KeyM147finalf06

KeyM147finalf06 - Alternative Hypothesis: The quart...

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MATH 147 FINAL EXAM Fall 2006 December 13, 2006 Answer Key Part I True/False 1. T 2. F 3. T 4. F 5 F 6. T 7. F 8. T 9. F 10 F 11 F Part II Fill in the blank 12. 515 ; 13. 2/3 14. -3 4.33 15. iii 16. 0.5 17. 0.7071 18. .2 19. 15 Part III Multiple Choice 20. D 21. A 22. D 23. C 24. E 25. B 26. C 27. A 28. E 29. D 30. C 31. A 32. B 33. D Part IV Problems 34. (a) 3/10 ; (b) 2/5 ; (c) 3/20 ; (d) 3/10 ; (e) 1/3 ; 35 (a.) The four possible progeny are: r/w, r/w, w/w, w/w; so that chance of red = 50% (b) EV = 1250; SE = 25; SU = 1.4; chance = 8.075% 36 SE ( for control ) = 12 / 50 ; SE ( for treatment ) = 12 / 50 ; SE ( for difference ) = 2 . 4 z = 1 . 75 p - value = 8 %; Therefore Do not reject Null hypothesis. The data do not support the hypothesis that there is a difference in the teaching methods. 37. Sample Average = 70 ; SE = 3 ; 95% C.I. : (70 +/- (2)(3) or (64 to 76) 38. (a) y - 187 18 = ( 0 . 7 ) x - 195 15 · ; (b) 191.2 ; (c) New Avg = 191.2; New SD = 12.8546; z = .6846 =.70; Chance = 24.195% ; 39. Null Hypothesis: The quart averages 128 oz.
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Unformatted text preview: Alternative Hypothesis: The quart averages less than 128 oz. Use t-test; Degrees of freedom = 9; SD + = . 3795; SE + = . 12; t = -2.667; p-value is between 1% and 2.5%; The result is significant; Reject the Null Hypothesis. In non statistical language, the data do support the claim that the containers are underfilled. 40. Null Hypothesis: There are the same number of each type of coin in the box. Alternate Hypothesis: There are not the same number of each type of coin in the box. χ 2 = 5 . 4; df = 3; 10 % < p-value < 30 % ; Conclusion: Do Not Reject Null Hypothesis i.e., The data do not support the hypothesis that ”There are not the same number of each type of coin in the box.” Updated 5/7/2007 1...
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This note was uploaded on 08/08/2008 for the course MATH 147 taught by Professor Klimko during the Fall '07 term at Binghamton.

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