1010 USP39-NF34 ANALYTICAL DATA INTERPRETATION AND TREATMENT (中英文)

Ratio = alternative procedure variance/current procedure variance = 45.0/25.0 = 1.8

Lower limit of confidence interval = ratio/F.05 = 1.8/2.168 = 0.83 Upper limit of confidence interval = ratio/F.95 = 1.8/0.461 = 3.90

Table 7. Example of Measures of Variance for Independent Runs (Specific to the Example in Appendix D) Procedure Alternative Current Variance (standard deviation) 45.0 (6.71) 25.0 (5.00) Sample Size 20 20 Degrees of Freedom 19 19

For this application, a 90% (two-sided) confidence interval is used when a 5% one-sided test is sought. The test is

one-sided, because only an increase in standard deviation of the alternative procedure is of concern. Some care must be exercised in using two-sided intervals in this way, as they must have the property of equal tails—most common intervals have this property. Because the one-side upper confidence limit, 3.90, is less than the allowed limit, 4.0, the study has demonstrated that the alternative procedure has acceptable precision. If the same results had been obtained from a study with a sample size of 15— as if 80% power had been chosen—the laboratory would not be able to conclude that the alternative procedure had acceptable precision (upper confidence limit of 4.47).

在这种情况下,当寻求单侧5%区间时,需要使用90%(双侧)置信区间。检验是单侧的,因为只有替代方法标准偏差的增加才是需要考虑的。此时使用双侧区间需要加以小心,因为他们必须具备等尾的特性-通常区间具备这一属性。因为单侧置信上限(3.90)小于允许限度(4.0),研究显示替代方法具有可接受的精密度。如果使用样本量15的研究获得了相同的结果,假设选择了80%效能,实验室不能做出替代方法具有可接受精密度的结论(此时置信上限为4.47)。

APPENDIX E: COMPARISON OF PROCEDURES—DETERMINING THE LARGEST ACCEPTABLE

DIFFERENCE, δ, BETWEEN TWO PROCEDURES

This Appendix describes one approach to determining the difference, δ, between two procedures

(alternative-current), a difference that, if achieved, still leads to the conclusion of equivalence between the two procedures. Without any other prior information to guide the laboratory in the choice of δ, it is a reasonable way to proceed. Sample size calculations under various scenarios are discussed in this Appendix.

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

Tolerance Interval Determination

Suppose the process mean and the standard deviation are both unknown, but a sample of size 50 produced a mean and standard deviation of 99.5 and 2.0, respectively. These values were calculated using the last 50 results generated by this specific procedure for a particular (control) sample. Given this information, the tolerance limits can be calculated by the following formula:

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

x ±KS

in which x is the mean; s is the standard deviation; and K is based on the level of confidence, the proportion of results to be captured in the interval, and the sample size, n. Tables providing K values are available. In this example, the value of K required to enclose 95% of the population with 95% confidence for 50 samples is 2.3828. The tolerance limits are calculated as follows:

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

99.5 ± 2.382 × 2.0

hence, the tolerance interval is (94.7, 104.3).

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

Comparison of the Tolerance Limits to the Specification Limits

8

There are existing tables of tolerance factors that give approximate values and thus differ slightly from the values reported here.

Assume the specification interval for this procedure is (90.0, 110.0) and the process mean and standard deviation have not changed since this interval was established. The following quantities can be defined: the lower specification limit (LSL) is 90.0, the upper specification limit (USL) is 110.0, the lower tolerance limit (LTL) is 94.7, and the upper tolerance limit (UTL) is 104.3. Calculate the acceptable difference, (δ), in the following manner:

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

A = LTL ? LSL for LTL < LSL

(A = 94.7 ? 90.0 = 4.7); B = USL ? UTL for USL < UTL (B = 110.0 ? 104.3 = 5.7); and δ = minimum (A, B) = 4.7

Figure 2. A graph of the quantities calculated above.

With this choice of δ, and assuming the two procedures have comparable precision, the confidence interval for the difference in means between the two procedures (alternative-current) should fall within ?4.7 and +4.7 to claim that no important difference exists between the two procedures.

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

Quality control analytical laboratories sometimes deal with 99% tolerance limits, in which cases the interval will widen. Using the previous example, the value of K required to enclose 99% of the population with 99% confidence for 50 samples is 3.390. The tolerance limits are calculated as follows:

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

99.5 ± 3.390 × 2.0

The resultant wider tolerance interval is (92.7, 106.3). Similarly, the new LTL of 92.7 and UTL of 106.3 would produce a smaller δ:

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的

检验。设定r=2,而n=10。

A = LTL ? LSL for LTL < LSL

(A = 92.7 ? 90.0 = 2.7); B = USL ? UTL for USL < UTL (B = 110.0 ? 106.3 = 3.7); and δ = minimum (A, B) = 2.7

Though a manufacturer may choose any δ that serves adequately in the determination of equivalence, the choice of a larger δ, while yielding a smaller n, may risk a loss of capacity for discriminating between procedures. 这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

Sample Size

Formulas are available that can be used for a specified δ, under the assumption that the population variances are known and equal, to calculate the number of samples required to be tested per procedure, n. The level of confidence and power must also be specified. [NOTE—Power refers to the probability of correctly concluding that two identical procedures are equivalent.] For example, if δ = 4.7, and the two population variances are assumed to equal 4.0, then, for a 5% level test9 and 80% power (with associated z-values of 1.645 and 1.282, respectively), the sample size is approximated by the following formula:

这是一种改进过的ESD检验,它可以从一个正态分布的总体当中发现预先设定数量(r)的异常值。对于仅检测1个异常值的情况,极端学生化偏离检验也就是常说的Grubb's检验。不建议将Grubb's检验用于多个异常值的检验。设定r=2,而n=10。

Thus, assuming each procedure has a population variance, s, of 4.0, the number of samples, n, required to conclude with 80% probability that the two procedures are equivalent (90% confidence interval for the difference in the true means falls between ?4.7 and +4.7) when in fact they are identical (the true mean difference is zero) is 4. Because the normal distribution was used in the above formula, 4 is actually a lower bound on the needed sample size. If feasible, one might want to use a larger sample size. Values for z for common confidence levels are presented in Table 8. The formula above makes three assumptions: 1) the variance used in the sample size calculation is based on a sufficiently large amount of prior data to be treated as known; 2) the prior known variance will be used in the analysis of the new

9

2

When testing equivalence, a 5% level test corresponds to a 90% confidence interval.

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