Stony Brook University 3
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Stony Brook University *
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Course
201
Subject
Statistics
Date
Jan 9, 2024
Type
Pages
5
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[\
I\:
\’\\\
Stony
Brook
University
“
Name:
\PxO
For
the
sole
use
of
students
at
Stony
Brook
University.
Copyright
©
2019
(Copyright
Dr.
Brenda
J
Anderson)
All
Rights
Reserved
Chapter
9
PRACTICE
PROBLEM
1:
Fifteen years
ago,
the
average
weight
of
a
select
subset
of
the
US
population
between
the
ages
of
30
and
50
was
166
|b.
(p=166).
A
researcher
would
like
to
determine
whether
there
has
been
any
change
in
this
figure
during
the
past
15
years.
A
sample
of
n
=
100
people
within
the
same
subset
and
between
30
and
50
years
old
was
randomly
chosen.
The
average
weight
for
the
sample
was
173
Ib
with
a
standard
deviation
of
23.
Can
the
researcher
conclude
that
there
is
a
significant
change
in
weight?
Use
a
two
tailed
test
with
alpha
=.05
1.1a
State
the
Null
Hypothesis
in
words:
M€
b
N0
dnong(
e
(]
T
p
‘
1.1b
State
the
Null
Hypothesis
in
mathematical
terms:
)
My
1.2a
What
is
your
alpha
level?
O5
df?
1o
\
1.2b
What
is
your
criterion
t
value?
_=
)
\)
Calculate
your
sample
statistic:
Under
these
conditions,
we
use
the
t-test.
The
strategy
for
testing
our
hypothesis
with
a
z-score
or
t-score
is
the
same.
And
that
similarity
is
very
evident
in
this
chapter,
where
we
assume
we
still
know
the
.
We
can
use
the
same
numerator,
M
-y,
but
we
have
to
replace
g,
as
the
denominator.
Just
as
M
estimates
|,
we'll
calculate
the
sample
standard
deviation,
s,
to
estimate.
0.
We'll
use
s
to
calculate
Sm_to
replace
o,
as
the
denominator.
13
|
:
[
oD
£
1
I
\“
\
1.4a
How
would
you
describe
this
to
your
aunt?
1.4b
How
would
you
report
this
in
a
research
paper?
»
g
b
|
Name:
A
S
.
!
For
the
sole
use
of
students
at
Stony
Brook
University.
Copyright
©
2019
(Copyright
Dr.
Brenda
J
Anderson)
All
Rights
Reserved
\V
Stony
Brook
University
|
[t(23)=
>,
p
05;
__
vt
the
Null
Hypothesis].
Is
this
a
“statistically
significant
difference”?
PRACTICE
PROBLEM
2:
Same
idea
but
starting
with
the
raw
data
for
the
sample.
“A
major
corporation
in
the
Northeast
noted
its
employees
averaged
5.8
absences
during
the
winter
season
(December
to
February)
last
year
(u
=
5.8).
The
distribution
is
normal.
In
an
attempt
to
reduce
absences,
the
company
offered
free
flu
shots
to
a
small
random
sample
of
employees.
For
a
sample
of
16
people
who
took
flu
shots,
the
number
of
absences
can
be
seen
in
the
box
below.
Use
a
1-tailed
test
and
alpha
.05,
2.1a
State
the
alternative
hypothesis:
i
Pl
shy
2.1b
State
the
Null
Hypothesis
in
words:
The
A
v
Y
redwu
S
2.1c
State
the
Null
Hypothesis
in
mathematical
terms:
_
|4,
2
M,
=
Au-
2.2
aWhat
is
your
alpha
level?
.
05
df?_Lb
-1
=15
2.2
b
What
is
your
criterion
t
value?
(the
direction/sign
is
determined
by
the
alternative
hypothesis)
—
1.
157
(note
that
in
the
t-table,
like
the
unit
normal
table,
only
positive
values
are
provided.
However,
the
t-distribution,
like
the
normal
distribution
has
negative
values.
When
conducting
a
one-tailed
test,
you
must
decide
whether
the
criterion
t-score
should
be
positive
or
negative.
Like
before,
draw
the
distribution.
Place
the
alpha
in
the
side
stated
by
the
alternative
hypothesis).
If
you
aren’t
sure,
see
the
help
item
(page
2)
in
the
chapter
folder.
Calculate
your
sample
statistic:
Under
these
conditions,
we
use
the
t-test.
