mont4e_sm_ch15_mind - = P( V 4| p = 0.5) = 0.1875 8....

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Mind-Expanding Exercises 15-43 Under the null hypothesis each rank has probability of 0.5 of being either positive or negative. Define the random variable X i as = negative is rank positive is rank X i 0 1 Then, 24 ) 1 2 )( 1 ( ) ( Then, 4 1 4 1 2 1 )] ( [ 2 1 0 2 1 1 ) ( ce independen by ) ( ) ( where 2 1 ) ( since 4 ) 1 ( 2 1 ) ( ) ( and 2 2 2 2 1 1 + + = = = + = = = + = = = = + + = + = + n n n R V X E X V i x V R V x E n n i i x E R E i x R i i i i n i i n i i 15-44 a) 5 4 ) ( 5 = > = + i i X X P p There appears to be an upward trend. b) V is the number of values for which . The probability distribution of V is binomial with n = 5 and p = 0.5. i i X X > + 5 c) V = 4 1. The parameter of interest is number of values of i for which . i i X X > + 5 trend upward an is there : 3 trend no is there : . 2 1 0 H H 4. α = 0.05 5. The test statistic is the observed number of values where or V = 4. i i X X > + 5 6. We reject H 0 if the P-value corresponding to V = 4 is less than or equal to α = 0.05. 7. Using the binomial distribution with n = 5 and p = 0.5, P-value
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Unformatted text preview: = P( V 4| p = 0.5) = 0.1875 8. Conclusion: do not reject H . There is not enough evidence to suspect an upward trend in the wheel opening dimensions at =0.05. 15-45 a) 32 sequences are possible b) Because each sequence has probability 1/32 under H , the distribution of W * is obtained by counting the sequences that result in each value of W * w * 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Prob*1/32 1 1 1 2 2 3 3 3 3 3 3 2 2 1 1 1 c) P( W * > 13) = 2/32 d) By enumerating the possible sequences, the probability that W * exceeds any value can be calculated under the null hypothesis as in part (c). This approach can be used to determine the critical values for the test. 15-19...
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This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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