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Final

# ADMS 2320 Formula Sheet for Final Exam Page 1.ppt

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School
York University
Department
Course
Professor
Marwan Hassan
Semester
Winter

Description
Describe a Single Population Compare 2 Populations Interval Nominal Interval Nominal p p x H :(p  p )0 z  , p  Mean (  ) Variability (F- test ) Case 1 0 1 2 H 0(p1 p2)D Case 2 Mean (  )  p(1 p) n H 1(p1 p2)0 H 1(p1 p2)D Is  known? Variability (c -test ) n Matched v1 n1-1, v2 n21 Pairs (t- test ) Independent (p 1 p ) ˆ 2 YES NO 2 ( n  1) s2 p(1 p) Samples 2 z  x   x   c  2 p z x D  D Are  known?  s1  1 z  t    /2 n t  LCL   2   1 1   / n s / n ( n  1) s 2 s / n  s2  F / 2,1, 2 p(1 p) ˆ    LCL  Rejection Region : D D YES NO  n n  x  z /2 / n) x  t / 2 / n ) c  / 2 , n  1Z   Z , Z  Z  s2   1 2  RejectionRegion: RejectionRegion:   df  n D UCL   1 F Z  Z , Z  Z t  t , t t ( n  1) s2 Z   Z  / 2r Z  Z  / 2  s2   / 2, 2,1 x  x x x   ,n1 ,n1UCL  2 s D  2  p  1 2 ,p  1 ,p  2 Z   /2 Z   /2  /2,n1t  /2,n1 c 1   / 2 , n  1 p  .5 H1:   .5 x D t  / 2 2 1 2 n Population s1 H 0 /1 2 1 n 1 n 2 n 1 n 2 n200, x 175, s 22,  .05 Rejection Region: A B D D 2 Rejection Region: Variances F  2 2 2 H 0  180 H1:  180 2 2 2 2 1 5 7 -2 4 t  t , t  t ( or  ) ? s2 H 0 /1 2 1 The Rejection Region is : X X 1,n1 X X ,n1 2 8 5 3 9  ,n1  ,n1 1 (p 1 p ) 2(p  p 1 2 t  .025 ,1991.972 or 2 2 2 2 . . . . . t   / 2,n1  / 2,n1 (F-test) RR : F  , F   , 1, 2 z  X X 1(/2),n1X /2,n1 H : 0, H : 0 F , 2, 1 p 11 p )ˆ1 p 21 p ) ˆ2 t  .025 ,199972 n  20, s  3800 ,   .05 x   D 14 1.4 0 D 1 D 1  The Value of Test Statistic is : 2 2 D n 10 RR : t .01 ,92.821 F  F or F  F  / 2, 1, 2 n n t 75 18 3.2141 H 0   5000 2 t  1.40  2.6888  / 2, 2, 1 1 2 22 / 200 H :  2 5000 2 SD 1  D   D) ]  2.71116465 / 10 Conclusion: Reject H . There is 1 n 1 n Conclusion:Do not Estimator for Case 1& 2 0 RR : X2  X.95,190.1170 SD  1.6465 Reject0H . There is not …  are NOT known enough evidence to infer that the 2 mean weight of new product is c  (20 1)380010.9744 Equal Unequal ˆ ˆ p1(1p )1 p 21p )ˆ 2 different from 180 lbs. 5000 Yes,  are known Variances Variances (p 1p )2z  / 2  n1 n2 Confidence Interval Estimator: Conclusion: Do not 175  1.972 (22 / 200 ) Reject H0. There is not (x  x ) (   ) (x  x )  (   ) enough evidence to infer
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