This preview shows pages 1–2. Sign up to view the full content.
36226 Summer 2010
Homework 8
Solutions
1.
(a) For the candidate to have more than 18500 valid signatures, it is necessary that a fraction
of
18500
19850
= 0
.
932 of the signatures are valid. Thus, we set
p
= true proportion of
signatures on the petition that are valid
H
0
:
p
≤
0
.
932
H
1
:
p >
0
.
932
Following the discussion we had in class, we can test this equivalently as
H
0
:
p
= 0
.
932
H
1
:
p >
0
.
932
NOTE: I don’t care which way you set up the hypotheses as long as you do the rest
correctly. This is the way I chose to do it because the problem says that we are interested
in testing whether more than 18500 of the signatures are correct. Thus, I feel that the
the generally accepted value is that that is not the case.
(b) Type I error: does not have enough signatures, but remains on the ballot.
Type II error: the candidate has enough signatures but he is removed from the ballot.
(c) Since
n
= 18500 is very large, we can invoke the CLT and use the sample for the large
sample test for a population proportion.
Test statistic:
Z
=
b
p

0
.
932
q
b
p
(1

b
p
)
n
.
Since the alternative is that the true proportion is greater than 0.932, we reject for large
values of
Z
.
To ﬁnd the test of size 0
.
05, we need to solve the following equation:
α
= 0
.
05 =
P
H
0
(reject
H
0
)
=
P
p
=0
.
932
(
Z > k
)
where
Z
∼
N(0
,
1) under
H
0
By the deﬁnition of the quantiles, we have
k
=
z
α
=
z
0
.
05
= 1
.
645
.
So, a test of size
α
= 0
.
05 rejects the null hypothesis if the test statistic
Z
is greater
than 1
.
645.
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 document was uploaded on 07/14/2011.
 Summer '09

Click to edit the document details