HOMEWORK5

# HOMEWORK5 - Homework 5 Stat 3022 Problem 6.57 1...

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Homework 5 Stat 3022 February 25, 2010 Problem 6.57 1. P(Z > -1.73)=pnorm(-1.73,lower.tail=F)=.958 2. P(Z < -1.73)=pnorm(-1.73)=.0418 3. P(Z < -1.73)+P(Z > 1.73)=2*P(Z < -1.73)=2*.0418=.0836 Problem 6.59 1. Regardless of the alternative we would reject the null hypothesis because 24 falls well inside the CI. 2. If the alternative is two-sided, the answer is yes. If the alternative is μ > 30 the answer is no. The answer is yes if the alternative is μ < 30 . Problem 6.68 For testing these hypotheses, we find Z = 10 . 2 - 8 . 9 2 . 5 /sqrt 6 = 1 . 27 . This is not significant because the p-value is .1020. Thus there is not enough evidence to conclude that these sonets were not written by our poet. Problem 6.95 1. Z = 508 - 505 100 /sqrt 100 = . 3 , so P=P(Z > .03)=.3821 2. Z = 508 - 505 100 /sqrt 1000 = . 95 , so P=P(Z > .95)=.1711 3. Z = 508 - 505 100 /sqrt 10000 = 3 , so P=P(Z¿3)=.0013 Problem 6.112 1

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1. α =P(Type 1 Error)=P( ¯ X > 26 when μ = 25 )= P( Z > 26 - 25 50 /sqrt 900 )=P(Z¿.6)=.2743 2. P(Type 2 Error when μ =28)=P( ¯ X 26 when μ = 28 )=P( Z 26 - 28 50 /sqrt 900 )=P(Z < - 1.2)=.1151 3. P(Type 2 Error when μ =30)=P( ¯ X < 26 when μ = 30 )= P( Z 26 - 30 50 /sqrt 900 )=P(Z¡- 2.4)=.0082 4. The sample size (n=900) is so large that the mean will be very close to
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