Statistical Methods for Biotechnology Products－USP Tests and Specifications

Statistical Methods for

Statistical Methods for

Biotechnology Products

Biotechnology Products

USP Tests and Specifications

by

Jen-pei Liu, PhD, Professor

Division of Biometry, Department of Agronomy National Taiwan University

and

Division of Biostatistics and Bioinformatics National Health Research Institutes

Outline

Outline



_Introduction



_{Sampling Plan and Acceptance Criteria}

Content uniformity of dosage units

USP/NF general chapter[905]

Dissolution Testing

USP/NF general chapter[711]

Disintegration Testing

USP/NF general chapter[701]

p2 p2

Outline

Outline



_{Probability of Passing USP Test}

Content Uniformity

Dissolution

Distintegration



_{Acceptance Limits}



_{In-house Specifications}



_Discussion

continued continued p2p2

Introducation

Introducation

Identity, strength, quality, and purity of a drug product.

Sampling plan and acceptance criteria required by USP XXII/NF XVII for

content uniformity testing [905] dissolution testing [711]

disintegration testing [701]

Approximate probabilities can be computed for passing USP testing. Acceptance limits also can be established to pass USP testing with a

high probability.

Tolerance limit approach can be used to set up the in-house specifications.

p3 p3

Introduction

Introduction

USP/NF Sampling Plan and Criteria

Multi-stage dependent mixed acceptance sampling plan

Discrete variables for attributes

Sample statistics for continuous variables

More than one acceptance criteria for some stages

Different acceptance criteria for different stages

p4 p4

Introduction

Introduction



_{Joint distributions of the attributes and}

individual units and sample statistics are

difficult to find.



_{Content uniformity:}

Joint distribution of each dosage unit and

coefficient of variation.



L

_h



Y

_hi



U

_h

and

CV

_h



c

_h



Pr



_{Dissolution testing:}

Joint distribution of each units and sample

mean.





h h h hi

L

and

Y

c

Y





Pr

p5 p5

Sampling Plan for Content Uniformity

Sampling Plan for Content Uniformity



_{Content uniformity of dosage units USP/NF}

general chapter [905]

1. Prepare a composite specimen of a sufficient number of dosage units to provide the sufficient amount of specimen.

2. Assay separately, accurately measured portions of the composite specimen.

3. Calculate the weight of active ingredient equivalent to one average dosage unit, by

(a) using the results obtained by assay procedure,

(b) using the results obtained by the special procedure.

p6 p6

Sampling Plan for Content Uniformity

Sampling Plan for Content Uniformity

4. Calculate the correction factor, F, by formula

P

A

F



in which A is the weight of the active ingredient equ

ivalent to one average dosage unit obtained by the a

ssay procedure, and P is the weight of active ingredi

ent equivalent to one average dosage unit obtained b

y the special procedure. If

,

10

100





A

P

A

The use of a correction factor is not valid.

continued

continued

p6 p6

Sampling Plan for Content Uniformity

Sampling Plan for Content Uniformity

5. A valid correction may be applied only if F is not

less than 1.03 or greater than 1.10, or not less than

0.90 or greater than 0.97. If F is between 0.97 and

1.03, no correction is required.

6. If F lies between 1.03 and 1.10, or between 0.90

and 0.97, calculate the weight of active ingredient

in each dosage unit by multiplying each of the

weights found using the special procedure by F.

continued

continued

p6 p6

If the average of the limits specified in the potency definition in the individual monograph in greater than 100%, proceed as follows:

1. If the average value of the dosage units tested is 100% or less, the requirements are the same as those given in Table 5.2.1. 2. If the average value if the dosage units tested is greater than or equal to the average of the limits specified in the potency definition in the individual monograph, the requirements are the same as those given in Table 5.2.1 except that the words “label claim” are replaced by the words “label claim

multiplied by the average of the limits specified in the potency definition in the monograph divided by 100.”

continued

continued

p7 p7

3. If the average value of the dosage units tested is

definition in the individual monograph, the

requirements are the same as those given in Table

5.2.1 except that the words “label claim“ are

replaced by the words “label claim multiplied by

the average value of the dosage units tested

(expressed as a percent of label claim) divided by 100.”

continued

continued

p7 p7

p8 p8

p11 p11

p10 p10

DisintegrationTesting

DisintegrationTesting

USP/NF general chapter [711]

USP/NF general chapter [711]

p9 p9

Probability of Passing USP Tests

Probability of Passing USP Tests



Let S

_i

and C

_ij

denote the events that the ith st

age of a K-stage test is passed and the event

that the jth criterion for the ith stage is met,

where j=1,…,m

_i

and i=1,…,K.

