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Final

ADMS 2320 - FINAL EXAM FORMULA SHEET 5 STARS!

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Department
Administrative Studies
Course Code
ADMS 2320
Professor
all

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Rejection Region:
Compare 2 Populations
Mean ( )
Nominal
Interval
Independent
Samples
Are known?
Matched
Pairs (t- test )
Equal
Variances Unequal
Variances
Variability (F- test )
2
2
2
1
1
1
21
21
21
21
n
x
p
ˆ
,
n
x
p
ˆ
,
nn
xx
p
ˆ
n
1
n
1
)p
ˆ
(1p
ˆ
)p
ˆ
p
ˆ
(
z
2
22
1
11
2121
n)p
ˆ
(1p
ˆ
n)p
ˆ
(1p
ˆ
)p(p)p
ˆ
p
ˆ
(
z
Case 1
YES NO
2
2
2
1
1,2,2/
2
2
2
1
2,1,2/
2
2
2
11
s
s
F
F
s
s
UCL
Fs
s
LCL
D
D
2/D
DD
DD
n
s
tx
n/s
x
t
1
)(
1
)(
)/(
d.f.
)(
)()(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
2
2
2
1
2
1
2/21
2
2
2
1
2
1
21
n
ns
n
ns
nsns
n
s
n
s
txx
n
s
n
s
xx
t
2 n2 n1 v
11
)(
2
)1()1(
11
)()(
21
2
2/21
21
2
22
2
11
2
21
2
2121
nn
stxx
nn
snsn
S
nn
s
xx
t
p
p
p
2
22
1
11
2/21 n)p
ˆ
1(p
ˆ
n)p
ˆ
1(p
ˆ
z)p
ˆ
p
ˆ
(
2 & 1 Case for Estimator
9704.96
5
2596.1
estimate toSize
2
2
2/
W
z
n
Sample
Case 2
Population
Variances
(F-test)
Describe a Single Population
Mean ( )
Is known?
NominalInterval
Variability (c- test )
YES NO
)/(
/
2/ nstx
ns
x
t
)/(
/
2/ nzx
n
x
z
21,2/1
2
21,2/
2
2
2
2
)1(
)1(
)1(
n
n
sn
UCL
sn
LCL
sn
c
c
c
n
pp
zp
n
x
p
n
pp
pp
z
)
ˆ
1(
ˆ
ˆ
ˆ
,
)1(
ˆ
2/
05.,22,175,200
sxn
180:180:10
HH
2/2/
,
:
ZZorZZ
ZZZZ
egionRejectionR
1,2/1,2/
1,1, ,
:
nn
nn
ttortt
tttt
egionRejectionR
:isegionRejectionRThe
:isStatisticTestofValueThe
ortt 972.1
199,025.
2141.3
200/22
180175
t
Conclusion: Reject H0. There is
enough evidence to infer that the
mean weight of new product is
different from 180 lbs.
972.1
199,025. tt
)200/22(972.1175
0677.3175
Confidence Interval Estimator:
LCL = 178.0667
UCL = 171.9323
We estimate with 95% confidence
the mean weight of new products
lies between 178.0667lbs and
171.9323 lbs.
21,2/
22 1,2/1
2
21,
22 1,1
2,
:
nn
nn
XXorXX
XXXX
egionRejectionR
2/2/
,
:
ZZorZZ
ZZZZ
egionRejectionR
05.,3800,20
sn
22
05000:
H
22
15000:
H
1170.10:219,95.
2XXRR
9744.10
5000
3800)120(
2
2
2
c
Conclusion: Do not
Reject H0. There is not
enough evidence to infer
that the standard
deviation will not exceed
$5,000.
1,2/1,2/
1,1, ,
:
nn
nn
ttortt
tttt
egionRejectionR
A B D D2
2 8 5 3 9
1 5 7 -2 4
. . . . .
4.1
10
14
n
D
xD
7111.2
1
/)( 22
2
n
nDD
SD
6465.1
D
S
2/2/
,
:
ZZorZZ
ZZZZ
egionRejectionR
5.:5.: 10
HpH
0:,0: 10 DD HH
821.2: 9,01.ttRR
6888.2
10/6465.1
04.1
t
Conclusion: Do not
Reject H0. There is not …
Yes, are known
vvvv ttortttttt ,2/,2/,, ,,
( v )
up) Round (Always 48207.481
03.
)2(.8.645.1)
ˆ
1(
ˆ22
2/
n
W
ppz
n
Sample Size to estimate p, (if p is unknown, take .5)
0)(:0)(: 211210
HH
50 n 0.43, S ,82.2x :ago years 10 75 n 0.38, S ,04.3x :Present .05, 222111
Variances = or :
1/:H 2
2
2
10
1/:H 2
2
2
11
RR :
8.1or 5747.
74.1
11 49,74,025.
74,49,025.
FF
F
F
2,1,2/
1,2,2/
1
FF
F
F
1/: 2
2
2
10
H
1/: 2
2
2
10
H
8837.
43.
38.
2
2
2
1s
s
F
.5747 1.8
Do not Reject H0, Variances are equal
1605.
25075
43).150(38).175(22
2
p
S
Value of Test Statistic :
RR :
658.1
25075,05.
tt
 
0078.3
50/175/11605.
0)82.204.3(
t
1.658 3.0x
Conclusion: Reject H0, there is sufficient evidence
the mean GPA now exceed the mean 10 years ago.
88.1575.233.2
96.1645.1282.1
03.005.01.
025.05.10.
ZZZ
ZZZ
0)(:
0)(:
211
210
ppH
ppH
DppH
DppH
)(:
)(:
211
210
08.)(:08.)(: 211210 ppHppH
300 n ,33x :A Defective 11
300 n ,84x :B Defective 22
11.300/33:
ˆ1p
28.300/84:
ˆ2p
2/2/
:egionRejectionR
ZZorZZZZ
ZZ
RR :
33.2
01.ZZ
1%: only if A is more than 8% smaller than B
85.2
300
)28..28(1
300
)11..11(1
.08)(.28)(.11
z
Value of Test Statistic :
or p-value =P(Z<2.85) = .5 .4978 = .0022
.0022 < .01() Reject H0
-2.85 -2.33
Conclusion: Reject H0, There is enough
evidence to conclude that management
should adopt process A.
are NOT known
25:,25:2
1
2
0
HH
05.,12,200 2
sn
2199,025.
22 199,975.
2XXorXX
 
)(
2/ nstxN
Total Number:
Under certain conditions, [np > 5
and n(1-p) > 5], is approximately
normally distributed, with = p
and 2 = p(1 - p)/n
nppzW /)
ˆ
1(
ˆ
2/
n1p1, n1(1 - p1),
n2p2, n2(1 - p2) are
all > or equal to 5

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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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