國
立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
Chapter 6
Concluding Remarks
In this study, we investigate several asymptotic and exact statistical pro-cedures for comparing two Poisson means in identifying the superiority and non-inferiority. Two types of Wald test are considered, and they give different forms with respect to the superiority and the non-inferiority test respectively.
The asymptotic power functions of the asymptotic procedures are derived and the correspondent asymptotic sample size formula are provided in the two testing problems. Consequently, the two asymptotic tests are compared in terms of the asymptotic power function and the required sample size. One concludes that the performances of the tests depend on the fraction of the group sizes. Moreover, the trends of asymptotic power function of testing superiority are consistent with that of testing non-inferiority. In this study, an exact test does not mean the use of the conventional p-value, which is de-noted as the standard p-value in Chapter 2. The test is exact in the sense that the calculation of the 𝑝-value or is based on the exact sampling distribution of the test statistic. In fact, in the Poisson problem, the calculation of the exact standard 𝑝−value is rather difficult because the null parameter space is unbounded. Two alternative procedures, in which the computation of a
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
p-value is taken either over a bounded space or a single point, are considered and proposed. The exact procedures under investigation are the confidence-set 𝑝-value and the estimated 𝑝-value. The definition and the computation of the exact 𝑝-values are introduced in details. The correspondent exact sam-ple sizes for power requirement are shown to be found numerically. In this study, intensive numerical studies are provided and it is concluded that the asymptotic tests tend to have inflated type I error rates. On the contrary, the exact procedures have adequate performance overall, and dominate the asymptotic tests. Moreover, the quick solutions based on the asymptotic sample size formulae are found to provide good approximations to the exact sample sizes for testing superiority or non-inferiority.
The confidence-interval 𝑝-value is the sum of the supremum over a 100(1−
𝛾)% confidence region of the nuisance parameter(s) and 𝛾. In which, the figure 𝛾 must be far less than the nominal level 𝛼. If not, the resultant 𝑝-value is easy to exceed 𝛼, and the test tends to give an insignificant conclusion.
The testing procedure becomes powerless and is meaningless. Further, from Table 4.1 and 4.2, we find that the confidence interval 𝑝-values with 𝛾 = 0.001 are more powerful than that with 𝛾 = 0.005 . It seems that the choice of 𝛾 affects the performance of the testing procedure. It is worthy to have more intensive investigations on the effect of 𝛾 in future study.
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
Bibliography
[1] Barnard, G. A.(1947) Significance Test for 2 × 2 Tables. Biometrika, 34, 123-138.
[2] Berger, R. L. and Boos, D. D.(1994) 𝑃 Values Maximized Over a Confi-dence Set for the Nuisance Parameter, Journal of the American Statistical Association, 89, 1012-1016.
[3] Casella, G. and Berger, R. L.(1990) Statistical Inference. Pacific Grove, CA: Wadsworth.
[4] Corinna, M. and Jochen, M. C.(2005) Power Calcualtion for Non-inferiority Trials Comparing Two Poisson Distributions. SAS Phuse Pa-pers , http://www.lexjansen.com/Phuse/2005/pk/pk01.pdf.
[5] Gail, M.(1974) Power Computations for Designing Comparative Poisson Trails. Biometrics, 30, 231-237.
[6] Gu, K., Ng, H. K., Tang, M. L. and Schucany, W. R.(2008) Testing the Ratio of Two Poisson Rates. Biometrical Journal, 50, 283-298.
[7] Krishnamoorthy, K. and Thomson, J.(2004) More Power Test for Two Poisson Means. Journal of Statistical Planning and Inference, 119, 23-35.
[8] Lehmann, E. L.(1986) Testing Statistical Hypotheses, 2nd edition, Wi-ley, New York.
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
[9] Lui, K. J.(2005) Sample Size Calculation for Testing Non-inferiority and Equivalence Under Poisson Distribution. Statistical Methodology, 2, 37-48.
