The function for calculating the probability of passing critical quality attributes (CQA) PPQ test .

PPQ_pp(Llim, Ulim, mu, sigma, n, n.batch, k)

Arguments

Llim

lower specification limit

Ulim

upper specification limit

mu

hypothetical mean of the attribute

sigma

hypothetical standard deviation of the attribute

n

sample size (number of locations) per batch

n.batch

number of batches for passing PPQ during validation

k

general multiplier for constructing the specific interval

Value

A numeric value of the passing/acceptance probability

References

Burdick, R. K., LeBlond, D. J., Pfahler, L. B., Quiroz, J., Sidor, L., Vukovinsky, K., & Zhang, L. (2017). Statistical Applications for Chemistry, Manufacturing and Controls (CMC) in the Pharmaceutical Industry. Springer.

See also

rl_pp.

Author

Yalin Zhu

Examples

if (FALSE) { PPQ_pp(Llim = 90, Ulim = 110, mu=105, sigma=1.5, n=10, k=3.1034) # One-sided tolerance interval with k=0.753 (95/67.5 one-sided tolerance interval LTL) PPQ_pp(sigma=0.03, mu=1.025, n=40, Llim=1, Ulim=Inf, k=0.753) sapply(X=c(0.1,0.5, 1,2,3,4,5,10), FUN = PPQ_pp, mu=97, n=10, Llim=95, Ulim=105, k=2.373) sapply(X=seq(0.1,10,0.1), FUN = PPQ_pp, mu=97, n=10, Llim=95, Ulim=105, k=2.373) sapply(X=c(0.1,0.5, 1,2,3,4,5,10), FUN = PPQ_pp, mu=100, n=10, Llim=95, Ulim=105, k=2.373) sigma <- seq(0.1, 4, 0.1) pp1 <- sapply(X=sigma, FUN = PPQ_pp, mu=97, n=10, Llim=95, Ulim=105, k=2.373) pp2 <- sapply(X=sigma, FUN = PPQ_pp, mu=98, n=10, Llim=95, Ulim=105, k=2.373) pp3 <- sapply(X=sigma, FUN = PPQ_pp, mu=99, n=10, Llim=95, Ulim=105, k=2.373) pp4 <- sapply(X=sigma, FUN = PPQ_pp, mu=100, n=10, Llim=95, Ulim=105, k=2.373) plot(sigma, pp1, xlab="Standard Deviation", main="LSL=95, USL=105, k=2.373, n=10", ylab="Probability of Passing", type="o", pch=1, col=1, lwd=1, ylim=c(0,1)) lines(sigma, pp2, type="o", pch=2, col=2) lines(sigma, pp3, type="o", pch=3, col=3) lines(sigma, pp4, type="o", pch=4, col=4) legend("topright", legend=paste0(rep("mu=",4),c(97,98,99,100)), bg="white", col=c(1,2,3,4), pch=c(1,2,3,4), lty=1, cex=0.8) mu <- seq(95, 105, 0.1) pp5 <- sapply(X=mu, FUN = PPQ_pp, sigma=0.5, n=10, Llim=95, Ulim=105, k=2.373) pp6 <- sapply(X=mu, FUN = PPQ_pp, sigma=1, n=10, Llim=95, Ulim=105, k=2.373) pp7 <- sapply(X=mu, FUN = PPQ_pp, sigma=1.5, n=10, Llim=95, Ulim=105, k=2.373) pp8 <- sapply(X=mu, FUN = PPQ_pp, sigma=2, n=10, Llim=95, Ulim=105, k=2.373) pp9 <- sapply(X=mu, FUN = PPQ_pp, sigma=2.5, n=10, Llim=95, Ulim=105, k=2.373) plot(mu, pp5, xlab="Mean Value", main="LSL=95, USL=105, k=2.373, n=10", ylab="Probability of Passing", type="o", pch=1, col=1, lwd=1, ylim=c(0,1)) lines(mu, pp6, type="o", pch=2, col=2) lines(mu, pp7, type="o", pch=3, col=3) lines(mu, pp8, type="o", pch=4, col=4) lines(mu, pp9, type="o", pch=5, col=5) legend("topright", legend=paste0(rep("sigma=",5),seq(0.5,2.5,0.5)), bg="white", col=c(1,2,3,4,5), pch=c(1,2,3,4,5), lty=1, cex=0.8) }