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q10-13sol - Math 218(Spring 2008 Solutions to Quizzes 1013...

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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 = 1 5 - 1 2 2 + 6 2 + 10 2 + 8 2 + 7 2 - 5(6 . 6) 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. (a) H 0 : μ = 12 ounces, H a : μ = 12 ounces. (b) z = X - μ 0 σ/ n . Reject H 0 if | z | > z . 015 = 2 . 17. (c) z obs = 11 . 85 - 12 . 2 / 10 = - 2 . 37 < - 2 . 17; hence reject H 0 . (d) p -value = 2 P ( z > 2 . 37) = 2( . 5 - . 4911) = . 0178. (e) Circle . 025, . 05, . 075, and . 10. Criterion: reject H 0 if p -value < α . (f) Say μ = 12 ounces when actually μ = 12 ounces. (g) H 0 : μ 12 ounces, H a : μ < 12 ounces. Q12. (a) H 0 : μ 15 minutes, H a : μ < 15 minutes. (b) t = X - μ 0 S/ n ; reject H 0 if t < - t 4 ,. 05 = - 2 . 132. (c) X = (6+10+11+14+14) / 5 = 11, S = 1 5 - 1 6 2 + 10 2 + 11 2 + 14 2 + 14 2 - 5(11) 2 = 11 = 3 . 3166, and thus t obs = 11 - 15 11 / 5 = - 2 . 6968 < - 2 . 132; reject H 0 and hence μ < 15 minutes. (d) With df = 4, p -value = P ( t < - 2 . 6968). From the t -table, P ( t < - 2 . 776) = P ( t > 2 .
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