HW8 - For 90 confidence za/2 = Z.05 = 1.645 Then the 90 6 cr is 16.3 1.645 x y25 ~ For 99 confidence za/2 =[14.33,18.27 = Z.005 = 2.576 Then the 99 cr

HW8 - For 90 confidence za/2 = Z.05 = 1.645 Then the 90 6...

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6.18 (a) (b) For 90% confidence, za/2 = Z.05 = 1.645. Then the 90% cr is 6 16.3 ± 1.645 x ~ = [14.33,18.27]. y25 For 99% confidence, za/2 = Z.005 = 2.576. Then the 99% cr is 6 16.3 ± 2.576 x ~ = [13.21,19.39]. y25 (b) lfn were increased to 100, the width ofthe cr would decrease by a factor of .)100/25 = 2. 6.15 (a) A 95% cr for J1. = 1l0.5 ± 1.96 x ~ = [108.64, 112.36]. Since this interval falls completely within the specification limits of [107.5,1l2.5], we can conclude that the specifications are met. (b) Only the lower specific limit is critical, i.e., we want J1. ~ 107.5 volts. Calculate the lower one-sided 95% confidence bound: u 3 J1. ~ x - Za . ~ = 100.5 - 1.645 x r;n = 108.94. . yn y10 Since this exceeds 107.5, the lower specification limit is met. (c) Similarly if we want J1. ::::; 112.5, calculate the upper one-sided 95% confidence bound: u 3 J1. ::::; x + ZQ ..;n = 1l0.5 + 1.645 x V15 = 112.06. Since this is smaller than 112.5, the upper specification limit is met. 6.16 (l)n-l Conf. Level: P(Xmin ::::; p, ::::; Xmax) = 1 - "2 For n = 10, (1) 10-1 Conf. Level: P(Xmin ::::; p, ::::; Xmax) = 1 -"2 = 0.998. Solutions for Section 6.3 6.17 (a) Ho : p = 0.54 vs. HI : p =f 0.54, where p is the actual proportion of voters who favor the congressman. (b) Ho : p = 0.05 VS. HI : p < 0.05, where p is the true proportion of overdue books. (c) Ho: p = 0.40 vs. HI : p < 0.40, where p is the true scrap rate. (d) Ho : p = ~ vs. HI : p =f ~, where p = P(prefer Gatorade). Ho : J1. = 3.4 vs. HI : J1. > 3.4, where J1. is the mean fat content per yogurt cup. i. Ho : J1. = 10,000 vs. HI : J1. > 10,000 ii. Ho : J1. = 10,000 vs. HI : J1. < 10,000, where J1. is the mean shear strength. (c) Ho: J1. = 25 vs. HI : J1..< 25, where J1. is the mean commuting time. - 96-
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(d) Ho : J.L = 0 vs. HI : J.L # 0, J.L is the mean difference in scores between GauraoiP Sport. 6.19 (a) Since it is more serious not to detect an unsafe food additive, the hypotheses_ It is not safe in the amount normally consumed vs. HI : It is safe. (b) Assuming it is more serious to not release an effective drug, the hypotheses ale": is not effective vs. HI : It is effective. . (c) Because of potential side effects, it is more important to reduce the chance r1 an inequivalent drug on the market. Then the hypotheses are Ho : It is not vs. HI : It is equivalent. (d) Because of possible unforeseen consequences of cloud seeding, it is more i!Jjl[_. 7 •. reduce the chance of accepting cloud seeding as a technique when it isn't effa:tiwe. ••••• the hypotheses are Ho : It is not effective vs. HI : It is effective. 6.20 (a) , a P(Reject HoIHo) = P(X = lip = 1/4) = 1/4 f3 P(Accept HoIHr) = P(X = Olp = 3/4) = 1/4 OC(P) - 97- - . P(XI + X2 f. 0 or 11Ho) = P(XI + X2 = 21Ho) = P(X1 = 1 and Xl = IiIRIIJ P(XI = 1IHo)P(X2 = 11Ho) = (1/4) x (1/4) = 1/16 P(XI + X2 = 0 or 11Hr) = 1 - P(XI + X2 = 21Hr) 1 - P(XI = 1 and X2 = 11Hr) 1 - (3/4) x (3/4) 7/16 (b) a = = (3 = 6.21
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- 98- Then .25 0.0 s u 0 -.2 0.00 .05 .10 .15 .20 P f3 = P(X ~ clJL = 1) = P(X ~ 0.5481JL = 1) ( 0.548-1 ) = P Z ~ 1/V9 IJL = 1 = P(Z ~ -1.355) = 0.087 .2 a = P(X > 21p =-0.02) = 1 - t (5?) (0.02)i(0.98)50-i = 1 - 0.922 = 0.078 i=O ' f3 = P(X ~ 2lp = 0.1) = t (5?) (0.1)i(0.9)50-i = 0.112 i=O ' Power(0.15) = P(X > 21p = 0.15) = 1 - t (5?) (0.15)i(0.85)50-i = 1 - 0.014 = 0.986. i=O ' Operating Characteristic Function 1.0 .8 .6 .4 (b) In a simulation, a Type I error was committed 5% of the time. In a separate simulation, _ a Type II error was committed 5% of the time. a and f3 were both very close to the .• risks calculated in (a).
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  • Spring '08
  • SAWYER
  • Statistics, Statistical hypothesis testing, Student's t-distribution

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