420
Spring STAT 2012
Name
ANSWERS
.
Section N1
Exam 1
Version A
Page
Points
Be sure to show all your work; your
partial credit might depend on it.
1
No credit will be given
without supporting work.
2
3
The exam is closed book and closed notes.
You are allowed to use a calculator and
one 8" x 11" sheet with notes.
4
5
6
7
Total
/ 55
___________________________________________________________________________
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.
Water flowing across farmland washes away soil. Researchers released water
across a test bed at different flow rates and measured the amount of soil washed
away. The following table gives the flow x (in liters per second) and the weight
y (in kilograms) of eroded soil.
Flow rate
Eroded soil
x
y
0.4
0.5
0.8
2.3
1.5
1.5
2.6
2.4
3.7
5.8
x = 9,
y = 12.5,
x 2 = 23.5,
y 2 = 47.19,
x y = 31.99,
( x x ) 2 = 7.3,
( y y ) 2 = 15.94,
( x x ) ( y y ) = 9.49.
Consider the model Y i = 0 + 1 x i + i .,
i = 1, 2, , 5,
where is are i.i.d. N ( 0, 2 ).
a) (4) Find the equation of the least-squares regression line.
SXY 9.49
1 =
=
= 1.3.
SXX
x=
7 .3
x = 9
n
5
= 1.8.
y=
y = 12.5
n
5
= 2.5.
0 = y 1 x = 2.41 1.3 1.8 = 0.16.
The least-squares regression line:
y = 0.16 + 1.3 x.
1.
(continued)
b) (7) Is the regression significant at a 5% level of significance? State the null
hypothesis and perform the test.
2
SSRerg = 1 SXX = 1.3 2 7.3 = 12.337.
SSTot = SYY = 15.94,
Source
SSResid = SSTot SSRerg = 15.94 12.337 = 3.603.
SS
df
MS
F
Regression
12.337
1
12.337
10.272273
Residual (Error)
3.603
n2=3
1.201
Total
15.94
n1=4
( Test Statistic )
F 0.05 ( 1, 3 ) = 10.13.
Reject H 0 : = 0.
( Critical Value )
( p-value 0.049142 )
OR
s e2 =
SSResid 3.603
=
= 1.201.
n2
3
s e = 1.201 1.0959.
Test Statistic:
T=
0
se
SXX
=
1.3 0
1.201
3.2050387.
7 .3
Rejection Region:
Reject H 0 if T < t 0.025 ( 5 2 = 3 df ) or T > t 0.025 ( 3 df )
t 0.025 ( 3 df ) = 3.182 Critical Values.
Do NOT Reject H 0 : = 0.
( p-value 2 0.024571 = 0.049142 )
1.
(continued)
c) (4) Construct a 90% prediction interval for the weight of eroded soil corresponding to
the flow rate x = 3.0.
y = 0.16 + 1.3 x = 0.16 + 1.3 3.0 = 4.06
x = 3.0
y t
se
2
1+
1
n
+
(x x ) 2
2
( xi x )
t 0.05 ( 5 2 = 3 df ) = 2.353
4.06 2.353 1.201
1+
1
+
5
( 3 .0 1 .8 ) 2
7 .3
4.06 3.048
2.
Consider the following salary data for 10
Y
x1
x2
tenure-track professors in the Department
48
48
49
1
3
1
3
2
5
49
59
3
2
4
2
59
62
4
5
1
4
66
72
78
4
2
5
5
8
6
of Philosophical Engineering at Anytown
State University in thousands of dollars per
year ( Y ), x 1 ( years at Anytown State
University ), and x 2 ( number of
publications in the last 5 years ).
Consider the model
Y = 0 + 1 x 1 + 2 x 2 + ,
where s are i.i.d. N ( 0, 2 ).
10 30 40
X X = 30 110 120 ,
40 120 200
T
Then
590
X T Y = 1850 ,
2480
and
T
(X X)
35
= 4 .
3
1
0.95 0.15 0.1
= 0.15 0.05
0 ,
0 .1
0
0.025
( y y ) 2 = 350,
and
( y y ) 2 = 1030.
a) (8) Perform the significance of the regression test at a 5% level of significance.
n 10,
p = = ( # of s ) = ( # of columns of matrix X ) = 3.
Source
SS
df
MS
F
Regression
680
p1=2
340
6.8
Error (Residual)
350
np=7
50
Total
1030
n1=9
H0 : 1 = 2 = 0
vs
H 1 : at least one of 1 , 2 is not zero.
Critical Value: F 0.05 ( 2 , 7 ) = 4.74.
Decision:
Reject H 0.
T
(X X)
2.
(continued)
b) (6) Construct a 90% confidence interval for 1 .
1
0.95 0.15 0.1
= 0.15 0.05
0
0 .1
0
0.025
Test H 0 : 1 = 6 vs. H 1 : 1 6
a 10% level of significance.
