409
Fall STAT 2009
Name
Version B
ANSWERS
.
AD1
AD2
AD3
34
45
23
Please circle your section.
Page
Possible
1
12
2
12
3
9
4
13
5
Earned
Exam 2
Be sure to show all your work; your
partial credit might depend on it.
Put your final answers at the end of
your work, and mark them clearly.
10
6
14
Total
No credit will be given
without supporting work.
The exam is closed book and closed notes.
You are allowed to use a calculator and
one 8" x 11" sheet with notes on it.
70
___________________________________________________________________________
Academic Integrity
The University statement on your obligation to maintain academic integrity is:
If you engage in an act of academic dishonesty, you become liable to severe disciplinary action. Such acts include cheating;
falsification or invention of information or citation in an academic endeavor; helping or attempting to help others commit
academic infractions; plagiarism; offering bribes, favors, or threats; academic interference; computer related infractions; and
failure to comply with research regulations.
Rule 33 of the Code of Policies and Regulations Applying to All Students gives complete details of rules governing
academic integrity for all students. You are responsible for knowing and abiding by these rules.
1.
Let X have a Binomial distribution with the number of trials n = 8 and with
probability of success p. We wish to test H 0 : p = 0.40 vs. H 1 : p > 0.40.
a) (4) Suppose we decided to use the rejection region Reject H 0 if X 6.
Find the significance level associated with this rejection region.
significance level = P ( Reject H 0 | H 0 is true ) = P ( X 6 | p = 0.40 )
= 1 P ( X 5 | p = 0.40 ) = 1 0.9502 = 0.0498.
b) (4) Find the power of the test from part (a) at p = 0.60.
Power = P ( Reject H 0 | H 0 is false ) = P ( X 6 | p = 0.60 )
= 1 P ( X 5 | p = 0.60 ) = 1 0.6846 = 0.3154.
c) (4) Suppose we observe X = 5. Find the p-value of this test.
p-value = P ( as extreme or more extreme than X = x observed | H 0 true )
= P ( X 5 | p = 0.40 ) = 1 P ( X 4 | p = 0.40 ) = 1 0.8263 = 0.1737.
Binomial distribution ( n = 8 ) CDF, P ( X x ):
p
n
x
0.20
0.30
0.40
0.50
0.60
0.70
8
0
1
2
3
4
5
6
7
0.1678
0.5033
0.7969
0.9437
0.9896
0.9988
0.9999
1.0000
0.0576
0.2553
0.5518
0.8059
0.9420
0.9887
0.9987
0.9999
0.0168
0.1064
0.3154
0.5941
0.8263
0.9502
0.9915
0.9993
0.0039
0.0352
0.1445
0.3633
0.6367
0.8555
0.9648
0.9961
0.0007
0.0085
0.0498
0.1737
0.4059
0.6846
0.8936
0.9832
0.0001
0.0013
0.0113
0.0580
0.1941
0.4482
0.7447
0.9424
Page 1 of 6
2.
Let X 1 , X 2 , , X n be a random sample from an exponential distribution with
the p.d.f.
f ( x ) = e x, x > 0.
We wish to test H 0 : = 25 vs. H 1 : > 25.
a) (8) If n = 3, find a uniformly most powerful rejection region with the significance
level = 0.05 that is based on the statistic
of
3
i =1 X i
3
i =1 X i .
That is, for which values
should H 0 be rejected?
( x 1 , x 2 ,..., x n ) =
L ( H 0 ; x 1 , x 2 ,..., x n )
L ( 25 ; x 1 , x 2 ,..., x n )
=
L ( H 1 ; x 1 , x 2 ,..., x n )
L ( ; x 1 , x 2 ,..., x n )
n
25 e
25 x i
e
xi
= i =1
n
n
n
25
= exp ( 25 ) x i .
i =1
i =1
( x 1 , x 2 ,..., x n ) k
Since > 25,
n
2 i =1 X i has a
Recall:
2(2n)
0.05 = = P ( Reject H 0 | H 0 is true ) = P (
n
x i c.
i =1
distribution.
3
X i c | = 25 )
i =1
3
= P ( 50
X i 50 c | = 25 ) = P ( 2 ( 6 ) 50 c ).
i =1
50 c =
2
0.95 ( 6 ) = 1.635.
c = 0.0327.
3
Reject H 0 if
x i 0.0327.
i =1
b) (4) Find the power of the test in part (a) if = 33.7.
Power = P ( Reject H 0 | H 0 is NOT true ) = P (
3
X i 0.0327 | = 33.7 )
i =1
3
= P ( 67.4
X i 2.204 | = 33.7 ) = P ( 2 ( 6 ) 2.204 ) = 0.10.
i =1
Page 2 of 6
3. (3) The power of a test will increase if ( Circle one. )
A
Either the sample size increases or the significance level increases.
B
Either the sample size increases or the significance decreases.
C
Either level the sample size decreases or the significance level increases.
D
Either the sample size decreases or the significance level decreases.
