Homework 04 - Answers and Points

Homework 04 - Answers and Points - 
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Unformatted text preview: 
 Homework
Assignment
4
 
 
 
 Psychology
60
 Spring
2010
 1) For
a
population
with
μ
=
50,
and
σ
=
10,
find
the
standard
error,
 σ X ,
when:

 σ 10 10 = = = 5
 a. (1
point)
n
=
4,







 σ X = 2 n 4 σ 10 10 = = =2
 b. (1
point)
n
=
25,





 σ X = 5 n 25 σ 10 10 = = = 1
 c. (1
point)
n
=
100,




 σ X = n 100 10 
 2) (2
point)
The
distribution
of
sample
means
is
not
always
distributed
normally.
Under
what
 circumstances
will
the
distribution
of
sample
means
not
be
normal?
 The
distribution
of
the
sample
means
will
be
non­normal
if
the
population
is
 non­normal
and
the
sample
size
is
smaller
than
about
35.

 
 3) For
a
population
with
σ
=
20:
 a. (1
point)
How
large
a
sample
would
be
needed
to
have
a
standard
error
less
than
10
 σ 20 , ∴ n > 4
 points?
 σ X < 10, n n b. (1
point)
How
large
a
sample
would
be
needed
to
have
a
standard
error
less
than
4
 σ 20 , ∴ n > 25 
 points?
 σ X < 4, n n c. (1
point)
How
large
a
sample
would
be
needed
to
have
a
standard
error
less
than
2
 σ 20 , ∴ n > 100 
 points?
 σ X < 2, n n 
 4) A
population
has
a
mean
of
μ
=
100
and
a
standard
deviation
of
σ
=
20.
Find
the
z‐score
for
 each
of
the
following
sample
means
obtained
from
this
population:
 a. (1
point)
 X =
102
for
sample
of
4
scores

 X−µ 102 − 100 102 − 100 2 zX = , zX = , zX = , zX = = 0.2 
 σX 10 σ/ n 20 / 4 b. (1
point)
 X =
102
for
sample
of
100
scores
 X−µ 102 − 100 102 − 100 2 zX = , zX = , zX = , zX = = 1 
 σX 2 σ/ n 20 / 100 c. (1
point)
 X =
95
for
sample
of
16
scores
 X−µ 95 − 100 95 − 100 −5 zX = , zX = , zX = , zX = = −1.0 
 σX 5 σ/ n 20 / 16 d. (1
point)
 X =
95
for
sample
of
25
scores
 zX = X−µ 95 − 100 95 − 100 −5 , zX = , zX = , zX = = −1.25 
 σX 4 σ/ n 20 / 25 
 5) A
random
sample
is
obtained
from
a
population
with
a
mean
of
μ
=
50
and
a
standard
 deviation
σ
=
12.
 a. (2
points)
For
a
sample
of
n
=
4
scores,
would
a
sample
mean
of
 X =
55
be
 considered
an
unlikely
sample
mean
or
a
fairly
typical
sample
mean?

Explain.

 X−µ 55 − 50 55 − 50 5 zX = , zX = , zX = , z X = = .833 ,

 σX 6 σ/ n 12 / 4 Fairly
typical
sample
mean.
A
mean
this
far
from
the
population
mean
would
occur
a
 large
proportion
of
the
time.

 
 b. (2
points)
For
a
sample
of
n
=
36
scores,
would
a
sample
mean
of
 X =
55
be
 considered
an
unlikely
sample
mean
or
a
fairly
typical
sample
mean?

Explain.
 X−µ 55 − 50 55 − 50 5 zX = , zX = , zX = , z X = = 2.50 ,

 σX 2 σ/ n 12 / 36 Fairly
atypical
sample
mean.
A
mean
this
far
from
the
population
mean
would
occur
a
 very
small
proportion
of
the
time.


