Formula Sheet

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22 Apr 2012
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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
1 n2 v21,- n1 v1
s
s
F
F
s
s
UCL
F
s
s
LCL
D
D
D
D
DD
DD
n
s
tx
ndf
ns
x
t
2/
1
/
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)(df
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?
Nominal Interval
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
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
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
12
22
n
D
D
n
SD
6465.1
D
S
2/2/
,:
ZZorZZ
ZZZZRR
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/
2,1,
1,2,
1
,
1
:RR
FFor
F
F
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 t
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/
,:RR
ZZorZZ
ZZZZ
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) =0.0022
0.0022 < .01() p-value< ,Reject H0
-2.85 -2.33
Conclusion: Reject H0, There is enough
evidence to conclude
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:
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