# chapter11 - 1(a H0 160 H1 > 160 x 0 158 160 = =-1.333 3/2 n...

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1 Chapter 11 11–1. (a) H 0 : μ 160 H 1 : μ > 160 Z 0 = x - μ 0 σ n = 158 - 160 3 / 2 = - 1 . 333 The ﬁber is acceptable if Z 0 > Z 0 . 05 = 1 . 645. Since Z 0 = - 1 . 33 < 1 . 645, the ﬁber is not acceptable. (b) d = μ - μ 0 σ = 165 - 160 3 = 1 . 67, if n = 4 then using the OC curves, we get β 0 . 05. 11–2. (a) H 0 : μ = 90 H 1 : μ < 90 Z 0 = x - μ 0 σ n = 90 . 48 - 90 p 5 / 5 = 0 . 48 Since Z 0 is not less than - Z 0 . 05 = - 1 . 645, do not reject H 0 . There is no evidence that mean yield is less than 90 percent. (b) n = ( Z α + Z β ) 2 σ 2 2 = (1 . 645 + 1 . 645) 2 5 / (5) 2 = 2 . 16 3. Could also use the OC curves, with d = ( μ 0 - μ ) = (90 - 85) / 5 = 2 . 24 and β = 0 . 05, also giving n = 3. 11–3. (a) H 0 : μ = 0 . 255 H 1 : μ 6 = 0 . 255 Z 0 = x - μ 0 σ/ n = 0 . 2546 - 0 . 255 0 . 0001 / 10 = - 12 . 65 Since | Z 0 | = 12 . 65 > Z 0 . 025 = 1 . 96, reject H 0 . (b) d = | μ - μ 0 | σ = | 0 . 2552 - 0 . 225 | 0 . 0001 = 2, and using the OC curves with α = 0 . 05 and β = 0 . 10 gives n 3. Could also use n ( Z α/ 2 + Z β ) 2 σ 2 2 = (1 . 96 + 1 . 28) 2 (0 . 0001) 2 / (0 . 0002) 2 = 2 . 62 = 3. 11–4. (a) H 0 : μ = 74 . 035 H 1 : μ 6 = 74 . 035 Z 0 = x - μ 0 σ/ n = 74 . 036 - 74 . 035 0 . 001 / 15 = 3 . 87 Since Z 0 > Z α/ 2 = 2 . 575, reject H 0 . (b) n = ( Z α/ 2 + Z β ) 2 σ 2 δ 2 = 0 . 712 ,n = 1 11–5. (a) H 0 : μ = 1 . 000 Z 0 = x - μ 0 σ n = 1014 - 1000 25 / 20 = 2 . 50 H 1 : μ 6 = 1 . 000 | Z 0 | = 2 . 50 > Z 0 . 005 = 1 . 96 , reject H 0 . 11–6. (a) H 0 : μ = 3500 Z 0 = x - μ 0 σ n = 3250 - 3500 p 1000 / 12 = - 27 . 39 H 1 : μ 6 = 3500 | Z 0 | = 27 . 39 > Z 0 . 005 = 2 . 575 , reject H 0 .

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2 11–7. (a) H 0 : μ 1 = μ 2 H 1 : μ 1 6 = μ 2 Z 0 = x 1 - x 2 q σ 2 1 n 1 + σ 2 2 n 2 = 1 . 349 Since Z 0 < Z α/ 2 = 1 . 96, do not reject H 0 . (d) d = 3 . 2, n = 10, α = 0 . 05, OC curves gives β 0, therefore power 1. 11–8. μ 1 = New machine, μ 2 = Current machine H 0 : μ 1 - μ 2 2, H 1 : μ 1 - μ 2 > 2 Use the t -distribution assuming equal variances: t 0 = - 5 . 45, do not reject H 0 . 11–9. H 0 : μ 1 - μ 2 = 0 , H 1 : μ 1 - μ 2 6 = 0 Z 0 = x 1 - x 2 q σ 2 1 n 1 + σ 2 2 n 2 = 2 . 656 Since Z 0 > Z α/ 2 = 1 . 645, reject H 0 . 11–10. H 0 : μ 1 - μ 2 = 0 , H 1 : μ 1 - μ 2 > 0 Z 0 = x 1 - x 2 q σ 2 1 n 1 + σ 2 2 n 2 = - 6 . 325 Since Z 0 < Z α = 1 . 645, do not reject H 0 . 11–11. H 0 : μ 1 - μ 2 = 0 , H 1 : μ 1 - μ 2 < 0 Z 0 = x 1 - x 2 q σ 2 2 n 1 + σ 2 2 n 2 = - 7 . 25 Since Z 0 < - Z α = - 1 . 645, reject H 0 . 11–12. H 0 : μ = 0 t 0 = x - μ 0 s/ n = - 0 . 168 - 0 8 . 5638 / 10 = - 0 . 062 H 1 : μ 6 = 0 | t 0 | = 0 . 062 < t 0 . 025 , 9 = 2 . 2622 , do not reject H 0 . 11–13. (a) t 0 = 1 . 842, do not reject H 0 . (b) n = 8 is not suﬃcient; n = 10. 11–14.
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## This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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chapter11 - 1(a H0 160 H1 > 160 x 0 158 160 = =-1.333 3/2 n...

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