Table Of ContentSTUDIES IN STATISTICAL INFERENCE,
SAMPLING TECHNIQUES AND DEMOGRAPHY
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Rajesh Singh, Jayant Singh, Florentin Smarandache
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ILQ
2009
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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)
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(ISBN-10): 1-59973-087-1
(ISBN-13): 978-1-59973-087-5
(EAN): 9781599730875
Printed in the United States of America
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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
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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.
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(cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) The Authors
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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)
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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.
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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
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(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)
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