Solution for Mid Term Test
Question 1:
a.
r
=
2
.
521
(7

1)
√
4
.
6667(0
.
0388)
= 0
.
9874
Since it is close to +1, we have a strong positive linear association between
T
and
S
.
b.
b
= 0
.
9874
q
0
.
0388
4
.
6667
= 0
.
09, and
a
= 1
.
792

0
.
09(4) = 1
.
432
Therefore
ˆ
S
= 1
.
432 + 0
.
09
T
For each year, the predicted sales increased by $0.09 millions.
c.
For year 2010, we have
T
= 16. Therefore
ˆ
S
= 1
.
432 + 0
.
09(16) = 2
.
872.
Hence predicted sales in 2010 is $2.872 millions.
No, we should not trust the result because we are extrapolating.
Question 2:
a.
min
= 25, and
max
= 130.
Q
1
= 2
.
75
th smallest number
=
between
(30
,
33) = 31
.
5
˜
x
= 5
.
5
th smallest number
=
between
(39
,
53) = 46
Q
3
= 8
.
25
th smallest number
=
between
(69
,
94) = 81
.
5
Hence the simple boxplot can be drawn.
b.
IQR
= 81
.
5

31
.
5 = 50
⇒
1
.
5
IQR
= 75
Inner fence = (31
.
5

75
,
81
.
5 + 75) = (

43
.
5
,
156
.
5)
Since all data are inside the inner fence, there is no outlier nor extreme outlier.
Question 3:
Question 4:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 AugustineWong
 Normal Distribution, Standard Deviation, smallest number

Click to edit the document details