Homework 06 - Answers and Points

Homework 06 - Answers and Points - 
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Unformatted text preview: 
 Homework
Assignment
6
 ‐ 
 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) Find
the
tcritical
values
for
a
two‐tailed
hypothesis
test
with
 α = 0.05 
for
the
follow
values
of
 df:
 a. (1
point)
df
=
2
 t critical = ± 4.303 
 b. (1
point)
df
=
8
 t critical = ± 2.306 
 c. (1
point)
df
=
20
 t critical = ± 2.086 
 d. (1
point)
df
=
40
 t critical = ± 2.021 
 2) The
following
scores
were
obtained
from
a
population
with
unknown
parameters.

 Scores:
6,


12,


0,


3,


4
 a. (1
point)
Compute
the
sample
mean
 ΣXi 25 X= ,X= , X =5
 n 5 b. (2
points)
Compute
the
estimate
of
the
population
variance
 SS s2 = , SS = Σ( Xi − X )2 
 n −1 SS = (6 − 5 )2 + (12 − 5 )2 + (0 − 5 )2 + ( 3 − 5 )2 + ( 4 − 5 )2 = 80 
 80 2 s2 = , s = 20 
 4 
 c. (2
points)
Compute
the
estimate
of
the
standard
error,
 s X 
 sX = s2 , sX = n 20 , 5 sX = 4 , sX = 2 
 
 3) To
evaluate
the
effect
of
a
treatment,
a
sample
of
n
=
9
is
obtained
from
a
population
with
a
 mean
of
μ
=
40,
and
the
treatment
is
administered
to
the
individuals
in
the
sample.
After
 treatment,
the
sample
mean
is
found
to
be
 X = 33 
 a. (6
points
total)
If
the
sample
has
a
variance
of
s2=81,
are
the
data
sufficient
to
 conclude
that
the
treatment
has
a
statistically
significant
effect
using
a
two‐tailed
test
 with
 α = 0.05 ?
 
 
 (2
points)
 s X = s2 , sX = n 81 , 9 sX = 9 , sX = 3 
 
 (2
points)
 t X = X−µ 33 − 40 −7 , tX = , tX = , t X = −2.33 
 sX 3 3 
 (1
point)
df
=
8,
 α = 0.05 ,
 t critical = ± 2.306 
 
 













(1
point)
|tobserved|
>
|tcritical|,

|­2.33|
>
|2.306|,
Reject
Null
Hypothesis
 
 b. (6
points
total)
If
the
sample
has
a
variance
of
s2=225,
are
the
data
sufficient
to
 conclude
that
the
treatment
has
a
statistically
significant
effect
using
a
two‐tailed
test
 with
 α = 0.05 ?
 (2
points)
 s X = s2 , sX = n 225 , 9 s X = 25 , s X = 5 
 
 (2
points)
 t X = X−µ 33 − 40 −7 , tX = , tX = , t X = −1.4 
 sX 5 5 
 (1
point)
df
=
8,
 α = 0.05 ,
 t critical = ± 2.306 
 

















 








(1
point)
|tobserved|
<
|tcritical|,
|­1.4|
<
|2.306|,
Fail
to
Reject
Null
Hypothesis
 
 
 c. (2
points)
Comparing
your
answers
in
A
and
B,
how
does
the
variance
you
calculate
 for
your
sample
affect
the
outcome
of
a
hypothesis
test?
 The
sample
variance
has
a
direct
effect
on
the
hypothesis
test;
a
larger
sample
 variance
leads
to
a
less
extreme
t
value,
which
makes
rejecting
the
null
less
likely
 
 
 4) Some
research
examining
the
effects
of
preschool
childcare
has
found
that
children
who
 spend
time
in
day
care,
especially
high‐quality
day
care,
perform
better
on
math
and
 language
tests
than
children
who
stay
home
with
their
mothers.
A
typical
result
in
this
area
 might
show
that
a
sample
of
n=25
children
who
attended
day
care
before
starting
school
had
 an
average
score
of
 X = 87 with
SS
=
1536
on
a
standardized
math
test
for
which
the
 population
mean
is
μ
=
81.

 a. (8
points
total)
Is
this
sample
sufficient
to
conclude
that
the
children
with
a
history
 of
preschool
daycare
are
statistically
significantly
different
from
the
general
 population?
Use
a
two
tailed
with
 α = 0.01 .

 
 SS 1536 (2
points)
 s 2 = , s2 = , s 2 = 64 
 n −1 24 
 (2
points)
 s X = 
 s2 , sX = n 64 , 25 s X = 2.56 , s X = 1.6 
 (2
points)
 t X = X−µ 87 − 81 6 , tX = , tX = , t = 3.75 
 sX 1.6 1.6 X 
 (1
point)
df
=
24,
 α = 0.01 ,
 t critical = ± 2.797 
 
 
















(1
point)


|tobserved|
>
|tcritical|,
|3.75
|
>
|2.797|,
Reject
Null
Hypothesis
 
 
 
 
 b. (2
points)
Compute
the
estimate
of
Cohen’s
d
to
measure
the
size
of
the
preschool
 effect.
 X−µ 6 est Cohen's d = , est Cohen's d = , est Cohen's d = 0.75, 
 64 s2 
 
 c. (2
points)
Write
a
sentence
showing
how
the
outcome
of
the
hypothesis
test
and
the
 measure
of
effect
size
would
appear
in
a
research
report.

 The
effect
of
day
care
on
children’s
mathematical
ability
was
investigated.
 Participation
in
day
care
statistically
significantly
improved
scores
on
a
standardized
 mathematics
test,
t(24)=3.75,
p
<
0.01,
d
=
0.75.

