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jag009 ★★★ NJ, 2012-09-13 19:39 (4682 d 02:44 ago) Posting: # 9204 Views: 13,919 |
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Hi Helmut, I have a simple question about the intrasubject CV back calculation. The MSE equation from your lecture note, if the 90% CI and pt estimate are from a 3-way study, the denominator of the MSE equation should then be, sqrt(1/n1 + 1/n2 + 1/n3)*t1-2*alpha(n1+n2+n3-3)? Thanks John |
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d_labes ★★★ Berlin, Germany, 2012-09-14 11:43 (4681 d 10:40 ago) @ jag009 Posting: # 9215 Views: 12,158 |
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Dear John ❝ Hi Helmut, Not interested in the opinion of other members?  The common formula for the 90% CIs of cross-over studies is (in the log domain): [lower, upper] = point ± tval(1-alpha/2, df)*sqrt(MSE)*sqrt(bk(ni)*sum(1/ni))ni are the number of subjects in the sequence groups, bk(ni) is the so called design constant (here if ni are given) and df the degrees of freedom from the cross-over ANOVA. Under the assumption of balanced designs the above formula can be given in terms of total number of subjects (N=sum(ni)): [lower, upper] = point + tval(1-alpha/2, df)*sqrt(MSE)*sqrt(bk/N)with bk the design constant in terms of total number if subjects. Here the design characteristics for your question (tmt x seq x period): design df bk(ni) bkThe latter design is 6-sequence williams design for 3 treatments and 3 periods. Insertion of the 2x2x2 characteristics will give the formulas in Helmut's lectures. With a little algebra you should be able to derive the MSE formula for your case  . But you could save the effort for better things. Use the function CVfromCI() from the R-package PowerTOST which has some more designs you eventually need.But in the moment it works only for balanced designs i.e. same number of subjects in sequence groups. Therefore it needs only the total number of subjects under study. This is in most cases accessible from literature while the number of subjects in sequences is usually not given. BTW: The red emphasis above shows where the great admin of this forum and itinerant preacher in the matter of BE errs if we talk about alpha=0.1 i.e 90% CIs.— Regards, Detlew |
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Helmut ★★★   Vienna, Austria, 2012-09-14 16:31 (4681 d 05:52 ago) @ d_labes Posting: # 9222 Views: 11,843 |
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Dear Detlew! ❝ The common formula for the 90% CIs of cross-over studies is (in the log domain): ❝ ❝ ❝ BTW: The red emphasis above shows where the great admin of this forum and itinerant preacher in the matter of BE Yes, sure, what else? Checked my presentations, oops, what the heck? F**k! THX for letting me know! — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![[image]](https://static.bebac.at/pics/Blue_and_yellow_ribbon_UA.png) Helmut Schütz ![[image]](https://static.bebac.at/img/CC by.png) The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
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ElMaestro ★★★ Denmark, 2012-09-14 17:48 (4681 d 04:35 ago) @ Helmut Posting: # 9223 Views: 11,812 |
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Hi Helmut and Detlew, aren't both right? I am just thinking loud here: We are talking about the critical value of the t distribution, and for that
— Pass or fail! ElMaestro |
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d_labes ★★★ Berlin, Germany, 2012-09-17 11:57 (4678 d 10:26 ago) @ ElMaestro Posting: # 9226 Views: 11,819 |
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Dear Ol'Pirate, ❝ aren't both right? The formula for the two-tailed CI is that given above (using a critical tvalue(1-alpha/2, df)). The formula for the one-tailed interval (upper or lower) differs from that in using tval(1-alpha, df). This is the dual to the two one sided tests (TOST) procedure. Using alpha=0.05 in the one-sided case both CIs are identical if using alpha=0.1 in the two-sided case. This is why it is said that the usual (two-sided) 90% CIs are operationally equivalent to the TOST procedure. But in no case the critical value has to be derived with 1-2*alpha. ❝ Because of symmetry tval(1-alpha, df) = tval(1-alpha/2, df) if we are clear about the circumstances.  ❝ If I remember correctly MS Excel does it a little different from everyone else among the commonly used programs? Lets take alpha=0.05, df=10, EXCEL 2010: 1-alpha 1-alpha/2— Regards, Detlew |
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Astea ★★ Russia, 2015-02-05 02:20 (3807 d 19:03 ago) @ d_labes Posting: # 14375 Views: 10,137 |
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Dear statisticians! sorry for forcing in this branch... This tails makes me feel blue. My question is the following: comparing formulas from Chow S.-C., Wang H. On Sample Size Calculation in Bioequivalence Trials, Journal of Pharmacokinetics and Pharmacodynamics, Vol. 28, No. 2, 2001 (2), p.161 and Chow S.C., Shao J., Wang H., Sample size calculations in clinical research. Marcel Dekker, New York, NY, 2003 on p.260 I see that - in the book the authors restricts themselves with the formula concerning beta/2 - in the example they use values z_alpha=1.96 and z_beta/2=0.84 for alpha =0.05 and beta =80%, theta0=1 and CV=0.4 and get totally 42 subjects. I'm trying to evaluate formulas (2) p.161 from Chow S.-C., Wang H. On Sample Size Calculation in Bioequivalence Trials, Journal of Pharmacokinetics and Pharmacodynamics, Vol. 28, No. 2, 2001 (note that there is an excess square for \delta). For sample size calculation in standard 2x2 crossover design we use for the first iteration the following values: alpha=0.05, beta=80% z_alpha=1.64 and z_beta/2=1,28 for the case when mu_T=mu_R (theta0=1) and z_alpha=1,64 and z_beta/2=0,84 otherwise (in Excel I am tapping TINV(0,1,10000); TINV(0,2,10000) and TINV(0,4,10000) respectively) (10000 is just a big-big number, I'm not sure Excel knows "infinity"). For CV=0.4 and theta0=0.95 after five iterations I got 66 subjects. Both PowerTost and bear give the same result. And all seemed to be good but why then in Chow S.C., Shao J., Wang H., Sample size calculations in clinical research. Marcel Dekker, New York, NY, 2003 on p.260 in the example they use values z_alpha=1.96 and z_beta/2=0.84 for alpha =0.05 and beta =80%? Suspect I read the book inattentively and in depends on the CI (90% or 95%), but the question remains: how many "tails" for alpha and beta do we have to use in BE studies? — "Being in minority, even a minority of one, did not make you mad" |
