# 410Hw13ans - STAT 410 Homework#13(due Friday December 4 by...

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STAT 410 Homework #13 Fall 2009 (due Friday, December 4, by 3:00 p.m.) 1. 8.2.3 Hint: Do NOT look at the answer at the back of the textbook. Reject H 0 if x 5 3 . γ ( θ ) = P ( Reject H 0 | θ ) = P ( X 5 3 | θ ) = P - 25 2 5 3 Z θ = 1 – Φ - 2 5 3 θ . 2. 5.5.4 H 0 : p = 1 / 2 H 1 : p = 1 / 4 Significance level = α = P ( Reject H 0 | H 0 is true ) = P ( X 3 | p = 1 / 2 ) = = - 3 0 10 10 2 1 2 1 k k k k C = 0.171875 . Power = P ( Reject H 0 | H 0 is not true ) = P ( X 3 | p = 1 / 4 ) = = - 3 0 10 10 4 3 4 1 k k k k C = 0.775875 .

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3. Alex wants to test whether a coin is fair or not. Suppose he decides to use 25 tosses. Let p denote the probability of obtaining heads. H 0 : p = 0.50 vs. H 1 : p 0.50. a) Find the Rejection Rule with the significance level α closest to 0.05. Let S denote the number of “successes” ( heads ) in 25 tosses. Decision rule: Reject H 0 if S a or S b . Want P( Type I error ) = 0.05. P( Type I error ) = P( Reject H 0 | H 0 true ) = P( S a or S b | p = 0.50 ) = CDF( a | p = 0.50 ) + ( 1 – CDF( b – 1 | p = 0.50 ) ) . Want CDF( a | p = 0.50 ) 0.025, ( 1 – CDF( b – 1 | p = 0.50 ) ) 0.025, CDF( b – 1 | p = 0.50 ) 0.975. CDF( 7 | p = 0.50 ) = 0.022, a = 7. CDF( 17 | p = 0.50 ) = 0.978, b – 1 = 17, b = 18. Decision rule: Reject H 0 if S 7 or S 18 . b) What is the actual value of the significance level α associated with the Rejection Rule obtained in part (a)? P( Type I error ) = P( Reject H 0 | H 0 true ) = P( S 7 or S 18 | p = 0.50 ) = CDF( 7 | p = 0.50 ) + ( 1 – CDF( 17 | p = 0.50 ) ) = 0.022 + ( 1 – 0.978 ) = 0.044 . c)
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410Hw13ans - STAT 410 Homework#13(due Friday December 4 by...

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