This preview shows pages 1–5. Sign up to view the full content.
ACTSC 433/833 – Winter 09
Solutions to Assignment 2
Question 1
(a)
j
y
j
r
j
s
j
1
1
5
1
2
2
4
1
3
4
5
1
4
5
3
1
5
8
2
1
(b)
S
10
(2) = (1 – 1/5)(1 – 1/4) = 0.6.
2
2
10
10
1
2
ˆ
Var(
(2))
(
(2))
()
11
0.6
54 43
0.048
j
j
jj
j
s
SS
rr s
=
=
−
⎛⎞
=+
⎜⎟
××
⎝⎠
=
∑
(c)
111
(4)
0.65
545
H
=++=
±
222
ˆ
Var(
(4))
0.1425
H
±
(d)
Using the results from part (b), a 95% symmetric linear confidence interval for
S
(2) is
given by (0.1706, 1). The interval of (0.1706, 1.0294) is also acceptable.
(e)
1.96 0.048
exp
0.2463
0.6ln(0.6)
U
==
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document A 95% logtransformed confidence interval for
S
(2) is given by (0.6
1/
U
, 0.6
U
) = (0.1257,
0.8818)
(f)
Using the results from part (c), a 95% symmetric linear confidence interval for
H
(4) is
given by (0, 1.3899). The interval of (
−
0.0899, 1.3899) is also acceptable.
Let
1.96 0.1425
exp
0.65
U
⎛⎞
=
⎜
⎜
⎝⎠
⎟
⎟
. A 95% logtransformed confidence interval for
H
(4) is
given by (0.65/
U
, 0.65
U
) = (0.2082, 2.0289).
Question 2
(a)
22
2
5 ... 50
ˆ
22.9
10
5
... 50
ˆ
777.9
10
μ
++
==
(b)
For the 25
th
percentile, we have (
n
+1)
p
=
11
×
0.25
=
2.75, which gives
g
=
2.
Interpolating between
x
(2)
and
x
(3)
, the smoothed estimate of the 25
th
percentile is given by
(3 – 2.75)
×
8 + (2.75 – 2)
×
10 = 9.5.
For the 75
h
percentile, we have (
n
+1)
p
=
11
×
0.7
=
8.25, which gives
g
=
8.
Interpolating between
x
(8)
and
x
(9)
, the smoothed estimate of the 75
th
percentile is given by
(9 – 8.25)
×
35 + (8.25 – 8)
×
50 = 38.75.
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.
This note was uploaded on 10/30/2009 for the course ACTSC 433 taught by Professor Johnny during the Winter '09 term at Waterloo.
 Winter '09
 johnny

Click to edit the document details