sol5S09 - PSTAT 120C: Solutions for Assignment #5 May 22,...

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Unformatted text preview: PSTAT 120C: Solutions for Assignment #5 May 22, 2009 1. The data has marginal totals: Favor Oppose Dont know Attends Every week 167 78 16 261 Church Every month 260 113 16 389 Services Never 258 81 14 353 685 272 46 1003 From this we can construct a table of the expected values, E i,j = r i c j /n : Favor Oppose Dont know Attends Every week 178.25 70.78 11.97 261 Church Every month 265.67 105.49 17.84 389 Services Never 241.08 95.73 16.19 353 685 272 46 1003 Thus the 2 statistic is X 2 = (167- 178 . 25) 2 178 . 25 + (78- 70 . 78) 2 70 . 78 + (16- 11 . 97) 2 11 . 97 + (260- 265 . 67) 2 265 . 67 + (113- 105 . 49) 2 105 . 49 + + (16- 17 . 84) 2 17 . 84 + (258- 241 . 08) 2 241 . 08 + (81- 95 . 73) 2 95 . 73 + (14- 16 . 19) 2 16 . 19 = 0 . 710 + 0 . 737 + 1 . 357 + 0 . 121 + 0 . 534 + 0 . 190 + 1 . 187 + 2 . 266 + 0 . 296 = 7 . 398 The critical value for = 0 . 05 chi-squared random variable with ( r- 1)( c- 1) = 4 degrees of freedom is 9.48773 which is bigger than our test statistic. Therefore, even though a somewhat higher proportion in the sample of those that never attend services favor the legislation, there is not a statistically significant difference between the three groups. There is not evidence thatis not a statistically significant difference between the three groups....
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This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.

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sol5S09 - PSTAT 120C: Solutions for Assignment #5 May 22,...

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