ex BABA BBAA.. 3 40-49 M Never Etc. 4 40-49 F Current 5 40-49 F Ex 6 40-59 F Never 7 50-59 M Current 8 50-59 M Ex 9 50-59 M Never 10 50-59 F Current 11 50-59 F Ex 12 50-59 F Never etc.
  • 59. Advantages  Tomake two study groups appear comparable with regard to specified factors, the power of the study can be increased by taking the stratification into account in the analysis.
  • 60. Disadvantages  The prognostic factor used in stratified randomization may be unimportant and other factors may be identified later are of more importance
  • 61. Mechanism Trial Type Mechanism No central registration office Randomization list sealed envelops Double blind drug trial Pharmacist will be involved Multi-centre trial Central registration office Single-centre trial Independent person responsible for patients registration and randomization
  • 62. An Example of Stratified Randomization  Patients will be stratified according to the following criteria:  1) Treatment center (Hospital A vs Hospital B vs Hospital C)  2) N-stage(N2 vs N3)  3) T-stage (T1-2 vs T3-4)
  • 63. What should be in the protocol?  A dynamic allocation scheme will be used to randomize patients in equal proportions within each of 12 strata. The scheme first creates time- ordered blocks of size divisible by three and then uses simple randomization to divide the patients in each block into three treatment arms, in equal proportion. The block sizes will be chosen randomly so that each block contains either 6 or 9 patients.
  • 64. Cont…  This procedure helps to ensure both randomness and investigator blinding (the block sizes are known only to the statistician), as recommended by Freedman et al. Randomization will be generated by the consulting statistician in sealed envelopes, labeled by stratum, which will be unsealed after patient registration.
  • 65. Adaptive Randomization  Number adaptive  Biased coin method  Baseline adaptive (MINIMIZATION)  Outcome adaptive
  • 66. Biased Coin Method  Advantages  Investigators can not determine the next assignment by discovery the blocking factor.  Disadvantages  Complexity in use  Statistical analysis cumbersome
  • 67. Minimization  Minimization is an well -accepted statistical method to limit imbalance in relative small randomized clinical trials in conditions with known important prognostic baseline characteristics.  It called minimization because imbalance in the distribution of prognostic factors are minimized
  • 68. Table 1 Some baseline characteristics of patients in a controlled trial of mustine versus talc in the control of pleural effusions in patients with breast cancer (Frientiman et al, 1983) Treatment Mustine (n=23) Talc(n=23) Mean age (SE) 50.3(1.5) 55.3(2.2) Stage of disease: 1 or 2 52% 74% 3 or 4 48% 26% Mean interval in 33.1(6.2) 60.4(13.1) month between BC diag. and effusion diag. (SE) Postmenopausal 43% 74%
  • 69. Minimization Factors Age ( years) <=50 Or >50 Stage of disease 1 or 2 Or 3 or 4 Time between diagnosis <=30 Or >30 of cancer and diagnosis of effusions(months) Menopausal Pre Or Post
  • 70. Table 2 Characteristics of the first 29 patients in a clinical trial using minimization to allocate treatment Mustine Talc Age <=50 7 6 >50 8 8 Stage 1 or 2 11 11 3 or 4 4 3 Time <=30m 6 4 Interval >30m 9 10 Menopausal Pre 7 5 Post 8 9
  • 71. Table 3 Calculation of imbalance in patient characteristics for allocating treatment to the thirtieth patient Mustine Talc (n=15) (n=14) Age >50 8 8 Stage 3 or 4 4 3 Time interval <=30m 6 4 Postmenopausal 8 9 Total 26 24
  • 72. Advantages  It can reduce the imbalance into the minimum level especially in small trial  Computer Program available (called Mini) and also not difficult to perform ‘by hand’  Minimization and stratification on the same prognostic factors produce similar levels of power, but minimization may add slightly more power if stratification does not include all of the covariance
  • 73. Disadvantages  It is a bit complicated process compare to the simple randomization
  • 74. Practical Considerations Study type Randomization Large studies Blocked Large, Multicentre studies Stratified by centre Small studies Blocked and Stratified by centre Large number of Minimization Prognostic factors Large studies Stratified analysis without stratified randomization