Also, let Pi be the probability of passing the

ith stage Then the probability of passing a

multiple-stage test is given by

p12 p12

Probability of Passing USP Tests

Probability of Passing USP Tests

 





























K



K i i k i i i j j k k k k

P

P

P

P

P

P

P

P

P

P

P

P

P

S

S

not

S

P

S

S

not

P

S

not

S

P

S

not

P

S

P

S

or

or

S

or

S

P

K

P

,

,

,

max

1

1

1

1

1

1

}

,

,

|

{

}

,

,

{

)

|

(

)

(

}

{

test}

stage

a

passing

{

2 1 1 1 1 1 1 1 3 2 1 2 1 1 1 1 1 1 1 2 1 1 2 1















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

























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

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

















 

       p13 p13

Probability of Passing USP Tests

Probability of Passing USP Tests



_Furthermore,



_{Therefore, a lower bound for passing the}

multiple K-stage test is given by

 

 

















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

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0

,

1

max

}

{

1 2 1 i m j ij im i i i i

m

C

P

C

and

and

C

and

C

P

S

P

P

i i



























0

,

1

)

(

max

1 i m j ij

m

C

P

i p13 p13

Probability of Passing the

Probability of Passing the

Content Uniformity Test

Content Uniformity Test

 _{The probability of passing the USP test content uniformly, denoted by PCU, is given}

by

};

0

,

1

)

(

)

(

,

1

)

(

)

(

max{

criteria}}

2

stage

USP

meet the

units

30

{all

criteria},

1

stage

meet USP

units

10

{first

max{

criteria}

2

stage

USP

meet the

units

30

all

or

criteria

1

stage

USP

meet

units

10

first

{

22 21 12 11















C

P

C

P

C

P

C

P

P

P

P

P

_CU p14 p14

Probability of Passing the

Probability of Passing the

Content Uniformity Test

Content Uniformity Test

claim}. label of 125% to 75 range the outside unit no and claim label of 115.0% to range the outside is unit no and calim label of 115.0% to 85.0 range the outside is ) S (S units 30 the of 1 than more not { }, 7.8% than kess is deviation standard relative the { claim}, label the of 115% to 85 range e within th is units 10 the of each { 6%} than less is deviation standard relative the { 2 1 22 21 12 11      C C C C where p14 p14

Therefore, a lower bound(LB) of PCU can be obtained by finding P(C11),

P(C12), P(C21), and P(C22).

Let Y be the content uniformity assay value, which is usually expressed as a percentage of label claim, Assume that Y follows a normal distribution with mean  and variance 2 [i.e., Y~ N(, 2)]. The distribution of the

inverse of the square of the sample coefficient of variance multiplied by sample size follows a noncentral F distribution, Therefore, given a set of values for  and 2 , P(C₁₁) and P(C₂₁) can be computed by the following

probability statements:

where , and are the sample mean and standard deviation for stage i, and k1=0.06 and k2=0.078. Let

Where n1=10 and n2=30. Then it follows that P(Cil)=P{Wi<ki}=P{Fi>ni/ki2}.

, 2 , 1 }, { ) (C ₁  P W  k i  P _{i i i} i i i s Y W  Y_i s_i 2 , 1 , 2 _  n W i F _{i i} p15 p15

110/07/16 Copyright by Jen-pei Liu, PhD 22

It can be verified that Fi=Fi(1, dfi, ) follows a no central

F distribution with 1 and dfi degrees of freedoms and

noncentrality parameter

where df1=9 and df2=29.

Bergum(1990) considered the approximation by a central F distribution to evaluate P(Ci1) based on the following

relationship:

where F(vi, dfi,0) is a central F distribution with vi

and dfi degree of freedom, and

, 2       







_i n_i







1

,

df

,



,

1

1

0

,

df

,

_{i i} i i i

F

v

F













i i i v





2 1 1 2    i



p16p16

In addition to the approximation by a central F distribution, a normal approximation to the distribution of the square root of Fi, as suggested by

Laubscher (1960), may be useful for the evaluation of P(Ci1). The method

is described as follows. Let

where i=1,2.Then

follows a standard normal distribution. As a result, P(Ci1) can be

computed by replacing Fi with ni / ki2 for a given set of values of and . p17 p17





2 2 3 2 2 2 2 i 1 1 2 1 df 1 2 1 1 2 df ) 1 df 2 ( i i i i i i i i i i i i F g g F g                 2 , 1 , 3 2 1 1    i g g g Z i i i 



