Bioequivalence and Bioavailability Forum

Anu ★ India, 2013-09-04 09:03 (4254 d 19:13 ago) Posting: # 11424 Views: 9,043	Sample Size [Power / Sample Size] Post reply
	Dear All, Kindly guide me how can I find the sample size from the following available information: A crossover study with 21 subjects. It was Average daily Urinary analysis study with phosphorus as a component. Bioequivalence assessment using pharmacodynamics: 1. Test-Reference= -0.69 2. 90 % CI = -1.07 to -0.30 which is in the limit of ( -20 to 20) % Thanks & Regards Anu

Dr_Dan
★★

Germany,
2013-09-04 11:52
(4254 d 16:24 ago)

@ Anu
Posting: # 11427
Views: 7,923

Sample Size

Post reply

Dear Anu,
I am not sure if you really know what you are talking about

❝ A crossover study with 21 subjects.

there you have a sample size

❝ Bioequivalence assessment using pharmacodynamics:

what kind of parameter? AUEC, Cmax?

❝ 1. Test-Reference= -0.69

ratio?

❝ 2. 90 % CI = -1.07 to -0.30 which is in the limit of ( -20 to 20) %

In bioequivalence testing the 90% confidence interval for the exponential of the difference in LSmeans between the Test and Reference product are calculated for the ln transformed parameters. The data you present do not fit.

Please reassess your question, maybe it would be helpful to provide more information and before you ask the same question a third time please ensure that you have the necessary basic understanding of bioequivalence assessment. Have a look into Helmut's lectures!
Kind regards
Dan

—
Kind regards and have a nice day
Dr_Dan

Helmut
★★★

Vienna, Austria,
2013-09-04 15:20
(4254 d 12:56 ago)

@ Anu
Posting: # 11431
Views: 8,079

Linear model and Fieller…

Post reply

Hi Anu,

please don’t fork; sounds like your previous thread.

❝ A crossover study with 21 subjects. It was Average daily Urinary analysis study with phosphorus as a component.

❝ Bioequivalence assessment using pharmacodynamics:

❝ 1. Test-Reference= -0.69

❝ 2. 90 % CI = -1.07 to -0.30 which is in the limit of ( -20 to 20) %

Are you sure you know what you are doing? I would not call P-excretion a PD measurement (though it might be an effect, you have still to deal with concentrations/amounts). Whether amount excreted follows a normal distribution or lognormal is subject of heated debates. Personally I tend towards the former.
You didn’t give units. Normal ranges of P in 24 h-urine are 0.8–1.6 mmol/L (2.5–4.9 mg/dL). If you assume a normal distribution, you should apply an additive model (untransformed data) and (!) state the acceptance range in absolute values – not percent (I have no idea, maybe –0.75 mg/dL, +0.75 mg/dL). So what was stated in the protocol? If you want to deal with percentages and still assume normal distributed data, you have to go with Fieller’s confidence interval (then the acceptance range would be 80–125%). For some references see this thread.

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jag009 ★★★ NJ, 2013-09-04 19:23 (4254 d 08:52 ago) @ Helmut Posting: # 11437 Views: 7,880	Linear model and Fieller… Post reply
	Hi Helmut, ❝ Are you sure you know what you are doing? I would not call P-excretion a PD measurement (though it might be an effect, you have still to deal with concentrations/amounts). Whether amount excreted follows a normal distribution or lognormal is subject of heated debates. Personally I tend towards the former. Study Maybe Anu got the terminology wrong? John

Anu
★

India,
2013-10-04 15:16
(4224 d 13:00 ago)

@ jag009
Posting: # 11610
Views: 7,622

Linear model and Fieller…

Post reply

Hi all,

Greetings!

Kindly check the link Urine Analysis

According to our data we carried out a Pilot study on 21 Subjects and the Results for Excretion were as follows:

Refernce Formulation (LSM,SD)=(12.225,0.171)
Test Formulation (LSM,SD)=(11.540,0.171)

PD measure...Test-Reference....90% CI ±20% limit of....TestReference 90% CI Average
daily
urinary
excretion ... -0.69...... -1.07 to -0.30 ..... -2.44 to 2.44

As for the calculation of Pivotol study sample size we require (CV %, Point Estimate (T/R)) from Pilot study usually.
I have no idea how to go for the sample size calculation for this study from pilot results so available. :confused:

How we can calculate sample size for Pivotal Study in this case?

Also in the given link dint get how they have calculated, PD equivalence was to be claimed if the 90% CI was completely contained within the critical reference range of (-3.47, 3.47).
The numbers in red. Which in our study is (-2.44 to 2.44) :confused:

How the calculations done?

