Homework 09 - Answers and Points

Homework 09 - Answers and Points - 
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Unformatted text preview: 
 Homework
Assignment
9
 Psychology
60
 Spring
2010
 
 ‐ 
 For
homework
problems
requiring
calculation,
you
must
show
all
of
your
work,
 including
intermediate
steps,
in
order
to
receive
credit.


 
 
 1) (4
points
total)
A
researcher
has
constructed
an
80%
confidence
interval
from
sample
data,
 and
found
the
interval
to
be
μ
=
45
±
8,
using
a
sample
of
n
=
25
scores.

 a. (1
point)
What
would
happen
to
the
width
of
the
interval
if
the
researcher
used
a
 larger
sample
size
(assuming
other
factors
are
held
constant)?
 The
width
of
this
interval
would
decrease
(the
standard
error
and
the
tconf
would
be
 smaller
leading
to
a
smaller
interval)
 
 b. (1
point)What
would
happen
to
the
width
of
the
interval
if
the
researcher
 constructed
a
90%
confidence
interval
rather
than
an
80%
confidence
interval?

 The
width
of
this
interval
would
increase
(the
tconf
for
a
90%
interval
is
larger
than
 the
tconf
for
an
80%
interval)
 
 c. (1
point)What
would
happen
to
the
width
of
the
interval
if
the
sample
variance
 increased
(assuming
other
factors
are
held
constant)?
 The
width
of
this
interval
would
increase
(the
standard
error
would
be
larger
leading
 to
a
larger
interval)
 
 d. (1
point)What
would
happen
to
the
width
of
the
interval
if
the
mean
of
the
sample
 were
twice
as
large
(assuming
other
factors
are
held
constant)?
 The
width
of
this
interval
would
be
unaffected
(neither
the
standard
error
nor
the
 tconf
would
change
as
a
result
of
the
mean
changing)
 
 
 
 2) (12
points
total)
A
researcher
obtains
a
sample
from
an
unknown
population
and
computes
 a
sample
mean
of
 X = 43 ,
and
a
standard
deviation
of
s
=
6.

 a. (3
points)
If
the
sample
has
n
=
16
scores,
what
is
the
80%
confidence
interval
to
 estimate
the
unknown
population
mean.
 sX = s2 62 36 , sX = , sX = , s X = 1.5 
 n 16 16 
 80% confidence interval with df = 15: t conf = 1.341 
 
 µ = X ± t conf × s X , µ = 43 ± 1.341 × 1.5, µ = 43 ± 2.0115, 

 
 40.9885 < µ < 45.0115 

 b. (3
points)
If
the
sample
has
n
=
36
scores,
what
is
the
80%
confidence
interval
to
 estimate
the
unknown
population
mean.

 
 sX = s2 , sX = n 62 ,s = 36 X 36 , s = 1
 36 X 
 80% confidence interval with df = 35 (actually df=30): t conf = 1.310 
 
 µ = X ± t conf × s X , µ = 43 ± 1.310 × 1.0, µ = 43 ± 1.310, 41.69 < µ < 44.31 
 
 
 c. (3
points)
If
the
sample
has
n
=
16
scores,
what
is
the
95%
confidence
interval
to
 estimate
the
unknown
population
mean.

 
 sX = s2 62 36 , sX = , sX = , s X = 1.5 
 n 16 16 
 95% confidence interval with df = 15: t conf = 2.131 
 
 µ = X ± t conf × s X , µ = 43 ± 2.131 × 1.5, µ = 43 ± 3.1965, 39.8035 < µ < 46.1965 
 
 
 d. (3
points)
If
the
sample
has
n
=
36
scores,
what
is
the
95%
confidence
interval
to
 estimate
the
unknown
population
mean.

 
 sX = s2 , sX = n 62 ,s = 36 X 36 , s = 1
 36 X 
 95% confidence interval with df = 35 (actually df=30): t conf = 2.042 
 
 µ = X ± t conf × s X , µ = 43 ± 2.042 × 1.0, µ = 43 ± 2.042, 40.958 < µ < 45.042 
 
 3) (5
points
total)
Standardized
measures
seem
to
indicate
that
the
average
level
of
anxiety
 has
increased
gradually
over
the
past
50
years.
In
the
1950s,
the
average
score
on
the
Child
 Manifest
Anxiety
Scale
was
μ
=
15.1.
Suppose
a
sample
of
n
=
16
children
today
produces
a
 mean
score
of
 X = 23.3 ,
and
SS
=
240.

 a. (1
point)
Based
on
the
sample,
make
a
point
estimate
of
the
population
mean
anxiety
 score
for
children
today.

