M8 Exam
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Course
110
Subject
Statistics
Date
Jan 9, 2024
Type
Pages
13
Uploaded by ChefOwl3912
M8:
Exam
-
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due
date
Points
50
Questions
5
Time
Limit
90
Minutes
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Instructions
You
may
only
have
the
following
items
when
taking
an
exam:
computer,
1-2
pieces
of
blank
scratch
paper,
a
pen/pencil,
and
a
calculator.
You
may
ONLY
use
the
equation
sheets
that
are
provided
WITHIN
the
exam.
The
use
of
printed
versions
will
be
considered
a
violation
of
the
Academic
Integrity
Policy.
Attempt
History
Attempt
Time
Score
LATEST
Attempt
1
81
minutes
38 out
of
50
Score
for
this
quiz:
38
out
of
50
Submitted
Oct
21
at
10:34am
This
attempt
took
81
minutes.
Question
1
8/10
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
=
table.pdf)
Suppose
we
have
independent
random
samples
of
size
nq
=
420
and
n,
=
510.
The
proportions
of
success
in
the
two
samples
are
p1=
.38
and
p,
=
43.
Find
the
99%
confidence
interval
for
the
difference
in
the
two
population
proportions.
Answer
the
following
questions:
1.
Multiple
choice:
Which
equation
would
you
use
to
solve
this
problem?
A.
F=F)—
2
[—+=—4+<
y—Yty
<(FH
—%)+
z
|—
+
—
g
n,
n;
n
n
B.
p_Pp,
4z
\lm(l-px)
4
P2(1-P3)
n,
nz
@
-
)
-
(M
—
1)
c
$i2
L
52
n,
n,
X
S
=
Sa
D.
d-t=<jy<d+
t==
2.
List
the
values
you
would
insert
into
that
equation.
3.
State
the
final
answer
to
the
problem
Your
Answer:
1.B
2.
P1=.38, P2=.43,
n1=420,
n2=510,
z=2.576
.38(1—.38)
N
43(1—.43)
38
—
.43
+
2.576\/
=
=0
—.05
%
2.576(.0010)|
3.
so
the
interval
is
-.047424
,
-.052576
From
table
6.1,
we
see
that
99%
confidence
corresponds
to
z=2.58.
Notice
that
the
sample
sizes
are
each
greater
than
30,
SO
wWe
may
use
eqn.
8.2:
i
2
P1(1"P1)+P2(1“P2)
P.
—
1
n,
B
38(1
=
38)
.43(1
—
43)
510
—.05
+
2.58(.03227)
So, the
interval
is
(.-.1333,.03326).
Question
2
10
/
10
pts
4
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
5
(https://iprevious.nursingabc.com/upload/images/Help_file
picture/Statistic
standard
normal
table
=
(https://previous.nursingabc.com/upload/images/Help_file
picture/standar:
t-table
=
(https://previous.nursingabc.com/upload/images/Help_file
picture/t-
table.pdf)
Your preview ends here
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4
In
certain
hospital,
nurses
are
required
to
constantly
make
rounds
to
check
in
on
all
of
the
patients.
The
nursing
supervisor
would
like
to
know
there
is
a
difference
between
the
number
of
rounds
completed
per
shift
k
the
nurses
on
the
day
shift
compared
to
the
nurses
on
the
night
shift.
So
the
nursing
supervisor
checks
the
records
of
77
day
shift
nurses
and
finc
that
they
complete
an
average
(a
mean)
of
32
rounds
per
shift
with
a
standard
deviation
of
4.1
rounds
per
shift.
The
nursing
supervisor
also
checks
the
records
of
72
night
shift
nurses
and
finds
that
they
complete
an
average
(a
mean)
of
26
rounds
per
shift
with
a
standard
deviation
of
6.3
rounds
per
shift.
a)
Find
the
95%
confidence
interval
for
estimating
the
difference
in
the
population
means
(1
-
Mo).
b)
Can
you
be
95%
confident
that
there
is
a
difference
in
the
means
of
tt
two
populations?
Answer
the
following
questions:
1.
Multiple
choice:
Which
equation
would
you
use
to
solve
this
problem?
A,
(x,-%)
512+322+<
<
(%
—%,)+
312+S22
1
2
,
n,
Hy
—
Ha
1
2
,
,
B.
P
Pk
Z\jm(l-m)_l_pz(l-pz)
n;
nz
3
-%)
-
(g
—
o)
C.