Thanks in advance

Anu

Helmut
★★★

Vienna, Austria,
2013-10-04 17:20
(4224 d 10:56 ago)

@ Anu
Posting: # 11611
Views: 7,668

Linear model and 80–120%

Post reply

Hi Anu,

❝ Refernce Formulation (LSM,SD)=(12.225,0.171)

❝ Test Formulation (LSM,SD)=(11.540,0.171)

Identical SDs? Your variability is unbelievably low (CV ~1.44%). In the linked study the CV was ~20.4% – which is already pretty low for phosphate excretion. Even if your values are the SEMs (not the SDs) I still doubt CVs of ~6.6%.

❝ As for the calculation of Pivotol study sample size we require (CV %, Point Estimate (T/R)) from Pilot study usually.

If you want to follow the method of the linked study you need CV% and the difference (!) T–R, not the ratio T/R.

❝ I have no idea how to go for the sample size calculation for this study from pilot results so available. :confused: How we can calculate sample size for Pivotal Study in this case?

You need the intrasubject CV from your study (untransformed data). If it is not given in the study report calculate it as CV_intra = √MSE/LSM_R.

❝ Also in the given link dint get how they have calculated, PD equivalence was to be claimed if the 90% CI was completely contained within the critical reference range of (-3.47, 3.47).

❝ The numbers in red. Which in our study is (-2.44 to 2.44) :confused:

❝ How the calculations done?

              n     T (SD)      R (SD)    80–120% of R   AR* for T–R  90% CI of T–R

linked study  49  16.01±3.40  17.35±3.39  13.88–20.82   –3.47, +3.47  –3.17, –0.80

your study    21  11.54±0.17  12.23±0.17   9.78–14.67   –2.45, +2.45  –1.07, –0.30

Note that both studies were imbalanced. In the imbalanced case generally the LSM of the difference ≠ the difference of treatment means.

               T–R   LSM

linked study  –1.34 –1.98 

your study    –0.69 –0.69

Check your calculations.

AR = (1±0.2)×R – R = [13.88, 20.82] – 17.35 = –3.47, +3.47

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Anu
★

India,
2013-10-07 11:59
(4221 d 16:17 ago)

@ Helmut
Posting: # 11614
Views: 7,575

Linear model and 80–120%

Post reply

Hi Helmut,

Thankyou so much for your esteemed reply.

❝ You need the intrasubject CV from your study (untransformed data). If it is not given in the study report calculate it as CV_intra = √MSE/LSM_R.

Calculated the Intrasubject CV(%) came out 5.83 % (yes its very low).
With the help of Fartssie Sample Size Calculation sheet and R-Software, I find sample size. For this study I have added the values of SD(0.17) & Difference (0.69) in the equivalence crossover and got sample size 5.

Am I right? Is there any other method kindly suggest.

❝ n T (SD) R (SD) 80–120% of R AR* for T–R 90% CI of T–R

❝ linked study 49 16.01±3.40 17.35±3.39 13.88–20.82 –3.47, +3.47 –3.17, –0.80

❝ your study 21 11.54±0.17 12.23±0.17 9.78–14.67 –2.45, +2.45 –1.07, –0.30

Thanx

I got it.

Thanks & Regards
Anu

Helmut
★★★

Vienna, Austria,
2013-10-07 16:14
(4221 d 12:02 ago)

@ Anu
Posting: # 11618
Views: 7,492

CVintra < CVtotal?

Post reply

Hi Anu,

❝ Calculated the Intrasubject CV(%) came out 5.83 % (yes its very low).

I’m still not sure whether your calculations are correct. You got a total CV of ~1.4% which is smaller than CV_intra? BTW, a CV_intra of 5.83% for phosphate excretion in urine is hardly to believe.

❝ With the help of Fartssie Sample Size Calculation sheet and R-Software, I find sample size. For this study I have added the values of SD(0.17) & Difference (0.69) in the equivalence crossover and got sample size 5.

Haven’t used FARTSSIE.xls for ages, but you need the CV, not the SD. However,

library(PowerTOST)

sampleN.TOST(logscale=FALSE, theta0=-0.69, theta1=-2.45, theta2=2.45, CV=0.0583)

+++++++++++ Equivalence test - TOST +++++++++++

            Sample size estimation

-----------------------------------------------

Study design:  2x2 crossover 

untransformed data (additive model)



alpha = 0.05, target power = 0.8

BE margins        = -2.45 ... 2.45

Null (true) diff. = -0.69,  CV = 0.0583



Sample size (total)

 n     power

4   1.000000

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The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes

Sample Size [Power / Sample Size]

Sample Size

Linear model and Fieller…

Linear model and Fieller…

Linear model and Fieller…

Linear model and 80–120%

Linear model and 80–120%

CVintra < CVtotal?