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Unformatted text preview: Practice Questions 7 Answers 1. D 2. A 3. B 4. C 5. A 6. C 7. B 8. B
0.5 0.5 2 z ( 1.96 ) 9. n = / 2 = = ( 32.67 ) = 1067.33 ; 1068 ME 0.03 2 2 10. 2 n =15, / 2 = 0.025. Then df =n 1 =14 and 14, 0.975 = 5.629 UCL= ( n  1) s 2
2 n1, 1 / 2 = 14s 2 2 = ( 46 ) s 2 = 850.783 s = 29.168 . 5.629 11.
nx = 55, sx = 5633, n y = 64, s y = 4793, df = nx + n y  2 = 117 tdf , / 2 = t117,0.025 t120,0.025 = 1.9799 , and 2 2 s 2 = [( nx  1) sx + ( n y  1) s y ]/( nx + ny  2) = 270,149,29. Then, p ( x  y)
12. a. tnx + n y  2, / 2 (s 2 p / nx s 2 / n y = (47,520 44,304) . 1.9799 (955.66) + p
= 3216 . 1892.11 or 1323.89 < x  y < 5108.11 ) The degrees of freedom are given by 2 2 2 s 2 1 2 s2 /( n1  1) + /(n2 ^ 19 1) n n1 2 s 2 2 s = 1 2 + n n1 2 b. The degrees of freedom are given by 2 2 2 s 2 2 s 1 /( n1  1) + 2 /(n2 ^ 9 1) n n1 2 s 2 2 s = 1 2 + n n1 2 c. The degrees of freedom are given by
2 2 2 s 2 1 2 s2 /( n1  1) + /(n2 ^ 17 1) n n1 2 s 2 2 s = 1 2 + n n1 2 d. The degrees of freedom are given by
2 2 2 s 2 1 2 s2 /( n1  1) + /(n2 ^ 12 1) n n1 2 s 2 2 s = 1 2 + n n1 2 13. C 14. D 15. A 16. B 17. A 18. a. C b. D 19. a. B b. C c. D d. A 20. a. A b.D c.C d. B 21. a. C b. B c. A 22. a. D b. B 23. B 24. H 0 : =14 vs. H1 : 14 Test statistic Z= (11.6  14) / (3.4 / 6.32) = 4.46 Since pvalue = 2P(Z < 4.46) . 0.0041, we reject H 0 at = 0.01. We conclude that there is sufficient evidence to disprove the claim that the average time to assemble an electronic component is 14 minutes (The population mean appears to be less than 14 minutes.) 25. H 0 : r 29 vs. H1 : < 29
Test statistics t = (26  29) / (6.2 / 3.83) = 1.874 Reject H 0 if t <  tn 1, = t14,0.10 = 1.345. Since t = 1.874, we reject H 0 at = 0.10. There is sufficient evidence to prove claim that the product will increase output per machine by 29 units per hour. 26.
H 0 : P = 0.80 vs. H1 : P < 0.80 ^ Test statistics Z = ( p  P0 ) / P0 (1  P0 ) / n = 2.09 Since pvalue = P(Z < 2.09) = 0.0183, we reject H 0 at = 0.025. There is sufficient evidence to doubt the manufacturer's claim. 27.
H 0 : P = 0.634 vs. H1 : P < 0.634 ^ Test statistic Z = ( p  P0 ) / P0 (1  P0 ) / n = 1.38 Since pvalue = P(Z < 1.38) = 0.0838, we reject H 0 at = 0.10. There is sufficient evidence to doubt the U.S. Postal Service claim 28. H 0 : . 60 vs. H1 : > 60
Test statistic t = (63.2  60) / (7.7/4.9) = 2.04 Since, t23,0.05 = 1.714 , we reject H 0 at . 0.05. There is sufficient evidence to disprove the accountant claim that he can complete a standard tax return in at most an hour. 29. H 0 : = 4.0 vs. H1 : < 4.0 Reject H 0 if Z = ( X 4.0) / (0.1/5) <  Z 0.10 = 1.28; that is, reject shipments for X < 4.0
1.28(0.02) = 3.9744 30.
H 0 : P = 0.023 vs. H1 : P > 0.023
Reject H 0 if ^ ( p  P0 ) / ( P0 ( 1  P0 ) / n ) > Z.05 = 1.645, that is, assembly line will be shut ^ down for maintenance if ( p 0.023) / 0.0106 > 1.645, or equivalently if sample proportion exceeds 0.0404 ...
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 Spring '08
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