Using StatXact to calculate tolerance intervals in studies ... - IOS Press

Model Assisted Statistics and Applications 4 (2009) 117–122 DOI 10.3233/MAS-2009-0112 IOS Press

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Using StatXact to calculate tolerance intervals in studies with Poisson outcomes Boris G. Zaslavsky FDA, 1401 Rockville Pike HFM-219, Rockville, MD 20852, USA Tel.: +1 301 827 8587; Fax: +1 301 827 5218; E-mail: [email protected]

Abstract. The mathematical theory of tolerance intervals for distributions with Poisson outcomes is well developed. One criticism is the relative difficulty of computing the Poisson tolerance intervals, particularly when approximate formulae are not applicable. We offer the computational algorithm for the exact upper and lower tolerance intervals founded on the estimate of the exact confidence limits for the Poisson rate. Thus the exact tolerance intervals may be calculated using StatXact. To illustrate the algorithm, we apply it to the analysis of transfusion – associated deaths. Keywords: Clinical trial, confidence probability, Poisson distribution, StatXact, tolerance limit

1. Introduction Tolerance intervals are broadly used in industry and clinical studies [3,8,9,15,16,21]. “Tolerance intervals . . . summarize uncertainty about values of a random variable, usually a future observation” [1]. Even though tolerance intervals may not be mentioned in clinical protocols, they are frequently used de facto in validation and process control [5,7,10,20]. Tolerance intervals with Poisson outcomes may be used in bioequivalence studies with count number of outcomes. Tolerance intervals for normally distributed variables can be calculated using available statistical tables [2]. For random effect models with normally distributed random effects and errors, a procedure for constructing two-sided tolerance intervals was developed by [14]. For the exponentiated scale family of distributions, the asymptotic approximation is available and bootstrap-based tolerance limits are obtained by [19]. For binary data, a convenient computational algorithm is described by [24]. The author does not know any statistical software able to calculate the tolerance intervals for Poisson data. The mathematical theory of tolerance intervals for distributions with Poisson outcomes is well developed [13,22, 23]. The approximate computational methods of this theory have been addressed [17]. In [22,23], exact calculation of the tolerance limits was reduced to calculation of the inverse of the Chi-square distribution. The goal of this research is to develop simple to understand and easy to implement algorithms for calculating exact tolerance intervals for Poisson data. We offer the computational formulae for the exact upper and lower tolerance intervals for the number of events founded on the estimate of the exact confidence limits for the Poisson rate. The formulae for upper, lower, and two–sided tolerance intervals are given. The computation of tolerance limits is implemented in StatXact [6]. The auxiliary formulae are given to enable the computation of all admissible values of tolerance parameter. ISSN 1574-1699/09/$17.00  2009 – IOS Press and the authors. All rights reserved

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B.G. Zaslavsky / Using StatXact to calculate tolerance intervals in studies with Poisson outcomes

2. Theoretical background 2.1. Probability distribution of Poisson rate −λ

i

Let X be a random variables taking nonnegative integer values (i = 0,1, . . ., m) with the probability of e i! λ . m e−λ λi is a cumulative Then the distribution of X is Poisson. For any nonnegative number m, P (m, λ) = i=0 i! distribution function (C.D.F) of X with m number of events. Xi be M independent identically distributed Poisson random variables. Then the distribution of T M = Let M X i=1 i is Poisson with expectation Mλ [22,23]. We define P = {P (λ)|0 < λ < ∞} as the family of Poisson distributions with expectation λ. 2.2. Uniformly Most Accurate (U.M.A.) confidence limits for the rate ∞ −t m Let m be known. By differentiating P (m, λ) with respect to λ, it is easily seen that P (m, λ) = λ e m!t dt [12]. Consequently, P (m, ∞) = 0, P (m, 0) = 1, P (m, λ) is a monotone decreasing function of λ and a monotone increasing function of m. Thus 1 − P (m, λ) is a C.D.F. of λ. Let γ be a confidence limit (0 0.5. If β 0.5, the above procedures is not applicable. Below we present auxiliary formulae that overcome the restriction β > 0.5. The following formulae are convenient for β 0.5: b(β, γ) = the least non-negative integer n such that θ γ (M, m) θ¯1−β (N, n) (8) b(β, γ) = the largest non-negative integer n such that θ γ (M, m) θ 1−β (N, n)

(8’)

(See the proof in Appendix 5).

4. Application: Calculation of bioequivalence boundaries In bioequivalence studies (BE), a test and reference products are compared to verify that they will reach the same therapeutic effect or that they are therapeutically equivalent [4,11]. Consider a two-arm trial of M subjects in the control arm and of N subjects in the treatment arm. In this section we assume that the treatment criterion is the number of events in each arm. For example, it may be a number of bleeding episodes in the population of hemophilia patients. We will use tolerance limits to gauge the difference between the new treatment and the standard, approved one. For example, consider a situation in which there are two different haemostatic medications. The confidence interval is characterized by the confidence limit γ and the

