varofsamplevar

Let zi xi for i 1 2 n so that e zi 0 since

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Unformatted text preview: = 1 2 : : : n so that E (Zi) = 0. Since V ar ;S 2 = E ;S 4 ; expression of E (S ) in terms of n and the moments. We can write 4 (2) 2 2 , we derive an Pn Z 2 Pn 2 S 2 = n i=1 n i(n; ( 1) i=1 Zi ) ; and by squaring S 4 ;Pn Z 2 2 ; 2 n ;Pn Z 2 (Pn Z )2 + (Pn Z )4 i=1 i i=1 i i=1 i i=1 i 2 (n ; 1)2 n ; ;Pn Z 2 (Pn Z )2 + E (Pn Z )4 2 E Pn Z 2 2 ; 2 n E n i=1 i i=1 i i=1 i i=1 i n2 = E (S 4 ) = n2 (n ; 1)2 Kentucky State University, Frankfort, KY, 40601, USA, Current address: eccho@math.snu.ac.kr, Seoul National University, Seoul, Korea, Eungchun Cho's work at Seoul National University was supported by The Korea Research Foundation and The Korean Federation of Science and Technology Socities Grant funded by Korea Government (MOEHRD, Basic Research Promotion Fund). y U.S. Bureau of Labor Statistics 2 Massachusetts Avenue, NE Washington, DC 20212 USA 1291 Section on Survey Research Methods – JSM 2008 Since Z1 : : : Zn are independent, we have ; Ei ;Z(ZZZ2j ) = 0 2 E i2 j = 2 ; E Zi 3 Zj = 0 ; E Zi 4 = 4 E Zi 2 Zj Zk = 0 for distinct i j k: Routine algebraic simpli cation with the expected values given above yields E n X 2!2 =n E Zi i=1 + n (n ; 1) 2 4 + n (n ; 1) 2 =n 0n ! 1 X 2 X !2 A Zi Zi E@ i=1 i=1 n X !4 i=1 n 4 =n Zi 4 + 3 n (n ; 1) ; 2 (3) 2 (4) 2 (5) 2 Substitution of (3), (4) and (5) into the expansion of E S 4 and simpli cation give ; E S4 and ; V ar S 2 = (n ; 1) 4 ; + n2 ; 2n + 3 n (n ; 1) 2 2 (6) ; = E S4 ; 22 ; (n ; 1) 4 + n2 ; 2n + 3 = n (n ; 1) 1 = n 4 ; n ; 3 22 n;1 1; ; 2 + 2 =n4 2 n(n ; 1) 2 2 2 2 ; 2 2 k To obtain an expression of the formula of V ar(S 2 ) in terms of and the moments 02 , substitute 2 4 into (1) and get ; V ar S 2 = = 0 2 0 4 ;2 ; 4 03 + 6 20 2 ;3 and 0 4 about zero, we 4 4 ; n 03 ; n nn;;31) 02 2 + ( n ; 3) n ; 3) + 4n(2n ; 1) 2 02 ; 2n(2n ; 1) ( ( 1 =n 0 3 0 4 (7) 4 3. Comparison with Without-replacement Samples ; Here we compare (1) with the variance of variance of without-r...
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