This preview shows page 1. Sign up to view the full content.
Second Edition
911
We need to solve this numerically for the smallest
n
that satisﬁes the inequality
(
n

1)
n
t
2
n

1
,.
025
·
64
≥
χ
2
n

1
,.
1
.
Trying diﬀerent values of
n
we ﬁnd that the smallest such
n
is
n
= 276 for which
(
n

1)
n
t
2
n

1
,.
025
·
64
= 306
.
0
≥
305
.
5 =
χ
2
n

1
,.
1
.
As to be expected, this is somewhat larger than the value found in a).
9.35 length = 2
z
α/
2
σ/
√
n
, and if it is unknown, E(length) = 2
t
α/
2
,n

1
cσ/
√
n
, where
c
=
√
n

1Γ(
n

1
2
)
√
2Γ(
n/
2)
and E
cS
=
σ
(Exercise 7.50). Thus the diﬀerence in lengths is (2
σ/
√
n
)(
z
α/
2

ct
α/
2
). A little
work will show that, as
n
→ ∞
,
c
→
constant. (This can be done using Stirling’s formula along
with Lemma 2.3.14. In fact, some careful algebra will show that
c
→
1 as
n
→ ∞
.) Also, we know
that, as
n
→ ∞
,
t
α/
2
,n

1
→
z
α/
2
. Thus, the diﬀerence in lengths (2
σ/
√
n
)(
z
α/
2

ct
α/
2
)
→
0
as
n
→ ∞
.
9.36 The sample pdf is
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

Click to edit the document details