[10] Ng, H. K. and Tang, M. L.(2005) Testing The Equality of Two Poisson Means Using The Rate Ratio. Satistics in Medicine, 24, 955-965.
[11] Przyborowski, H. and Wilenski, H.(1940) Homogeneity of Results in Testing Samples from Poisson Series. Biometrika, 31, 313-323.
[12] Pirie, W. R. and Hamdan, M. A.(1972) Some Revised Continuity Cor-rections for Discrete Distributions. Biometrics, 28, 3, 693-701.
[13] R¨ohmel, J and Mansmann, U.(1990) Unconditional Non-Asymptotic One-Sided Tests for Independent Binomial Proportions When the Interest Lies in Showing Non-Inferiority and/or Superiority. Biometrical Journal, 41, 149-170.
[14] Shiue, W. and Bain, L. J.(1982) Experiment Size and Power Compar-isons for Two-Sample Poisson Tests. Applied Statistics, 31, 130-134.
[15] Storer, B. E. and Kim, C.(1990) Exact Properties of Some Exact Test Statistics for Comparing Two Binomial Proportions. Journal of the Amer-ican Statistical Association, 85, 409, 146-155.
[16] Song, J. X.(2009) Sample Size for Simultaneous Testing of Rate Differ-ences in Non-inferiority Trials With Multiple Endpoints. Computational Statistics and Data Analysis, 53, 1201-1207.
[17] Thode, H. C.(1997) Power and Sample Size Requirements for Tests of Differences Between Two Poisson Rates. The Statistican, 46, 227-230.
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
Appendix
A.1
Theorem 1. Let 𝛿0 be the true value of 𝛿, and 𝜌 = 𝑛1/𝑛2 ∈ (0, 1) be the sample size fraction of the first group to the second group. As 𝑛1, 𝑛2 → ∞,
𝑍𝑅𝜎 − 𝜇 → 𝑁 (0, 1) and 𝑍𝑑 𝑈− 𝜇 → 𝑁 (0, 1).𝑑
Under 𝜆1 = 𝜆2 = 𝜆, define the testing statistic 𝑍𝑅=
𝑌¯1− ¯𝑌2
√𝜆˜0(𝑛1
1 + 𝑛1
2) ,
where ˜𝜆0 = 𝑌𝑛1+𝑌2
1+𝑛2. By C. L. T, we have that 𝑍𝑅= 𝑌¯1− ¯𝑌2
√𝜆1
𝑛1 + 𝜆𝑛2
2
= 𝑌¯1− ¯𝑌2
√ 𝜆(𝑛1
1 +𝑛1
2)
→ 𝑁 (0, 1).𝑑
And,
𝑌1 𝑛1
= ˆ𝜆1 → 𝜆,𝑝 𝑌2 𝑛2
= ˆ𝜆2 → 𝜆,𝑝 𝑌1+ 𝑌2
𝑛1+ 𝑛2
→ 𝜆,𝑝
√𝑌1+𝑌2
𝑛1+𝑛2
√𝜆
→ 1.𝑝
‧
Therefore, the 𝑍𝑅 is valid.
Alternative, if 𝜆1 > 𝜆2 is true, the 𝛿0 = 𝜆1− 𝜆2 is defined. The following
‧
Subsequently, we can find that
𝑌¯1− ¯𝑌2−𝛿0
Given 𝜌, 𝛿0 and 𝜆2, the required sample sizes of the second group satisfied the power is greater than 1 − 𝛽0 is
‧
that is, the type I error is controlled at 𝛼 asymptotically.
Under 𝛿 = 𝛿0, the variance of the MLE of 𝛿, is given by 𝜎2ˆ
𝛿 = 𝜆1 𝑛1 + 𝜆2
𝑛2,
since ¯𝑌1 and ¯𝑌2 are independent. Then the estimated standard error satisfies 𝑠𝑒(ˆ𝛿)
𝜎𝛿ˆ
→ 1, provided 𝜆𝑝 1, 𝜆2 > 0.
