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Final

ADMS 2320 - FINAL EXAM FORMULA SHEET 5 STARS!

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Department
Administrative Studies
Course
ADMS 2320
Professor
All Professors
Semester
Fall

Description
Describe a Single Population Compare 2 Populations Under certain conditions, [np > 5 n 1 1 1 (1 -1p ), and n(1-p) > 5], is approximately W  z p(1 p)/ n n p , n (1 - p ) are Interval norma2ly distributed, with  = Nominal  /2 Interval 2 2 2 2 Nominal and  = p(1 - p)/n all > or equal to 5 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 H 1(p1 p2) 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) Samples LCL   2 z  x   x   c  2 p z  /2 Are  known?  s2  F /2,1, 2 z  t   n x D  D  1 1   / n s / n ( n  1) s 2 t  YES NO s12 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  Z   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 s D 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 x D t  /2 F  1 2 2 p  ,p 1 ,p 2  /2  /2  /2,n1  /2,n1 c 1   / 2 , n  1 1 n D Population s2 H 0 /1 12 n  n n n n200, x 175, s 22,  .05 RejectionRegion: 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  .025 ,1991.972 or X X 2 orX X 2 . . . . .  / 2,n1  / 2,n1 (F-test) F  t  .025 ,199972 1/2,n1 /2,n1  D 14 H 0 D0, 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 :   5000 2 n 10 1.4001 ,9 F  F  / 2 , 1, 2 n1 n 2 t 75 180 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 Estimatorfor Case1& 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 )ˆ p (1p )ˆ c  5000 10.9744 Yes,  are known Equal Unequal (p 1p )2z  / 2 1 1  2 2 different from 180 lbs. Variances Variances n n Confidence Interval Estimator: Conclusion: Do not 1 2 Reject H0. There is not 175  1.972 (22 / 200 ) enough evidence to infer (x1x )2(  )1 2 (x1 x )2 (   )  Rejection Region : Z  Z 175  3.0677 LCL = 178.0667 t  t   UCL = 171.9323 that the standard  1 1   s2 s2  deviation will not exceed s 2    1  2  Z  Z  Z  Z  /2 or Z  Z  /2 We estimate with 95% confidence $5,000. pn n   n n  the mean weight of new products 2  1 2   1 2  lies between 178.0667lbs and n  202 , s 12, 2 .05 2 2 2 2 Defective A: x 1 33, n 301 171.9323 lbs. H 0  25, H :1  25 2 (n 11)s 1n 1)2 2 s1 s2 X X 2 orX X 2 S p (x1 x )2 t  / 2  Defective B: x 284, n 3002 Total Number: N x  / 2n) .975,199 .025,199 n 1n 2 n1 n2 1%: only if A is more than 8% smaller than B Sample Size to estimate    (s 2 n  s / n ) 2 p :33/300 .11 p :84 /300 .28 Rejection Region : (x x )t s2  1  1  d.f.  1 1 2 2 1 2 2 2 1 2  /2 p   2 2 2 2  z / 2   1.96  25  Z  Z  , Z  Z  n1 n 2 ( v ) (s1 n1)  (s2 n 2 H 0(p 1 )2.08 H 1(p 1p )2.08 n        96 .04  97 Value of Test Statistic :  W   5  Z  Z  / 2or Z  Z  / 2 vn1n22 n 11 n21 (.11.28)(.08) Sample Size to estimate p , (if p is unknown, take .5) t  t  , v , t  t  , v, t  t  / 2, vor t  t  / 2, v z 2.85 .11(1.11) .28(1.28) 2 2  .05, Present:x  1.04,S 0.31, n 75 1 10 yearsago:x  2.82,S 0.42, n 50 2   z /2 p(1 p) ˆ   1.645 .8(.2)  300 300 n       H 0( 1 ) 2 0 H 1( 1 ) 2 0 2 (751).38 (501).43 2  W   .03  2 2 2 2 Sp .1605     Variances = or  : H0: 1  2 H : 1  11 2 75502 RR : ZZ 2.33 1 1 RR : .01 -2.85 -2.33 n  481.07  482 (Always Round up) RR : F   .5747 or F  F.025,74,49 t  t.05,755021.658 F.025,49,7474 Value of Test Statistic : or p-value =P(Z 5. If les
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