 
 6) (3
points)
A
random
sample
is
obtained
from
a
normal
population
with
a
mean
of
μ
=
80
and
 a
standard
deviation
of
σ
=
8.
Which
of
the
following
outcomes
is
more
likely:
A
sample
mean
 greater
than X 
=
86
for
a
sample
of
n
=
4
scores,
or
a
sample
mean
greater
than
 X =
84
for
a
 sample
of
n
=
16
scores.
Explain
and
defend
your
answer.
 X−µ 86 − 80 86 − 80 6 zX = , zX = , zX = , z X = = 1.50 
 σX 4 σ/ n 8/ 4 X−µ 84 − 80 84 − 80 4 , zX = , zX = , z X = = 2.00 
 σX 2 σ/ n 8 / 16 Obtaining
a
sample
mean
greater
than
84
for
a
sample
of
16
scores
will
happen
less
 frequently
than
obtaining
a
sample
mean
greater
than
86
for
a
sample
of
4
scores
as
if
 evident
from
the
z‐scores
showing
how
far
out
in
the
distribution
each
case
is
(NOTE:
If
the
 population
is
normal
(implying
necessarily
a
normal
sampling
distribution
of
the
sample
 mean),
one
can
find
the
exact
proportion
difference;
but,
even
if
drawn
from
a
non‐normal
 population,
the
second
mean
represents
a
more
extreme
case.
)
 zX = 
 7) For
a
normal
distribution
with
μ
=
50
and
σ
=
8:
 a. (2
points)
What
proportion
of
scores
in
the
population
have
values
between
46
and
 54?
 x−µ 46 − 50 −4 x−µ 54 − 50 4 z= , z= , z= , z = −0.5 ,









 z = , z= , z = , z = 0.5 
 σ 8 8 σ 8 8 p(−0.5 < z < 0.5 ) = 2 × (0.1915 ) = 0.383 
 
 b. (2
points)
What
proportion
of
samples
of
size
n
=
4
from
this
population
will
have
 sample
means
between
46
and
54?,

 zX = X−µ 46 − 50 46 − 50 −4 , zX = , zX = , zX = = −1.00 
 σX 4 σ/ n 8/ 4 X−µ 54 − 50 54 − 50 4 , zX = , zX = , z X = = +1.00 
 σX 4 σ/ n 8/ 4 p(−1.0 < z X < +1.0 ) = 2 × (0.3413) = 0.6826 
 zX = 
 c. (2
points)
What
proportion
of
samples
of
size
n
=
16
from
this
population
will
have
 sample
means
between
46
and
54?
 X−µ 46 − 50 46 − 50 −4 zX = , zX = , zX = , zX = = −2.00 
 σX 2 σ/ n 8 / 16 X−µ 54 − 50 54 − 50 4 , zX = , zX = , z X = = +2.00 
 σX 2 σ/ n 8 / 16 p(−2.0 < z X < +2.0 ) = 2 × (0.4772 ) = 0.9544 
 zX = 
 8) The
average
age
for
licensed
drivers
in
the
US
is
μ
=
42.6
years
old
with
a
standard
deviation
 σ
=
12.
The
distribution
is
approximately
normal.
 a. (2
point)
A
researcher
obtained
a
sample
of
n
=
36
drivers
who
received
parking
 tickets.
The
average
age
for
these
drivers
was X 
=
41.7.

Is
this
sample
mean
unlikely
 to
occur
by
chance
alone
or
is
it
a
fairly
typical
sample
mean
with
a
sample
of
n
=
36
 from
this
population?

 X−µ 41.7 − 42.6 41.7 − 42.6 −0.9 zX = , zX = , zX = , zX = = −0.45 
 σX 2 σ/ n 12 / 36 Fairly
typical
sample
mean.
A
mean
this
far
from
the
population
mean
would
occur
a
 large
proportion
of
the
time.
 
 b. (2
points)
The
same
researcher
obtained
a
sample
of
n
=
16
drivers
who
received
 speeding
tickets.
The
average
age
for
these
drivers
was X 
=
34.4.

Is
this
sample
mean
 unlikely
to
occur
by
chance
alone
or
is
it
a
fairly
typical
sample
mean
with
a
sample
of
 n
=
16
from
this
population?

 X−µ 34.4 − 42.6 34.4 − 42.6 −8.2 zX = , zX = , zX = , zX = = −2.733 
 σX 3 σ/ n 12 / 16 Fairly
atypical
sample
mean.
A
mean
this
far
from
the
population
mean
would
occur
a
 very
small
proportion
of
the
time.
 
 
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