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Unformatted text preview: Practice Exam Questions and Solutions for the Final Exam; Fall, 2008 Statistics 301, Professor Wardrop Part A, Chapters 7, 12 and 15 Chapter 7 1. An observational study yields the following “collapsed table.” Group S F Total 1 72 228 300 2 88 212 300 Total 160 440 600 Below are two component tables for these data. Complete these tables so that Simpson’s Para dox is occurring or explain why Simpson’s Paradox cannot occur for these data. For the latter, you must provide computations that jus tify your answer. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 30 30 60 1 42 198 240 2 120 2 180 Tot 180 Tot 420 2. An observational study yields the following “collapsed table.” Group S F Total 1 100 100 200 2 156 244 400 Total 256 344 600 Below are two component tables for these data. Complete these tables so that Simpson’s Para dox is occurring or explain why Simpson’s Paradox cannot occur for these data. For the latter, you must provide computations that jus tify your answer. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 70 30 100 1 30 70 100 2 80 2 320 Tot 180 Tot 420 3. An observational study yields the following col lapsed table. Group S F Total 1 41 59 100 2 35 65 100 Total 76 124 200 Below are two component tables for these data. Complete these tables so that Simpson’s Para dox is occurring or explain why Simpson’s Paradox cannot occur for these data. For the latter, you must provide computations that jus tify your answer. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 5 15 20 1 36 44 80 2 40 2 60 Tot 60 Tot 140 4. An observational study yields the following col lapsed table. Group S F Total 1 100 150 250 2 42 58 100 Total 142 208 350 Below are two component tables for these data. Complete these tables so that Simpson’s Para dox is occurring or explain why Simpson’s Paradox cannot occur for these data. For the latter, you must provide computations that jus tify your answer. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 57 93 150 1 43 57 100 2 60 2 40 Tot 210 Tot 140 5. In a ‘Chapter 7’ problem, you are given the fol lowing information: radicalBigg ˆ p 1 ˆ q 1 n 1 = 0 . 025 , and 1 ˆ p 2 ˆ q 2 n 2 = 0 . 005 . Calculate the halfwidth of the 90% confidence interval for p 1 p 2 . 6. In a ‘Chapter 7’ problem, you are given the fol lowing information: 1 . 96 radicalBigg ˆ p 1 ˆ q 1 n 1 = 0 . 0679 , and 1 . 645 radicalBigg ˆ p 2 ˆ q 2 n 2 = 0 . 0409 . Calculate the halfwidth of the 80% confidence interval for p 1 p 2 . Chapter 12 7. A sample of size 40 yields the following sorted data. Note that I have deleted x (39) (the sec ond largest number). This fact will NOT pre vent you from answering the questions below. 14.1 46.0 49.3 53.0 54.2 54.7 54.7 54.7 54.8 55.4 57.6 58.2 58.3 58.7 58.9 60.8 60.9 61.0 61.1 63.0 64.3 65.6 66.3 66.6 67.0 67.9 70.1 70.3 72.1 72.4 72.9 73.5 74.2 75.3 75.4 75.9 76.5 77.0 x (39) 88.9 (a) Construct the density scale histogram for these data using the class inter vals: [10 , 40] , [40 , 50] , [50 , 60]...
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This note was uploaded on 10/23/2009 for the course STAT STATS 301 taught by Professor Professorwardrop during the Fall '08 term at Wisconsin.
 Fall '08
 ProfessorWardrop
 Statistics

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