stat 302 case study 6
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DESCRIPTION
Purdue North CentralStat 302Case Study 6This is the 6th case study required for the Stat 302 class at PNC taught by Professor K.TRANSCRIPT
STAT 302
Case Study 6
4
6
4
1
3
6
3
1
2
6
2
1
1
6
1
1
6
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
5
9
5
U
C
L
=
1
2
.
4
7
2
L
C
L
=
4
.
7
1
8
X
b
a
r
C
h
a
r
t
o
f
C
o
l
1
4
6
4
1
3
6
3
1
2
6
2
1
1
6
1
1
6
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
5
9
5
U
C
L
=
1
2
.
4
7
2
L
C
L
=
4
.
7
1
8
X
b
a
r
C
h
a
r
t
o
f
C
o
l
1
The X bar chart shows that the process is in control. We are basing this off of our assessment of the WECO rules.
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
2
5
0
U
C
L
=
1
2
.
1
7
3
L
C
L
=
4
.
3
2
8
X
b
a
r
C
h
a
r
t
o
f
C
o
l
1
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
2
5
0
U
C
L
=
1
2
.
1
7
3
L
C
L
=
4
.
3
2
8
X
b
a
r
C
h
a
r
t
o
f
C
o
l
1
Based on the Waco rules we have determined that this Xbar chart is out of control. Between the points of 5,6, and 7 there are more than 3 standards deviations on either side.
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
2
7
1
U
C
L
=
1
2
.
0
0
0
L
C
L
=
4
.
5
4
3
X
b
a
r
C
h
a
r
t
o
f
C
o
l
2
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
2
7
1
U
C
L
=
1
2
.
0
0
0
L
C
L
=
4
.
5
4
3
X
b
a
r
C
h
a
r
t
o
f
C
o
l
2
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
5
4
9
U
C
L
=
1
2
.
4
6
3
L
C
L
=
4
.
6
3
5
X
b
a
r
C
h
a
r
t
o
f
C
o
l
3
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
5
4
9
U
C
L
=
1
2
.
4
6
3
L
C
L
=
4
.
6
3
5
X
b
a
r
C
h
a
r
t
o
f
C
o
l
3
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
4
8
9
U
C
L
=
1
2
.
5
1
4
L
C
L
=
4
.
4
6
3
X
b
a
r
C
h
a
r
t
o
f
C
o
l
4
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
4
8
9
U
C
L
=
1
2
.
5
1
4
L
C
L
=
4
.
4
6
3
X
b
a
r
C
h
a
r
t
o
f
C
o
l
4
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
5
5
8
U
C
L
=
1
2
.
5
4
5
L
C
L
=
4
.
5
7
2
X
b
a
r
C
h
a
r
t
o
f
C
o
l
5
2
8
2
5
2
2
1
9
1
6
1
3
1
0
7
4
1
1
3
1
2
1
1
1
0
9
8
7
6
5
4
S
a
m
p
l
e
S
a
m
p
l
e
M
e
a
n
_
_
X
=
8
.
5
5
8
U
C
L
=
1
2
.
5
4
5
L
C
L
=
4
.
5
7
2
X
b
a
r
C
h
a
r
t
o
f
C
o
l
5