Paper, pencil, brain ;-) [Software]

posted by yuvrajkatkar – Pune, Maharashtra (India), 2009-09-21 17:10 (6109 d 00:34 ago) – Posting: # 4224
Views: 32,291

Dear Sir,

Suppose Geometric Least Square Mean Ratio (T/R)=1.20
Lower Confidence Limit=1.08
Upper Confidence Limit=1.33
Number of Subjects (N)=42
Degrees of Freedom=N-2=40;

TINV(.10,40)=1.683851

Based on Lower Confidence Limit
Residual SUMSQ(1.08,1.20)=2.6064
MSresidual=2.6064/40=.06516
and SEdifference=sqrt(2*MSresidual/42)=0.055703296
CVintra=100*SQRT(EXP(MSresidual)-1)=26%

Based on Upper Confidence Limit
Residual SUMSQ(1.33,1.20)=3.2089
MSresidual=3.2089/40=0.0802225
and SEdifference=sqrt(2*MSresidual/42)=0.061807112
CVintra=100*SQRT(EXP(MSresidual)-1)=29%

I want to confirm,
Whether I am right or wrong?

I need your reply.

Best Regards,
Yuvraj

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