4. (6) A baseball player claims that he hits better during night games than during day
games. Last season, the players performance was as follows:
Day
Games
Night
Games
At bat, n i ( attempts )
240
160
Base hits, Y i ( successes )
72
56
Find the p-value of the test H 0 : p Day = p Night vs. H 1 : p Day < p Night .
Y
72
p1 = 1 =
= 0.30.
n 1 240
Y
56
p2 = 2 =
= 0.35.
n 2 160
Y + Y 2 = 72 + 56 = 128 = 0.32 .
p= 1
n 1 + n 2 240 + 160 400
0.30 0.35
Test Statistic:
Z=
P-value:
= 1.05.
Left tailed.
1
1
0.32 0.68
+
240 160
P-value = ( area of the left tail ) = P( Z 1.05 ) = 0.1469.
Page 3 of 6
5.
Assume that the population of adult hippopotamus weights is normally distributed,
and the overall (population) standard deviation of the weights is 400 pounds. A
random sample of n = 16 adult hippos had an average weight of 5,700 pounds.
We wish to test H 0 : = 5,500 pounds vs. H 1 : > 5,500 pounds, where
is the overall average weight of adult hippo.
a) (5) Find the p-value of this test.
is known.
Test Statistic:
Right tailed test.
Z=
X 0
=
n
5,700 5,500
= 2.00.
400
16
p-value = ( area of the right tail ) = P ( Z 2.00 ) = 0.0228.
b) (8) Find the power of this test if the true value of the overall average weight of adult
hippo is = 5,678, and we use a 5% level of significance.
is known.
Test Statistic:
Rejection Region:
Z=
X 0
Right - tailed.
X > 5,500 + 1.645
400
16
n
=
Show all work.
X 5,500
.
400
16
Reject H 0 if Z > z = 1.645.
= 5,664.5.
Power = P ( Reject H 0 | H 0 is NOT true ) = P ( X > 5,664.5 | = 5,678 )
5,664.5 5,678
= P Z >
= P ( Z > 0.135 ) = 0.5537.
400
16
Page 4 of 6
6. (4) Let X 1 , X 2 , , X n be a random sample from the distribution with probability
density function
f ( x; ) = 2 2 x 3 e x
2
> 0.
x>0
H 0 : = 5 vs. H 1 : = 3.
Find the most powerful rejection region for testing
( You do not have to simplify. )
Reject H 0 if
( x 1 , x 2 ,..., x n ) =
Since
L ( H 0 ; x 1 , x 2 ,..., x n )
L ( H 1 ; x 1 , x 2 ,..., x n )
25
=
9
( x 1 , x 2 ,..., x n )
( x 1 , x 2 ,..., x n ) k
7.
n
n
5 x i2
25 2 x 3 e
i
i =1
=
n
3 x i2
23 2 x 3 e
i
i =1
k.
n
exp 2 x i2 ,
i =1
n
x i2
i =1
c.
Suppose the lifetime of a particular brand of light bulbs is normally distributed.
A random sample of 27 light bulbs yields an average lifetime of 728 hours and a
sample standard deviation of 67 hours.
H 0 : = 750 hours vs. H 1 : 750 hours.
a) (6) Find the p-value of the test
Test Statistic:
is unknown.
T=
X 0
s
n
=
728 750
= 1.7062.
67
27
n 1 = 26 degrees of freedom
Two tailed test.
p-value = 2 ( area of the tail ) = 2 P ( T ( 26 ) 1.7062 )
2 0.05 = 0.10.
Page 5 of 6
7.
b) (6) Test
x = 728,
n = 27,
(continued)
H 0 : = 56 hours vs. H 1 : > 56 hours
( n 1 ) s 2
=
2
2
Test Statistic:
0
Rejection Region:
s = 67.
at a 5% level of significance.
( 27 1 ) 67 2
=
= 37.2175.
56 2
Right tailed.
2
Reject H 0 if 2 >
n 1 = 26 degrees of freedom.
2 ( 26 ) = 38.88.
0.05
= 0.05
Reject H 0 if 2 > 38.88.
The value of the test statistic does not fall into the Rejection Region.
Do NOT Reject H 0 at = 0.05.
c) (8) Find the power of the test in part (b) at = 84 hours.
( n 1 ) s 2
2 =
Test Statistic:
02
2
Reject H 0 if 2 > ( n 1 ) =
( 27 1 ) s 2
56
2
> 38.88
=
Show all work.
( 27 1 ) s 2 .
56 2
2 ( 26 ) = 38.88.
0.05
s 2 > 4689.526,
Recall: If X 1 , X 2 , , X n are i.i.d. N ( , 2 ), then
s > 68.48.
( n 1 ) S 2
2
is
2 ( n 1 ).
Power = P ( Reject H 0 | H 0 is not true ) = P ( S 2 > 4689.526 | = 84 )
= P(
( n 1 ) S 2
2
>
( 27 1 ) 4689.526 = 84 )
|
2
84
= P ( ( 26 ) > 17.28 ) 0.90.
2
Page 6 of 6