final_f05sol

# final_f05sol - 17.0026 = Pr(Z<.36.50/sqrt.5.5/100 =...

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Solutions to Final Exam, ST 370 Online/Distance, Fall 2005 1. True: sd(c+X1,c+X2, . .., c+Xn) = sd(X1,X2, . .., Xn) 2. statistic: by definition, a statistic is calculated from the sample 3. The middle 50% of the data is spread out over 45.7 grams 4. 0.931 = [1-(1-.9)(1-.8)]*.95 5. 0.0473 = int(1,2) 3exp(-3x) dx = [-exp(-3x)](1,2) = -exp(-6)+exp(-3) 6. Experiment 7. irrigation regime 8. growth 9. 10 10. 30 11. t > 1.895 (use t because s is used in place of sigma, > comes from the direction in H1) 12. 1.935 = (2.23-2.1)/[.19/sqrt(8)] 13. Type I Error (FRN - Falsely Reject Null ) 14. (2.07,2.39) = 2.23+-2.365*.19/sqrt(8) 15. Pr(B|A)=Pr(B) 16. Z < -2.33 (use Z with proportions problems, < comes from the direction of H1)
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Unformatted text preview: 17. .0026 = Pr(Z < (.36-.50)/sqrt(.5*.5/100)) = Pr(Z < -2.80) 18. Accept the null hypothesis since 0.14434 > .05 19. 6.80 = .4*2+.2*5+.3*10+.1*20 20. 0.21 = Pr(Z > (140-132)/10)= Pr(Z>.80)=1-.7881 21. 0.85 = Pr((116-132)/10 < Z < (145-132)/10)=Pr(-1.6<Z<1.3)=.9032-.0548 22. 140.4 = 10*0.84+132 23. 0.14 = Pr(Z < (130-132)/[10/sqrt(30)])=Pr(Z < -1.095) 24. 0.6769 (from p. 527, Table A) 25. 0.1485 = 0.55*0.27 26. 1/9 (1=c*int(0,3)x^2 dx=c*[x^3/3](0,3)=c*9 ==> c=1/9) 27. The approximate normal distribution of xbar 28. correlation coefficient 29. n=11 (n=(1.645*.02/.01)^2=10.824, round to 11) 30. (0.063,0.113) 44/500=.088 0.88+-1.96*sqrt(.088*(1-.088)/500)...
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## This note was uploaded on 01/16/2012 for the course ST 370 taught by Professor Nail during the Spring '08 term at N.C. State.

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