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STUDIES IN STATISTICAL INFERENCE, SAMPLING TECHNIQUES AND DEMOGRAPHY (cid:2) (cid:2) Rajesh Singh, Jayant Singh, Florentin Smarandache (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) ILQ 2009 1 This bbook can be ordered in a paper bound reprint from: BBooks on Demand ProQuest Information & Learning (University of Microfilm International) 300 N. Zeeb Road P.O. Box 1346, Ann Arbor MI 48106-1346, USA (cid:2)(cid:2) Copyright 2009 by InfoLearnQuest (Ann Arbor) and the Authors Many books can be downloaded from the following Digital Library of Science: http://www.gallup.unm.edu/~smarandache/eBooks(cid:3)otherformats.htm(cid:2)(cid:2)(cid:2)(cid:2) Peer Reviewers: Prof.(cid:2)Mihaly(cid:2)Bencze,(cid:2)Department(cid:2)of(cid:2)Mathematics,(cid:2)Áprily(cid:2)Lajos(cid:2)College,(cid:2)Bra(cid:3)ov,(cid:2)Romania.(cid:2) Dr.(cid:2)Sukanto(cid:2)Bhattacharya,(cid:2)Department(cid:2)of(cid:2)Business(cid:2)Administration,(cid:2)Alaska(cid:2)Pacific(cid:2) University,(cid:2)U.S.A.(cid:2) (cid:2) (cid:2) (ISBN-10): 1-59973-087-1 (ISBN-13): 978-1-59973-087-5 (EAN): 9781599730875 Printed in the United States of America 2 Contents Preface: 4 1 Optimum Statistical Test Procedure: 5 2 A Note on Testing of Hypothesis: 21 3 Improvement in Estimating Population Mean using Two Auxiliary Variables in Two-Phase Sampling: 26 4 Improved Exponential Estimator for Population Variance Using Two Auxiliary Variables: 36 5 Structural Dynamics Of Various Causes Of Migration In Jaipur: 45 3 Preface This volume is a collection of five papers. Two chapters deal with problems in statistical inference, two with inferences in finite population, and one deals with demographic problem. The ideas included here will be useful for researchers doing works in these fields. The following problems have been discussed in the book: Chapter 1. In this chapter optimum statistical test procedure is discussed. The test procedures are optimum in the sense that they minimize the sum of the two error probabilities as compared to any other test. Several examples are included to illustrate the theory. Chapter 2. In testing of hypothesis situation if the null hypothesis is rejected will it automatically imply alternative hypothesis will be accepted? This problem has been discussed by taking examples from normal distribution. Chapter 3. In this section improved chain-ratio type estimator for estimating population mean using some known values of population parameter(s) has been discussed. The proposed estimators have been compared with two-phase ratio estimator and some other chain ratio type estimators. Chapter 4. In this section we have analysed exponential ratio and exponential product type estimators using two auxiliary variables are proposed for estimating unknown population variance S2 . Problem is extended to the case of two-phase sampling. y Chapter 5. In this section structural dynamics of various causes of migration in Jaipur was analysed. Reasons of migration from rural to urban areas and that of males and females are studied. (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) The Authors (cid:2) 4 Optimum Statistical Test Procedure Rajesh Singh Department of Statistics, BHU, Varanasi (U.P.), India [email protected] Jayant Singh Department of Statistics Rajasthan University, Jaipur, India [email protected] Florentin Smarandache Chair of Department of Mathematics, University of New Mexico, Gallup, USA [email protected] Introduction Let X be a random variable having probability distribution P(X/(cid:2)), (cid:2)(cid:4)(cid:3). It is desired to test H :(cid:2)(cid:4)(cid:3) against H :(cid:2)(cid:4)(cid:3) (cid:6)(cid:3)(cid:5)(cid:3) . Let S denote the sample space of 0 0 1 1 0 outcomes of an experiment and x(cid:6)(x ,x ,(cid:5)(cid:5)(cid:5),x )denote an arbitrary element of S. A test 1 2 n procedure consists in diving the sample space into two regions W and S – W and deciding to reject H if the observed x(cid:4)W. The region W is called the critical region. The function 0 (cid:7)((cid:2))(cid:6)P ( x(cid:4)W) = P (W), say, is called the power function of the test. (cid:2) (cid:2) 5 We consider first the case where (cid:3)0 consists of a single element, (cid:2)0 and its complement (cid:3) also has a single element (cid:2) . We want to test the simple hypothesis 1 1 H : (cid:2)(cid:6)(cid:2) against the simple alternative hypothesis H : (cid:2)(cid:6)(cid:2) . 