The
strategy
for
testing
our
hypothesis
with
a
z-score
or
t-score
is
the
same.
And
that
similarity
is
very
evident
in
this
chapter,
where
we
assume
we
still
know
the
u
We
can
use
the
same
numerator,
M
-,
but
we
have
to
replace
a,,
as
the
denominator.
Just
as
M
estimates
u,
we'll
calculate
the
sample
standard
deviation,
s,
to
estimate
o.
We'll
use
s
to
calculate
sy
to
replace
@,
as
the
denominator.
"\%
;
\\
Stony
Brook
University
Name:
For
the
sole
use
of
students
at
Stony
Brook
University.
Copyright
®
2019
(Copyright
Dr.
Brenda
J
Anderson)
All
Rights
Reserved
X
X2
To
calculate
SS,
s2,
s
and
finally
0
)
Sy,
you
first
need
to
find
¥
X
and
0
(T
X)?
i
1
~
2
SS=
-
2
3
<
3
2
=
———
-
35S
§2
=
3
4
4
/
\
4
r
\
]
AL
5
6
e
6
S
J
oAk
6
Sm
—
E
-
.
\
\
A\
\k:
\
9
S
\
=9
2
/’-—,—
5
Y.
X=5%
XX=
193
-
(05
M
=105
Calculate
the
t-score
(A7
L
c.l
.
a
S@lol
20
-
1
pf
e
M-
pu
T
=T
(,05
Sm
To
make
your decision
about
the
null
hypothesis,
you
need
to
compare
the
statistic
from
your
sample
to
the
criterion
score.
When
we
report
the
t-score,
we
list
the
degrees
of
freedom
in
parentheses
after
the
“t”
then
place
an
“="
sign
and
list
the
value
of
the
score
you
calculated.
Finally,
if
your
score
is
more
extreme
than
the
criterion
score,
then
state
the
following,
p
<.05,
because
more
extreme
scores
have
smaller
tails
in
the
distribution.
If
your
sample
statistic
is
closer
to
O (in
the
center)
than
the
criterion
score,
then
it
is
highly
likely
your
sample
came
from
the
original
t-distribution.
In
that
case,
you
don’t
have
good
enough
evidence
that
the
null
is
false,
and
thus
you
fail
to
reject
the
null
hypothesis.
In
that
case,
you
state
p
>.05.
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4]
W\
.
e
E\‘[\\\
Stony
Brook
University
Name:
For
the
sole
use
of
students
at
Stony
Brook
University.
Copyringr.
Brenda
J
Anderson)
All
Rights
Reserved
2.4a
Draw
a
conclusions
and
report
the
statistics
by
stating
a
conclusion
about
the
effect
and
then
reporting
the
statistics
between
brackets
following
the
sentence
as
shown
here
tC)=_p<
.05;
Reject/Fail
to
Reject
the
Null
Hypothesis].
To
start
you
off,
make
a
statement
that
answers
the
following
question:
Did
flu
shots
reduce
absences?
2.4b
Why
do
you
think
it
is
important
to
list
the
df?
Will
t-scores
have
the
same
probability
when
df
changes?
11
2.5a
What
is
the
estimated
effect
size?
Use
Cohen’s
d
(covered
in
video
3
and
slides
79-82)
2.5b
What
is
the
effect
size?
Use
r?
3a
In
both
examples,
we
had
to
estimate
the
standard
error
of
the
mean,
because
we
didn’t
have
the
population
standard
deviation.
What
did
we
substitute
in
the
formula
to
obtain
this
estimate?
S
When
we
knew
the
population
standard
deviation
(chapters
7
&
8),
we
could
calculate
the
Oy.
That
told
us
with
some
certainty
the
distance
between
the
typical
sample
mean
and
the
population
mean.
Now,
we
have
to
estimate
the
@,
which
leaves
us
less
certain
about
distance
between
sample
means
and
the
population
mean.
In
other
words,
we
don’t
know
the
width
of
our
distribution
of
means.
4
3b
To
be
cautious,
we
judge
samples
means
against
a
distribution
that
is
broader
than
the
normal
distribution.
This
distribution
is
called
__
4
-l
)b
uih
O
3¢
This
type
of
distribution
gets
wider
when
we
have
Lowe
11,
and
narrow
when
we
have
3d
One
way
to
compare
the
normal
distribution
and
the
new
distribution
is
to
look
at
the
difference
in
criterion
scores.
