Math
112
Spring
’09
Quiz
3
Name____________________________________________
STOP
and
read
the
directions!
40
points
total,
8
points
each
problem
For
each
problem
except
the
last
one
calculate
and
find
the
following
in
this
order:
1.
2.
3.
4.
5.
6.
7.
Claim
and
alpha
Set
up
your
Ho
and
Ha
Find
your
test
statistic
If
you
are
asked
to
do
the
traditional
method
then
find
Critical
Value
now.
Graph
and
shade.
Test
your
claim.
What
is
your
decision
&
why,
reject
Ho,
fail
to
reject
Ho?
Interpret.
Please
write
neatly
use
pencil
and
follow
the
order
above.
Good
Luck!
I.
Claim:
the
mean
life
span
of
desktop
PCs
is
less
than
7
yrs.
The
sample
data:
n=21,
x = 6.8 years, s = 2.4years. The sinificance level is α = 0.05 Answer:
1. Claim : µ < 7
€
α = 0.05
2. H o : µ = 7yrs
H a : µ < 7yrs (claim)
3. Test Statistics :
x-µ
−0.2
t* =
=
= −0.3819
s
0.5237
n
€
tcdf(
‐99,‐0.3819,
20)=0.3533
6. Is p < α ? No !
Fail to reject Ho!
7. There is not enough evidence to support the claim that the mean
life span of desktop PCs is less than 7yrs.
€
1
II.
When
109,
857
arrests
for
federal
offenses
were
randomly
selected,
it
was
found
that
32,
116
of
them
were
drug
offenses.
Test
the
claim
that
more
than
29%
of
federal
crimes
were
for
drug
offenses.
Use
alpha
0.05
significance
level.
Use
the
traditional
method
to
solve
this
problem.
1. Claim : That more than 29% of federal
crimes were for drug offenses, i.e. P > 0.29
α = 0.05
2. H 0 : p = 0.29
H a : p > 0.29 (claim)
32116
pˆ − p 109857 − 0.29
3. z =
=
= 1.712
pq
(0.29)(0.71)
n
109857
4. CV = zα = 1.645
Extra
:
Normcdf=(1.712,99)=0.0434
€
6. Is the test statistic to the right of CV? Yes!
Reject Ho!
7. There is enough evidance to reject the claim that more than 29% of
federal crimes were for drug offenses.
€
NOTE
!
FOR
THIS
YOU
MUST
ROUND
OUT
TO
AT
LEAST
5
DECIMAL
PLACES
TO
GET
THE
SAME
SOLUTION
AS
I.
IF
YOU
ROUND
YOUR
TEST
STATISTIC
TO
FOUR
DECIMAL
PLACES
YOU
WILL
FAIL
TO
REJECT
Ho.
2
III. With individual lines at the checkout, a store manager finds that the standard
deviation for the waiting times on Monday mornings is 5.7min. After switching to a
single waiting line, he finds that for n=29 customers, the waiting times have a
standard deviation of 4.9min. Use a 0.025 significance level to test the claim that
with a single line, waiting time vary less than with individual lines.
Summary and Hint:
σ = 5.7min Monday waiting time individual lines;
s = 4.9 n = 29 single line
Claim : That with a single line, waiting time vary less than with individual lines.
*If it is a left tail test find critical value 1- alpha, if it is a right tail test than it's just alpha in the right tail.
Answer:
€
1. Claim : With a single line, waiting time vary less than with individual lines.
i.e σ < 5.7min, with α = 0.025
2. H 0 : σ = 5.7min
H a : σ < 5.7min (Claim)
(n −1)s2 28 × 4.9 2
3. Test statistic χ =
=
= 20.7
σ2
5.7 2
2
2
4. The critica value left tail for χ 0.025,28
= 15.308
5. Graph is given in class.
6. Is the test statistics to the left or smaller( because is a left tail test) than CV ? NO!
7. There is not enough evidence to support the claim that with single line, waiting time
wary less than with individual lines.
€
3
IV.
A
researcher
finds
that
of
1,000
people
who
said
that
they
attend
a
religious
service
at
least
once
a
week,
31
stopped
to
help
a
person
with
a
car
trouble.
Of
1,200
interviewed
who
had
not
attended
a
religious
service
at
least
once
a
month,
22
stopped
to
help
a
person
with
car
trouble.
At
the
0.05
significance
level,
test
the
claim
that
the
two
proportions
are
equal.
Summary:
Test
the
claim
that
the
two
proportions
are
equal.
Population
ONE:
Attend
religions
service
at
least
once
a
week,
n=1000,
x=31
Population
TWO
Not
attend
religions
services
at
lease
once
a
month,
n=1200,
x=22.
Answer:
1 Claim : The proportion of people who attend religious services and not attend are the same
Let p1 = attend religious services
p 2 = not attend religiouse services
Claim : p1 = p2
2. H 0 : p1 = p2
H a : p1 ≠ p2
31
22
pˆ1 =
= 0.0310, pˆ 2 =
= 0.01833, p = 0.0241, q = 0.9759
1000
1200
( pˆ − pˆ 2 ) − ( p1 − p2 )
(0.0310 − 0.01833) − 0
3. Test statistic : z = 1
=
=
pq pq
(0.0241× 0.9759) (0.0241× 0.9759)
+
+
n1
n2
1000
1200
z=
0.01267
0.01267
=
= 1.93
0.02352 0.02352 0.00657
+
1000
1200
5.
€
Normcdf(1.93,99)=0.02687
multiply
this
by
two
because
it
is
two
tail
p=0.05257
4
6. Is p < α ? NO!
Fail to reject Ho!
7. There is not enough evidence to reject the claim that the two
proportions of people who attend religiouse services at least once a week
and not attend religious services at least once a month are equal.
€
V.
Construct
and
find
a
90%
confidence
interval
for
the
above
problem
(IV).
Work
it
out
step
by
step
and
interpret.
Answer:
E = zα
€
€
2
pˆ1qˆ1 pˆ 2qˆ 2
0.03004 0.018
+
= 1.645 ×
+
= 0.01104 n1
n2
1000
1200
( pˆ1 − pˆ 2 ) − E < ( p1 − p2 ) < ( pˆ1 − pˆ 2 ) − E
0.01267 − 0.01104 < ( p1 − p2 ) < 0.01267 + 0.01104
0.0016 < ( p1 − p2 ) < 0.0237
The interval does not contain "0"
We are 90% confident that people who attend religious services at least once a week differ
from the group of people who don't attent religious services at least once a week.
Note that the differance is very low. At a higher cofidence interval there would not be a differance
between the two groups.
€
5