Then according to the asymptotic normality of the MLE, Z’s easily derived that for
𝑍𝑈 − 𝛿0 𝜎2ˆ
𝛿
→ 𝑁 (0, 1).𝑑
Hence the asymptotic power of Z can be derived too as 𝛽¯𝑍𝑈(𝛿0,𝜆1,𝜌,𝑛2) = 𝑃 (𝑍𝑈 ≥ 𝑧𝛼 ∣ 𝛿 = 𝛿0)
the required size of the second group should satisfies 𝑧𝛼− 𝛿0
√𝜆2(1+𝜌)+𝛿0
𝑛2𝜌
≤ −𝑧𝛽0.
‧
The necessary asymptotic sample size is thus 𝑛∗2,𝑍𝑈 ≥( 𝑧𝛼+ 𝑧𝛽0
The sampling distribution of√
𝑛2( ¯𝑌1− ¯𝑌2) can be derived straight forward
‧
and the polled variance estimate is,𝑠2𝑝 = (𝑛1− 1)𝑠21+ (𝑛2− 1)𝑠22
By Slustky Theorem, the T-test statistic has an asymptotical standard nor-mal distribution, 𝑧𝛼. Consequently, the T-test has asymptotical level 𝛼, i.e.
𝑃 (𝑇 ≥ 𝑡(𝑛1+𝑛2−2,𝛼)∣ 𝐻0) ≈ 𝑃 (𝑍 ≥ 𝑧𝛼) ≤ 𝛼.
‧
Then the asymptotic power of the T-test can be derived as, 𝛽¯𝑇(𝛿0, 𝜆2, 𝜌, 𝑛2) = 𝑃 (𝑇 ≥ 𝑡(𝑛1+𝑛2−2,𝛼) ∣ 𝛿0) the required size of the second group should satisfies
𝑧𝛼
‧
The necessary asymptotic sample size is thus
𝑛∗2 ≥
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
and (𝑡1, 𝑡2) ∈ 𝐴. Multiplying (2) by −1, we have −𝑞𝑡1+ 𝑝𝑡2 = −1. Adding the left hand side of the equation by 𝑞𝑝(𝑡1+ 𝑡2) − 𝑝𝑞(𝑡1 + 𝑡2), then
𝑞(𝑡1(𝑝 − 1) + 𝑡2𝑝) − 𝑝(𝑡2(𝑞 − 1) + 𝑡1𝑞) = −1, or 𝑞𝑡′1− 𝑝𝑡′2 = −1, (3) where 𝑡′1 = 𝑡1(𝑝 − 1) + 𝑡2𝑝 > 0 and 𝑡′2 = 𝑡2(𝑞 − 1) + 𝑡1𝑞 > 0, and (𝑡′1, 𝑡′2) ∈ 𝐴.
That is, for any 𝑝, 𝑞, such that (𝑝, 𝑞) = 1, there exist (𝑡1, 𝑡2), (𝑡′1, 𝑡′2) ∈ 𝐴 such that (2), (3) are true.
Similarly, when 𝑠1 < 0 and 𝑠2 > 0, we can find (𝑡∗1, 𝑡∗2) and (𝑡∗∗1 , 𝑡∗∗2 ) in 𝐴 such that 𝑞𝑡∗1− 𝑝𝑡∗2 = 1 and 𝑞𝑡∗∗1 − 𝑝𝑡∗∗2 = −1. Hence, (𝑞∑ 𝑌1𝑖− 𝑝∑ 𝑌2𝑖) has positive mass at 1, −1 and their multiples. The support is the set of integers and the space 𝑏 = 1. So, ˆ𝛿 has equal spacings with 𝑏 = 𝑚1. The continuity corrected 𝑍-test and 𝑇 -test are
𝑍𝐶 =
ˆ𝛿 − 2𝑚1
𝑠𝑒(ˆ𝛿) , 𝑇𝐶 =
𝛿 −ˆ 2𝑚1 𝑠𝑝√
1 𝑛1 +𝑛1
2
.