2

Let p1 be the probability that an assay result is in the range 85 to 115% of

label claim. Then

Let p2 be the probability that an assay result is in the range 75 to 85% of

label claim or 115 to 125% of label claim. Then

Therefore, we have

P(C12)=P{all 10 units are between 85 and 115% of label claim}

= (p1)10

P(C22)=P{ all 30 units are between 85 and 115% of label claim}

+P{29 units are between 85 and 115% of label claim and 1 unit is between 75 and 85% of label claim or 115 and 125% of label

claim} =(p1)30+30(p1)29 p2 p18 p18



85 115



1  P Y  p } 125 115 { } 85 75 { 2  P Y   P Y  p

p19 p19

Probability of Passing the Dissolution Test

Let Y be the dissolution assay result, which is usually expressed as a percentage of label claim, Assume that Y follows a normal distribution with mean and variance . Also, let

and denote the sample mean of 24 assay values. Then the probability of passing USP dissolution test is given by

P{passing USP test}= P(S1 or S2 or S3},

where

S₁= { the first six pass the stage 1 criteria}, S2 = { the first 12 pass the stage 2 criteria},

S3 = { all 24 pass the stage 3 criteria}.

p20 p20



15



, ₂



25 15



1  P Y  Q  p  P Q   Y  Q  p 24 Y



 2

To obtain a lower bound, we may consider the following.

So a lower bound(LB) for the probability of passing the dissolution test, given and , is given by





 



24



1 2 23 1 2 2 22 1 2 1 24 24 24 3 3 2 1 24 276 } 24 { 25} than less is units no and 15 an greater th es assay valu 24 of 22 least at { } { 1 25} than less is unit no and 15 an greater th es assay valu 24 of 22 least at { } { 25} than less is unit no and 15 an greater th are es assay valu 24 of 22 least at and { } { or or p p p p p Q Z P Q Q P Q Y P Q Q P Q Y P Q Q Q Y P S P S S S P                          



2

 



24



1 2 23 1 2 2 22 1 2 1 24 276 } 24 {Z Q p p p p p P LB          p21 p21

p22 p22

p23 p23

 _{Disintegration Testing}

Let Y be the disintegration time. Again we assume that Y follows a normal distribution with mean and variance . Also, let

p = P{0 < Y < UL},

where UL denotes the specified limit. Since the disintegration test involves only one acceptance criterion at both stages of the sampling plan, the exact probability can be computed. Let

C11 = {all six units disintegrate completely},

C12= {one unit fails to disintegrate completely},

C13= {two units fail to disintegrate completely},

C21 = {11 of 12 additional units disintegrate completely},

C22 = {all 12 additional units disintegrate completely}.

 _ 2

p24 p24

 _{Then the exact probability of passing the disintegration test} is given as follows:













. ) 1 ( 87 ) 1 ( 6 1 2 6 p 1 1 6 1 11 12 } {C } C | {C } {C } C | C {C } {C pass} { 2 16 17 6 2 4 12 5 12 11 6 13 13 22 12 12 22 21 11 p p p p p p p p p p p p p P P P P P P                                                     p24 p24

 _{It can easily be verified that if the desired probability of passing the disintegr}

ation test is 0.5, p is approximately about 0.831. If, in addition, the specified time limit, UL, is 30 min, it follows that

where Z is a standard normal variable and Z(0.169) is the 16.9% upper quant ile of a standard normal distribution. Therefore,

Hence the contour for and is a linear decreasing function of given by

where 0.957=Z(0.169) 831 . 0 )} 169 . 0 ( { } 30 { } 30 { } {            Z Z P Y P Y P UL Y P p     p25 p25



0.169



0.957 30 _{ } Z  





 30



957

.

0

 _ 2

p26 p26

In-house Specifications

In-house Specifications



_{Definition – Specifications}

Specific intervals that sample mean and standard

deviation must be contained to meet the USP

requirements.

Example:

Dissolution specifications at 4 hours after

encapsulation

Mean: (35%, 60%)

SD: (0, 11%)

p37 p37

In-house Specifications

In-house Specifications



_{Source of variation}

- Lab-to-lab

- day-to-day

- analyst-to-analyst

- location-to-location

.

.

