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Homework6_solution

Homework6_solution - Homework#6 Solutions 9.1(a With no...

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Homework #6 Solutions 9.1 (a) With no prior estimate of p, 062 . 663 ) 5 . 0 )( 5 . 0 ( 05 . 0 575 . 2 ˆ ˆ 2 * * 2 2 = = = q p E z n α or 664 Therefore, 664 subscribers should be surveyed. (b) With a prior estimate of p=0.30, 972 . 556 ) 7 . 0 )( 3 . 0 ( 05 . 0 575 . 2 ˆ ˆ 2 * * 2 2 = = = q p E z n α or 557 Therefore, 557 subscribers should be surveyed. (c) If a 40% nonresponse rate is anticipated, we need to end up with 557 surveys, after nonresponse. So 0.6N=557 or N=928 surveys must be mailed. Such a nonresponse rate may cause a bias in the results if the nonresponse rate is related to the proportion of interest, i.e. subscribers with income above \$100,000 may be less likely to respond to the survey. 9.2 (a) 5 . 0 : 0 = p H vs 5 . 0 : 1 p H . The alternative should be two-sided because there is no prior knowledge of whether heads should be more or less likely than tails. (b) The test statistic is 34 . 1 10000 ) 5 . 0 )( 5 . 0 ( 5 . 0 5067 . 0 ˆ 0 0 0 = - = - = n q p p p z The P-value is 180 . 0 0901 . 0 2 )) 34 . 1 ( 1 ( 2 = × = Φ - = P Since 05 . 0 = α P , do not reject 0 H and conclude that heads and tails are equally likely. (c) A 95% CI is given by ] 5165 . 0 , 4969 . 0 [ 10000 ) 4933 . 0 )( 5067 . 0 ( 96 . 1 5067 . 0 ˆ ˆ 96 . 1 ˆ = ± = ± n q p p

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9.4 (a) With a prior estimate of p=0.01, 96 . 9507 ) 99 . 0 )( 01 . 0 ( 002 . 0 96 . 1 ˆ ˆ 2 * * 2 2 = = = q p E z n α or 9508 Therefore, 9508 parts should be sampled.
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