A.4
Theorem 4. Let 𝐶𝛾∗ = 𝐶𝛾,0∩ Ω02 be the truncated confidence set. Then 𝑃 ((𝜆1, 𝜆2) ∈ 𝐶𝛾∗ ∣ 𝜆1, 𝜆2) ≥ 1 − 𝛾, for all (𝜆1, 𝜆2) ∈ Ω02.
First, we express 𝐶𝛾∗ as the following form,
𝐿1 ≤ 𝜆1 ≤ min(𝑈1, 𝜆2), 𝐿2 ≤ 𝜆2 ≤ 𝑈2.
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
Note that the two intervals are build on two independent statistics. Then, at any (𝜆1, 𝜆2) such that 𝜆1 ≤ 𝜆2,
𝑃 (𝐿1 ≤ 𝜆1 ≤ min(𝑈1, 𝜆2), 𝐿2 ≤ 𝜆2 ≤ 𝑈2∣𝜆1, 𝜆2)
= 𝑃 (𝐿1 ≤ 𝜆1 ≤ min(𝑈1, 𝜆2)∣𝜆1, 𝜆2)𝑃 (𝐿2 ≤ 𝜆2 ≤ 𝑈2∣𝜆1, 𝜆2).
And
𝑃 (𝐿1 ≤ 𝜆1 ≤ min(𝑈1, 𝜆2)∣𝜆1, 𝜆2)
= 𝑃 (𝐿1 ≤ 𝜆1 ≤ 𝑈1, 𝑈1 < 𝜆2∣𝜆1, 𝜆2) + 𝑃 (𝐿1 ≤ 𝜆1 ≤ 𝜆2, 𝑈1 ≥ 𝜆2∣𝜆1, 𝜆2)
= 𝑃 (𝐿1 ≤ 𝜆1 ≤ 𝑈1, 𝑈1 < 𝜆2∣𝜆1, 𝜆2) + 𝑃 (𝐿1 ≤ 𝜆1, 𝑈1 ≥ 𝜆2∣𝜆1, 𝜆2).
Since under Ω02, 𝜆1 ≤ 𝜆2, {𝑈1 ≥ 𝜆2} is a subset of {𝑈1 ≥ 𝜆1}, and 𝑃 (𝐿1 ≤ 𝜆1, 𝑈1 ≥ 𝜆2∣𝜆1, 𝜆2)
= 𝑃 (𝐿1 ≤ 𝜆1, 𝑈1 ≥ 𝜆1, 𝑈1 ≥ 𝜆2∣𝜆1, 𝜆2)
= 𝑃 (𝐿1 ≤ 𝜆1 ≤ 𝑈1, 𝑈1 ≥ 𝜆2∣𝜆1, 𝜆2).
Consequently, for 𝜆1 ≤ 𝜆2,
𝑃 (𝐿1 ≤ 𝜆1 ≤ min(𝑈1, 𝜆2)∣𝜆1, 𝜆2) = 𝑃 (𝐿1 ≤ 𝜆1 ≤ 𝑈1∣𝜆1, 𝜆2), and
𝑃 (𝐿1 ≤ 𝜆1 ≤ min(𝑈1, 𝜆2), 𝐿2 ≤ 𝜆2 ≤ 𝑈2∣𝜆1, 𝜆2)
= 𝑃 (𝐿1 ≤ 𝜆1 ≤ 𝑈1∣𝜆1, 𝜆2)𝑃 (𝐿2 ≤ 𝜆2 ≤ 𝑈2∣𝜆1, 𝜆2)
≥ 1 − 𝛾.
A.5
Theorem 5. Let 𝑆 be a test statistic that depends on the data only through the two sufficient statistics (𝑌1, 𝑌2) in comparing two Poisson means. Suppose
‧
𝑆 satisfies the convexity condition. Then given 𝑠0, the supremum of 𝑃 (𝑆 ≥ 𝑠0∣𝜆1, 𝜆2) occurs at a boundary point of the parameter space.