Var ( 1 ) = C 11 s 2 = 0.05 50 = 2.5.
4 1.895
t 0.05 ( 7 df ) = 1.895.
4 2.996
2 .5
( 1.004 , 6.996 )
6 IS covered by the 90% confidence interval for 1 .
Do NOT Reject H 0
at = 0.10.
OR
46
= 1.265.
Test Statistic:
T=
Critical Values:
t 0.05 ( 7 df ) = 1.895.
Do NOT Reject H 0
2 .5
at = 0.10.
c) (4) Construct a 90% prediction interval for the salary of a tenure-track professor who has
been at Anytown State University for 4 years and published 6 times in the last 5 years.
X 0T = [ 1 4 6 ]
X 0T
Y 0 = 1 35 + 4 4 + 6 3 = 6 9 .
0.95 0.15 0.1 1
0.25
0.15 0.05
4 = [ 1 4 6 ] 0.05 = 0.25.
C X 0 = [ 1 4 6 ]
0
0 .1
0 .0 5
0
0.025 6
[ 1 + X 0T C X 0 ] s 2 = ( 1 + 0.25 ) 50 = 62.5.
t 0.05 ( 7 ) = 1.895.
69 1.895
62.5
69 14.98
3. (7) Suppose a complete second-order model
Y = 0 + 1 x1 + 2 x2 + 3 x3 + 4 x4 + 5 x5 + 6 x6 + 7 x7 + 8 x8 +
was fit to n = 36 data points.
> sum( lm( y ~ 1 )$residuals^2 )
[1] 204
> sum( lm( y ~ x1 + x3 + x5 + x6 + x7 )$residuals^2 )
[1] 138
> sum( lm( y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 )$residuals^2 )
[1] 108
Test H 0 : 2 = 4 = 8 = 0 at a 5% level of significance.
Full Model:
Y = 0 + 1 x1 + 2 x2 + 3 x3 + 4 x4 + 5 x5 + 6 x6 + 7 x7 + 8 x8 +
dim ( V ) = p = 9.
SSResid Full = 108.
Null Model:
Y = 0 + 1 x1
dim ( V0 ) = q = 6.
+ 3 x3
+ 5 x5 + 6 x6 + 7 x7
SSResid Null = 138.
+
Numerator d.f. = dim ( V ) dim ( V0 ) = 9 6 = 3.
Denominator d.f. = n dim ( V ) = 36 9 = 27.
SS
DF
MS
F
Diff.
SSResid Null SSResid Full
dim ( V ) dim ( V0 )
Full
SSResid Full
n dim ( V )
Null
SSResid Null
n dim ( V0 )
SS
DF
MS
F
Diff.
30
3
10
2.5
Full
108
27
4
Null
138
30
Test Statistic
Critical Value: F 0.05 ( 3 , 27 ) = 2.96.
F = 2.5 < 2.96 = F 0.05 ( 3 , 27 )
Decision:
Do NOT Reject H 0
at = 0.05.
4.
Suppose that the price of a birthday present a child receives from his parents ( X )
and the price of a birthday present a child receives from his grandparents ( Y ) have
a bivariate normal distribution with
X = $51,
X = $16,
Y = $38,
Y = $10,
= 0.40.
a) (5) What is the probability that the total value of the presents is over $100?
That is, find P ( X + Y > 100 ).
X + Y has Normal distribution,
E ( X + Y ) = X + Y = 51 + 38 = 89,
2
2
2
2
Var ( X + Y ) = X + 2 XY + Y = X + 2 X Y + Y
= 16 2 + 2 0.40 16 10 + 10 2 = 484.
P ( X + Y > 100 ) = P ( Z >
100 89
) = P ( Z > 0.50 ) = 0.3085.
484
b) (5) What is the probability that the grandparents present is more expensive than the
parents present? That is, find P ( X < Y ).
P ( X < Y ) = P ( X Y < 0 ).
X Y has Normal distribution,
E ( X Y ) = X Y = 51 38 = 13,
2
2
2
2
Var ( X Y ) = X 2 XY + Y = X 2 X Y + Y
= 16 2 2 0.40 16 10 + 10 2 = 228.
P(X Y < 0) = P(Z <
0 13
) = P ( Z < 0.86 ) = 0.1949.
228
4.
X = $51, X = $16, Y = $38, Y = $10, = 0.40.
(continued)
c) (5) Suppose the grandparents present is worth $47. What is the probability that the
parents present is worth over $60? That is, find P ( X > 60 | Y = 47 ).
Given Y = 47, X has Normal distribution
with mean
X +
and variance
X
( y Y ) = 51 + 0.40 16 ( 47 38 ) = 56.76
Y
10
2
(1 2 ) X = (1 0.40 2 ) 16 2
P ( X > 60 | Y = 47 ) = P ( Z >
= 215.04.
60 56.76
) = P ( Z > 0.22 ) = 0.4129.
215.04