Jfi+§z_2.
ny
n;
Jot
oo
iz
d
s
52
2.
List
the
values
you
would
insert
into
that
equation.
2
Qtata
the
final
ancwer
tn
thea
nrnhlam
Your
Answer:
1.
A
2.n1=77,n2=72,s1=4.1,
s2=6.3,
x1=32,
x2=26,
z=1.960
2
2
|(32
—
26)
—
1.960\/%
+
%5
<
—
p2
<
(32—
26)
+
1.9604
3.14.2806
<
p1
—
pa
<
7.7194)
interval
is
positive
so
we
can
be
confident
that
there
is
a
difference
in
the
means
of
the
2
populations
<
4
When
we
look
back
at
table
6.1,
we
see
that
95%
confidence
corresponds
to
z=1.96.
If
we
say
that
the
day
shift
nurses
corresponds
to
population
1
and
the
night
shift
nurses
corresponds
to
population
2,
then:
n{=77,
no=72,
s1=4.1,
$,=6.3,
X1=32,
X,=26
a)We
will
use
equation
8.1:
S
_
’51:
5,7
L
-
5,°
X.—X>)—2Z|—+
<
—
B
L
(Xy—
X))+
Z
+
(%,
_)
.
i
My
—
U2
(
1
2)
n,
4.1°
o
6.3°
77
72
(32—
26)—
1.96
<py—ps<(32-26)+1
4281
<pu,—u.<
17719
1.
b)
Since
the
entire
confidence
interval
is
positive,
we
can
be
95%
sure
that
there
is
a
difference
in
the
means
of
the
two
populations.
Question
3
8/10
pts
You
may
find
the
following
files
helpful
in
throughout
the
exam:
Statistics_Equation_Sheet
=
(https://previous.nursingabc.com/upload/images/Help_file_
picture/Statistics
standard
normal
table
/=
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
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t-table
>
table.pdf)
A
head
librarian
supervises
a
number
of
libraries
in
a
large
county.
He
wants
to
know
if
full-time
library
workers
and
part-time
library
workers
re-
shelve
books
at
the
same
rate.
So, he
checks
the
records
of
45
full-time
library
workers
and
finds
that
they
re-shelve
an
average
of
166
books
per
hour
with
a
standard
deviation
of
9.3
books
per
hour.
The
records
of
45
part-time
library
show
that
they
re-shelve
an
average
of
159
books
per
hour
with
a
standard
deviation
of
12.2
books
per
hour.
Using
a
level
of
significance
of
a=.01,
is
there
enough
evidence
to
indicate
a
difference
in
the
mean number
of
books
re-shelved
by
full-time
workers
compared
to
part-time
workers?
Answer
the
following
questions:
1.
Multiple
choice:
Which
equation
would
you
use
to
solve
this
problem?
A
(x,-%)
S‘2+Szz+<
<
(%,
—%,)
+
S‘2+s22
1
2
n
,
Hi
—
H2
1
2
,
,
B.
S
N
Jm(l—m)
4
P2(-p2)
n,
n;
@
-x)
-
(M
—
1)
C.
Jfi.flfi
ny
n,
P
TR
Y
D
\/H
Ha
\/H
2.
List
the
values
you
would
insert
into
that
equation.
3.
State
the
final
answer
to
the
problem
Your
Answer:
1.C
2.
x1=166, x2=159,
s1=9.3
,s2=12.2,
n1=45
,
n2=45
.01/2=.005
z=2.576,
z=-2.576
166—159)—0
LRI
D
T
—3.040
9.32
+12.22
1/5.30
45
45
3.
z-score
is
>
null
hypothesis
so
we
will
reject
The
null
hypothesis
is
that
there
is
no
difference
between
the
mean
number
of
books
re-shelved
by
the
full-time
and
part-
time
workers:
Ho:pq-u2=0
Hq
:
gy
-
g2
#0.
Since
this
is
a
two-tailed
test,
we
must
find
the
z
that
satisfies:
P(Z<z)=.01/2=.005
and
P(Z
>
2)=.01/2=.005.
In
the
standard
normal
table,
z=-2.576
and
z=2.576.
We
will
reject
the
null
hypothesis
if
the
z-score
is
less
than
-2.576
or
the
z-score
is
greater
than
2.576.
We
now
find
the
z-score:
(f1—%)—-0
(166
—159)
—
0
51%
.