To obtain estimates of variance component 

an estimate of standard deviation

p37 p37

In-house Specifications

In-house Specifications

p37p37

























2 2 2 2 2 2 2 2 U 2 2 L 2 2

1

,

1

,

/

,

/

,

1

,

/

{

}

{c

}

c

and

U































d

N

c

N

N

b

z

N

a

z

s

N

N

Y

Z

where

P

z

Z

z

P

d

s

P

b

Y

a

P

d

s

b

Y

a

P

L U L U L

























































In-house Specifications

In-house Specifications



 and 

2

are usually unknown and need to be

estimated based on the sample. Furthermore,

based on the sample mean and variance (or

standard deviation), it is desirable to have some

idea with a certainty (i.e., probability) regarding

the range where the mean and variance of a future

sample.

p37 p37

In-house Specifications

In-house Specifications

These ranges are referred to as the prediction intervals for the samp le mean and standard deviation (Hahn, 1970). On the other hand, it is als o important to estimate an interval such that

100% of the population lies within the interval with

100% confidence. This interval is known as the tolerance interval [see, e. g., Bowker (1947), Hahn (1970), and Graybill (1976)]. In addition, differ ent sources of variation need to be taken into account when constructing t he confidence interval for the total variability and the prediction interval for the standard deviation of a future sample. In the following we introdu ce the concept of the use of the prediction interval and the tolerance

interval for the development of an in-house dissolution specification thro ugh a real example.

p37 p37

p38 p38

p39 p39

 _{Suppose that an experiment was conducted to develop an in-house}

dissolution specification for the tablets of a drug product. A total of

I batches of granulation were each made into J tablet strengths. K

dissolution assays were performed at 2, 4, 8, and 14 h on each of IJ batch-by-strength combinations. The purpose of the study is to

recommend in-house specifications for the mean of the percent of the amount of dissolved active ingredient. At a particular time point, a recommended in-house specification can be obtained under the following two-way cross-classification mixed model:

p40 p40

 

K

k

J

j

I

i

e

BS

S

B

Y

_{ijk i i ij ijk}

1,...,

,

,...,

1

,

,...,

1

,



















 where Y_ijk denotes the assay result of the kth tablet in the ith b

atch for the jth strength, Sj is the fixed effect for the jth streng

th, Bi is the random effect for the ith batch, (BS)ij is the rando

m effect for the jth strength made from the ith batch, and eijk i

s the random error in observing Yijk. In the model above, it is

assumed that Bi’s are i.i.d. normal with mean zero and varian

ce ,

(BS)ij's are i.i.d. normal with mean zero and variance

,

and eijk's are i.i.d normal with mean zero and variance

p40 p40

 

~



0,



] ., . [ 2 2 BS ij BS i e BS N   )] , 0 ( ~ ., . [ 2 2 e ijk e i e e N   )] , 0 ( ~ ., . [ 2 2 B i B i e B N  

Thus a approximate two-side symmetric prediction interval for any future sample mean of the jth

strength over the I' batches at a particular time point is given by (Lp,Up) where p41 p41  

 

   

 

       2 2 2 2 2 2 . 2 1 2 , 1 1 1 1 1 1 , ˆ 1 ) ( 1 } ) ( 1 ] [ 1 1 { var ˆ 1 1 1 ˆ ) ( 2 1                       _                  



     JK MSB I JK BS MS J J I d d df MSB JK BS MS JK J BS MS MSB JK MSE BS MS K MSE K Y Y I Y I I t Y L U ij ij ij I i ij j ij df j p p    



1



100%

Recall that is a random sample of sizes I chosen from a normal distribution with mean , and variance which can be estimated unbiasedly by and ,respectively.

Therefore, as pointed out by Wallis (1951), an approximate symmetrical tolerance interval with a confidence level of 1 - , which contains at least (1 - ) 100% of the sample mean . for the jth strength based on K tablets over a population of /random batches, is given by (TL,TU), where

Hence a (1 - )  100% prediction interval for the future mean based on batches at a particular time point can be obtained as (LTP,UTP), where

Where

An unbiased estimator of is given by

p42 p42



Y ...,1j, YIj



j S   _ij2 j Y_ ˆij2 ij Y              _ I I I t Y L U_TP( _TP) , 1 ˆ_i2 1 1 2 1 

 

_{L j}





_ij U T Y g I T  _{ }  , , ˆ            I i i J j K k ijk i _JK Y Y _I Y Y 1 1 1 , 1 2 i  JK MSB i  2 

Furthermore, a symmetrical tolerance interval with

confidence level of 1 -  , which contains be obtained as (TTL,TTU), where

Therefore, an approximate (1 - )100% confidence interval for y is given by (LY, UY), where

