Dr. Hackney STA Solutions pg 153

# Dr. Hackney STA Solutions pg 153 - Second Edition 9-11 We...

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Second Edition 9-11 We need to solve this numerically for the smallest n that satisﬁes the inequality ( n - 1) n t 2 n - 1 ,. 025 · 64 χ 2 n - 1 ,. 1 . Trying diﬀerent values of n we ﬁnd that the smallest such n is n = 276 for which ( n - 1) n t 2 n - 1 ,. 025 · 64 = 306 . 0 305 . 5 = χ 2 n - 1 ,. 1 . As to be expected, this is somewhat larger than the value found in a). 9.35 length = 2 z α/ 2 σ/ n , and if it is unknown, E(length) = 2 t α/ 2 ,n - 1 cσ/ n , where c = n - 1Γ( n - 1 2 ) 2Γ( n/ 2) and E cS = σ (Exercise 7.50). Thus the diﬀerence in lengths is (2 σ/ n )( z α/ 2 - ct α/ 2 ). A little work will show that, as n → ∞ , c constant. (This can be done using Stirling’s formula along with Lemma 2.3.14. In fact, some careful algebra will show that c 1 as n → ∞ .) Also, we know that, as n → ∞ , t α/ 2 ,n - 1 z α/ 2 . Thus, the diﬀerence in lengths (2 σ/ n )( z α/ 2 - ct α/ 2 ) 0 as n → ∞ . 9.36 The sample pdf is
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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