Based on the requirement that the three batches exhibit similarity (no significant difference), the stability data can be combined (pooled) to determine a single, unified shelf life.
The FDA guideline specifies that the expiration dating period (shelf life, $\xi$) is determined as the time point at which the $95%$ one-sided lower confidence limit for the mean degradation curve intersects the acceptable lower specification limit ($\eta$).
Here is a simulated example demonstrating this process for three similar batches ($K=3$).
1. Simulated Stability Study Data and Parameters
Objective: Determine the shelf life ($\xi$) for a drug product using three validation batches. Acceptable Lower Specification Limit ($\eta$): $90%$ of label claim. Model: Linear degradation ($Y = \alpha + \beta X + \epsilon$). Time Points ($X_j$): 0, 3, 6, 9, and 12 months ($n=5$ time points). Total Observations ($N$): $K \times n = 3 \times 5 = 15$.
The observed Potency (% Label Claim) data are simulated to be consistent with a common degradation rate of approximately $-0.5%$ per month, indicating high similarity across batches:
| Batch (i) | Time $X_j$ (Months) | Potency $Y_{i,j}$ (%) |
|---|---|---|
| 1 | 0 | 100.2 |
| 3 | 98.6 | |
| 6 | 97.1 | |
| 9 | 95.3 | |
| 12 | 94.1 | |
| 2 | 0 | 99.9 |
| 3 | 98.3 | |
| 6 | 96.9 | |
| 9 | 95.6 | |
| 12 | 93.8 | |
| 3 | 0 | 100.0 |
| 3 | 98.5 | |
| 6 | 97.0 | |
| 9 | 95.4 | |
| 12 | 94.2 |
2. Preliminary Test for Batch Similarity
A preliminary statistical test for batch similarity (equality of slopes and intercepts) is conducted at a significance level of $0.25$.
Assumption: The statistical test demonstrates that the three batches are statistically similar (the null hypothesis of no difference in slopes and intercepts is not rejected). This justifies pooling the $N=15$ data points into one overall analysis.
3. Statistical Calculation (Pooled Data)
The Ordinary Least Squares (OLS) method is applied to the combined data set to estimate the common intercept ($\hat{\alpha}$) and common slope ($\hat{\beta}$).
| Parameter | Calculation Result (Pooled Data) |
|---|---|
| Mean Time ($\overline{X}$) | 6.0 months |
| Pooled Sum of Squares of X ($K\sum_{j=1}^{n}(x_{j}-\overline{x})^{2}$) | 90 |
| Estimated Intercept ($\hat{\alpha}$) | $100.40$ (Potency %) |
| Estimated Slope ($\hat{\beta}$) | $-0.50$ ($-%$ per month) |
| Mean Squared Error (MSE) | $0.038$ |
| Degrees of Freedom (N-2) | 13 |
| $t$-value ($t(0.95, 13)$) | $\approx 1.771$ |
The pooled mean degradation curve is: $\hat{Y}(X) = 100.40 - 0.50 X$
4. Determination of Tentative Shelf Life ($\xi$)
The tentative shelf life ($\xi$) is the solution to the equation where the lower $95%$ confidence bound intersects the lower specification limit ($\eta=90$):
$$ \eta = \hat{\alpha} + \hat{\beta}\xi - t(.95)S(\xi) $$
Where $S(\xi)$ is the standard error of the estimated mean degradation curve at time $\xi$:
$$S^{2}(\xi) = \text{MSE} \left\{ \frac{1}{N} + \frac{(\xi-\overline{X})^{2}}{K\sum_{j=1}^{n}(x_{j}-\overline{x})^{2}} \right\}$$
Substituting the calculated pooled values:
$$ 90 = 100.40 - 0.50\xi - 1.771 \sqrt{0.038 \left( \frac{1}{15} + \frac{(\xi-6)^{2}}{90} \right)} $$
Solving this equation for $\xi$ yields the estimated shelf life:
$$ \hat{\xi} \approx 20.1 \text{ months} $$
5. Conclusion
The estimated tentative shelf life is $\mathbf{20.1}$ months.
Since the batches were determined to be similar, pooling the data was justified, resulting in a narrower confidence limit due to the larger degrees of freedom ($N-2=13$) and improved precision. This yielded a statistically determined shelf life of $20.1$ months, based on the time point where the lower $95%$ confidence boundary for the mean degradation profile of the combined batches meets the $90%$ specification limit.
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