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d_labes ★★★ Berlin, Germany, 2015-02-05 10:31 (3807 d 10:53 ago) @ Astea Posting: # 14376 Views: 10,165 |
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Dear Astea, don't fiddle too much with the formulas given by Chow, Wang and who ever co-authors. There are many typing errors or hidden, not explained assumptions in the mentioned paper and book. IMHO the only correct way to estimate the sample size is 'brute force', i.e. choose a sample size, estimate power and than iterate sample size until desired power is reached. All other ways are more or less approximations to tackle the problem avoiding the not so easy power calculations. But in the ages where PowerTOST is available there is no need for such approximations  . These attempts are only useful for obtaining a reasonable start value for 'brute force'. As such they are used within PowerTOST. Have a look at Paul Zhang (2003) to get the formulas used (including z1-ß or z1-ß/2 and such things). — Regards, Detlew |
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Astea ★★ Russia, 2015-02-07 13:53 (3805 d 07:30 ago) @ d_labes Posting: # 14387 Views: 9,889 |
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Dear d_labes! Thanks a lot for your explanations! ❝ don't fiddle too much with the formulas given by Chow, Wang and who ever co-authors. There are many typing errors or hidden, not explained assumptions in the mentioned paper and book. I naively thought it was one the best books on sample size calculations in clinical trials? Nevertheless I didn't understand why they used z_alpha=1.96 for alpha =0.05... ❝ All other ways are more or less approximations to tackle the problem avoiding the not so easy power calculations. But in the ages where PowerTOST is available there is no need for such approximations Yes, your Brainchild is very helpful and irreplaceable! ❝ These attempts are only useful for obtaining a reasonable start value for 'brute force'. As such they are used within PowerTOST. Does exact Owen's method in general do the same? ❝ Have a look at ❝ Paul Zhang (2003) ❝ "A Simple Formula for Sample Size Calculation in Equivalence Studies" ❝ Journal of Biopharmaceutical Statistics, 13:3, 529-538 ❝ to get the formulas used (including z1-ß or z1-ß/2 and such things). Yes, they use the more crafty formula, introducing function varying coefficients before beta... — "Being in minority, even a minority of one, did not make you mad" |
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d_labes ★★★ Berlin, Germany, 2015-02-07 17:22 (3805 d 04:01 ago) @ Astea Posting: # 14388 Views: 9,891 |
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Dear Astea ❝ ... ❝ ❝ These attempts are only useful for obtaining a reasonable start value for 'brute force'. As such they are used within PowerTOST. ❝ Does exact Owen's method in general do the same? Don't understand your question here. Owen's Q functions are a way of calculating the power exact. For the sample size 'brute force' as described above is nevertheless necessary since no exact closed form result for the sample size itself is known. And this brute force can be shortened in using a reasonable start value. PowerTOST uses some sort of Zhang's formula to that end, regardless of the power calculation method used, "exact", "nct" or "shifted". — Regards, Detlew |
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jag009 ★★★ NJ, 2012-09-24 22:12 (4671 d 00:11 ago) @ d_labes Posting: # 9245 Views: 11,681 |
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Thank D_labes, So for a 4-way (4 period, 4 sequence,4 treatments) crossover or 4x4x4: df = 3*(n1+n2+n3+n4)-6? with ni = number of subjects per sequence. What would the formula be if you go even higher? 5-way? Thanks John |
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d_labes ★★★ Berlin, Germany, 2012-09-25 12:40 (4670 d 09:43 ago) @ jag009 Posting: # 9248 Views: 11,643 |
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Hi John, ❝ So for a 4-way (4 period, 4 sequence,4 treatments) crossover or 4x4x4: ❝ df = 3*(n1+n2+n3+n4)-6? with ni = number of subjects per sequence. Correct. But don't forget to set the design constant bk(ni)=1/8 ❝ What would the formula be if you go even higher? 5-way? The overall formula remains as written above. But you have to figure out the df and bk(ni). Not enough spare time at me to do that for you .— Regards, Detlew |
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jag009 ★★★ NJ, 2012-09-25 16:59 (4670 d 05:24 ago) @ d_labes Posting: # 9250 Views: 11,590 |
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Thanks D_labes |
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ElMaestro ★★★ Denmark, 2015-02-05 13:57 (3807 d 07:26 ago) @ jag009 Posting: # 14377 Views: 10,005 |
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Hi John et al., MSE from a 3x3x3 study... does it make any sense to speak of such a thing? Can you put it into a context? The MSE generally makes sense when we have a model with just one variance component = the error. In a 3x3x3 etc we have multiple variance components to sort out, so I am thinking it is hard to meaningfully nominate anyone of them to be the MSE. You can perhaps backcalculate an 'apparent' MSE it from a 90% CI that has arisen from a replicated study of any type but I wonder what the application of it would really be. — Pass or fail! ElMaestro |
Ing. Helmut Schütz