Homework 07 - Answers and Points

Homework 07 - Answers and Points - 
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Unformatted text preview: 
 Homework
Assignment
7
 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
sample
of
n
=
16
individuals
participates
in
a
repeated‐measures
study
that
 produces
a
sample
mean
difference
of
 X D = 9 
with
SS
=
960.

 a. (1
point)
Calculate
the
variance
for
the
sample
of
difference
scores.
 SS 960 2 2 2 sD = D , sD = , sD = 64 
 df 15 b. (1
point)
Briefly
explain
what
the
measured
by
the
variance
of
difference
scores
 The
variance
of
the
difference
scores
is
a
measure
of
the
consistency
of
the
effect.

 
 c. (1
point)
Calculate
the
estimated
standard
error
for
the
sample
mean
difference
 sXD = 2 sD 64 , sXD = , sXD = 4 , sXD = 2 
 n 16 
 d. (1
point)
Briefly
explain
what
is
measured
by
the
estimated
standard
error
in
this
 situation
 The
estimated
standard
error
here
measures
the
standard
deviation
of
a
sampling
 distribution
of
means
of
samples
of
16
difference
scores.
That
is,
it
measures
typical
 amount
that
a
mean
of
a
sample
of
16
difference
scores
will
differ
from
the
population
 mean
of
difference
scores
 
 
 2) (9
points
total)
The
following
data
are
from
a
repeated‐measures
study
examining
the
effect
of
 a
treatment
by
measuring
a
group
of
n
=
4
participants
before
and
after
they
receive
treatment.
 
 Participant
 
 Before
 After
 
 A
 
 
 7
 
 10
 
 B
 
 
 6
 
 13
 
 C
 
 
 9
 
 12
 
 D
 
 
 5
 
 8
 
 a. (2
points)
Calculate
the
difference
scores
and
 X D 
 (Either
way
of
subtracting
is
fine)
 Participant
 
 Before
 After
 
 Difference
Score
(Before
–
After)
 
 A
 
 
 7
 
 10
 
 ‐3
 
 B
 
 
 6
 
 13
 
 ‐7
 
 C
 
 
 9
 
 12
 
 ‐3
 
 D
 
 
 5
 
 8
 
 ‐3
 
 
 XD = ΣX D -16 = = -4 
 n 4 
 b. (2
points)
Compute
the
SS
for
the
difference
scores
 
 ( X Di − X D ) 
 
 ( X Di − X D )2 
 
 Difference
Score

 
 

 ‐3
 
 
 ‐3
‐
‐4
=
1
 
 
 1
 

 ‐7
 
 
 ‐7
‐
‐4
=
‐3
 
 
 9
 

 ‐3
 
 
 ‐3
‐
‐4
=
1
 
 
 1
 

 ‐3
 
 
 ‐3
‐
‐4
=
1
 
 
 1
 
 
 SSD
=
12

 
 c. (1
point)
Compute
the
sample
variance
for
the
difference
scores
 
 SS 12 2 2 2 sD = D , sD = , sD = 4 
 df 3 
 d. (1
point)
Compute
the
estimated
standard
error
of
the
difference
scores
 
 sXD = 2 sD , sXD = n 4 , s = 1
 4 XD 
 e. (1
point)
Compute
 t XD 
 
 t XD = XD -4 , t XD = , t XD = -4 
 sXD 1 
 f. Is
there
a
statistically
significant
treatment
effect?
Use
a
non‐directional
hypothesis
test
 with
 α = 0.05 
 
 (1
point)

df
=
3,
 α = 0.05 ,
 t critical = ± 3.182 
 
 
(1
point)
|tobserved|
>
|tcritical|,

|­4|
>
3.182,
Reject
Null
Hypothesis
 
 3) (13
points
total)
A
researcher
uses
a
matched‐samples
design
to
investigate
whether
single
 people
who
own
pets
are
generally
happier
than
single
people
without
pets.
A
mood
inventory
 questionnaire
is
administered
to
a
group
of
20‐
to
29‐year
old
non‐pet
owners
and
a
similar
age
 group
of
pet
owners.
The
pet
owners
are
matched
one‐to‐one
with
the
non‐pet
owners
for
 income,number
of
close
friendships
and
general
health.
The
data
are
as
follows.
 