 The
best
point
estimate
is
the
observed
sample
mean,
so
the
best
point
est
is
μ
=
23.3
 

 b. (4
points)
Construct
a
90%
confidence
interval
estimating
the
population
mean
of
 children
today
 
 s2 = SS 2 240 2 ,s= , s = 16 













 s X = df 15 s2 16 , sX = , s = 1
 n 16 X 
 90% confidence interval with df = 15: t conf = 1.753 
 
 µ = X ± t conf × s X , µ = 23.3 ± 1.753 × 1.0, µ = 23.3 ± 1.753, 21.547 < µ < 25.053 
 
 4) (7
points
total)
A
researcher
would
like
to
determine
how
physical
endurance
is
affected
by
 a
common
herbal
supplement.
The
researcher
measures
endurance
for
a
sample
of
n
=
9
 participants.
Each
individual
is
then
given
a
30‐day‐suppy
of
herbs
and,
1
month
later,
 endurance
is
measured
again.

For
this
sample,
endurance
increased
by
an
average
of
 X D = 6 
 with
SSD
=
216.
 
 a. (1
point)
Based
on
the
sample,
make
a
point
estimate
of
the
population
mean
 difference
in
endurance
for
individuals
taking
the
herbs.

 
 The
best
point
estimate
is
the
observed
sample
mean,
so
the
best
point
est
is
μD
=
6
 
 b. (4
points)
Construct
a
95%
confidence
interval
estimating
the
population
mean
 difference
in
endurance
for
individuals
taking
the
herbs
 
 2 sD = SSD 216 2 , s2 = , sD = 27 









 s XD = df 8 2 sD , sXD = n 27 , s XD = 3, s XD = 1.732 
 9 
 95% confidence interval with df = 8: t conf = 2.306 
 
 µ D = ( X D ) ± t conf × s XD , µ D = (6 ) ± 2.306 × 1.732, µ D = (6 ) ± 3.994, 2.006 < µ D < 9.994 
 
 c. (2
points)
Describe
what
this
confidence
interval
implies
about
whether
the
herbs
 have
an
effect
on
endurance
 Because
zero
is
not
contained
in
the
95%
confidence
interval,
we
have
(at
least)
95%
 confidence
that
the
true
population
mean
difference
score
is
not
0,
implying
that
 there
is
some
effect
of
herbs.
In
fact,
we
have
95%
confidence
the
true
mean
 difference
score
is
between
about
2
and
10,
so
it
is
within
that
range
that
we
believe
 the
true
value
of
μD
to
be.

 
 
 5) (11
points
total)
A
developmental
psychologist
wants
to
determine
whether
infants
display
 any
color
preferences.
A
stimulus
consisting
of
four
color
patches
(red,
green,
blue
and
 yellow)
is
projected
onto
the
ceiling
above
a
crib.
Instants
are
placed
in
the
crib,
on
at
a
time,
 and
the
psychologist
records
how
much
time
each
infant
spends
looking
at
each
of
the
four
 colors.
The
color
that
receives
the
most
attention
during
a
100‐second
test
period
is
 identified
as
the
preferred
color
that
infant.
The
preferred
colors
for
a
sample
of
60
infants
 are
shown
in
the
following
table:
 
 Red
 Blue
 Yellow
 12
 
 Green
 20
 10
 18
 a. (2
points)
What
are
the
null
and
alternative
hypotheses
for
this
experiment?
 H0:
There
is
no
preference
in
the
population
for
red,
green,
blue
or
yellow


 or
H0:
The
percent
of
preference
in
the
population
for
these
colors
is
each
25%
 
 H1:
There
is
some
preference
in
the
population
for
red,
green,
blue
or
yellow


 or
H1:
The
percent
of
preference
in
the
population
for
these
colors
is
NOT
each
25%
 
 
 b. (2
points)
What
is
the
critical
value
of
the
relevant
test‐statistic?
 
 2 df = (c − 1), df = ( 4 − 1), df = 3, for α = 0.05 χ critical = 7.81 
 
 
 c. (2
points)
What
are
the
expected
values
for
each
of
the
colors
in
this
experiment?
 
