STAT 415 - Homework 8 Solution

STAT 415 - Homework 8 Solution - Hmewwk 8 folzdfim Ho p=...

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Unformatted text preview: Hmewwk 8 folzdfim Ho: p= 0.07 us. (4.: > 0.07 84‘7' A: :2. f 209 M 2 ix Jan/1x o. (33/20? > 2.27 (a) E: 2‘17) (£41; rfjed‘ Ho (6) 5-2.27 < 2.316, 8.1—8 The value of the tat statistic is ‘3 I” g 0.70 — 0.75 z = (\(msstyaso = ‘2'280' (3) Since 2 = —2.280 < —1.645, reject Ho. (b) Since z = -2.280 > —2.326, do not reject Ho. (c) p-value a: P(Z S -2.280) = 00113. Note that 0.01 < p-value (0.05. 34’? 1551.7. ‘— ‘ ‘2 c — A _D( 9/ Page ’2 ( 8.!.l9. (4) Ho: (= 0.034 v.5, H.: No.0” (5) z 2 2.326 (0) Z = 20/300-009’] = 1. 711 > 2.526. Hence reject Ho. ’0. 03’} x 0. qsa/aoo 8. I. [q Q97 coincidence 131 -5 121 54% 400 TnfEfi/fid for m"ff I 0 75 1104: included. 0.138’lx 0.869 amaq '[ 8qu ’Ioo ( 7' H' 7 11.46 ...
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