 
 
 5) In
a
two‐sample
study,
one
sample
has
SS
=
48
and
a
second
sample
has
SS=32.

 a. (4
points)
Assuming
n=5
for
both
samples,
find
each
of
the
sample
variances
and
 compute
the
pooled
variance
(because
the
samples
are
the
same
size,
you
should
find
 that
the
pooled
variance
is
exactly
halfway
between).
 SS 48 SS 32 2 s12 = , s12 = , s12 = 12 












 s2 = , s12 = , s12 = 8 
 n −1 4 n −1 4 SS1 + SS1 48 + 32 80 s 2 ed = , s 2 ed = , s 2 ed = , s 2 ed = 10 
 pool pool pool pool df1 + df2 4+4 8 
 
 
 b. (4
points)
Now
assume
that
n=5
for
the
first
sample
and
n=9
for
the
second
sample.
 Again
calculate
the
two
sample
variances
and
the
pooled
variance
(you
should
find
 that
the
pooled
variance
is
closer
to
the
variance
of
the
second
sample).

 SS 48 SS 32 2 s12 = , s12 = , s12 = 12 













 s2 = , s12 = , s12 = 4 
 n −1 4 n −1 8 SS + SS1 48 + 32 2 80 s 2 ed = 1 , s 2 ed = , s pooled = , s 2 ed = 6.667 
 pool pool pool df1 + df2 4+8 12 
 
 6) Two
separate
samples
receive
two
different
treatments.
The
first
sample
has
n=9
with
 SS=710,
and
the
second
sample
has
n=6
with
SS
=
460.

 a. (2
points)
Compute
the
pooled
variance
for
the
samples
 SS + SS1 710 + 460 2 1170 s 2 ed = 1 , s 2 ed = , s pooled = , s 2 ed = 90 
 pool pool pool df1 + df2 8+5 13 b. (2
points)
Calculate
the
estimate
standard
error
for
the
sample
mean
difference
 s( X1 − X2 ) = s 2 ed pool n1 + s 2 ed pool n2 , s( X1 − X2 ) = 90 90 + , s( X1 − X2 ) = 10 + 15 , s( X1 − X2 ) = 5 
 9 6 
 c. (3
points
total)
If
the
sample
mean
difference
is
10
points,
is
this
enough
evidence
to
 reject
the
null
hypothesis
using
a
two‐tailed
test
with
 α = 0.05 
 
 ( X − X2 ) 10 (1
point)
 t( X1 − X2 ) = 1 , t( X1 − X2 ) = , t( X1 − X2 ) = 2 
 s( X1 − X2 ) 5 
 (1
point)
df
=
(n1‐1)+(n2‐1)=13,
 α = 0.05 ,
 t critical = ± 2.160 
 
 (1
point)
|tobserved|
<
|tcritical|,
|2.00
|
<
|2.160|,
Fail
to
Reject
Null
Hypothesis
 
 d. (3
points
total)
If
the
sample
mean
difference
is
13
points,
is
this
enough
evidence
to
 reject
the
null
hypothesis
using
a
two‐tailed
test
with
 α = 0.05 
 
 
 ( X − X2 ) 13 (1
point)
 t( X1 − X2 ) = 1 , t( X1 − X2 ) = , t( X1 − X2 ) = 2.60 
 s( X1 − X2 ) 5 
 (1
point)
df
=
(n1‐1)+(n2‐1)=13,
 α = 0.05 ,
 t critical = ± 2.160 
 
 (1
point)
|tobserved|
>
|tcritical|,
|2.60
|
>
|2.160|,
Reject
Null
Hypothesis
 
 
 7) (9
points
total)
An
independent‐measures
research
study
was
used
to
compare
two
 treatment
conditions
with
n=12
participants
in
each
treatment.
The
first
treatment
has
a
 mean
of
 X1 = 55 and
variance
 s12 = 8 ;
the
second
treatment
has
a
mean
of
 X2 = 52 and
 2 variance
 s2 = 4 .
Do
these
data
indicate
a
statistically
significant
difference
between
the
two
 treatments?
Use
a
two‐tailed
test
with
 α = 0.05 .

 
 (3
points
for
pooled
variance)
 NEED
SS
to
get
pooled
variance
using
formula.

 
 SS1 SS SS2 SS 2 s12 = , 8 = 1 ∴ SS1 = 88 ,

 s2 = , 4 = 2 ∴ SS2 = 44 
 n1 − 1 11 n2 − 1 11 
 SS + SS1 88 + 44 2 132 s 2 ed = 1 , s 2 ed = , s pooled = , s 2 ed = 6 
 pool pool pool df1 + df2 11 + 11 22 
 OR:
because
n1
=
n2,
pooled
variance
will
be
the
simple
average
of
the
sample
 variances,
(4+8)/2,

(12/2)

=
6
 
 
 
 
 (2
points)
 s( X1 − X2 ) = s 2 ed pool n1 + s 2 ed pool n2 , s( X1 − X2 ) = 6 6 + , s( X1 − X2 ) = 0.5 + 0.5 , s( X1 − X2 ) = 1 
 12 12 
 (2
points)


 t( X1 − X2 ) = ( X1 − X2 ) 55 − 52 , t( X1 − X2 ) = , t( X1 − X2 ) = 3 
 s( X1 − X2 ) 1 
 (1
point)


df
=
(n1‐1)+(n2‐1)=22,
 α = 0.05 ,
 t critical = ± 2.074 
 
 (1
point)

|tobserved|
>
|tcritical|,
|3.00
|
>
|2.074|,

Reject
Null
Hypothesis
 ...
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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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