q10-13sol

Q10-13sol - Math 218(Spring 2008 Solutions to Quizzes 1013 TA Wei Lin Q10(a X =(2 6 10 8 7/5 = 33/5 = 6.6(b S = 1 22 62 102 82 72 5(6.6)2 = 8.8 =

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Math 218 (Spring 2008) Solutions to Quizzes 10–13 TA: Wei Lin Q10. (a) X = (2 + 6 + 10 + 8 + 7) / 5 = 33 / 5 = 6 . 6. (b) S = r 1 5 - 1 ± 2 2 + 6 2 + 10 2 + 8 2 + 7 2 - 5(6 . 6) 2 ² = 8 . 8 = 2 . 9665. (c) α = (1 - . 95) / 2 = . 025 and t 4 ,. 025 = 2 . 776. Thus, the confidence interval for μ is X ± t 4 ,. 025 S n = 6 . 6 ± 2 . 776 8 . 8 5 , i.e., 2 . 9172 μ 10 . 2828. (d) First use z -score to approximate the t -score. z . 025 = 1 . 96. We want z . 025 S n < . 5 S and thus n > (2 z . 025 ) 2 = (2(1 . 96)) 2 = 15 . 3664. Hence, n 16. Try n = 16: t 15 ,. 025 S 16 = 2 . 131 S 16 = . 5328 S > . 5 S , failed. Try n = 17: t 16 ,. 025 S 17 = 2 . 120 S 17 = . 5142 S > . 5 S , failed. Try n = 18: t 17 ,. 025 S 18 = 2 . 110 S 18 = . 4973 S < . 5 S . Therefore, n 18 moviegoers. (e) μ X ± z . 025 S 5 = 6 . 6 ± 1 . 96 8 . 8 5 , i.e., 3 . 9998 μ 9 . 2002. Q11.
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This note was uploaded on 05/04/2008 for the course MATH 218 taught by Professor Haskell during the Fall '06 term at USC.

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Q10-13sol - Math 218(Spring 2008 Solutions to Quizzes 1013 TA Wei Lin Q10(a X =(2 6 10 8 7/5 = 33/5 = 6.6(b S = 1 22 62 102 82 72 5(6.6)2 = 8.8 =

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