Consider the probability function of the Poisson distribution, 𝑝𝑜𝑖(𝑥∣𝜆), then
∂
∂𝜆𝑃 (𝑋∣𝜆) = 𝑃 (𝑋 − 1∣𝜆) − 𝑃 (𝑋∣𝜆), for 𝑥 = 1, 2, ⋅ ⋅ ⋅ .
Given one the test statistic 𝑆 and one observation (𝑦10, 𝑦20), then the p-value is
‧
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
and
∂𝑃𝑆(𝜆1, 𝜆2)
∂𝜆1
= (𝑝𝑜𝑖(𝑦1 = 𝑎 − 1∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎∣𝜆1)) ∑
𝑦2≤ℎ∗(𝑎)
𝑝𝑜𝑖(𝑦2∣𝜆2) +(𝑝𝑜𝑖(𝑦1 = 𝑎∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎 + 1∣𝜆1)) ∑
𝑦2≤ℎ∗(𝑎+1)
𝑝𝑜𝑖(𝑦2∣𝜆2) +(𝑝𝑜𝑖(𝑦1 = 𝑎∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎 + 1∣𝜆1)) ∑
𝑦2≤ℎ∗(𝑎+1)
𝑝𝑜𝑖(𝑦2∣𝜆2) + ⋅ ⋅ ⋅
+(𝑝𝑜𝑖(𝑦1 = 𝑁1∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑁1∣𝜆1)) ∑
𝑦2≤ℎ∗(𝑁1)
𝑝𝑜𝑖(𝑦2∣𝜆2)
= (𝑝𝑜𝑖(𝑦1 = 𝑎 − 1∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎∣𝜆1))𝑝𝑜𝑖(𝑦2 = 0∣𝜆2)
+(𝑝𝑜𝑖(𝑦1 = 𝑎∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎 + 1∣𝜆1))(𝑝𝑜𝑖(𝑦2 = 0∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 1∣𝜆2)) +(𝑝𝑜𝑖(𝑦1 = 𝑎 + 1∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎 + 2∣𝜆1))
(𝑝𝑜𝑖(𝑦2 = 0∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 1∣𝜆2) + 𝑝𝑜𝑖(𝑦=2∣𝜆2)) +(𝑝𝑜𝑖(𝑦1 = 𝑎 + 2∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑎 + 3∣𝜆1))
(𝑝𝑜𝑖(𝑦2 = 0∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 1∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 2∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 3∣𝜆2)) + ⋅ ⋅ ⋅
+(𝑝𝑜𝑖(𝑦1 = 𝑁1− 1∣𝜆1) − 𝑝𝑜𝑖(𝑦1 = 𝑁1∣𝜆1))
(𝑝𝑜𝑖(𝑦2 = 0∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 1∣𝜆2) + 𝑝𝑜𝑖(𝑦2 = 2∣𝜆2) + ⋅ ⋅ ⋅ + 𝑝𝑜𝑖(𝑦2 = ℎ∗(𝑁1)))
=
𝑁1
∑
𝑦1=𝑎
(𝑝𝑜𝑖(𝑦1∣𝜆1) − 𝑝𝑜𝑖(𝑁1∣𝜆1))𝑝𝑜𝑖(𝑦2 = ℎ∗(𝑦1)∣𝜆2)
> 0,
where 𝑁1 → ∞, such that 𝑝𝑜𝑖(𝑁1∣𝜆1) .
= 0. Moreover, the space of the null hypothesis 𝐻02 is a compact set, then the supremum of the 𝑃𝜃 is maximum can be shown in the Poisson problem.
‧
Theorem 6. 𝑍𝑅, 𝑍𝑈 satisfy the convexity condition.
For the test statistic
‧
‧
國立 政 治 大 學
‧
N a tio na
l C h engchi U ni ve rs it y
hence 𝑍𝑅is increasing in 𝑌1 and decreasing in 𝑌2, then it can be provided 𝑍𝑅(𝑌1, 𝑌2) ≤ 𝑍𝑅(𝑌1+ 1, 𝑌2) and 𝑍𝑅(𝑌1, 𝑌2) ≤ 𝑍𝑅(𝑌1, 𝑌2− 1), hence the 𝑍𝑅
satisfies the convexity condition.
