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Final

# Formula Sheet Final Exam Page 1.ppt

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York University
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Summer

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Describe a Single Population Compare 2 Populations Interval Nominal Interval Nominal p p x H :(p  p )0 H :(p  p )D z  , p  Mean (  ) Variability (F- test ) Case 1 0 1 2 0 1 2 Case 2 Mean (  )  p(1 p) n H1:(p1 2 )0 H 1(p1 p2)D Is  known? Variability (c -test ) n Matched 2 2 Pairs (t- test ) Independent s1  1 (p 1 p )ˆ 2 YES NO 2 ( n  1) s p(1 p)ˆ Samples LCL   2 z  x   x   c  2 p z  /2 Are  known? s 2 F  /2,1, 2 z  t   n xD  D  1 1   / n s / n ( n  1) s 2 t  YES NO  s1  p(1 p) ˆ    x  z ( / n) LCL  2 Rejection Region : s / n UCL   2 F /2, 2,1  n 1 n 2   /2 x  t / 2 / n ) c  / 2 , n  1Z   Z  , Z  Z  D D  s2  RejectionRegion: RejectionRegion: 2 Z   , Z   t ,n1, t ,n1 ( n  1) s Z   Z  / 2r Z  Z  / 2 sD s2 H 0 /1 12 x 1 x 2 x 1 x 2 Z  Z or Z  Z UCL  2 H : p  .5 H :   .5 xD t  /2 F  1 2 2 p  ,p 1 ,p 2  /2  /2t  /2,n1t /2,n1 c 1   / 2 , n  1 1 nD Population s2 H 0 /1 12 n  n n n n200, x 175, s 22,  .05 Rejection Region: 2 Variances 1 2 1 2 A B D D Rejection Region: Rejection Region: H 0  180 H1:  180 X X 1,n1 X X ,n1 1 5 7 -2 4 t   ,n1 t  ,n1 ( or  ) ? (p  p )  (p  p ) The Rejection Region is : 2 8 5 3 9 t  t or t  t 1 z  1 2 1 2 t  t.025 ,1991.972 or X X 1/2,n1 X /2,n1 . . . . .  / 2,n1  / 2,n1 (F-test) F  t  .025 ,199.972  D 14 H0:D0, H 1 D0 F  / 2 , 2 , 1 p 11 p )ˆ1 p 21 p ) ˆ2 n  20, s  3800 ,   .05 xD  1.4 RR : t  t  2.821  The Value of Test Statistic is : H :  2 5000 2 n 10 1.401 ,9 F  F  / 2 , 1, 2 n 1 n 2 t 75 18 3.2141 0 2  D ( D) /n t   2.6888 22 / 200 H1:  2 5000 2 SD n1  2.7111 1.6465 / 10 Conclusion: Reject H0. There is 2 2 Conclusion:Do not Estimator for Case 1& 2 enough evidence to infer that the RR : X  X .95,190.1170 SD  1.6465 Reject0H . There is not …  are NOT known mean weight of new product is 2 (20 1)3800 p (1p ) p (1p ) c  50002 10.9744 Yes,  are known Equal Unequal (p1p )2 z  / 2 1 1  2 2 different from 180 lbs. Variances Variances n1 n 2 Confidence Interval Estimator: Conclusion: Do not Reject H0. There is not 175  1.972 (22 / 200 ) enough evidence to infer (x1 x 2( 1) 2 (x 1
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