ISM_chapter10 - Chapter 10: Hypothesis Testing 10.1 10.2...

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201 Chapter 10: Hypothesis Testing 10.1 See Definition 10.1. 10.2 Note that Y is binomial with parameters n = 20 and p . a. If the experimenter concludes that less than 80% of insomniacs respond to the drug when actually the drug induces sleep in 80% of insomniacs, a type I error has occurred. b. α = P (reject H 0 | H 0 true) = P ( Y 12 | p = .8) = .032 (using Appendix III). c. If the experimenter does not reject the hypothesis that 80% of insomniacs respond to the drug when actually the drug induces sleep in fewer than 80% of insomniacs, a type II error has occurred. d. β (.6) = P (fail to reject H 0 | H a true) = P ( Y > 12 | p = .6) = 1 – P ( Y 12 | p = .6) = .416. e. β (.4) = P (fail to reject H 0 | H a true) = P ( Y > 12 | p = .4) = .021. 10.3 a. Using the Binomial Table, P ( Y 11 | p = .8) = .011, so c = 11. b. β (.6) = P (fail to reject H 0 | H a true) = P ( Y > 11 | p = .6) = 1 – P ( Y 11 | p = .6) = .596. c. β (.4) = P (fail to reject H 0 | H a true) = P ( Y > 11 | p = .4) = .057. 10.4 The parameter p = proportion of ledger sheets with errors. a. If it is concluded that the proportion of ledger sheets with errors is larger than .05, when actually the proportion is equal to .05, a type I error occurred. b. By the proposed scheme, H 0 will be rejected under the following scenarios (let E = error, N = no error): Sheet 1 Sheet 2 Sheet 3 N N . N E N E N N E E N With p = .05, α = P ( NN ) + P ( NEN ) + P ( ENN ) + P ( EEN ) = (.95) 2 + 2(.05)(.95) 2 + (.05) 2 (.95) = .995125. c. If it is concluded that p = .05, but in fact p > .05, a type II error occurred. d. β ( p a ) = P (fail to reject H 0 | H a true) = P( EEE , NEE , or ENE | p a ) = . ) 1 ( 2 3 2 a a a p p p + 10.5 Under H 0 , Y 1 and Y 2 are uniform on the interval (0, 1). From Example 6.3, the distribution of U = Y 1 + Y 2 is < = 2 1 2 1 0 ) ( u u u u u g Test 1 : P ( Y 1 > .95) = .05 = α . Test 2 : α = .05 = P ( U > c ) = 2 ) 2 ( c du u = 2 = 2 c + .5 c 2 . Solving the quadratic gives the plausible solution of c = 1.684.
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202 Chapter 10: Hypothesis Testing Instructor’s Solutions Manual 10.6 The test statistic Y is binomial with n = 36. a. α = P (reject H 0 | H 0 true) = P (| Y – 18| 4 | p = .5) = P ( Y 14) + P ( Y 22) = .243. b. β = P (fail to reject H 0 | H a true) = P (| Y – 18| 3 | p = .7) = P (15 Y 21| p = .7) = .09155. 10.7 a. False, H 0 is not a statement involving a random quantity. b. False, for the same reason as part a. c. True. d. True. e. False, this is given by α . f. i. True. ii. True. iii. False, β and α behave inversely to each other. 10.8 Let Y 1 and Y 2 have binomial distributions with parameters n = 15 and p . a. α = P (reject H 0 in stage 1 | H 0 true) + P(reject H 0 in stage 2 | H 0 true) ) , 6 ( ) 4 ( ) 3 , 6 ( ) 4 ( 1 2 1 3 0 1 1 2 1 1 i Y Y Y P Y P Y Y Y P Y P i + + = + + = = ) ( ) 6 ( ) 4 ( 1 2 3 0 1 i Y P i Y P Y P i + = = = .0989 (calculated with p = .10). Using R , this is found by: > 1 - pbinom(3,15,.1)+sum((1-pbinom(5-0:3,15,.1))*dbinom(0:3,15,.1)) [1] 0.0988643 b. Similar to part a with p = .3: α = .9321. c. β = P (fail to reject H 0 | p = .3) = ) ( ) 5 ( ) 5 , ( 1 3 0 2 2 1 3 0 1 i Y P i Y P Y Y i Y P i i = = = + = = = = .0679. 10.9 a. The simulation is performed with a known p = .5, so rejecting H 0 is a type I error. b. - e. Answers vary. f. This is because of part a.
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This note was uploaded on 03/18/2009 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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ISM_chapter10 - Chapter 10: Hypothesis Testing 10.1 10.2...

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