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predetermined BE limit is characterized by the tolerance limit β . Let m events be observed among the subjects treated with an approved drug. The trial succeeds if the number of events in the test arm does not exceed the upper tolerance limit b(β, γ). For example, let M = N =18, γ = β = 0.95, and m = 18. From formula (5) follows that θ 0.95 (18, 18) = 1.483 < θ0.95 (18, 35 + 1) = 1.485. Thus the trial is successful if 35 or fewer events were observed in the treatment arm. In a superiority trial, the number of events in the test arm must be less than the lower tolerance limit b(β, γ). For example, from formula (6) follows that θ 0.95 (18, 18) θ 0.95 (18, 6 − 1). Thus the new treatment is superior if the number of events in the test arm is less than six. 5. Examples Example 1. The occurrence of transfusion – associated deaths has very small probability and the number of transfusions per year is very high. Therefore, the probability distribution of such an event would usually follow a Poisson distribution. Over three years from 1976 through 1978, the United States FDA received reports of 70 fatalities associated with transfusion [18]. To compare the safety standards of blood transfusion from 1979 through 1985 to the level of 1976–1978, we calculate the (95%, 95%) and (95%, 99%) upper tolerance limits for the period of 1976–1978. If the standards did not deteriorate, the number of deaths should be below the upper tolerance limit. Given γ = 0.95 (ˆ γ = 0.90) and β = 0.95 ( βˆ = 0.90), we have θ 0.95 (3, 70) = 28.47 < θ0.95 (1,39) = 29.33; if γ = 0.95 and β = 0.99 ( βˆ = 0.98) then θ 0.95 (3, 70) = 28.47 < θ0.99 (1,43) = 29.23. Thus the upper tolerance limits are 38 for (95%, 95%) and 42 for (95%, 99%). Comparing these limits with the reported data: 44 (1979), 46 (1980), 36 (1981), 32 (1982), 52 (1983), 37 (1984), 38 (1985), we can see that 1979, 1980, 1983 fatalities exceeded the most stringent (95%, 95%) tolerance limit. The 1980 and 1983 fatalities exceeded more liberal (95%, 99%) tolerance limit. Example 2. Over 4 years from 1976 trough 1979, 10 fatalities resulted from blood transfusion to a wrong person in a surgical suite. Over 6 years from 1980 trough 1985 this error happened 12 times [18]. To evaluate statistical significance of this improvement, we calculate the lower (95%, 95%) tolerance limit. Considering that θ0.95 (4, 10) = 1.356 > θ0.95 (6, 4 − 1) = 1.292, it follows that a statistically significant improvement would be 3 fatalities or less. Thus the observed improvement was not statistically significant. 6. Conclusion and discussion Although the theory of tolerance interval for Poisson data is well developed, the exact computational algorithm is not easily understood and implemented [22]. Our approach reduces the calculation of tolerance limits for the number of Poisson outcomes to the calculation of confidence intervals for the rates of Poisson events and resolving an inequality for these rates. In this form, the computational issues can be easily resolved by StatXact (CYTEL Software Corporation, 2001). For convenience of numerical calculations, we provide the formulae reducing the calculation of upper tolerance limits to the calculation of lower tolerance limits and vice versa. Acknowledgment I thank Tammy Massie and Ghanshyam Gupta for helpful comments and suggestions. Appendix 1 Proof of formula (5’). Following [22,23], the upper β, γ tolerance limit L(β, γ) for P is an integer exceeding a β -quantile of the Poisson distribution with confidence probability γ: Probabilityλ (L(β, γ) P −1 (β, N λ)) γ for all λ. The U.M.A. upper β, γ tolerance limit b(β, γ) is the smallest of all L(β, γ). Since P (n, N λ) is a decreasing function of λ and an increasing function of n, b(β, γ) is defined as (5’) [13, p. 126][22,23].


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Appendix 2 Proof of formula (5). Let k be a solution of inequality (5’). By definition (2), P (x − 1, N θ β (N, x)) = β for all nonnegative integers x and positive integers N . If x − 1 = k, from formula (5’) follows that P (k, N θ¯γ (M, m)) P (k, N θβ (M, k + 1)). Since P (k, N λ) is a monotone decreasing function of the second argument, formulae (5’) and (5) are equivalent.

Appendix 3 Proof of formula (6’). Following [22], the lower β, γ tolerance limit L(β, γ) for P is defined as Probabilityλ (L(β, γ) P −1 (β, N λ)) γ for all λ. The U.M.A. lower β, γ tolerance limit is the largest of all L(β, γ). Given λ, Probability(T k n) = 1 − P (n − 1, N λ) β is a probability bound on the smallest number of events in the future study. Thus from (2) follows that the U.M.A. β, γ lower tolerance limit b(β, γ) for P is defined by formula (6’) [13, p. 126].

Appendix 4 Proof of formula (6). Let us recall that P (k − 1, N θ β (N, k − 1)) = 1 − β for any positive integer k. Therefore, P (k − 1, N θγ (M, m) P (k − 1, N θβ (N, k − 1)). Since P (k − 1, N λ) is a monotone decreasing function of λ, we get Eq. (6).

Appendix 5 ˜ γ) = the least nonnegative integer k such Proof of formula (8). Let β˜ = 1 − β. By definition ¯b(1 − β, ˜ ¯ ¯ that P (k, N θγ (M, m)) 1 − β = P (k, N θβ˜(N, k)). Since P (n, N λ) is a monotone decreasing function of λ, θ¯γ (M, m) θ¯β˜ (N, k). ˜ γ) = the largest nonnegative integer n such Proof of formula (8’). β˜ = 1 − β. By definition b(1 − β, ˜ that 1 − P (n − 1, N θγ (M, m)) 1 − β. By formulae (1) and (2’), 1 − P (n − 1, N θ γ (M, m)) 1 − β˜ = 1 − P (n − 1, N θβ˜ (N, n)). Since P (n − 1, N λ) is a monotone decreasing function of λ, θ γ (M, m) θβ˜ (N, n). References [1] [2] [3] [4] [5]

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