A.7
The derivation of restricted MLE of 𝜆1 and 𝜆2 on 𝐻03.
The constrained MLE maximizes the following likelihood function, 𝐿(𝜆1, 𝜆2) = 𝑌1ln 𝜆1− 𝑛1𝜆1+ 𝑌2ln 𝜆2− 𝑛2𝜆2, subject to 𝜆1 = 𝜆2− Δ0.
The likelihood function can be rewritten to as the following function of 𝜆2,
𝐿(𝜆2) = 𝑌1ln(𝜆2− Δ0) − 𝑛1(𝜆2− Δ0) + 𝑌2ln 𝜆2− 𝑛2𝜆2,
taking the derivative of the likelihood function 𝐿(𝜆2) with respect to 𝜆2, we have
∂𝐿(𝜆2)
∂𝜆2 = 𝑌1
𝜆2− Δ0 − 𝑛1+ 𝑌2
𝜆2 − 𝑛2 = 0.
The RMLE of 𝜆2 satisfies
(𝑛1+ 𝑛2)𝜆22− [(𝑛1+ 𝑛2)Δ0+ 𝑌1+ 𝑌2]𝜆2+ 𝑌2Δ0 = 0, and hence we have possible multiple solutions of 𝜆2,
[(𝑛1+ 𝑛2)Δ0 + 𝑌1+ 𝑌2] ±√[(𝑛1+ 𝑛2)Δ0+ 𝑌1+ 𝑌2]2− 4(𝑛1+ 𝑛2)𝑌2Δ0
2(𝑛1+ 𝑛2) ,
The solution with negative squared term leads to a negative RMLE of 𝜆1 and thus is not a valid RMLE. In the following, we have the RMLEs of 𝜆2 and 𝜆1 on 𝜆1 = 𝜆2− Δ0. Define
ˆ𝜆1 = 𝑌1
𝑛1, ˆ𝜆2 = 𝑌2
𝑛2, ˜𝜆0 = 𝑌1+ 𝑌2
𝑛1+ 𝑛2 = 𝜌 1 + 𝜌
ˆ𝜆1+ 1 1 + 𝜌
ˆ𝜆2.
‧
And the testing statistic is defined as follows,𝑍𝑅∗ = 𝑌¯1− ¯𝑌2 + Δ0
√˜
𝜆1
𝑛1 + ˜𝜆𝑛2
2
‧
Hence, we have as follows
𝑍𝑅∗ = 𝑌¯1− ¯𝑌2+ Δ0
‧
Therefor, we can derive that
𝑃 (𝑍𝑅∗ ≥ 𝑧𝛼∣𝜆1 = 𝜆2− Δ0) ≤ 𝛼, following the limit converge form (Jun Shao, 1998)
˜𝜆1 → 𝑞𝑝 1(𝜆1, 𝜆2), ˜𝜆2 → 𝑞𝑝 2(𝜆1, 𝜆2),
Then, the limit distribution can be derived as follows
𝑌¯1− ¯𝑌2+Δ0−𝛿∗0
‧
Next, the above equation can be rewritten as follows
𝑌¯1− ¯𝑌2+Δ0−𝛿∗0
Similarly, the asymptotic distribution of 𝑍𝑈∗ can be derived as follows:
𝑍𝑈∗− 𝛿0∗
‧
‧
Similarly, the power function of 𝑍𝑈∗ be can fund as follows:
𝛽¯𝑍𝑈 ∗(𝛿∗0, 𝜆2, 𝑛2, 𝜌, Δ0) = 1 − Φ(𝑧𝛼− 𝜇∗),
Theorem 8. 𝑍𝑅∗ satisfy the convexity condition.
Firstly, we execute the partial derivative of 𝑍𝑅∗ wrt ˆ𝜆1:
‧
‧
The following can be derived,
˜𝜆1
‧
And, we have follows as,
√