 Expected
values
=
n/c
 
 Red
 Green
 Blue
 Yellow
 60/4
=
15
 60/4
=
15
 60/4
=
15
 60/4
=
15
 
 
 
 d. (5
points)
Do
the
data
indicate
any
statistically
significant
preference
among
the
four
 colors?
Test
using
 α = 0.05 
and
explain
what
this
means
for
the
original
research
 question.
 
 ⎛ ( f o − f e )2 ⎞ 2 χ observed = ∑ ⎜ ⎟ fe ⎝ ⎠ ⎛ (12 − 15 )2 ⎞ ⎛ (20 − 15 )2 ⎞ ⎛ (10 − 15 )2 ⎞ ⎛ (18 − 15 )2 ⎞ 2 χ observed = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ 15 15 15 15 ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎛ 9 ⎞ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎛ 9 ⎞ 2 χ observed = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 15 ⎠ ⎝ 15 ⎠ ⎝ 15 ⎠ ⎝ 15 ⎠ 2 χ observed = 0.6 + 1.66 + 1.66 + 0.6 2 χ observed = 4.533 
 
 
 
 2 2 χ observed = 4.533, χ critical = 7.81 
 2 2 χ observed < χ critical , Fail to reject H 0 The
data
do
not
provide
evidence
for
any
color
preference
for
infants
at
this
age.
 
 6) (12
points
total)
A
local
county
is
considering
a
budget
proposal
that
would
allocate
extra
 funding
toward
the
renovation
of
city
parks.
A
survey
is
conducted
to
measure
public
 opinion
concerning
the
proposal.
A
total
of
150
individuals
responded
to
the
survey:
50
who
 live
within
the
city
limits,
and
100
who
live
in
the
surrounding
suburbs.
The
local
county
 wants
to
determine
if
there
is
any
difference
in
preference
for
people
who
live
in
the
city
and
 people
who
live
in
the
surrounding
suburbs.
The
frequencies
of
responses
are
as
follows:
 
 Opinion
 
 Favor
 Oppose
 City
 35
People
 15
People
 Suburb
 55
People
 45
People
 
 a. (2
points)
What
are
the
null
and
alternative
hypotheses
for
this
study?
 
 H0:
Region
living
and
opinion
on
funding
are
independent

 
 H1:
Region
living
and
opinion
on
funding
are
not
independent
 
 
 b. (2
points)
What
is
the
critical
value
of
the
relevant
test‐statistic?
 
 2 df = (r − 1)(c − 1), df = (2 − 1)(2 − 1), df = 1, for α = 0.05 χ critical = 3.84 
 
 
 c. (3
points)
What
are
the
expected
values?
 
 ff fe = c r 
 n 
 
 Expected
 Opinion
 
 Favor
 Oppose
 City
 60*50/150
=
20
 Suburb
 
 
 
 90*50/150
=
30
 90*100/150
=
60
 60*100/150
=
40
 
 
 d. (5
points)
Do
the
data
indicate
a
statistically
significant
difference
in
the
preference
 of
those
people
living
in
the
city
and
those
people
living
in
the
suburbs?
Test
using
 α = 0.05 
and
explain
what
this
means
for
the
original
research
question.

 
 
 χ 2 observed ⎛ ( f o − f e )2 ⎞ = ∑⎜ ⎟ f ⎝ ⎠ e ⎛ ( 35 − 30 )2 ⎞ ⎛ (15 − 20 )2 ⎞ ⎛ (55 − 60 )2 ⎞ ⎛ ( 45 − 40 )2 ⎞ 2 χ observed = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ 30 20 60 40 ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎛ 25 ⎞ 2 χ observed = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 30 ⎠ ⎝ 20 ⎠ ⎝ 60 ⎠ ⎝ 40 ⎠ 
 2 χ observed = 0.833 + 1.25 + 0.4166 + 0.625 2 χ observed = 3.125 
 
 
 2 2 χ observed = 3.125, χ critical = 3.84 2 2 χ observed < χ critical , Fail to reject H 0 
 The
data
do
not
provide
evidence
that
opinions
are
related
to
the
region
in
which
 individuals
live.

 
 ...
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This note was uploaded on 10/16/2011 for the course PSYCH 60 taught by Professor Parris during the Spring '11 term at UCSD.

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