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| Title: | 簡單順序假設波松母數較強檢定力檢定研究 -兩兩母均數差
More Powerful Tests for Simple Order Hypotheses in Poisson Distributions -The differences of the parameters |
| Authors: | 孫煜凱
Sun, Yu-Kai |
| Contributors: | 劉惠美
Liu, Hui-Mei
孫煜凱
Sun, Yu-Kai |
| Keywords: | 波松分配
條件檢定
精確性檢定
較強力檢定
信賴區間
蒙地卡羅法
Poisson distribution
Conditional test
Exact test
More powerful test
Confidence set
Monte carlo method |
| Date: | 2017 |
| Issue Date: | 2017-08-10 09:42:49 (UTC+8) |
| Abstract: | 波松分配（Poisson Distribution）常用在單位時間或是區間內，計算對有興趣之某隨機事件次數（或是已知事件之頻率），例如：速食餐廳的單位時間來客數，又或是每段期間內，某天然災害的發生次數，可以表示為某一特定事件X服從波松分配，若lambda為單位事件發生次數或是平均次數，我們稱lambda為此波松分配之母數，記作Poisson(lambda)，其中lambda屬於實數。
今天我們若想要探討由兩個服從不同波松分配抽取的隨機變數，如下列所述：令X={(X1,X2)}為一集合，其中Xi為X(i,1),X(i,2),...,X(i,ni)~Poisson(lambda(i)),i=1,2。欲探討兩波松分配之均數是否相同或相差小於某個常數d時，考慮以下檢定:H0:lambda2-lambda1<=d與H0:lambda2-lambda1>d，對於此問題可以使用的檢定方法有Przyborwski和Wilenski(1940)提出的條件檢定（Conditional test,C-test）或K.Krishnamoorthy與Jessica Thomson(2002)提出的精確性檢定（Exact test,E-test），其中的精確性檢定為一個非條件檢定（Unconditional Test）；K.Krishnamoorthy與Jessica Thomson比較條件檢定與精確性檢定的p-value皆小於顯著水準(apha)，而精確性檢定的檢定力不亞於條件檢定，因此精確性檢定比條件檢定更適合上面所述之假設問題。
Roger L.Berger（1996）提出一個以信賴區間的p-value所建立的較強力檢定，而目前只用於檢定兩二項分配（Binomial Distribution）的機率參數p是否相同為例，然而Berger在文中提到，較強力檢定比非條件檢定有更好的檢定力，而且要求的計算時間較少，可以提升檢定的效率。
本篇論文我們希望在固定apha與d時檢定的問題，建立一個兩波松分配均數顯著水準為apha的較強力檢定。
利用Roger L.Berger與Dennis D.Boos(1994)提出以信賴區間的p-value方法，建立波松分配兩兩母均數差的較強力檢定；研究發現此較強力檢定與精確性檢定的p-value皆小於apha，然而我們的檢定的檢定力皆不亞於精確性檢定所計算得出的檢定力，然而其apha及虛無假設皆需要善加考慮以本篇研究來看，當檢定為單尾檢定時，若apha<0.01，我們的較強力檢定沒有辦法找到比精確性檢定更好地拒絕域，換言之，此時較強力檢定與精確性檢定的檢定力將會相等。
Poisson Distribution is used to calculate the probability of a certain phenomenon which attracted by researcher. If we want to test two random variable in an experiment .Therefore ,let X={(X1,X2)} be independent samples ,respectively ,from Poisson distribution ,also X(i,1),X(i,2),...,X(i,ni)~Poisson(lambda(i)),i=1,2.
The problem of interest here is to test:
H0:lambda2-lambda1<=d and H0:lambda2-lambda1>d,
where 0<apha<1/2 ,and let Y1 equals sum of X1 and Y2 equals sum of X2, where apha ,lambda,d be fixed.
In this problem of hypothesis testing about two Poisson means is addressed by the conditional test.However ,the exact method of testing based on the test statistic considered in K.Krishnamoorthy,Jessica Thomson(2002) also commonly used.
Roger L.Berger ,Dennis D.Boos(1994) give a new way to calculate
p-value,which replace the old method ,called it a valid p-value .In 1996, Roger L.Berger used the new way to propose a new test for two parameter of binomial distribution which is more powerful than exact test. In the other hand, Roger L.Berger also explain the unconditional test is more suitable than the conditional test.
In this paper,we propose a new method for two parameter of Poisson distribution which revise from Roger L.Berger’s method. The result we obtain that our new test is really get a much bigger rejection region.We found when the fixed increasing ,the set of more powerful test increasing, and when the fixed power increasing ,the required sample size decreasing. |
| Reference: | 1. C.J Clopper,E.S.Pearson(1934)“The Use of Confidence or Fiducial Limits Illustrated in the Case of the Binomial”.
2. K.Krishnamoorthy,Jessica Thomson(2002),“A more powerful test for comparing two Poisson means”.
3. K.Krishnamoorthy,Jessica Thomson(2002),“Hypothesis Testing About Proportions in Two Finite Populations”.
4. Roger L. Berger (1989),“Uniformly more powerful tests for hypotheses concerning linear inequalities and normal means”, Journal of the American Statistical Association,84(405), 192-199.
5. Roger L. Berger ,Dennis D.Boos(1994)“P Values Maximized Over a Confidence Set for the Nuisance Parameter”.
6. Roger L. Berger(1994)“Power Comparison of Exact Unconditional Tests for Comparing Two Binomial Proportions”.
7. Roger L. Berger(1996),“More Powerful Test From Confidence Interval p values”. |
| Description: | 碩士
國立政治大學
統計學系
104354020 |
| Source URI: | http://thesis.lib.nccu.edu.tw/record/#G0104354020 |
| Data Type: | thesis |
| Appears in Collections: | [統計學系] 學位論文
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