sample size for parallel design [Software]

posted by yuvrajkatkar – Pune, Maharashtra (India), 2009-10-03 09:44 (5737 d 16:29 ago) – Posting: # 4289
Views: 28,594

Dear Sir,

Thanks a lot, Sir.

Ok Please see the Chow and Liu Sample size formula for parallel design

Suppose CV=.15
Assume n=32
n            32         32         32
alpha        0.05       0.05       0.05
beta         0.2        0.2        0.2
power        0.8        0.8        0.8
Theta        0.95       1          1.05
LN(Theta)2  -0.10259    0          0.09758
CV           0.15       0.15       0.15
CV^2         0.0225     0.0225     0.0225
talpha,2n-2  1.998971   1.998971   1.998971
tbeta,2n-2   1.669804   1.669804   1.669804

       2CV^2 (talpha,2n-2+tbeta/2, 2n-2)^2
  n >= -----------------------------------
                 0.2-|Theta|^2

Using this formula, I have got n= 2.001728, 3.028481, 5.913866

and using the below SAS code

proc power;
  Pairedmeans
  test  = Equiv_ratio
  dist  = lognormal
  CV    = 0.15
  meanratio = 0.95 1 1.05
  lower = 0.8
  upper = 1.25
  alpha = 0.05
  power = 0.8
  npairs= .
  corr  = 0.0
  ;
run;

I have got n= 12, 10, 11 per group
It means n=24, 20, 22 for parallel design

Which one is the right answer?


Edit: Reformated using BBCode (code); tabs substituted by blanks. Please see the Policy. [Helmut]

Best Regards,
Yuvraj

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