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Unformatted text preview: Chapter 8 odd number solutions 7. A larger sample size would be needed to reduce the margin of error. Section 8.3 can be used to show that the sample size would need to be increased to n = 246. Solving for n , shows n = 246 15. 90% confidence df = 64 t .05 = 1.669 19.5 1.669 19.5 1.08 or 18.42 to 20.58 95% confidence df = 64 t .025 = 1.998 19.5 1.998 19.5 1.29 or 18.21 to 20.79 21. liters of alcoholic beverages t .025 = 2.093 df = 19 95% confidence interval: t .025 130 2.093 130 30.60 or 99.40 to 160.60 liters per year 27. Planning value a. b. c. d. Sampling 5403 college graduates to obtain the $100 margin of error would be viewed as too expensive and too much effort by most researchers. 43. a. Margin of Error = 95% Confidence Interval: .53 .0253 or .5047 to .5553 b. Margin of Error = 1.96= .0234 95% Confidence Interval: .31 .0234 or .2866 to .3334 c. Margin of Error = 1.96= .0110 95% Confidence Interval: .05 .0110 or .039 to .061 d. The margin of error decreases asgets smaller. If the margin of error for all of the interval estimates must be less than a given value (say .03), an estimate of the largest proportion should be used as a must be less than a given value (say ....
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This note was uploaded on 06/06/2011 for the course MGSC 291 taught by Professor Rollins during the Fall '09 term at South Carolina.
 Fall '09
 Rollins

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