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Solutions to Tutorial 2
1.
(a)
ˆ
Y
=1
0
.
2+
4
.
00
X
(
SE
)(
0
.
6633)
(0
.
4690)
MSE
=2
.
199289
,R
2
=0
.
9009
,F
=72
.
73
see code
(R code)
for the plot.
Yes, the linear regression function Fts the data well.
(b)
ˆ
Y
=10
.
2+4
.
00
∗
1=14
.
2
(c)
b
1
±
t
(0
.
975
,
8)
∗
s
(
b
1
)=4
±
2
.
306
∗
0
.
469 = [2
.
918486
,
5
.
081514]
The interval does not include 0, indicating that we have (big) conFdence that
β
1
is di±erent from 0, i.e.
β
1
is not zero, and thus the linear association is signiFcant.
(d) We need to test
H
0
:
β
1
v.s.H
1
:
β
1
6
Note that

t

=

b
1
−
0
s
(
b
1
)

=8
.
528
>t
(1
−
α/
2
,n
−
2) = 2
.
306
We reject
H
0
. In other words, there is
signiFcant
linear association.
(e)
b
0
±
t
(0
.
975
,
8)
∗
s
(
b
0
)=10
.
2
±
2
.
306
∗
0
.
6633 = [8
.
67043
,
11
.
72957]
The interval does not include 0, indicating we have (big) conFdence that
β
0
is
di±erent from 0. Even if no shipment, there are still broken ampules (due to the
other reasons)
(f) We need to test
H
0
:
β
1
v.s.
H
1
:
β
1
6
Note that

t

=

b
1
−
2
s
(
b
1
)

=4
.
264392
(1
−
α/
2
−
2) = 2
.
9
We reject
H
0
. In other words, the past experience is not applicable to the data
1
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View Full Document2. The absolute value of the coeﬃcient
−
1
.
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 Fall '09
 XIAYingcun
 Linear Regression, Regression Analysis

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