M8 Problem Set
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Course
110
Subject
Mathematics
Date
Jan 9, 2024
Type
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7
Uploaded by MinisterOctopusPerson886
M8:
Problem
Set
Due
No
due
date
Points
5
Questions
5
Time
Limit
None
Attempt
History
Attempt
Time
Score
LATEST
Attempt
1
11,690
minutes
5outof
5
Score
for
this
quiz:
5
out
of
5
Submitted
Sep
16
at
10:02pm
This
attempt
took
11,690
minutes.
Question
1
0/0
pts
A
company
would
like
to
determine
if
there
is
a
difference
in
the
number
of
days
that
employees
are
absent
from
the
East
Side
Plant
compared
to
the
West
Side
Plant. So,
the
company
takes
a
sample
of
54
employees
from
the
East
Side
Plant
and
finds
that
these
people
missed
an
average
of
5.3
days
last
year
with
a
standard
deviation
of
1.3
days.
A
sample
of
41
employees
from
the
West
Side
plant
revealed
that
these
people
were
absent
an
average
of
6.8
days
last
year
with
a
standard
deviation
of
1.8
days.
a)
Find
the
96%
confidence
interval
for
estimating
the
difference
in
the
population
means
(Jq
-
Y2).
b)
Can
you
be
96%
confident
that
there
is
a
difference
in
the
means
of
the
two
populations?
c)
Interpret
the
results.
Your
Answer:
A.
Iny
=54,ny
=41,8;
=
1.3,8,
=
1.8,71
=
5.3,72
=
6.8
_
32
7
—
=
~
(B1
—Z2)
—
24/
—+
=
<pp
—p2
<
(B1
—Z2)
+
24/
—
+
—
ni
N2
|(5.3
—
6.8)
—
2.05\/
L&
4
L8
<
1
—
pp
<
(5.3
—
6.8)
+
2.054
|—2.181
<
p1
—
pg
<
—.8192
B.
96%
confident
that
there
is
a
difference
in
the
mean
of
the
two
population.
C.
Due
to
the
confidence
interval
being
negatve,
96%
confident
that
there
is
a
difference
in
the
mean
of
the
two
population.
Therefore,
people
on
the
West
Side
will
be
missing
more
than
people
from
the
East
Side
Plant.
4
4
Solution.
When
we
look
back
at
table
6.1,
we
see
that
96%
confidence
corresponds
to
z=2.05.
If
we
say
that
the
East
Side
Plant
corresponds
to
population
1
and
the
West
Side
Plant
corresponds
to
population
2,
then:
n4{=54,
n,=41,
s1=1.3,
s,=1.8,
x%
=
5.3,
Xxd
=
6.8,
a)
We
will
use
eqn.
8.1:
L
51
8550
o
s
T
x—x)—z’—+—+<
—h
<(X—%)+z
|
—+
—
(x;
—x;
n
",
W
—Hy
<
(¥
—X;)
n
n,
132
182
132
182
+—
<y
—p;
<(53-68)+
2.05
|—
+
O
—08)
203
feep
Sy
et
A
-2181
<
J;
—
|,
<—.8192
b)
Notice
that
the
entire
96%
confidence
interval
is
negative
(it
is
never
positive
or
zero).
Therefore,
we
can say
that
we
are
96%
confident
that
there
is
a
difference
in
the
two
population
means.
c)
Since
the
entire
confidence
interval
is
negative,
we
can
be
96%
confident
that
(u1
-
y2)
is
negative.
This
means
that
on
average,
people
from
the
West
Side
Plant
will
be
absent
more
days
than
people
from
the
East
Side
Plant.
Question
2
0/0pts
Seven
students
take
a
speed
reading
course
in
order
to
improve
their
reading
ability.
The
reading
scores
of
the
seven
before
and
after
taking
the
course
are
given
below:
Person
1
2
3
4
5
6
7
x
Test
Score
(Before)
350
371
360
400
450
389
410
y
Test
Score
(After)
370
366
365
413
440
420
419
Note
that
n
=
7.
We
will
define
dq
=
x4
-
y4.
After
doing
the
appropriate
calculations,
we
find
that
.
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d
=—9
s;,=
14.12.
Find
the
80
%
confidence
interval
for
mean
of
the
differences,
Jg.