Where 2 (/2,df_Y) is the (/2)th upper quantile of a central

Chi-square distribution with degrees of freedom dfy. where

p43 p43





_i TL TU T Y g I T ( )  _  , , ˆ









2 / 1 2 1 2 2 Y 2 / 1 2 1 2 2 Y df , 1 ˆ df , df , ˆ df                Y Y Y Y y y U L        

 

_ˆ 2 2 _d2 df_Y  _{Y Y}        2 2 2 2 2 1 1 1 ) ( 1 1 1 1 1 1 1 1 ) ( 1 1 )] ( [ 1 ] [ 1 ˆ                                    MSB JK I BS MS JK I J I MSB K K K IJ d MSB JK BS MS JK J MSE K K BS MS MSB JK MSE BS MS K MSE Y Y 

Suppose that a future experiment is planned with K' total of J' str engths made from a total of I' batches. Under the same model, the va riance of a single assay from this future experiment is also .

Let be an unbiased estimator of from a future experiment. Let df' be the degrees of freedom associated with .Since

are independent and each is approximately distributed as a central ch i-square random variable with dfy and degrees of freedom, respect

ively,

follows approximately a central F distribution with and dfy degre

es of freedom. p44 p44 2 Y  2 ˆ_Y  2 Y  2 ˆ_Y  2 2 2 2 _{df ˆ} and ˆ df Y Y Y Y Y Y       2 2 ˆ ˆ Y Y F    y f d  y f d 

 As indicated by Hahn (1970), a (1- ) 100% upper prediction lim

it for the future standard deviation y of the assay can be obtained a

s

Note that is usually unknown and needs to be

estimated. An adhoc estimate of can be obtained simply by replac ing I, J, and K with F, J', and K'. Since the mean squares from the c urrent study are used to estimate the degrees of freedom for the futur e experiment, strictly speaking, the numerator and denominator in F

are not truly independent. Further research is necessary in this area.

p44 p44 y f d  y f d 







₂



1/2 f d , f d ,    _{Y y} F F U  

Example for In-house Specifications

Example for In-house Specifications



_{An experiment to recommend in-house dissolution}

specifications for the tablets of a pharmaceutical

compound.



_{Batch: 3}

Strength: 200mg, 300mg, 400mg

Time: 2,4,8,14 hours

# of Assays: 12 per combination

p45 p45

p46 p46

p47 p47

p48 p48

p49 p49

p50 p50

Discussions

Discussions

Bergum (1990) perform a simulation study for examining the accuracy of approximation (Table 5.6.1).

Tsong et al (1992) found that the sampling plan and

acceptance criteria for dissolution testing are rather liberal: a high fraction of units < Q and sample mean slightly greater than Q.

They proposed two modified sampling plans and acceptance criteria:

I: Change Q-15% in USP to Q-5%

II: A new three-stage sampling plan based only on sampling means and standard deviations.

Results of their simulation study: USP test >I> II.

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p52 p52

p53 p53

USP XXIII (2000)

USP XXIII (2000)



_Dissolution

Stage NO. Tested

Pass if

S1

6

Min(X) > Q+5%

S2

6

 Q

Min(X) > Q-15%

S3

12  Q

Min(X) > Q-15%

No more 2 units < Q-5%

Y Y

New Dissolution Testing

New Dissolution Testing

Tsong et al (2004)

 _HO: P(YQ)  P vs. H_a: P(YQ) > P; Y is N(, 2)

HO: +Z1-P  Q vs. Ha: +Z1-P  Q

 _{one-sided tolerance limit}

 _{Group sequential procedure to multiple-stage sampling with O’Brien-Flemin}

g boundaries

 _{N1=6, n2=6, n3=12}

 ₁ = 0.0009, ₂ = 0.00548, and _{3 = 0.04439}  C(₁) = 3.7498, C(₂) = 2.5406, C(_{3) = 1.6626}  _{Acceptance limits at each stage}

- Q > Ai, i=1,2,3. Ai = C(i) si/ni – siZ0.1

Recap



_{Description of USP/NF sampling plan and}

acceptance criteria for content uniformity,

dissolution, and disintegration testing



_{Multi-stage dependent mixed acceptance sampling}

plan



_{Discrete variables for attributes}



_{Sample statistics for continuous variables}



_{More than one acceptance criteria for some stages}



_{Different acceptance criteria for different stages}



_{Sampling plan depends upon the results from the}

 _{Probability of passing}

USP test

 _{Exact probability is difficult}

Good Approximation is available

 _{In-house Specifications}

Tolerance limits approach under a two-way

cross-classification mixed model (always tighter for passing the USP/NF testing).