In
chapter
8,
an
alpha
=
.05
and
a
2-tailed
test
gave
us
a
criterion
z-score
of
A
,
regardless
of
sample
size.
Ffl‘:\x
Stony
Brook
University
d
Name:
For
the
sole
use
of
students
at
Stony
Brook
University.
Copyright
©
2019
(Copyright
Dr.
Brenda
J
Anderson)
All
Rights
Reserved
3e
When
working
with
t-distributions,
how
many
subjects
would
we
need
to
have
a
t-
criterion
score
of
+1.96?
(@2
3f
If
we
had
only
20
subjects
in
our
sample,
our
criterion
t-score
would
be
2
:
That
is
nearer/farther
lcircle
one)
from
the
center
of
the
distribution
than
the
criterion
score
for
an
infinite
number
of
sample
subjects.
4.
1magine
we
have
a
sample
mean
of
24
with
a
standard
deviation
of
10
and
a
sample
size
of
25.
The
population
has
a
mean
of
20)The
population
is
normally
distributed.
We
don’t
know
the
population
standard
deviation
so
we
have
to
estimate
the
standard
error
of
the
mean
(sw).
This
produces
a
t-score
because
it
belongs
to
a
wider
distribution
called
the
t-distribution.
Would
you
reject
or
fail
to
reject
the
null
hypothesis?
Assume
you
are
using
a
2-tailed
test
and
alpha
.05.
4.1a
What
is
your
criterion
t-score?
2
:
4.1b
What
is
your
sample
t-score?
4.1c
Would
you
reject
the
null
hypothesis?
_\Q
LV
Now
imagine
that
you
found
the
population
standard
deviation.
It
was
also
10.
Since
this
allows
us to
calculate
the
true
@,,,
we
can
calculate
the
z-score
instead
of
the
t-scores.
Z-scores
are
normally
distributed,
and
therefore
you
can
use
the
unit
normal
table.
In
this
case,
would
you
reject
or
fail
to
reject
the
null
hypothesis?
4.2a
What
is
your
criterion
z-score?
-\
O\.s
A
4.2b
What
is
your
sample
z-score?
VL
e
3
T
L
4.2¢
Would
you
reject
the
null
hypothesis?
A
=
\J
4.3
The
two
conditions
above
differed.
In
the
first,
we
didn’t
know
o,
and
in
the
second,
we
did.
If
your
rejection
of
the
null
was
different
for
the
two
conditions,
explain
why.
Use
your
own
words.
You
may
use
figures.
(if
you
have
trouble,
refer
back
to
video
2,
and
slides
37-75).
0
AU
AN
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The following table lists the approximate number of adults (18+) who are obese (BMI ≥ 25) in North America in 2014 by country and gender.
Female Obese
Male Obese
Female Not Obese
Male Not Obese
Canada
11 M
12 M
7 M
6 M
Mexico
40 M
37 M
21 M
22 M
U.S.A.
101 M
113 M
61 M
44 M
(adapted from http://apps.who.int/gho/data/node.main.A897A?lang=en)
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Answer in decimal form. Round to 3 decimal places as needed.
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Passenger car occupant 13,100
| 4,500
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Source: Virginia Transportation Research Council, "Research Report: The Impact of Red Light Cameras
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Directions: Perform an appropriate significance test to determine whether or not the reduction in the
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RED
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F3
F4
F5
F6
F7
F8
F9
F10
28
F11
#3
2$
&
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7
8
9.
