Preliminaries

Without loss of generality, suppose $$n$$ outcomes of the Critical Quality Attribute (CQA) are normally distributed, which is denoted by $$X_i \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(\mu, \sigma^2)$$, where $$i=1,\dots, n$$, then the distributions of sample mean and standard deviation are as known: $\begin{equation} \bar{X} \sim \mathcal{N}(\mu, \frac{\sigma^2}{n}) \end{equation}$ and $\begin{equation} \label{diststd} \dfrac{(n-1)S^2}{\sigma^2} \sim \chi^2 (n-1). \end{equation}$ Moreover, sample mean and sample standard deviation are independent under normal distribution assumption.

Denote the lower and upper specification limits as $$L$$ and $$U$$, respectively. The prediction or tolerance interval can be expressed by $\begin{equation} \left[ Y_1, Y_2 \right] = \left[ \bar{X} - kS, \ \bar{X} + kS \right] , \end{equation}$ where $$k$$ is a specific multiplier for the interval. For example, for prediction interval, $$k=t_{1-\alpha/2,n-1}\sqrt{1+\frac{1}{n}}$$.

Specification test for one release batch

The outcome at release can be any one of the sample, so $$X_{rl} \sim \mathcal{N}(\mu, \sigma^2)$$, then the probability of passing PPQ at release should be

$\begin{equation} \begin{split} \Pr(\text{Passing Specification for Release}) & = \Pr(L \le X_{rl} \le U) \\ & = \Phi (U) - \Phi(L) \end{split} \end{equation}$

This probability is very easy to calculate using software, such as pnorm() in R.

Test for PPQ Batches

$\begin{equation} \label{probpass} \begin{split} \Pr(\text{Passing a Single PPQ Batch}) & = \Pr(L \le Y_1 \le Y_2 \le U) \\ & = \int_{L}^{U} \int_{L}^{y_2}f_{Y_1,Y_2}(y_1, y_2) dy_1 dy_2 \end{split} \end{equation}$

Now it is essential to obtain the bivariate joint distribution of the lower and upper prediction/tolerance interval, that is, find joint probability density function (PDF) $$f_{Y_1,Y_2}(y_1,y_2)$$.

Since $$Y_1=\bar{X} - k S$$ and $$Y_2=\bar{X} + k S$$, we can use another bivariate PDF $$f_{\bar{X},S}(x,s)$$ to calculate $$f_{Y_1,Y_2}(y_1,y_2)$$ by using Jacobian transformation.

Solve $$\bar{X}$$ and $$S$$ as $$x=\dfrac{y_1+y_2}{2}$$ and $$s=\dfrac{y_2-y_1}{2k}$$, then Jacobian of the transformation is $\begin{equation} |J|= \left| \begin{array}{cc} \dfrac{\partial x}{\partial y_1} & \dfrac{\partial x}{\partial y_2}\\ \\ \dfrac{\partial s}{\partial y_1} & \dfrac{\partial s}{\partial y_2} \end{array} \right| = \left| \begin{array}{cc} \dfrac{1}{2} & \dfrac{1}{2}\\ \\ -\dfrac{1}{2k} & \dfrac{1}{2k} \end{array} \right| = \dfrac{1}{2k}. \end{equation}$

Thus, () can be calculated as

$\begin{equation} \label{extend} \begin{split} \int_{L}^{U} \int_{L}^{y_2}f_{Y_1,Y_2}(y_1, y_2) dy_1 dy_2 & = \int_{L}^{U} \int_{L}^{y_2}f_{\bar{X},S}\left(\dfrac{y_1+y_2}{2}, \dfrac{y_2-y_1}{2k}\right) |J| dy_1 dy_2 \\ & = \dfrac{1}{2k} \int_{L}^{U} \int_{L}^{y_2}f_{\bar{X}}\left(\dfrac{y_1+y_2}{2}\right) f_{S}\left( \dfrac{y_2-y_1}{2k}\right) dy_1 dy_2. \end{split} \end{equation}$ The second equation follows from normal sample mean and standard deviation being independent.

Similarly, we can obtain the PDF of sample standard deviation $$f_S(s)$$. By (), let $$v= \dfrac{(n-1)s^2}{\sigma^2}$$, then Jacobian of the transformation is $\begin{equation} |J| = \left|\dfrac{dv}{ds}\right| = \dfrac{2(n-1)s}{\sigma^2}. \end{equation}$ Thus, $\begin{equation} \label{Spdf} \begin{split} f_S(s) & = f_V\left(\dfrac{(n-1)s^2}{\sigma^2}\right) |J| \\ & = \dfrac{2(n-1)s}{\sigma^2}f_V\left(\dfrac{(n-1)s^2}{\sigma^2}\right) \end{split} \end{equation}$

Plug () in (), we can get the final results. $\begin{equation} \begin{split} & \Pr(\text{Passing a Single PPQ Batch}) \\ = & \dfrac{1}{2k} \int_{L}^{U} \int_{L}^{y_2}f_{\bar{X}}\left(\dfrac{y_1+y_2}{2}\right) \dfrac{2(n-1)\dfrac{y_2-y_1}{2k}}{\sigma^2}f_V\left\{\dfrac{(n-1)\left[\dfrac{y_2-y_1}{2k}\right]^2}{\sigma^2}\right\} dy_1 dy_2 \\ = & \dfrac{n-1}{2k^2 \sigma^2}\int_{L}^{U} \int_{L}^{y_2}f_{\bar{X}}\left(\dfrac{y_1+y_2}{2}\right) f_V\left\{\dfrac{(n-1)(y_2-y_1)^2}{4k^2\sigma^2}\right\} (y_2-y_1) dy_1 dy_2, \end{split} \end{equation}$ where $$\bar{X} \sim \mathcal{N}(\mu, \frac{\sigma^2}{n})$$ and $$V \sim \chi^2 (n-1)$$. Then this quantity can be easily calculated by software, such as functions dnorm(), dchisq() and integrate() in R.

We can also calculate the probability of passing $$m$$ PPQ batches, then under the assumption of independence and similar expected performance across batches, the probability will be $\Pr(\text{Passing } m \text{ batches }) = \{\Pr(\text{Passing a Single PPQ Batch}) \}^m$