50 a accepting h and concluding the mean average age

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50. a. Accepting H 0 and concluding the mean average age was 28 years when it was not. b. Reject H 0 if z -1.96 or if z 1.96 0 28 / 6/ 100 x x z n µ σ = = Solving for x , we find at z = -1.96, x = 26.82 at z = +1.96, x = 29.18 Decision Rule: Accept H 0 if 26.82 < x < 29.18 Reject H 0 if x 26.82 or if x 29.18 At µ = 26, 26.82 26 1.37 6/ 100 z = = β = .5000 +.4147 = .0853 At µ = 27, 26.82 27 .30 6/ 100 z = = −
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Hypothesis Testing 9 - 21 β = .5000 +.1179 = .6179 At µ = 29, 29.18 29 .30 6/ 100 z = = β = .5000 +.1179 = .6179 At µ = 30, 29.18 30 1.37 6/ 100 z = = − β = .5000 -.4147 = .0853 c. Power = 1 - β at µ = 26, Power = 1 - .0853 = .9147 When µ = 26, there is a .9147 probability that the test will correctly reject the null hypothesis that µ = 28. 51. a. Accepting H 0 and letting the process continue to run when actually over - filling or under - filling exists. b. Decision Rule: Reject H 0 if z -1.96 or if z 1.96 indicates Accept H 0 if 15.71 < x < 16.29 Reject H 0 if x 15.71 or if x 16.29 For µ = 16.5 16.29 16.5 1.44 .8/ 30 z = = − β = .5000 -.4251 = .0749
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Chapter 9 9 - 22 16.5 16.29 c β c. Power = 1 - .0749 = .9251 d. The power curve shows the probability of rejecting H 0 for various possible values of µ . In particular, it shows the probability of stopping and adjusting the machine under a variety of underfilling and overfilling situations. The general shape of the power curve for this case is .00 .25 .50 .75 1.00 15.6 15.8 16.0 16.2 16.4 Possible Values of u Power 52. 0 .01 4 15 2.33 16.32 50 c z n σ µ = + = + = At µ = 17 16.32 17 1.20 4/ 50 z = = − β = .5000 - .3849 = .1151 At µ = 18 16.32 18 2.97 4/ 50 z = = − β = .5000 - .4985 = .0015 Increasing the sample size reduces the probability of making a Type II error.
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