HW3-Solution - 1 Accept null H X2 = = =12.456>...

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Unformatted text preview: 1. Accept null H. X2 = = =12.456 > X2U,0.01=21.67 95% C.I. (196,1384) X2U,0.025=19.02 X2L,1-0.025=2.7 2. Reject All null H. X2 = 3. Accept null H df1=11 df2=14 F= =22.03 > X2U,0.01/2 , X2U,0.1/2 , X2U,0.05/2 = 1.847 >F(1-0.01/2),11,14 =1/ F(0.01/2),14,11 1/5.05 =0.198 99% C.I. FL(0.01/2),11,14= 1/4.6 , FU(0.01/2),14,11 = 5.05 (1.847* FL,1.847* FU)= (0.401, 9.329) 4. Accept null H df1=7 df2=5 F= 7.2 a. b. c. d. e. 7.3 2 a. 0.005 = 21.95 = 1.68 > F(1-0.01/2),7,5 =1/ F(0.01/2),5,7 = 1/ 9.52 = 0.105 P(Y 52.62) = 0.001 P(Y 34.38) =0.1 P(Y 14.61) = 1- 0.95 = 0.05 P(Y 10.52) = 1-0.995= 0.005 P(10.52 Y 34.38) = P(10.5<Y) – P(34.38<Y)=(1-0.1)-(0.005)=0.895 2 b. 0.025 = 17.53 2 c. 0.975 = 2.180 2 d. 0.995 = 1.344 2 e. 0.95 = 2.733 7.11 a. b. c. d. e. 7.23 b. 95% C.I. on Mean Brand 1- x=38.79, Sx=1.9541 df=9 - t0.05/2* Sx / , x + t0.05/2* Sx / Brand 2- y =40.69, Sy =5.579 df=9 ( x 0.05, df1 5, df 2 15 . F=2.9 0.025, df1 15, df 2 15 F=2.86 0.01, df1 10, df 2 12 F=4.3 0.001, df1 15, df 2 5 F=25.91 0.005, df1 8, df 2 13 F= 5.08 )= (37.39, 40.19) )= (36.68, 44.66) ( y - t0.05/2* Sy / , y + t0.05/2* Sy / 95% C.I. on variance Brand 1- Sx=1.9541 df=9 X2U,0.025=19.02 , X2L,1-0.025=2.7 ( , )= (1.344, 3.567) Brand 2-Sy =5.579 df=9 ( , )= (3.838, 10.186) c. Reject null H for F= S2x/ S2y=0.1227 < F(1-0.01/2),9,9 =1/ F(0.01/2),9,9 = 1/5.35 Accept null H for u Use t ’ test Chpt 6.2 page 301 ...
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This note was uploaded on 12/19/2010 for the course AMS 315 taught by Professor Staff during the Fall '08 term at SUNY Stony Brook.

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HW3-Solution - 1 Accept null H X2 = = =12.456>...

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