- On 25 May 2006 at 11:55:15, shekar (shekar_stats.aaa.yahoo.co.in) sent the message

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Dear All,

I Thanks in advance, if anyone could clarify my

question. In the Statistical Analysis of one study, We came acrose

this situation.

The Anova for LnAUCt as Dependent variable, and Sequence, Period , Sub

(seq) and Treatment as independent variables carried out.

the results are as follows:

R-Square Coeff Var Root MSE lnAUCt

Mean

0.853861 4.139193 0.199158

4.811522

Source DF Type III SS Mean

Square F Value Pr > F

seq 1 0.00000001

0.00000001 0.00 0.9997

sub(seq) 30 6.95082384

0.23169413 5.84 <.0001

per 1 0.00039561

0.00039561 0.01 0.9211

trt 1 0.00131702

0.00131702 0.03 0.8566

Tests of Hypotheses Using the Type III MS for sub(seq)

as an Error Term

Source DF Type III SS Mean

Square F Value Pr > F

seq 1 5.3387176E-9

5.3387176E-9 0.00 0.9999

as observed in the above result, the type III S.S and M.S.S for Seq is

5.3387176E-9 and Probability is 0.9999

Could any one tell me, whether this result is acceptable.

What could be the interpretation?

explain it. - On 25 May 2006 at 16:35:13, "ogwal sidney" (sidneyogwal.aaa.hotmail.com) sent the message

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The following message was posted to: PharmPK

Where statistics have failed it does not mean automatically that the

results are bad. Have you tried other statistical tests besides

ANOVA??? You do have several choices

s.o.o - On 26 May 2006 at 10:47:48, "Silva, Nuno" (nmens.-at-.ff.ul.pt) sent the message

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The following message was posted to: PharmPK

For bioequivalence studies, the ANOVA sequence effect must be tested

against Subject nested within Sequence (Subj(Seq)). Therefore, and

according to your results, p value for sequence effect is 0.9999, which

means that your study sequence effect is not statistically significative

(p<0.05).

Best regards

Nuno Silva

Faculty of Pharmacy, U.L.

Biopharmaceutics and Pharmacokinetics Dept.

Av. Prof. Gama Pinto

1649 003 Lisbon / Portugal

Tel: (+351) 217 946 406

Fax: (+351) 217 937 703

E-mail: nmens.-at-.ff.ul.pt - On 28 May 2006 at 17:41:47, "Pereira, Luis" (Luis.Pereira.-a-.mcphs.edu) sent the message

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The following message was posted to: PharmPK

Dear Shekar

There's an important piece of information missing in your posting

pertaining to the pooled variance for the error term in your analysis

of variance. Without a full picture it's hard to tell you whether

your conclusions, rather than your results, are acceptable.

You may be aware of an unfortunate misconception that still lingers

around about the use of p-value. Many people still believe that it

indicates the strength or the magnitude of a given comparison, when

in fact it doesn't. A p-value of 0.00001 does not imply that the

magnitude of the difference between lets say treatments is larger

than when the p-value is 0.049. For example, a p-value of 0.6

relative to the F ratio for the formulations does not provide any

information about bioequivalence if the mean square error is let's

say 0.4. Likewise, a p-value of 0.0000005 for the formulations

doesn't convey any information about bioequivalence if the p-value

for the sequence (carryover) effect is let's say 0.001. Testing

separately the carryover effect will allow the use of the proper

corrected p-value and adequate bioequivalence testing.

So I suggest that you put all the pieces of the puzzle together and

be ware for the misleading p-value.

A short but significant communication by Brian P. Smith, "It's time"

The AAPS Journal 2005, 7(3), E655, may lead you in the way.

Cheers

Luis

--

Luis M. Pereira, Ph.D.

Assistant Professor, Biopharmaceutics and Pharmacokinetics

Massachusetts College of Pharmacy and Health Sciences

179 Longwood Ave, Boston, MA 02115

Phone: (617) 732-2905

Fax: (617) 732-2228

Luis.Pereira.aaa.bos.mcphs.edu - On 21 Aug 2006 at 15:35:15, "gang li" (gangli01.-a-.gmail.com) sent the message

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Hi,

I think the problem is that your model is overfitted. The treatment

is uniquely determined by the combination of sequence and period if

the design is a crossover design. But you included sequence, period

and treatment all in the predictors. Notice that the F-value for the

sequence is close to zero which comfirmed this. Hope this can help you!

Best Regards,

Gerry Li - On 24 Aug 2006 at 08:47:05, "gang li" (gangli01.aaa.gmail.com) sent the message

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Hi,

I think the problem is that your model is overfitted. The treatment

is uniquely determined by the combination of sequence and period if

the design is a crossover design. But you included sequence, period

and treatment all in the predictors. Notice that the F-value for the

sequence is close to zero which comfirmed this. Hope this can help you!

Best Regards,

Gerry - On 7 Dec 2006 at 22:04:44, (efoconnor.aaa.cox.net) sent the message

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The following message was posted to: PharmPK

----Would anynone out there consider it prudent to proceed with an

ANOVA without verifying the suitability of the data beforehand. - On 8 Dec 2006 at 14:17:21, "Nadeem Irfan Bukhari" (nadeemirfan_bukhari.-at-.imu.edu.my) sent the message

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The following message was posted to: PharmPK

Does verification mean the distribution of data?

In my opinion, the repeated ANOVA is based on the CI, which already

taking into account of the error in Mean.

Nadeem - On 8 Dec 2006 at 14:47:52, jfinsi (jamil.finsi.aaa.gmail.com) sent the message

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The following message was posted to: PharmPK

----That is obvious, indeed the credibility of anova test is gained

thru verifying the dependent and the factor elements. For instance:

The Mean Square Error (MSE) of an estimate is the variance of the

estimate plus the square of its bias; therefore, if an estimate is

unbiased, then its MSE is equal to its variance, as it is the case in

the ANOVA table.

--

Jamil Finsi - On 9 Dec 2006 at 01:17:51, "Zhang Jin" (zjhzpl26.aaa.hotmail.com) sent the message

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The following message was posted to: PharmPK

There're several assumptions need to be checked before drawing the

conclusions.

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