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- Question 12** The following table lists the approximate number of adults (18+) who are obese (BMI ≥ 25) in North America in 2014 by country and gender. Female Obese Male Obese Female Not Obese Male Not Obese Canada 11 M 12 M 7 M 6 M Mexico 40 M 37 M 21 M 22 M U.S.A. 101 M 113 M 61 M 44 M (adapted from http://apps.who.int/gho/data/node.main.A897A?lang=en) Based on this data, what is the probability that a randomly selected male in North America will be obese? Answer in decimal form. Round to 3 decimal places as needed. Your Answer:arrow_forwardThe table below shows the number of deaths in the U.S. in a year due to a variety of causes. For these questions, assume these values are not changing from year to year, and that the population of the United States is 312 million people. Cause Deaths Passenger car occupant 13,100 | 4,500 Motorcycle driver Tornado 553 Skydiving 56 g) People sometimes claim skydiving is less dangerous than driving. Does the data support this claim?arrow_forwardBackground: To prevent crashes caused by running red lights, many states are installing cameras at dangerous intersections. These cameras are used to take photographs of the license plates of vehicles that run a red light. The Virginia Department of Transportation (VDOT) obtained data on the number of crashes per year caused by running a red light at 13 intersections in Fairfax County, Virginia. Source: Virginia Transportation Research Council, "Research Report: The Impact of Red Light Cameras (Photo-Red Enforcement) on Crashes in Virginia", June 2007 Directions: Perform an appropriate significance test to determine whether or not the reduction in the number of crashes was statistically significant. 1. Click on the Data button below to display the data. Copy the data into a statistical software package and click the Data button a second time to hide it. Data Before After 3.7 1.26 0.1 0 4.55 1.69 2.4 1.94 2.09 3.24 2.5 2.72 0.83 0.24 3.05 1.67 3.21 0.33 1.08 0.18 1.25 0.99 7.25 5.02 191…arrow_forward
- Background: To prevent crashes caused by running red lights, many states are installing cameras at dangerous intersections. These cameras are used to take photographs of the license plates of vehicles that run a red light. The Virginia Department of Transportation (VDOT) obtained data on the number of crashes per year caused by running a red light at 13 intersections in Fairfax County, Virginia. Source: Virginia Transportation Research Council, "Research Report: The Impact of Red Light Cameras (Photo-Red Enforcement) on Crashes in Virginia", June 2007 Directions: Perform an appropriate significance test to determine whether or not the reduction in the number of crashes was statistically significant. RED LIGHT PHOTO ENFORCED 1. Click on the Data button below to display the data. Copy the data into a statistical software package and click the Data button a second time to hide it. Data Before After 3.5 1.46 0.47 0.1 0.39 0 4.55 1.69 2.6 2.04 2.39 3.14 2.5 2.62 0.83 0.14 3.25 1.57 3.41…arrow_forwardWhich one of these graphs does not represent a normal distribution? to search F2 F3 F4 F5 F6 F7 F8 F9 F10 28 F11 #3 2$ & 2 4 7 8 9. E R Y U w/arrow_forwardplease answer all questionarrow_forward
- “Radon: The Problem No One Wants to Face” is the title of an article appearing in Consumer Reports. Radon is a gas emitted from the ground that can collect in houses and buildings. At certain levels, it can cause lung cancer. Radon concentrations are measured in picocuries per liter (pCi/L). A radon level of 4 pCi/L is considered “acceptable.” Radon levels in a house vary from week to week. In one house, a sample of 8 weeks had the following readings for radon level (in pCi/L): 1.9 2.8 5.7 4.2 1.9 8.6 3.9 7.2 Find the mean, median, and mode. Find the sample standard deviation, coefficient of variation, and range. Interpretation: Based on the data, would you recommend radon mitigation in this house? Explain.arrow_forwardBackground: To prevent crashes caused by running red lights, many states are installing cameras at dangerous intersections. These cameras are used to take photographs of the license plates of vehicles that run a red light. The Virginia Department of Transportation (VDOT) obtained data on the number of crashes per year caused by running a red light at 13 intersections in Fairfax County, Virginia. RED LIGHT PHOTO ENFORCED Source: Virginia Transportation Research Council, "Research Report: The Impact of Red Light Cameras (Photo-Red Enforcement) on Crashes in Virginia", June 2007 Directions: Perform an appropriate significance test to determine whether or not the reduction in the number of crashes was statistically significant. 1. Click on the Data button below to display the data. Copy the data into a statistical software package and click the Data button a second time to hide it. Data Before After 3.7 1.26 0.37 0.1 0.39 0 4.55 1.69 2.4 1.94 2.09 3.24 2.5 2.72 0.83 0.24 3.05 1.67 3.21…arrow_forwardA certain virus affects 0.5% of the population. A test used to detect the virus in a person is positive 90% of the time if the person has the virus (true positive) and 15% of the time if the person does not have the virus (false positive). Fill out the remainder of the following table and use it to answer the two questions below based on a total sample of 100,000 people. Virus No Virus Total Positive Test Negative Test Total 100,000 a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest hundredth of a percent and do not include a percent sign. % b) Find the probability that a person does not have the virus given that they test negative. Round your answer to the nearest hundredth of a percent and do not include a percent sign.arrow_forward
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