Your
Answer:
Sd
Sd
‘d—t%
<
g
<d+t%
a
14.12
B
14.12
‘
9
—
L441B
<y
<
9+1.44—fi‘
=9
—
7.69
<
pg
<
—9+7.69)
‘—16.69
<
pg
<
—1.31\
Solution.
When
we
look
at
the
student’s
t
chart
for
80%
confidence
(the
80%
Is
found
along
the
bottom
row
of
the
chart)
and
df=7-1=6
(the
6
is
found
in
the
leftmost
column)
we
find
that
t=1.44.
Then
Sd
vn
=
S4
=
d—t—Sreiigad
¥
t
Vn
14.12
14.12
—9-
144
——
<
p,
<
-9+
144
——
V7
V7
—9-7.69
<
py
<—9+
7.69
—16.69
<
py
<
—1.31
Question
3
0/0
pts
A
company
wants
to
buy
new
accounting
software
for
its
employees.
It
must
decide
between
software
A
and
Software
B.
The
company
will
decide
between
the
two
software
packages
based
on
how
quickly
its
employees
can
learn
how
to
use
the
software.
So,
seven
employees
are
chosen
to
test
the
software
packages.
The
number
of
hours
that
it
takes
each
employee
to
learn
each
software
package
is
given
by:
Employee
1
2
3
4
5
6
7
x
Hours
to
learn
Software
A
27
33
22
40
37
38
25
y
Hours
to
learn
Software
B
30
36
25
4
37
35
23
Note
that
.
We
will
define
,
d4
=
x4
-
y4.
After
doing
the
appropriate
calculations,
we
find
that.
d=-1143
s;=
2.80.
Find
the
80
%
confidence
interval
for
mean
of
the
differences,
Mg.
Your
Answer:
d—t=%
d+t—=<
‘
\/fi<,uld<
+
7
~
~
{44280
_
2.80
‘
1.143
—
14428
<y
<
1.143+1.44fi‘
|—1.143
—
1.523
<
pg
<
—1.143
+
1.523|
‘—2.67
<
pg
<
.381
Solution.
When
we
look
at
the
student’s
t
chart
for
80%
confidence
(the
80%
is
found
along
the
bottom
row
of
the
chart)
and
df=7-1=6
(the
6
is
found
in
the
leftmost
column)
we
find
that
t=1.44.
Then
Sd
vn
3
Sa
™
d—t—<|.ld<d+t
Vn
2.80
2.80
—1.
43—
1A~
iy
<1143
144
——
V7
V7
—1.143
-
1.523
<
py;
<
—1.143
+
1.523
—2.67
<
p,
<
.381
Question
4
0/0
pts
Consider
the
following
dependent
random
samples:
Observations
1
2
3
4
5
6
7
8
9
10
x-values
153
160
163
168
171
158
157
166
170
155
y-values
158
162
165
163
17.7
161
163
17.0
173
152
a)
Do
hypothesis
testing
to
see
if
ug
<
0
at
the
a
=.025.
Your
Answer:
HO:Hd:OH1:Md<O
t
=
d
=
—23
—
—1.99
Left
tailed
test,
DOF=10-1=9,
|
g25
=
2.262
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Solution.
Since
we
are
testing
whether
or
not
ud
<
0,
then
our
null
and
alternate
hypothesis
will
be
set
as
follows:
Ho:
ug
=0
Hi:pg
<0
Note
n=10.
If
we
calculate
d4
=
x4
—V4,
...,
d10
=
X410
—
Y10-
Then
find
the
mean
of
the
d’s,
we
get
d*
=
-.23.
The
standard
deviation,
Syq
=
.3653.
-l
d
t=
=
=
(sa/yn)
(3653
/,10)
=159
This
is
a
left-tailed
test.
So,
for
DOF=10-1=9
we
have
t.qo5
=
2.262.
Since
this
is
a
left-tailed
test,
we
will
reject
the
null
hypothesis
if
the
test
statistic
t
(-1.99)
is
less
than
negative
Ta
(-2.262).
Since
t;s
not
less
than
-t.go5,
we
do
not
reject
the
null
hypothesis.
Question
5
513
pts
As
a
reminder,
the
questions
in
this
review
quiz
are
a
requirement
of
the
course
and
the
best
way
to
prepare
for
the
module
exam.
Did
you
complete
all
questions
in
their
entirety
and
show
your
work?
Your
Answer:
Yes
Quiz
Score:
5
out
of
5