§p*
J9.32
12.22
J
n,
o
15
*
715
= 3.06.
Since
the
z-score
is
above
2.576,
we
reject
the
null
hypothesis.
State
the
null
and
alternative
hypothesis.
State
the
critical
value.
Question
4
6/10
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
(https://iprevious.nursingabc.com/upload/images/Help_file
picture/Statistics
standard
normal
table
=
(https://previous.nursingabc.com/upload/images/Help_file
picture/standardn
Your preview ends here
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t-table
=
(https://previous.nursingabc.com/upload/images/Help_file
picture/t-
table.pdf)
Consider
the
following
dependent
random
samples
Observations
1
2
3
4
5
6
x-values
155
140
142
159
162
169
y-values
169
150
151
145
173
180
a)
Determine
the
difference
between
each
set
of
points,
Xx;
-
y;
b)
Do
hypothesis
testing
to
see
if
uyy
<
0
at
the
a
=
.05.
Your
Answer:
n=6
t=.05=2.447
fpr
6-1=5
(L
tail)
me
5,—
/20
t<t.05
so
accept
null
hypothesis
Since
we
are
testing
whether
or
not
ud
<
0,
then
our
null
and
alternate
hypothesis
will
be
set
as
follows:
Ho:
Mlg=0
H4y:pg<0
There
are
6
data
points.
So,
n=6.
This
is
a
left-tailed
test.
Note
that
for
t.o5
=
-2.015
for
6-1
=
5.
We
find
the
mean
in
the
usual
way:
Then
using
the
mean,
d
=
-6.833,
and
the
standard
deviation,
sq=
10.34,
that
we
found
above:
—6.833
=
—1.619.
d
TGl
(1038706
Since
t
>
t.p5,
we
do
not
reject
the
null
hypothesis.
State
the
null
and
alternative
hypothesis.
State
the
critical
value.
Double
check
the
Sd
value
and
the
d-bar
value.
Question
5
6/10
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
(https://previous.nursingabc.com/upload/images/Help_file_picture/Statistics
standard
normal
table
=
(https://previous.nursingabc.com/upload/images/Help_file
picture/standardn
t-table
=
(https://previous.nursingabc.com/upload/images/Help_file_
picture/t-
A
new
energy
drink
is
supposed
to
improve
a
person’s
time
in
the
one
mile
run.
The
times,
in
seconds,
of
eight
runners
with
and
without
the
drink
are
given
below:
Runner
1
2
3
4
5
6
7
8
x-time
(before)
260
280
275
295
267
247
250
241
y-time
(after)
255
270
281
275
260
243
255
239
Find
the
90
%
confidence
interval
for
mean
of
the
differences,
ug.
Answer
the
following
questions:
1.
Multiple
choice:
Which
equation
would
you
use
to
solve
this
problem?
A
(x,-%)
S‘2+Szz+<
<
(%,
—%,)
+
S‘2+s22
1
2
n
,
Hi
—
H2
1
2
,
,
B.
Pl
-
PZ
+
ZJPl(l'p1)+Pz(1’pz)
ny
nz
@
-x)
-
(M
—
1)
C.
Jfi.flfi
ny
n;
Tt
i
e
d
g
¢
2
D
\/E
Ha
\/H
2.
List
the
values
you
would
insert
into
that
equation.
3.
State
the
final
answer
to
the
problem
Your
Answer:
1.D
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2.
n=8,
d1=x1-y1
mean=-4.625
standard
deviation=11.236
dof=8-1=7
t=2.998
_
_
11.236
.
11.236
‘
4.625
—
2.998120
<
pi;
<
—4.625
+
2.998
1128
3.|—16.535
<
pg
<
7.285
Note
that
n=8.
We
will
define
,
d;
=
x;
-
y;.
After
doing
the
appropriate
calculations,
we
find
that
d
=4.625
s4=
8.314.
When
we
look
at
the
student’s
t
chart
for
90%
confidence
(the
90%
is
found
along
the
bottom
row
of
the
chart)
and
DOF=8-
1=7
(the
df=7
is
found
in
the
leftmost
column)
we
find
that
t=1.895.
Then
D.
-
Sd
-
Sd
d-t
—=
<
<d+t—
e
Vn
8.314
8.314
4625-
1895
——
<
<
4625+
1.895
——
v
V8
4625
-
557
<
py
<
4.625+
5.57
Quiz
Score:
38
out
of
50
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