0 0 1 1 Let L = L(X/H) and L = L(X/H) be the likelihood functions under H and H 0 0 1 1 0 1 respectively. In the Neyman – Pearson set up the problem is to determine a critical region W such that (cid:7)((cid:2)0)(cid:6)P(cid:2)0(W)(cid:6)(cid:9)L0dx(cid:6)(cid:8), an assigned value (1) w and (cid:7)((cid:2)1)(cid:6)P(cid:2)1(W)(cid:6)(cid:9)L1dx is maximum (2) w compared to all other critical regions satisfying (1). If such a critical region exists it is called the most powerful critical region of level (cid:2). By the Neyman-Pearson lemma the most powerful critical region W0 for testing (cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:5)(cid:3) against (cid:2)(cid:7)(cid:4)(cid:5)(cid:6)(cid:5)(cid:7) is given by W (cid:6){x:L(cid:10)kL } 0 1 0 provided there exists a k such that (1) is satisfied. For this test (cid:7)((cid:2) )(cid:6)(cid:8) and (cid:7)((cid:2) )(cid:11)1 as n (cid:11)(cid:8). 0 1 But for any good test we must have (cid:7)((cid:2) )(cid:11)0 and (cid:7)((cid:2) )(cid:11)1 as n (cid:11)(cid:12) because complete 0 1 discrimination between the hypotheses H and H should be possible as the sample size 0 1 becomes indefinitely large. 6 Thus for a good test it is required that the two error probabilities (cid:8) and (cid:13) should depend on the sample size n and both should tend to zero as n (cid:11)(cid:12). We describe below test procedures which are optimum in the sense that they minimize the sum of the two error probabilities as compared to any other test. Note that minimizing ((cid:8)(cid:14)(cid:13)) is equivalent to maximising 1 - ((cid:8)(cid:14)(cid:13)) = (1(cid:5)(cid:13))(cid:5)(cid:8) = Power – Size. Thus an optimum test maximises the difference of power and size as compared to any other test. Definition 1 : A critical region W will be called optimum if 0 (cid:9)Ldx(cid:5) (cid:9)L dx(cid:10)(cid:9)Ldx(cid:5)(cid:9)L dx (3) 1 0 1 0 w0 w0 w w for every other critical region W. Lemma 1: For testing H0: (cid:2)(cid:6)(cid:2)0 against H1 : (cid:2)(cid:6)(cid:2)1 the region W (cid:6){x:L (cid:10)L } is optimum. 0 1 0 Proof : W is such that inside W , L (cid:10)L and outside W , L (cid:15)L . Let W be any other 0 0 1 0 0 1 0 critical region. (cid:9)(L (cid:5)L )dx(cid:5)(cid:9)(L (cid:5)L )dx 1 0 1 0 W0 W = (cid:9) (L (cid:5)L )dx(cid:5) (cid:9)(L (cid:5)L )dx, 1 0 1 0 W0(cid:2)Wc W(cid:2)W0c by subtracting the integrals over the common region W (cid:2) W. 0 7 (cid:10) 0 since the integrand of first integral is positive and the integrand under second integral is negative. Hence (3) is satisfied and W is an optimum critical region. 0 Example 1 : Consider a normal population N((cid:2),(cid:16)2)where (cid:16)2 is known. It is desired to test H0: (cid:2)(cid:6)(cid:2)0 against H1 : (cid:2)(cid:6)(cid:2)1 , (cid:2)1 (cid:17)(cid:2)0. L( x (cid:2))= (cid:22)(cid:22)(cid:24) 1 (cid:19)(cid:19)(cid:21)ne(cid:5)i(cid:18)(cid:6)n1(x2i(cid:16)(cid:5)(cid:2)2)2 (cid:23)(cid:16) 2(cid:25)(cid:20) (cid:18)(x (cid:5)(cid:2) )2 (cid:5) i 1 L e 2(cid:16)2 1 (cid:6) L0 (cid:5)(cid:18)(xi(cid:5)(cid:2)0)2 e 2(cid:16)2 The optimum test rejects H o L if 1 (cid:10)1 L 0 L i.e. if log 1 (cid:10)0 L 0 (cid:18)(x (cid:5)(cid:2) )2 (cid:18)(x (cid:5)(cid:2) )2 i.e. if (cid:5) i 1 (cid:14) i 0 (cid:10)0 2(cid:16)2 2(cid:16)2 i.e. if 2(cid:2) (cid:18)x (cid:5)n(cid:2)2(cid:5)2(cid:26) (cid:18)x (cid:14)n(cid:2)2 (cid:10)0 1 i 1 0 i 0 8 (cid:27) (cid:28) i.e. if (cid:27)2(cid:2) (cid:5)2(cid:2) (cid:28)(cid:18)x (cid:10)n(cid:2)2(cid:5)(cid:2)2 1 0 i 1 0 (cid:18) x (cid:2) (cid:14)(cid:2) i.e. if i (cid:10) 1 0 n 2 (cid:2) (cid:14)(cid:2) i.e. if x(cid:10) 1 0 2 Thus the optimum test rejects H 0 (cid:2) (cid:14)(cid:2) if x(cid:10) 1 0 2 We have " (cid:2) (cid:14)(cid:2) (cid:31) (cid:8)(cid:6)PH x(cid:10) 1 0(cid:29) 0! 2 (cid:30) = PH0 "x(cid:5)(cid:2)0 (cid:10) n(cid:27)(cid:2)1(cid:5)(cid:2)0(cid:28)(cid:29)(cid:31) !(cid:16)/ n 2(cid:16) (cid:29)(cid:30) x(cid:5)(cid:2) Under H, 0 follows N(0,1) distribution. o (cid:24)(cid:16) (cid:21) (cid:22) (cid:19) (cid:23) n(cid:20) (cid:8) (cid:8)(cid:6)1(cid:5)#(cid:22)(cid:22)(cid:23)(cid:24) n(cid:27)(cid:2)21(cid:16)(cid:5)(cid:2)0(cid:28)(cid:19)(cid:19)(cid:20)(cid:21) where # is the c.d.f. of a N(0,1) distribution. " (cid:31) " (cid:2) (cid:14)(cid:2) (cid:31) x(cid:5)(cid:2) (cid:5) n(cid:27)(cid:2) (cid:5)(cid:2) (cid:28)(cid:29) (cid:13)(cid:6)PH1 !x(cid:15) 1 2 0(cid:29)(cid:30) = PH1 (cid:16) 1 (cid:15) 2(cid:16)1 0 (cid:29) ! n (cid:29)(cid:30) 9