 Matched
Pair
 Non­Pet
 Pet



 
 A
 
 
 11
 
 13
 
 B
 
 
 9
 
 8
 
 C
 
 
 11
 
 14
 
 D
 
 
 13
 
 13
 
 E
 
 
 6
 
 12
 
 F
 
 
 9
 
 11
 
 a. Is
there
a
statistically
significant
difference
in
the
mood
scores
for
non‐pet
owners
versus
 pet
owners?
Use
a
non‐directional
hypothesis
test
with
 α = 0.05 
 
 
 (2
points)
For
finding
difference
scores
and
mean
 
 
 
 (Was
listed
as
Before
/
After
in
original
HW,
sorry)
 Pet
 13
 8
 14
 13
 12
 11
 
 
 
 
 
 
 
 Difference
(NP
–
Pet)
 ‐2
 1
 ‐3
 0
 ‐6
 ‐2
 Difference
(NP
–
Pet)
 
 ( X Di − X D ) 
 
 
 ( X Di − X D )2 
 
 
 
 
 
 
 
 
 
 
 
 
 Matched
Pair
 A
 
 
 B
 
 
 C
 
 
 D
 
 
 E
 
 
 F
 
 
 Non­Pet
 11
 
 9
 
 11
 
 13
 
 6
 
 9
 
 ‐2
 1
 ‐3
 0
 ‐6
 ‐2
 ‐2
‐
‐2
=
0
 1
‐
‐2
=
3
 ‐3
‐
‐2
=
‐1
 0
‐
‐2
=
2
 ‐6
‐
‐2
=
‐4
 ‐2
‐
‐2
=
0
 
 
 
 
 
 
 
 
 
 
 
 
 0
 9
 1
 4
 16
 0
 
 SSD
=
30
 ΣX D −12 = = -2 
 n 6 XD = 
 
 (2
points)
for
finding
SS
 
 
 
 
 
 
 
 (1
point)
 2 sD = 
 
 
 
 
 
 
 
 
 
 
 
 SSD 30 2 2 , sD = , sD = 6 
 df 5 
 (1
point)

 
 sXD = 2 sD , sXD = n 6 , s = 1
 6 XD (1
point)

 
 (1
point)

 
 
(1
point)

 
 t XD = XD -2 , t XD = , t XD = -2 
 sXD 1 df
=
5,
 α = 0.05 ,
 t critical = ± 2.571 
 |tobserved|
<
|tcritical|,


|­2|
<
2.571,
Fail
to
Reject
Null
Hypothesis
 
 b. (2
points)
Write
a
sentence
describing
the
outcome
of
the
hypothesis
test
as
it
would
 appear
in
a
research
report
 
 A
study
was
run
to
investigate
whether
single
people
who
own
pets
are
generally
happier
 than
single
people
without
pets.
No
evidence
for
a
difference
in
happiness
was
found,
 t(5)=‐2,
p
>
0.05.

 
 c. (2
points)
Why
does
the
matching
of
subjects
in
this
way
increase
the
power
of
this
test?
 
 Individual
differences
in
happiness
that
are
attributable
to
income,
number
of
close
 friendships
and
general
health
are
removed
from
the
analysis
as
a
consequence
of
the
 matching.

 
 4) (2
points)

A
repeated‐measures
study
and
an
independent‐measures
study
both
produced
a
t
 statistic
with
df
=
16.

How
many
individuals
participated
in
each
study?
 
 For
the
repeated
measures
study,
df
=
(nD
‐1),
thus
there
were
17
difference
scores,
which
 means
17
individuals
participated.

 
 For
the
independent
measures
study,
df
=
(n1‐1)
+
(n2
‐1),
or
(ntotal
–
2),
thus
there
were
18
 individuals
total.

 
 5) (2
points)
A
matched‐subjects
study
and
an
independent‐measures
study
both
produced
a
t
 statistic
with
df
=
16.

How
many
individuals
participated
in
each
study?
 
 For
the
matched
subjects
study,
df
=
(nD
‐
1),
thus
there
were
17
difference
scores;
difference
 scores
are
between
different
individuals,
thus
there
were
34
participants.
 
 For
the
independent
measures
study,
df
=
(n1‐1)
+
(n2
‐1),
or
(ntotal
–
2),
thus
there
were
18
 individuals
total.
 
 6) (1
point)
A
researcher
conducts
a
research
study
comparing
two
treatment
conditions
and
 obtains
30
scores
in
each
treatment.

If
the
researcher
used
a
repeated‐measures
design,
then
 how
many
individuals
participated
in
the
research
study?
 
 Repeated
measures
studies
have
measures
for
each
subject
twice,
thus
30
people
 participated
in
this
study,
giving
60
total
scores,
30
in
each
treatment.

 
 
 7) (2
points)

Briefly
explain
the
advantages
and
disadvantages
of
using
a
repeated‐measures
 design
as
opposed
to
an
independent‐measures
design.
 
 The
main
advantage
to
a
repeated
measures
design
is
that
variability
due
to
individual
 differences
is
removed
prior
to
analysis
because
individuals
are
compared
to
themselves.
 The
main
disadvantage
is
carry‐over
and
order
effects;
that
is,
the
effect
of
one
treatment
can
 leak
into
another,
and
that
one
treatment
always
comes
before
or
after
another.

In
 independent
measures
designs
one
does
not
have
to
worry
about
order
and
carry
over
 effects,
but
variability
due
to
individual
differences
is
measured
together
with
the
effect
of
 some
treatment
or
manipulation,
leading
to
less
power
to
detect
the
effect.

 
 ...
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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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