Statistical Tools for Similarity Assessment of Quality Attributes

Transcription

1 Statistical Tools for Similarity Assessment of Quality Attributes Aili Cheng, Ph.D, Director, Pfizer Statistics FDA-PQRI Conference on Evolving Product Quality Sept 16-17,

2 Comparability vs. Similarity Comparability: Usually, process change by the same sponsor Historical data available (e.g. development, process qualification and control ) Acceptance criteria are relatively easy to set Similarity: Change in manufacturer Limited (no historical) data available May require more rigorous analytical and statistical evaluations This talk will focus on similarity 2

3 Outline Background Statistical challenges Possible statistical approaches Pros and cons of each method Comments / questions 3

4 Background Objectives Biosimilar vs. US-licensed Reference Product For global development: non-us-licensed comparator vs. USlicensed Reference Product Importance of Analytical Studies in Biosimilar Development Analytical studies provide the foundation for an assessment of the proposed protein product for submission Include analytical studies that demonstrate that the biological product is highly similar to the reference product notwithstanding minor differences in clinically inactive components Source: FDA Guidance for Industry Quality Considerations in Demonstrating Biosimilarity to Reference Protein Product,

5 Statistical Challenges in Determining Similarity Limited number of lots Different testing methods result in different data types What difference is meaningful and what is not?--- acceptable similarity margin The plurality of candidates of critical quality attributes within specific developments, as well as the usually low number of drug batches available had been identified as the most limiting factors, rendering the use of statistical routines usually performed on basis of clinical patientdata inappropriate most of the time. Source: EMA Concept Paper on the need for a reflection paper on statistical methodology for the comparative assessment of quality attributes in drug development,

6 Possible Statistical Solutions +/- 3 Standard Deviation (SD) Tolerance Interval Equivalence testing 6

7 +/- 3 Standard Deviation Calculate Mean+/-3SD range based on the Reference lots Expect the comparator lots or biosimilars to fall within the range Pros Simple to understand Easy to implement Consistent with quality control principle Cons Assumes normal distribution Limited sample size in practice Concern is on the tails, so usual sample size for means may be too small here 7

8 Relationship Between Sample Size and SD 99.73% σ = lots: 96.4, 106.3, 97.8, , 100.5, 102.0, , Mean=102.5, SD=4.6 Estimated +/-3SD range: (88.67, ) 5 lots: 101.1, 94.7, 104.1, 92.7, Mean= 99.6, SD=5.6 Estimated +/-3SD range: (82.8, 116.4) 99.73% of the sample values are expected to fall within (100-3*4.5, 100+3*4.5), i.e. (86.5, 113.5) 8

9 Relationship Between Sample Size and SD For example, the estimated SD from a sample size of 10 can differ from the true SD by 45% with 95% chance See details in : Robert W. Burnett, CLINICAL CHEMISTRY, Vol. 21, No. 13,

10 Tolerance Interval Interval within which we expect a stated proportion of the population to lie with some confidence. +/- 3 std dev can be restated as a TI where we are xx% confidence that 99.7% is within the stated limits where xx% depends on the sample size Alternatively, any TI can be calculated on the reference data and the comparator be expected to fall within this limit Non-parametric tolerance interval is available if data does NOT follow normal distribution, but large sample size is generally required 10

11 Relationship between Tolerance Parameters and Sample Size k2 Multiplier Univariate TI = μ ± k 2 σ where k 2 is a function of confidence, coverage and sample size where μ and σ are estimated reference mean and variability, respectively 8 7 Figure 3 k2 multiplier for various sample sizes and confidence/coverage TI 20 confidence/coverage [0.90/0.95] [0.90/0.99] [0.95/0.99] (Target) Sample Size (N lots)

12 +/- 3SD and Tolerance: An Example 10 lots: 96.4, 106.3, 97.8, , 100.5, 102.0, , 104.8, Mean=102.5, SD=4.6 Mean+/-3SD Mean+/-k 2 *SD, k 2 = % confidence, 99.7% coverage (88.7, 116.3) (81.4,123.5) The width of the tolerance interval is dependent upon the sample size, confidence level, and coverage level +/-3SD range does not take sample size into account +/-3SD range in this case is same as TI with 37.6% confidence and 99.7% coverage When sample size is big enough, these two intervals are close to each other. 12

13 Probability of Failing Assuming Normal Distribution Mean Difference between reference and comparator mean+/- 3σ (%) 90/99 TI* (%) 0 σ 0.3 (1.5**) 0.08 (0.4) 0.50 σ 0.6 (3) 0.2 (1) 1.0 σ 2.2 (11) 0.9 (4) 2 σ 16 (58) 8.5 (36) *90/99 TI was based on n = 20 lots **Probability of failing at least one in the next 5 lots Calculation assumes that SD is known. 13

14 Testing for Equivalence Equivalence compares the expected difference, confidence interval on the mean difference, (fat point) to similarity limits (L1, L2) Similarity Limits are determined prior to experimentation H 0 = Non equivalence Not Equivalent H A = Equivalence Fat Point Confidence Interval On Average Difference ( ) L 1 L 2 L 1 0 L 2 H 0 = Non equivalence Not Equivalent Conclude: If Confidence Interval is within Similarity Limit then reference and comparator are Equivalent 1-2 Reference - Comparator 14

15 Equivalence Testing Pros Assess if the mean difference is within acceptable margin Relies on comparisons of means which (in theory) follow a normal distribution Cons How to determine the equivalence margin? Bigger variability require bigger sample size Determination of confidence level and power Multiple comparisons 15

16 Summary Focus on Individual lots +/- 3SD range Tolerance Interval Focus on mean Equivalence Testing Be careful of assumptions required by different statistical tools 16

17 Acknowledgement Kim Vukovinsky Beverly Ingram Andrew Rugaiganisa Brad Evans Greg Steeno Debbie Kraus 17

18 Relationship Between Sample Size and SD True SD=4.5 True SD=2 True SD=0.5 4 = 100, = 4.5 = 2.0 = Relative Frequency Distribution Estimates for n = Distribution Estimates 1 for n = Distribution Estimates for n = 10 Distribution Estimates for n = 20 18